Factorial/Exponential Identity, Infinity
Ross A. Finlayson
I like to think about infinity, and the mathematical work around
infinity, because I enjoy deriving novel results and especially
extending them to show them non-trivial.
I derived this some time ago:
lim n->oo n! / (2 (n/2)! 2^n) = 1 = (2 (n/2)! 2^n) / n!
What it is there is an identity relating the factorial to the binary
exponent. What it says there is that
lim n-> oo n! / (2(n/2)!) = 2^n
So on the one side of the equation is something along the theoretical
lines of (oo * (oo-1) * (oo-2) * ... * ((oo/2)+1) / 2 = 2^oo. That's
just trying to explain that partial factorial of half of the integers,
the big ones, each multiplied together, divided by two, is equal in
the limit to, basically, aleph one. I know that's some incorrect
usage in standard theories, that's why I qualify it and call it
theoretical. Please note that a large part of my words correctly use
standard theories and standardly-defined theoretical terms.
I'm interested if someone can say if this is already well-known. I
haven't seen it elsewhere.
One thing of interest with regards to this is the concept of "half an
infinity". This is among consideration of a scalar infinity. An
other place I've come across a "half a scalar infinity" is in the
impulse function, which I think at zero evaluates to half an infinity.
Another thing that is interesting is that it evaluates to equal to one
so that it is equal to its own reciprocal.
I'm interested in your opinion about this, probably. Greetings.
Ross F.
William Elliot
> I like to think about infinity, and the mathematical work around
> infinity, because I enjoy deriving novel results and especially
> extending them to show them non-trivial.
>
> I derived this some time ago:
> lim n->oo n! / (2 (n/2)! 2^n) = 1 = (2 (n/2)! 2^n) / n!
By Sterling's approximation, n! has some resemblance to n^n in the large.
Thus by my reckoning lim n->oo n! / (2 (n/2)! 2^n) = oo
Also when n = 2
4 = (2 (n/2)! 2^n)/n! /= 1
> What it is there is an identity relating the factorial to the binary
> exponent. What it says there is that
>
> lim n-> oo n! / (2(n/2)!) = 2^n
>
Mathematical that makes no sense as
lim... is a constant and 2^n a variable.
David C. Ullrich
wrote:
>On Sun, 27 Jul 2003, Ross A. Finlayson wrote:
>
>> I like to think about infinity, and the mathematical work around
>> infinity, because I enjoy deriving novel results and especially
>> extending them to show them non-trivial.
>>
>> I derived this some time ago:
>> lim n->oo n! / (2 (n/2)! 2^n) = 1 = (2 (n/2)! 2^n) / n!
>
>By Sterling's approximation, n! has some resemblance to n^n in the large.
>Thus by my reckoning lim n->oo n! / (2 (n/2)! 2^n) = oo
Um, of course the formula he gives is wrong, but you haven't shown
that. (_If_ I have Stirling's formula right then
sqrt(n) (2n)!/(4^n (n!)^2)
has a finite limit, although your "reckoning" would show it also
tends to infinity.)
>Also when n = 2
> 4 = (2 (n/2)! 2^n)/n! /= 1
So?
>> What it is there is an identity relating the factorial to the binary
>> exponent. What it says there is that
>>
>> lim n-> oo n! / (2(n/2)!) = 2^n
>>
>Mathematical that makes no sense as
> lim... is a constant and 2^n a variable.
But saying that n! "has some resemblance to" n^n
does make mathematical sense?
************************
David C. Ullrich
Ross A. Finlayson
Hi,
I think the identity is correct, actually. It's obvious that n! > 2^n
in the large. That starts at n=4 where 24>16, for n=5 120>32, and for
all following integers. What happens though is you divide n! by
(n/2)! and 2.
About the limit and having the variable on both sides of the equation,
the idea there is that each side of that equality could be multiplied
or divided by the corresponding sides of some other equality with n
diverging.
lim_n->oo n!/(2(n/2)!) = lim_n->oo 2^n
There's only one variable, n. If it was x on the left and y on the
right then it wouldn't be true uness x=y.
The expresion n! has a value less than n^n for any integer n greater
than one.
I'm interested in this Stirling function(s) and approximation(s), I am
not familiar with them and read about them.
Assume the identity is correct. What good is it? I derived it from
assuming that half of the infinite binary strings have equal numbers
of zeros and ones.
Consider the rational binary numbers of these forms:
.0101010101...
.1010101010...
I think those are 1/3 and 2/3 but I don't know. Let's see:
1/4 + 1/16 + 1/64 + 1/256 + ...
1/2 + 1/8 + 1/32+ 1/128 + ...
How do I evaluate those binary rational numbers? I want to know
reduced fractional forms for those and all other rationals between
zero and one with equal numbers of zeros and ones in their binary
expansion.
Hi.
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> lim_n->oo n!/(2(n/2)!) = lim_n->oo 2^n
>
> There's only one variable, n. If it was x on the left and y on the
> right then it wouldn't be true uness x=y.
Wrong! The variable on each side is a bound (or dummy) variable
which has n meaning outside of the limit statement, so changing the
variable within either limit statement has no effect on the meaning
of that limit statement.
Michael Hartley
> assuming that half of the infinite binary strings have equal numbers
> of zeros and ones.
This assumption, depending on how you define 'half' in this context, is incorrect.
>
> Consider the rational binary numbers of these forms:
>
> .0101010101...
> .1010101010...
>
> I think those are 1/3 and 2/3 but I don't know. Let's see:
Yes.
>
> 1/4 + 1/16 + 1/64 + 1/256 + ...
> 1/2 + 1/8 + 1/32+ 1/128 + ...
>
> How do I evaluate those binary rational numbers? I want to know
> reduced fractional forms for those and all other rationals between
> zero and one with equal numbers of zeros and ones in their binary
> expansion.
Suppose you have .a1a2a3a4...ana1a2..., a binary number which repeats after n bits.
Then it equals (.a1a2a3...an)(1 + 1/2^n + 1/2^2n + ...)
The first bracket is a finite binary string and you can evaluate it.
Call the second a(n).
Then 2^n . a(n) = 2^n + a(n), so a(n) = 2^n / (2^n - 1).
Yours, Mike H...
>
> Hi.
>
> Ross
William Elliot
Newsgroups: sci.math
Subject: Re: Factorial/Exponential Identity, Infinity
>William Elliot <ma...@xx.com> wrote:
>>On Sun, 27 Jul 2003, Ross A. Finlayson wrote:
>
>>> lim n->oo n! / (2 (n/2)! 2^n) = 1
>
>>By Sterling's approximation, n! has some resemblance to n^n in the
>>large. Thus by my reckoning lim n->oo n! / (2 (n/2)! 2^n) = oo
>Um, of course the formula he gives is wrong, but you haven't shown
>that. (_If_ I have Stirling's formula right then
> sqrt(n) (2n)!/(4^n (n!)^2)
>has a finite limit, although your "reckoning" would show it also
>tends to infinity.)
Sterling's approximation: n! = (n/e)^n sqr(2n.pi)
2n! 2^2n = 2(n/e)^n 2^2n sqr(2n.pi)
(2n)! = (2n/e)^2n sqr(4n.pi) = (n/e)^2n 2^2n sqr(4n.pi)
(2n)! / (2n! 2^2n) = (sqr 1/2)(n/e)^n
I reckon so.
----
Ross A. Finlayson
That's fair. I concede that and say that what you can do is use those
terms as an equality and then take the limit after moving the
variables to one side of the equation.
I went through my proof about the identity again and found a mistake.
It's trivial, now my opinion and belief is that:
lim_n->oo n! ( (n/2)! 2^n) = 1
Notice there is no longer the lone 2 in the denominator.
Let me describe my rationale for this statement and derivation of this
identity.
The binomial probability distribution describes how many of a given
sequence of coin tosses will be heads and how many tails. The coin
toss is an event with random results that has equal probability of
achieving either of two possible results.
There is a function that determines with some exactitude the
probability of a given number of successes for a given number of
trials. The function is often denoted or symbolized as the number of
trials over the number of successful results, contained in
parentheses. This notation doesn't easily transfer to ASCII or other
text that doesn't contain mathematical symbols. The function thus
describes for a sequence of true/false, yes/no results the probablity
that a given fraction of the results will be one or the other.
The derivation considers all of the infinite sequences of binary
values, zero and one, 0 and 1. The first thing to notice is among all
of the sequences A, at a given index i in the sequence half of the
sequences have a value of zero and the other half have a value of 1.
Here it is convenient to consider the infinite bit sequences as being
representative of the expansions of the reals between zero and one
inclusive, yet the same should hold true for all reals.
Consider the first element of the sequence, i=1, it is so that for all
of the values less than one half that the value for each A_n of
element i is equal to zero, and for the other half of the values it is
equal to one.
.000... = 0
...
.100... = .011... = 1/2
...
.111... = 1
Another example is to consider for any finite and even integer the
enumeration of all the sequences of that length to see that half of
the values of A_n_i are zero and the other half one, to show it true
for all finite cases.
00
01
10
11
Notice that in the first column there are two zeros and two ones, and
in the second column two zeros and two ones.
So what can be understood from this is that for any given sequence A_n
of the sequences A, that for any index i that the probability of the
value of A_n at i being either value is equal and is equal to one
half.
The binomial probability function is as follows: the probability P of
k successes in n trials with p probability of success and q
probability of failure is:
P = (n! / k!(n-k)!) p^k q^(n-k)
We have that p = q = 1/2.
In the case of the infinite bit sequences, n->oo.
The value k = n/2. This is because we are concerned with the
sequences of length n where half of the elements are successes. Thus
k = n-k = n/2. Thusly, k! = (n-k)! = (n/2)!. In multiplying p^k and
q^(n-k), p = q = 1/2 and k = n-k = n/2 thus that (p^k)(q^(n-k)) =
(1/2)^(n/2) * (1/2)^(n/2) = (1/2)^n.
P = lim n->oo (n!/ (n/2)! (n/2)!) (1/2)^n = n! / (2 (n/2)! 2^n)
Here's the major assumption: half of the sequences of A contain equal
numbers of zeros and ones. Thus the probability of an infinite
sequence having equal numbers of zeros and ones is 1/2, that is to
say, P = 1/2.
1/2 = lim n->oo (n!/ (n/2)! (n/2)!) (1/2)^n = n! / (2 (n/2)! 2^n)
At this point I am thinking that I made an error in that I forgot to
cancel the 2 from each side, or perhaps I multiplied too many 1/2's,
so I correct that to get:
lim n->oo n! / ((n/2)! 2^n)) = 1
As the expression evaluates to equal to one, so does its reciprocal.
Notice that I make a huge assumption: that half of the sequences
contain equal numbers of zeros and ones. This is not only that half
the values at a given index are zero and the other half one, it is
saying that of all the sequences that half of the sequences have equal
numbers of ones and zeros, and that the entire population describes
the distribution. This is true for all finite even values of n.
Ross
Martin Cohen
> Virgil <vmh...@comcast.net> wrote in message news:<vmhjr2-197209....@blade.randori.com>...
>
> I went through my proof about the identity again and found a mistake.
> It's trivial, now my opinion and belief is that:
>
> lim_n->oo n! ( (n/2)! 2^n) = 1
>
> Notice there is no longer the lone 2 in the denominator.
>
.................
>
> Ross
Not quite right. If you apply Stirling's theorem in the form
n! ~ sqrt(2*pi*n)*(n/e)^n, you get
(2n)!/(4^n*(n!)^2) ~ 1/sqrt(pi*n)
Martin Cohen
Virgil
[snip]
> 1/2 = lim n->oo (n!/ (n/2)! (n/2)!) (1/2)^n = n! / (2 (n/2)! 2^n)
>
> At this point I am thinking that I made an error in that I forgot to
> cancel the 2 from each side, or perhaps I multiplied too many 1/2's,
> so I correct that to get:
>
> lim n->oo n! / ((n/2)! 2^n)) = 1
Assuming that we evaluate only for even n, so that (n/2)! has an
unambigious meaning, where f(n) = n! / ((n/2)! 2^n)), one gets
n f(n)
-- ------
2 1/2 = 0.5
4 3/4 = 0.75
6 15/8 = 1.875
8 105/16 = 6.5625
10 945/32 = 29.53125
12 10395/64 = 162.421875
14 135135/128 = 1055.7521875
etc.
And f(100) is a number 64 digits long.
I see no evidence that the indicated limit has any finite value.
>
> As the expression evaluates to equal to one, so does its reciprocal.
>
> Notice that I make a huge assumption: that half of the sequences
> contain equal numbers of zeros and ones. This is not only that half
> the values at a given index are zero and the other half one, it is
> saying that of all the sequences that half of the sequences have equal
> numbers of ones and zeros, and that the entire population describes
> the distribution. This is true for all finite even values of n.
>
> Ross
I have no idea whare Ross is trying to go with this, but I, for one,
do not wish to follow him there.
Ross A. Finlayson
> ...
>
> I have no idea whare Ross is trying to go with this, but I, for one,
> do not wish to follow him there.
That's too bad.
I've gone through my inductive chain of logic to assert a claim, or
"proof" as they're called, a couple times, and think it's good. I
admit it's a suprising result. If you can show an error I'd be
interested to know of it.
One issue is that of probabilistic distribution over an infinite set.
I say things like "half the integers are multiples of two". One way
to consider it is this: select a hundred different integers between
one and a hundred inclusive. Fifty of a hundred, or 1/2, are even
numbers. Thus it's fair to say that as random as a method of
selection could be over a uniform probability distribution over the
numbers from one to a hundred, the probability of selecting an even
number would be 0.5.
It's so for a hundred, it's so for ten million, it's so for all
positive integer multiples of two. Thus, I say that it is true for
all integers, agreeing with the asymptotic density result of number
theory, where the asymptotic density of evens in the integers is one
half.
Of all the infinite bit strings half of them have equal numbers of
ones and zeros. Thus, the probability of selecting one of those at
random from some uniform probability distribution over the strings is
also equal to one half. It's not just a sample of the population
that's considered to determine projections about the population, it's
the entire population.
I'm not quite sure where all those sequences with equal numbers of
ones and zeros are in that set of those sequences, where for any two
sequences a and b where a is having equal numbers of ones and zeros
that a<b or a>b. That's not true for any two sequences c and d, for
example .100... = 0.011..., 1/2.
That is to say, the sequences with equal numbers of zeros and ones, F,
for flat, are not uniformly distributed among those sequences in their
order. For example, starting from the first value .000..., there
would be infinitely many sequences with a higher valuation than zero
and not being an element of F where each is less than any element of
F. F does not alternate in the reals. That implies that rational and
irrational sequences of F are indefinite neighbors, besides that they
are at some point dense in the reals.
I really wouldn't put this forward if I didn't think it was a correct
line of reasoning, I don't lie to sci.math. That's one reason why
mathematics is so great, there are definite right and wrong answers,
where right and left are irrelevant. Then again, I think the
political right and left are extremist and not representative of the
moderate polity.
Anyways I'm interested in these sequences because I think they display
in their interrelations some characteristics of the reals that are
valuable to know and that have meaning and utility.
Plus, I have Virgil as a punching bag on sci.math, and he has me. I
can probably take a punch, and have some knowledge of nerve junctures
and the mechanics of joint action. But that's mostly machismo.
Anyways I want people to find errors where they exist in my logical
development of the factorial/exponential identity. Virgil makes a
point about how the result of the expression appears to diverge, yet
the same would be true for a horseshoe shaped function, or saddle,
what-have-you. Until it reaches the point of inflection the function
is increasing and increasingly increasing.
So, go through the proof of the identity and verify it or contradict
it.
Ross
Martin Cohen
Not quite right. If you apply Stirling's theorem in the form
n! ~ sqrt(2*pi*n)*(n/e)^n, you get
(2n)!/(4^n*(n!)2) ~ 1/sqrt(pi*n)
This means that
n! ( (n/2)!^2 2^n) ~ 1/sqrt(pi*n/2)
or n! ( (n/2)! 2^n) ~ (n/2)! /sqrt(pi*n/2)
Martin Cohen
Ross A. Finlayson
> [...]
>
> Not quite right. If you apply Stirling's theorem in the form
> n! ~ sqrt(2*pi*n)*(n/e)^n, you get
>
> (2n)!/(4^n*(n!)^2) ~ 1/sqrt(pi*n)
>
> Martin Cohen
Hi,
I read some more about Stirling's approximation. Of note is
Stirling's approximation which gives an approximate value for n!,
gamma(n+1), for large values of n.
MathWorld gives Stirling's approximation as
n! ~= sqrt(2pi) n^(n+1/2)e^(-n)
Consider it as n -> oo, with having
n! = (n/2)! 2^n
Divide each side of the approximation by the corresponding side of the
other equation.
1 ~= ( sqrt(2pi) n^n n^(1/2) ) / ( e^n (n/2)! 2^n )
1 ~= ( sqrt(n2pi) n^n ) / ( (n/2)! (2e)^n )
lim n->oo ( sqrt(n2pi) n^n ) / ( (n/2)! (2e)^n) ) ~= 1
Does lim n->oo n! / ( (n/2)! 2^n ) = 1? Why or why not?
Ross
Virgil
Martin Cohen <martin_...@west.raytheon.com> wrote:
> > Virgil <vmh...@comcast.net> wrote in message
> > news:<vmhjr2-337808.21070429072003@[63.218.45.211]>...
> >
> >>...
> >>
> >>I have no idea whare Ross is trying to go with this, but I, for one,
> >>do not wish to follow him there.
> >
> >
> > That's too bad.
> >
> > I've gone through my inductive chain of logic to assert a claim, or
> > "proof" as they're called, a couple times, and think it's good. I
> > admit it's a suprising result. If you can show an error I'd be
> > interested to know of it.
Ross has claimed that if f(n) = n!/( (n/2)! 2^n) )
then lim_{n -> +oo} f(n) = 1, and has rejected my suggestion that he
reconsider.
Ross, consider this, which you may easily verify yourself:
f(6) = 15/8 and f(2n+2)/f(2n) = n + 1/2 for all positive integers n.
Thus f(2n+2) > f(2n) > 1 for every integer n > 2,
and f(2n) gets further away from 1 as n increases,
and, in fact, diverges towards +oo.
Ross A. Finlayson
f(n) = n! / ( (n/2)! 2^n )
f(2n+2) = (2n+2)! / ( (n+1)! 2^(2n+2) )
f(2n) = (2n)! / ( n! 2^(2n) )
f(2n+2)/f(2n) = (2n+2)! n! 2^(2n) / (2n)! (n+1)! 2^(2n+2)
= (2n+1)(2n+2) / ( (n+1) 4 )
= (2n+1)2/4
= n + 1/2
Yes, n + 1/2 is divergent, showing f(2n+2) infinitely greater than
f(2n). That seems to be a counterexample.
(2n+2) / 2n = 1 + 1/n
lim (1 + 1/n) = 1
That doesn't really help here because f(x)/f(y) != f(x/y), f(1) = 1/ 2
gamma(3/2) = 1/2gamma(1/2) = 1/2sqrt(pi) != n+1/2
Considering (n+1/2) / (n+1), in the limit it equals one.
Here's one consideration: find a similar form for some other
identity.
If Stirling's identity is lim n! e^n / n^n sqrt(n 2pi) = 1, with
having
f(n) = n! e^n / n^n sqrt(n)sqrt(2pi)
Then for n+1 and n
f(n+1) = (n+1)! e^(n+1) / (n+1)^(n+1) sqrt(n+1) sqrt(2pi)
f(n+1)/f(n) = ( (n+1)! e^(n+1) n^n sqrt(n) sqrt2pi ) / n! e^n
(n+1)^(n+1) sqrt(n+1) sqrt(2pi) )
= ( (n+1) e n^n sqrt(n) ) / (n+1)^(n+1) sqrt(n+1)
= e n^n sqrt(n) / (n+1)^n sqrt(n+1)
That appears to go to zero. That is to say, f(n) is infinitely larger
than f(n+1). That is because (n+1)^n >> n^n. I think that to show
its convergence there in that way that what you want there is to show
that it would go to one.
Is that similar in example to the proferred counterexample?
I'm more interested in contradictions to the steps of the proof, yet I
still have to figure out how to respond to the proferred
counterexample, unless you do.
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> > Ross has claimed that if f(n) = n!/( (n/2)! 2^n) )
> > then lim_{n -> +oo} f(n) = 1, and has rejected my suggestion that he
> > reconsider.
> >
> > Ross, consider this, which you may easily verify yourself:
> >
> > f(6) = 15/8 and f(2n+2)/f(2n) = n + 1/2 for all positive integers n.
> >
> > Thus f(2n+2) > f(2n) > 1 for every integer n > 2,
> > and f(2n) gets further away from 1 as n increases,
> > and, in fact, diverges towards +oo.
>
>
>
> f(n) = n! / ( (n/2)! 2^n )
>
> f(2n+2) = (2n+2)! / ( (n+1)! 2^(2n+2) )
> f(2n) = (2n)! / ( n! 2^(2n) )
>
> f(2n+2)/f(2n) = (2n+2)! n! 2^(2n) / (2n)! (n+1)! 2^(2n+2)
> = (2n+1)(2n+2) / ( (n+1) 4 )
> = (2n+1)2/4
> = n + 1/2
>
> Yes, n + 1/2 is divergent, showing f(2n+2) infinitely greater than
> f(2n). That seems to be a counterexample.
My point is that f(2n+2) is (n+1/2) times greater than f(2n). There
is no integer n for which (n+1/2) is infinite.
>
> (2n+2) / 2n = 1 + 1/n
> lim (1 + 1/n) = 1
But this is irelevant. What one concludes from my result is that
f(2n) > f(2)*n! for all n > 1.
And since n! -> +oo, amd f(1) <> 0. f(2n) -> +oo.
>
> I'm more interested in contradictions to the steps of the proof, yet I
> still have to figure out how to respond to the proferred
> counterexample, unless you do.
>
> Ross
My counterexample proves your argument contains at least one error.
An acknowledgement of that fact would be in order.
The problem of where those errors occur is all yours.
Ross A. Finlayson
Hi,
I understood your point, thanks for clearing up the part about how
n+1/2 is the quotient of f(2n+2)/f(2n).
What we're looking for in this as a test of possible convergence is
that that quotient should be equal to or have a limit evaluation of
one.
What I want to make clear is the application of that as a test or as a
counterexample that the same reasoning would imply the divergence of
Stirling's identity, which is proven by different means.
So I think I have a proof that doesn't depend on successive terms
demonstrating equality, and I look to a well-known proof of a similar
identity and see that its successive terms don't approach equality. I
want to resolve that as a line of argument against the result, and
start by saying it would invalidate another unrelated result, which is
known to be true and accurate. I'm saying the successive term
quotient argument would be invalid for the same reaons it is invalid
against the well-known result, Stirling's identity.
Now, I haven't verified Stirling's identity, to be frank it is beyond
my current knowledge and would require weeks.
Here's something I like to consider, where I assume
lim n-> oo n! / ( (n/2)! 2^n) = 1
It's that
(n!)^2 / ( ((n/2)!)^2 2^2n ) = 1
(n!)^4 / ( ((n/2)!)^4 2^4n ) = 1
(n!)^8 / ( ((n/2)!)^8 2^8n ) = 1
etcetera, from
n! / ( (n/2)! 2^n) = ( (n/2)! 2^n) / n!
That also leads me to think
sqrt(n!) / ( sqrt((n/2)!) 2^(n/2) ) = 1
that is
x E Z | lim n->oo ( n! ^(2^x) ) / ( ((n/2)!)^(2^x) (2^n)^(2^x) ) = 1
It's a good argument, the quotient argument, I don't yet know how to
counter it directly, so I counter it indirectly by showing the
counterexample against a well-known good result. If the argument
affects the F/E identity proof, then it would affect the Stirling
identity proof.
So my response is "I understand that and am considering it, and don't
consider it to affect the rationale of the proof, tell it to
Stirling."
I'm happy because it, the argument, doesn't represent a contradiction
to the progression of the proof.
About the factorial/exponential identity proof, I think there are gaps
in the proof, that is to say, I haven't fully demonstrated all the
aspects of the proof. It's only been put forth on this thread on
sci.math within the past several days, I imagine it will get further
review.
So, please attack the proof with logical, rational arguments so I can
see if it holds.
Ross
Ross A. Finlayson
squared, and, the 2 is exponentiated by n-1.
That's pretty simple, I should have caught that.
We have:
P = ( n! / ( (n-k)! k! ) ) p^k q^(n-k)
k = n-k = n/2 ->
P = ( n! / ((n/2)!)^2) p^(n/2) q^(n/2)
P = p = q = 1/2 ->
1/2 = n! / ( ((n/2)!)^2 (2^(n/2)) (2^(n/2)) )
1/2 = n! / ( ((n/2)!)^2 2^n )
1 = n! / ( (n/2)!^2 2^(n-1) )
lim n->oo n! / ( (n/2)!^2 2^(n-1) ) = 1 = lim n->oo ( (n/2)!^2 2^(n-1)
) / n!
Those aren't acceptable errors. My error was that I said the
incorrect "(n/2)! (n/2)! = 2 (n/2)!" instead of the correct
"(n/2)!(n/2)! = (n/2)!^2".
In consideration of Stirling's identity,
lim n->oo n! / ((n/e)^n sqrt(n2pi)) = 1
there is that:
lim n->oo ( (n/2)!^2 2^(n-1) e^n) / ( n^n sqrt(n) sqrt(2pi) ) = 1
lim n->oo ( (n/2)!^2 (2e)^n ) / ( 2 n^n sqrt(n) sqrt(2pi) ) = 1
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> I've found more errors in my derivation. The (n/2)! term is to be
> squared, and, the 2 is exponentiated by n-1.
>
> That's pretty simple, I should have caught that.
>
> We have:
>
> P = ( n! / ( (n-k)! k! ) ) p^k q^(n-k)
>
> k = n-k = n/2 ->
>
> P = ( n! / ((n/2)!)^2) p^(n/2) q^(n/2)
>
> P = p = q = 1/2 ->
The above, P = p = q, looks like you are claiming that in 2*n tosses
of a fair coin, the probability of getting exactly the same number
of heads as tails is 1/2.
Infortunately, this only is true for n = 1. The probability of an
equal number of heads and tails strictly decreases as n increases.
>
> 1/2 = n! / ( ((n/2)!)^2 (2^(n/2)) (2^(n/2)) )
The above is only true when n = 2, so the rest of the derivation,
which depends on it being true for all n, is also wrong.
Ross A. Finlayson
Hi,
I claim that in all the possible infinite sequences of coin tosses
that half of them have equal numbers of heads and tails, P=1/2, where
there is a one half probability of each coin toss being heads, p=1/2
and an equal probability of being tails, p=1/2.
Consider the case with two coin tosses. Here are the possible
results:
00
01
10
11
Notice that two of the sequences, one half of those four, have equal
numbers of ones and zeros, those being 01 and 10.
Consider with four coin tosses:
0000
0001
0010
0011 -
0100
0101 -
0110 -
0111
1000
1001 -
1010 -
1011
1100 -
1101
1110
1111
Only 3/8 of the sequences have equal numbers of ones and zeros,
exactly n! / ((n/2)!^2 2^n).
Uh oh. I need to prove that for infinite sequences that half of the
possible sequences have equal numbers of ones and zeros. Damnit!
Let me tell you where I got that idea, it was while we were talking
about "Cantor & Infinity", I have this notion that it is true. I
reread this and see that I did not explain it there. That would be
insufficient foundation to presume that then. I would have to
actually prove that to use it.
I write a program to look at some of the values that it calculates. I
notice some patterns in the values. The program calculates P = f(n)
= n! / (n/2)!^2 2^n for n being powers of two. I have the program
multiply P by n, n f(n). I notice that the value of n f(n) about
doubles in even powers of two. I determine a function on n to divide
this away. I'm observing something like this:
f(1024) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59537
f(4096) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59567
f(16304) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59574
f(65536) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59576
That seems to imply to me that the limit of that function in the
extreme is somewhere slightly greater than 1.595. Yet, over all of
the variables as opposed to even powers of two there is quite the
range of values, the odd powers of two evaluate to around 1.128...
with a different function. Oh well, I'll probably find a bunch more
errors. Anyways, I'm interested in any known constants that have a
rational functional relation to around 1.595.
If you have some information in relation to the above please do note
it.
I still need to figure out the whole "half the sequences have equal
numbers of zeros and ones" thing, as I still think it's true, but I
need to figure it out.
Ross
Ross A. Finlayson
> notice some patterns in the values. The program calculates P = f(n)
> = n! / (n/2)!^2 2^n for n being powers of two. I have the program
> multiply P by n, n f(n). I notice that the value of n f(n) about
> doubles in even powers of two. I determine a function on n to divide
> this away. I'm observing something like this:
>
> f(1024) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59537
> f(4096) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59567
> f(16304) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59574
> f(65536) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59576
>
> That seems to imply to me that the limit of that function in the
> extreme is somewhere slightly greater than 1.595. Yet, over all of
> the variables as opposed to even powers of two there is quite the
> range of values, the odd powers of two evaluate to around 1.128...
> with a different function. Oh well, I'll probably find a bunch more
> errors. Anyways, I'm interested in any known constants that have a
> rational functional relation to around 1.595.
>
> If you have some information in relation to the above please do note
> it.
>
> I still need to figure out the whole "half the sequences have equal
> numbers of zeros and ones" thing, as I still think it's true, but I
> need to figure it out.
>
Hi,
I let the program run overnight. It calculated another result:
f(262144) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 408.5165... /
256
I found the Inverse Symbolic Calculator and entered the value but it
is not yet precise enough, it is still converging, around 1.59577....
It would probably take a few computer years to get many more
calculated results. I definitely need to write a better algorithm for
it to run more efficiently, I should start with generating the
factorials only once and then using them to generate the higher
factorials. I attach the Java file I wrote yesterday, pretty it is
not. What are good extended precision C and Java libraries?
Ross
public class SeqBitCount{
public java.util.Vector vecn;
public java.util.Vector vecP;
public java.util.Vector vecPtimesn;
public java.util.Vector vecPtimesnbyL;
public SeqBitCount(){
vecn = new java.util.Vector();
vecP = new java.util.Vector();
vecPtimesn = new java.util.Vector();
vecPtimesnbyL = new java.util.Vector();
}
public static void main(String[] av){
SeqBitCount c = new SeqBitCount();
c.test4();
c.outputVectors();
}
/*
Calculate value of P for n
*/
public java.math.BigDecimal P(java.math.BigInteger n){
java.math.BigInteger nbang =
this.factorial(n);
//java.lang.System.out.println("n! = " + nbang.toString());
java.math.BigInteger two_to_n =
this.powerOfTwo(n);
//java.lang.System.out.println("2^n = " + two_to_n.toString());
java.math.BigInteger halfnbang1 =
this.factorial(n.shiftRight(1));
// this.factorial(n.divide(new java.math.BigInteger("2")));
//java.lang.System.out.println("(n/2)! = " + halfnbang1.toString());
java.math.BigInteger halfnbang2 =
new java.math.BigInteger(halfnbang1.toByteArray());
java.math.BigInteger halfnbangsquared =
halfnbang1.multiply(halfnbang2);
//java.lang.System.out.println("(n/2)!^2 = " +
halfnbangsquared.toString());
java.math.BigInteger denominator=halfnbangsquared.multiply(two_to_n);
return( new java.math.BigDecimal(nbang).divide(
new java.math.BigDecimal(denominator),
//scale
24,
// java.math.BigDecimal.ROUND_UNNECESSARY) );
java.math.BigDecimal.ROUND_DOWN) );
}
/*
Calculate factorial of n
*/
public java.math.BigInteger factorial(java.math.BigInteger n){
// unit
java.math.BigInteger u =
java.math.BigInteger.ONE;
// factorial value
java.math.BigInteger f =
new java.math.BigInteger(u.toByteArray());
// iterator
java.math.BigInteger i =
new java.math.BigInteger(n.toByteArray());
while(i.getLowestSetBit()>32 || i.intValue()!=0){
f=f.multiply(i);
i=i.subtract(u);
}
return(f);
}
/*
Calculate 2^n for n
*/
public java.math.BigInteger powerOfTwo(java.math.BigInteger n){
// unit
java.math.BigInteger u =
java.math.BigInteger.ONE;
// power value
java.math.BigInteger p =
new java.math.BigInteger(u.toByteArray());
// iterator
java.math.BigInteger i =
new java.math.BigInteger(n.toByteArray());
// two
//java.math.BigInteger two = new java.math.BigInteger("2")
while(i.getLowestSetBit()>32 || i.intValue()!=0){
p=p.shiftLeft(1);
i=i.subtract(u);
}
return(p);
}
/*
test0: generate a bunch of values of P and P*n for n=2^i
*/
public void test0(){
// test value
java.math.BigInteger t =
new java.math.BigInteger(java.math.BigInteger.ONE.toByteArray());
java.math.BigDecimal _P;
java.math.BigDecimal _Ptimesn;
for(int i=0;i<256;i++){
//multiply by 2
t=t.shiftLeft(1);
vecn.addElement(t);
_P=this.P(t);
vecP.addElement(_P);
_Ptimesn= _P.multiply(new java.math.BigDecimal(t));
vecPtimesn.addElement(_Ptimesn);
}
}
/*
test1: generate a bunch of values of P
*/
public void test1(){
// test value
java.math.BigInteger t =
new java.math.BigInteger(java.math.BigInteger.ONE.toByteArray());
java.math.BigDecimal _P;
for(int i=0;i<256;i++){
//multiply by 2^32
t=t.shiftLeft(32);
java.lang.System.out.println("For n = " + t.toString() );
_P=this.P(t);
java.lang.System.out.println("P = " + _P.toString() + "\n\n");
java.lang.System.out.println("P*n = " +
_P.multiply(new java.math.BigDecimal(t))
);
}
}
/*
test2 generate value of P and compare to 1/n
*/
/*
test3: generate a bunch of values of P and P*n for n=i
*/
public void test3(){
// test value
java.math.BigInteger t =
new java.math.BigInteger(java.math.BigInteger.ZERO.toByteArray());
java.math.BigDecimal _P;
java.math.BigDecimal _Ptimesn;
for(int i=0;i<256;i++){
//add two
t=t.add(java.math.BigInteger.ONE);
t=t.add(java.math.BigInteger.ONE);
vecn.addElement(t);
_P=this.P(t);
vecP.addElement(_P);
_Ptimesn= _P.multiply(new java.math.BigDecimal(t));
vecPtimesn.addElement(_Ptimesn);}
}
/*
test4: generate a bunch of values of P and P*n for n=2^2i, and also
the values of P*n / 2^((log_2 n)/2-1)
*/
public void test4(){
// test value
java.math.BigInteger t =
new java.math.BigInteger(java.math.BigInteger.ONE.toByteArray());
java.math.BigDecimal _P;
java.math.BigDecimal _Ptimesn;
java.math.BigInteger _L;
java.math.BigDecimal _PtimesnbyL;
// change 16 to 18 to get higher values, it runs for hours, at sixteen
it runs for less than an hour
for(int i=0;i<16;i+=2){
//multiply by 2
t=t.shiftLeft(2);
vecn.addElement(t);
_P=this.P(t);
vecP.addElement(_P);
_Ptimesn= _P.multiply(new java.math.BigDecimal(t));
vecPtimesn.addElement(_Ptimesn);
_L = new java.math.BigInteger(new Long(0x0000000000000001 <<
((i+2)/2-1) ).toString() );
//java.lang.System.out.println("L="+_L.toString());
_PtimesnbyL = _Ptimesn.divide(
new java.math.BigDecimal(_L),
24,
java.math.BigDecimal.ROUND_DOWN
);
vecPtimesnbyL.addElement(_PtimesnbyL);
}
}
public void outputVectors(){
java.util.Enumeration vecnenum = vecn.elements();
java.util.Enumeration vecPenum = vecP.elements();
java.util.Enumeration vecPtimesnenum = vecPtimesn.elements();
java.util.Enumeration vecPtimesnbyLenum = vecPtimesnbyL.elements();
while(vecnenum.hasMoreElements()){
System.out.println("n = " + vecnenum.nextElement());
System.out.println("P = " + vecPenum.nextElement());
System.out.println("P*n = " + vecPtimesnenum.nextElement());
if(vecPtimesnbyLenum.hasMoreElements()){
System.out.println("P*n/L = " + vecPtimesnbyLenum.nextElement());
}
}
}
}
Richard Carr
:Date: 3 Aug 2003 02:38:11 -0700
:From: Ross A. Finlayson <r...@tiki-lounge.com>
:Newsgroups: sci.math
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:> I write a program to look at some of the values that it calculates. I
:> notice some patterns in the values. The program calculates P = f(n)
:> = n! / (n/2)!^2 2^n for n being powers of two. I have the program
:> multiply P by n, n f(n). I notice that the value of n f(n) about
:> doubles in even powers of two. I determine a function on n to divide
:> this away. I'm observing something like this:
:>
:> f(1024) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59537
:> f(4096) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59567
:> f(16304) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59574
:> f(65536) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59576
:>
:> That seems to imply to me that the limit of that function in the
:> extreme is somewhere slightly greater than 1.595. Yet, over all of
:> the variables as opposed to even powers of two there is quite the
:> range of values, the odd powers of two evaluate to around 1.128...
:> with a different function. Oh well, I'll probably find a bunch more
:> errors. Anyways, I'm interested in any known constants that have a
:> rational functional relation to around 1.595.
:>
Did you consider sqrt(8/pi)?
Richard Carr
:Date: Sun, 3 Aug 2003 08:16:16 -0400
:From: Richard Carr <ca...@cpw.math.columbia.edu>
:Newsgroups: sci.math
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:
Also, if f(n)=(n*n!)/((n/2)!^2 2^n 1/2sqrt(n)), then f(2n) will tend to
sqrt(8/pi) anyway - you don't need to only look at f(2^(2n)).
Maybe if you use the gamma function to define (n/2)! when n is odd you can
also get that f(n) tends to sqrt(8/pi).
Ross A. Finlayson
Hey, great.
The Inverse Symbolic Calculator at
http://www.cecm.sfu.ca/projects/ISC/ISCmain.html says that 2sqrt(8/pi)
= 1.5957691216057307117597842397376....
Please explain how you arrived at that value. I had noticed that the
values for f(2n) approached that value, I just noticed the doubling in
f(2^n).
f(n) = (n * n!) / ( (n/2)!^2 2^n 2^((log_2 n)/2-1) )
Replacing 2^((log_ 2n)/2-1) with sqrt(n)/2 confused me for a second as
I though you had written 1/2sqrt(n), instead it's 1/2 sqrt(n).
f(n)=(n*n!)/((n/2)!^2 2^n 1/2 sqrt(n))
f(n) = ( sqrt(n) * n! ) / ( (n/2)!^2 2^(n-1) )
A writer directed me to Abramowitz and Stegun's "Handbook of
Mathematical Functions" I have a copy, 6'th ed., formula 6.1.18 on
page 256 is a "duplication formula" for gamma(x) = (x-1)!, y! =
gamma(y+1):
gamma(2x) = sqrt(2Pi) 2^(2x - 1/2) gamma(x) gamma(x+1/2)
I set x=n/2.
f(x) = ( sqrt(2x) (2x)! ) / ( x!^2 2^(2x) 2^(2x-1) )
f(x) = ( sqrt(2x) gamma(2x+1) / ( gamma(x+1)^2 2^(2x) 2^(2x-1) )
f(x) = ( sqrt(2x) 2x gamma(2x) / ( (x gamma(x))^2 2^(2x) 2^(2x-1) )
f(x) = ( sqrt(2x) 2x sqrt(2Pi) 2^(2x - 1/2) gamma(x) gamma(x+1/2) ) /
( x^2 gamma(x)^2 2^(2x) 2^(2x-1) )
f(x) = 2 sqrt(2x) sqrt(2Pi) gamma(x+1/2) / ( x gamma(x) 2^2x 2^(2x-1)
)
f(x) = 4 sqrt(xPi) gamma(x+1/2) / ( x gamma(x) 2^(4x-1) )
f(x) = ( sqrt(Pi) gamma(x+1/2) ) / ( sqrt(x) gamma(x) 2^(4x-3) )
Formula 6.1.12 shows a form for gamma(x+ 1/2) in terms of gamma(1/2)
and a production of the odd integers less than 2x over 2^x. It's
possible to show that production as a function of the factorial and a
summation of the sequence of the powers of two. Anyways I am looking
at the other functions for something to reduce gamma(x+1/2)/ gamma(x),
I think gamma(1/2) = 3/2 gamma(3/2) because gamma(x) = x gamma(x-1),
6.1.9 gives gamma(3/2) = 1/2 sqrt(Pi), gamma(1/2) = 3/4 sqrt(Pi). As
gamma(z) = z gamma(z-1), z!/(z-y)! gamma(z-y) = gamma(z). Let's see
on that:
gamma(x) = x gamma(x-1)
gamma(x-1) = (x-1)gamma(x-2)
gamma(x-2) = (x-2)gamma(x-3)
gamma(x) = x (x-1) (x-2) gamma(x-3)
So z = x + 1/2, y = x-1, gamma(x + 1/2 - x + 1) = ((x+1/2)!/x!)
gamma(3/2) = (gamma(x+3/2)/ gamma(x) ) / gamma(3/2), is that right?
f(x) = ( sqrt(Pi) gamma(x+1/2) ) / ( sqrt(x) gamma(x) 2^(4x-3) )
f(x) = ( sqrt(Pi) gamma(x+3/2) / ( sqrt(x) gamma(x)^2 gamma(3/2)
2^(4x-3) )
f(x) = gamma(x+3/2) / ( sqrt(x) gamma(x)^2 2^(4x-4) )
Here I plan to return the variable x to n/2.
f(n) = gamma( n/2 + 3/2) / ( sqrt(n/2) gamma(n/2)^2 2^(2n-4) )
I'm not sure if that's right, I just wrote it right now and haven't
reviewed it many times.
Please explain your reasoning and thought process about this.
I'm thinking here that lim n->oo f(n) = sqrt(8/Pi)
( sqrt(n) * n! ) / ( (n/2)!^2 2^(n-1) ) = 2 sqrt(2)/sqrt(Pi)
(sqrt(n) * n! ) / ( (n/2)!^2 2^n ) = sqrt(2)/sqrt(Pi)
(sqrt(n Pi) n! ) / ( (n/2)!^2 sqrt(2) 2^n) ) = 1
I haven't proven that, only observed it.
Stirling's equation has e in it, that is something for a different
time.
Do you have a proof of this identity?
Thanks,
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> > I write a program to look at some of the values that it calculates. I
> > notice some patterns in the values. The program calculates P = f(n)
> > = n! / (n/2)!^2 2^n for n being powers of two. I have the program
> > multiply P by n, n f(n). I notice that the value of n f(n) about
> > doubles in even powers of two. I determine a function on n to divide
> > this away. I'm observing something like this:
> >
> > f(1024) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59537
> > f(4096) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59567
> > f(16304) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59574
> > f(65536) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59576
Ross has a very odd way of doing math.
Originally he proposed that f(n) -> 1 as n -> +oo.
I showed him that f(n) was a decreasing function, at least for even
n, whose limiting value must be between 0 and 1 as n -> +oo, so he
now proposes a limit greater than 1.
Richard Carr
:Date: 3 Aug 2003 09:47:33 -0700
:From: Ross A. Finlayson <r...@tiki-lounge.com>
:Newsgroups: sci.math
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:Richard Carr <ca...@cpw.math.columbia.edu> wrote in message news:<Pine.LNX.4.56.03...@cpw.math.columbia.edu>...
No! sqrt{8/pi} not 2sqrt{8/pi} since sqrt{8/pi}>1 it is clear that
2sqrt{8/pi}>2.
:
:Please explain how you arrived at that value. I had noticed that the
:values for f(2n) approached that value, I just noticed the doubling in
:f(2^n).
The number looked a bit like sqrt{8/pi}.
:
:f(n) = (n * n!) / ( (n/2)!^2 2^n 2^((log_2 n)/2-1) )
:
:Replacing 2^((log_ 2n)/2-1) with sqrt(n)/2 confused me for a second as
:I though you had written 1/2sqrt(n), instead it's 1/2 sqrt(n).
Aren't they the same?
:
:f(n)=(n*n!)/((n/2)!^2 2^n 1/2 sqrt(n))
:
Certainly for even n it is. You can get it using inequalities connected
with Stirling's formula (you can also use Gosper's formula although
without having any kind of expression for the error term that wouldn't
quite be good enough. With inequalities on either side though you can use
a sandwich theorem to get the required result. One would expect that it is
likely to hold also for odd n if one took the time to analyze it.
:( sqrt(n) * n! ) / ( (n/2)!^2 2^(n-1) ) = 2 sqrt(2)/sqrt(Pi)
:
:(sqrt(n) * n! ) / ( (n/2)!^2 2^n ) = sqrt(2)/sqrt(Pi)
:
:(sqrt(n Pi) n! ) / ( (n/2)!^2 sqrt(2) 2^n) ) = 1
:
:I haven't proven that, only observed it.
:
:Stirling's equation has e in it, that is something for a different
:time.
Ah...
:
:Do you have a proof of this identity?
I actually used Stirling, I'm afraid to sandwich the limit between
sqrt{8/pi}-g(n) and sqrt{8/pi}+h(n) where g(n),h(n)->0 as n->infinity (at
least or even n).
:
:Thanks,
:
:Ross
:
Ross A. Finlayson
I think I can further reduce the equation to get a closed form in
relation to gamma. The idea here is to reduce the function to a
constant.
Euler gives a form for Gamma: this "Handbook of Mathematical
Functions" is 1046 pages long, a math test is one single-sided page of
equation and formulae. Chaper Six starting on page 255 is: "6: Gamma
Function and Related Functions". Euler's Formula is:
Gamma(z) = lim n->oo n!n^z / (n+z)!/(z)!
Gamma(z) = lim n->oo n! n^z z! / (n+z)!
The function gamma(z) is also given in Euler's infinite product form:
Gamma(z) = 1 / z e^(gamma z) II_n=1^oo [ (1+z/n) e^(-z/n)] , |z|<oo
The Euler-Mascheroni constant, called gamma, is approximately
.57721565... . I think I can write gamma(x) for the function and
gamma for the constant.
What is x for Gamma(x) = gamma?
So, it appears that
n! / (n/2)!^2 2^(n-1) = 2 / sqrt( n Pi)
The expression 2/ sqrt( n Pi) there goes to zero but when divided by
n!/ ( (n/2)!^2 2^(n-1) ) that goes to one.
So what that implies is P = 2 / sqrt(n Pi), with that many reals
between zero and one for n = |N| having equal numbers of ones and
zeros.
I wonder how many values of the reals in the unit interval have only
one one, it is the same quantity as how many have only one zero. The
same holds true for any other value of N in question, exactly as many
numbers have a given finite number of ones as there are different
numbers with that many zeros. For example, there is only and exactly
one value with no ones: zero, and only one value with no zeros: one.
For having one zero, or one one the value of that P is:
(n! / (n-1)! 1!) (1/2) (1/2)^(n-1) = n! / ( (n-1) 2^n )
For having one less than half of the values being zeros or ones, the
next most probably case after having equal numbers of zeros and ones,
the expression would be similar to this:
(n! / (n/2-1)! ( (n/2+1)! 2^n )
n! / ( k! (n-k)! 2^n )
Let's see, sum P_k=0^n = 1, sum P_k=0^(n/2)-1 = (1-2P(n/2))/2 = sum
P_k=((n/2)+1)^n. What that is saying is that the probability of there
being any number of ones less than n/2 is the same as that of there
being the same number of zeros and also of there being more ones than
n/2.
I still think about there being half the numbers with equal numbers of
ones and zeros.
So, anyways this expression is how many of the unit interval reals are
having one zero, it is equal to how many have one one, like one half,
.100..., one quarter, .0100..., etcetera, 2^(-n):
n! / ((n/2-1)!(n/2+1)! 2^n)
That is similar to:
gamma(n+1) / (gamma(n/2) gamma(n/2+2) 2^n)
A number with one zero is one half= .011...., a different sequence
than .100..., another such sequence is 3/4= .1011..., 7/8=.11011...,
etcetera, 1-2^(-n).
See, now I figure I should write another program for the new function
and observe its values at exponents of two again and see if there is a
pattern.
That Gosper's formula, it is an approximation for n! at large values
that is more accurate for low finite values of n than Stirling's
formula. I don't know if Gosper's formula is true in the limit as n
-> oo as Stirling's formula is.
Also, I want to prove my formula is another approximation of n! so I
can call it the Factorial/Exponential identity, except it has Pi in
it, and I haven't proven it.
So anyways I should make a program that gives me some values for small
values of n for P(n,k) to visually determine relations for when n ->
oo. Then, if I can determine a pattern, it might help to determine a
finite form, thus deriving a family of identities.
Here's one way to start, let's notice that two of the values have no
ones or no zeros, zero and one.
n! / ( n! 0! 2^n ) = 1 / 2^n
n! / n! = 2^n / 2^n
1=1
The idea there is that there are 2^n possible sequences, as is so for
any finite number of n, and that two of the sequences have all zeros
or all ones, and in consideration of one of them at a time, then 1 /
2^n is the probability of selecting one and the probability of
selecting zero from the unit interval.
Anyways that's the easy case, we can say with certitude that there is
only one sequence with all zeros, and only one sequence with all ones.
For having one or more zeros with one or more ones there are
infinitely many.
Then, when we have P, for example having P = 2 / sqrt( n Pi) for
k=n/2, then the number of sequences where 2^n is the count of all the
sequences is P * 2^n or 2^(n+1) / sqrt(n Pi).
For what value of k will P(n,k) * 2^n = n? There might exist one, as
there exist values of k where P(n,k) * 2^n is less than one, for
example 1, and values where P(n,k) * 2^n is greater than n, for
example our 2^(n+1) / sqrt(n Pi): 1 < n < 2^(n+1) / sqrt(n Pi). Heh
heh heh heh! Wait a second: is 2^(n+1) / sqrt(n Pi) greater than n?
It is for n = 2, 3, 4, ..., where 1 is less than n for n = 2, 3, 4,
.... Anyways if there was a value of k for which there are n or n/2
sequences, that is where P(n,k) = n/ 2^n or n/ 2^(n+1), I wonder what
that would be.
n! k!(n-k!) = n
Here with having one one:
n! /(n-1)! 1! 2^n = n/2^n
What that implies is that of the 2^n sequences, n many of them have
one one and n many of them have one zero. The sequences that have one
one are the set of numbers 2^(-n), the sequences that have one zero
are the sequences 1-2^(-n).
How about having two zeros or two ones? It might seem possible to
consider that there would be n^2 of them, but it might also be much
larger than that. For the x'th element of 2^(-x), there is a first
one in the sequence, then n-x other elements in the sequence to flip
the bit. So, the sum of those for each x over n many of the sequences
would be the amount that have two ones.
The first of these numbers would be .1100..., 3/4. The next would be
.10100..., 5/8, then .100100..., 9/16, .1000100..., 17/32, etcera,
((2^(n-1)+1) / 2^n). Then, the next ones would be .01100..., 3/8,
.010100..., 5/16, .0100100..., 9/32, etcetera. The numbers with two
zeros start with one quarter, .0011..., then .01011..., 3/8,
.011011..., 7/16, etcetera.
n! / (n-2)! 2! 2^n = x/2^n
n^2-n / 2 = x
All of these numbers are rationals that have denominators with powers
of two, we haven't gotten to any with infinite ones and infinite
zeros except as we have been discussing them and about half of them.
In base three,
n! / k! (n-k)! 1/3^k 2/3^(n-k) = x / 3^n
For k = 0, x = 1, for zero and one.
(n! / (n-1)! 1!) 1/3 (2/3)^(n-1) = x / 3^n
n 2^(n-1) = x
For k = 1, x=n 2^(n-1), implying (n 2^(n-1))/3^n of the sequences are
having one zero and many ones or twos in the trinary.
Ross
Ross A. Finlayson
a new thing to consider are those sequences having 1/4, 1/8, 1/16,
etcetera ones, and then also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8,
3/8, 1/9, 2/9, 4/9, etcetera many ones, or zeros.
Rational numbers with one third ones are similar to
.100100100(100)... = 1/2 + 1/16 + 1/128 + ... = sum i=0^oo 2^(-3x-1)
.010010010(010)... = 1/4 + 1/32 + 1/256 + ... = sum i=0^oo 2^(-3x-2)
.001001001(001)... = 1/8 + 1/64 + 1/512 + ... = sum i=o^oo 2^(-3x-3)
They start with any finite sequence with one third ones and terminate
with any repeating finite sequence of one third ones.
There are also irrational numbers of the form.
These would be the expressions of the probability of a sequence having
1/3, 1/5, 2/5, etcetera many ones, as n->oo.
n! / ( (n/3)! (2n/3)! 2^n )
n! / ( (n/5)! (4n/5)! 2^n )
n! / ( (2n/5)! (3n/5)! 2^n )
n! / ( (n/6)! (5n/6)! 2^n )
n! / ( (n/7)! (6n/7)! 2^n )
n! / ( (2n/7)! (5n/7)! 2^n )
n! / ( (n/8)! (7n/8)! 2^n )
n! / ( (3n/8)! (5n/8)! 2^n )
n! / ( (n/9)! (8n/9)! 2^n )
n! / ( (2n/9)! (7n/9)! 2^n )
n! / ( (4n/9)! (5n/9)! 2^n )
Then, for each of those, consider having integer x more or less than,
for example, one third.
n! / ( (n/3+x)! (2n/3-x)! 2^n)
n! / ( (n/3-x)! (2n/3+x)! 2^n)
For example, this would be the kind of sequences with a finite
sequence with one more or one less one than one third and a
terminating sequence with one third ones.
.110(100)...
.000(100)...
What would these probabilities tell us? Well, it might be possible to
easily derive what their limits are as they are evaluated for large n
as multiplied by other expressions and then to prepare a whole bunch
of forms with various fractions so that if/when the (2 sqrt2) / sqrt(n
Pi) case for even numbers of ones and zeros is proven that all the
other identities could be proven at once, giving an infinite family of
expressions for n! as n->oo in terms of 2^n and ((qn)+-x)! and
(((1-q)n)-+x)! for rational q.
One: the count of sequences with all zeros, zero.
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> Starting with sequences with equal numbers of zeros and ones, I think
> a new thing to consider are those sequences having 1/4, 1/8, 1/16,
> etcetera ones, and then also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8,
> 3/8, 1/9, 2/9, 4/9, etcetera many ones, or zeros.
>
> Rational numbers with one third ones are similar to
>
> .100100100(100)... = 1/2 + 1/16 + 1/128 + ... = sum i=0^oo 2^(-3x-1)
> .010010010(010)... = 1/4 + 1/32 + 1/256 + ... = sum i=0^oo 2^(-3x-2)
> .001001001(001)... = 1/8 + 1/64 + 1/512 + ... = sum i=o^oo 2^(-3x-3)
>
> They start with any finite sequence with one third ones and terminate
> with any repeating finite sequence of one third ones.
>
> There are also irrational numbers of the form.
Any n-ary expansion that eventually becomes periodic (merely
repeating some finite sequence of digits) is rational, regardless of
the base.
Ross A. Finlayson
> Starting with sequences with equal numbers of zeros and ones, I think
> a new thing to consider are those sequences having 1/4, 1/8, 1/16,
> etcetera ones, and then also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8,
> 3/8, 1/9, 2/9, 4/9, etcetera many ones, or zeros.
>
ones or zeros.
n! / ( (n/4)!(3n/4)! 2^n )
I problem here with getting an identity that converges to a value is
that (n/4)!(3n/4)! >> (n/2)!(n/2)!. I thought I might be able to use
Euler's formula for Gamma.
Gamma(z) = lim n->oo n! n^z / ((z+1)(z+2)...(z+n))
What I did was set x to n/2 and z to n/4, so that z+x = 3n/4.
Gamma(n/4) = (n/2)! (n/2)^(n/4) / ( (3n/4)!/(n/4)!
(n/4 -1)! = (n/2)! (n/2)^(n/4) (n/4)! / (3n/4)!
(n/2)! (n/2)^(n/4) (n/4) = (3n/4)!
Here what I want is a relation between (n/2)!^2 and (3n/4)!(n/4)!
(3n/4)!(n/4)! = (n/2)! (n/2)^(n/4) (n/4) (n/4)!
(3n/4)!^2 (n/4)!^2 = (n/2)!^2 (n/2)^(n/2) (n/4)^2 (n/4)!^2
(3n/4)! (n/4)! = (n/2)!^2 (n/2)^(n/2) (n/4)^2 (n/4)! / (3n/4)!
(3n/4)! (n/4)! / (n/2)!^2 = (n/2)^(n/2) (n/4)^2 (n/4)! / (3n/4)!
So I think that where I multiplied n!/((n/2)!^2 2^n) by n/(2^(log_2
(n/2-1)) that I could multiply n!/ (n/4)!(3n/4)! 2^n by n/(2^(log_2
(n/2-1))) and also (n/2)^(n/2) (n/4)^2 (n/4)! / (3n/4)!, however, the
values I have computed show it going to zero instead of any positive
constant, which is what is more or less derived from the case with
(n/2), equal numbers of ones and zeros.
n= 4, f(n)= 0.66666666666666666666666666666666666666666666666666666666
n= 16, f(n)= 2.98810325476992143658810325476992143658810325476992143658
n= 64, f(n)= 0.26720147600790400813975210460541880927282821458127736128
n= 256, f(n)= 0.00000000407055224032997121708258424900273493808866341188
n= 1024, f(n)= 0.00000000000000000000000000000000000000000005319865455386
n= 4096, f(n)= 0.00000000000000000000000000000000000000000000000000000000
n= 16384, f(n)= 0.00000000000000000000000000000000000000000000000000000000
n= 65536, f(n)= 0.00000000000000000000000000000000000000000000000000000000
I don't know if that's the correct way to repace the variables in the
Gamma function. If x = n/2 and y = n/4 =x/2, then:
Gamma(x/2) = limit n->oo (2x)! (2x)^(x/2) / ((5x/2)! /(x/2)!)
(x/2-1)! = (2x)! (2x)^(x/2) (x/2)! / (5x/2)!
(5x/2)! (x/2-1)! = (2x)! (2x)^(x/2) (x/2)!
(5x/2)! = (2x)! (2x)^(x/2) (x/2)
What I want there is an expresion for (x/2)!, (3x/2)! and x!.
(x/2)! = (5x/2)! (x/2-1)! / ( (2x)! (2x)^(x/2) )
On the right side of the equation there I'm looking for factorials of
fractions of x, where all the fractions there are higher than one,
where the fractions of the factorial terms are to sum to one.
(n/4)! = (5n/4)! (n/4-1)! / ( n! n^(n/4) )
That expression doesn't have terms (3n/4)! and (n/2)!, just (n/4)!. I
could multiply both sides by (3n/4)! / (n/2)!^2.
(n/4)!(3n/4)! / (n/2)!^2 = (5n/4)! (3n/4)! (n/4-1)! / ( n! n^(n/4)
(n/2)!^2 )
So that if correct allows me to multiply the thing by (5n/4)! (3n/4)!
(n/4-1)! / ( n! n^(n/4) (n/2)!^2) and get the thing to multiply by n/
2^(log_2(n/2-1)) to get a constant. Let's see, in my program here I
have x=n/2, y=n/4, z=3n/4, so 5n/4 is n+y, I have to write a new
function to generate n^(n/4) except I can use ((n/4)^(n/4))^4 or
(y^y)^4.
Calculating (n/4-1)! is a hassle, the other results are pretty much
precalculated except for (n+y)!. I could use a better algorithm to
calculate intermediate values, I precalculated n!, 2^n, and n^n for
2x, x=1...32, and 2^x for x=1...20. That's about a megabyte.
n= 4, f(n)= 0.000014112573173691905592530497270628348207985458404601827860
n= 16, f(n)= 0.000000000000000000000000000000000000000000000007226596805738
n= 64, f(n)= 0.000000000000000000000000000000000000000000000000000000000000
n= 256, f(n)= 0.000000000000000000000000000000000000000000000000000000000000
n= 1024, f(n)= 0.000000000000000000000000000000000000000000000000000000000000
n= 4096, f(n)= 0.000000000000000000000000000000000000000000000000000000000000
n= 16384, f(n)= 0.000000000000000000000000000000000000000000000000000000000000
I guess n^(n/4) does not equal ((n/4)^(n/4))^4, for example 8^2 = 64
while (2^2)^4 = 256. I need to write a new function to generate
n^(n/4). In the meantime as y is less than Integer.MAX_VALUE I can
use the pow() function without writing a new function for the use of
big integers, the value in the denominator would be less.
n= 4, f(n)= 1.875000000000000000000000000000000000000000000000000000000000
n= 16, f(n)= 0.696873106062412261962890625000000000000000000000000000000000
n= 64, f(n)= 0.707269486896323441061911497431205955411422732924867762503011
n= 256, f(n)= 45.830017329992275093956651137014803794959323615450153685895410
n= 1024, f(n)= 51109458999.835860517712702106878661662829103279886830226237808560868620
n= 4096, f(n)= 5044451663733146794724285287208666268270792829347.414806442674833401752204405871217611766651003570779252566509
n= 16384, f(n)= 30614458091559570281379975788497681477287936472086313154343024159912605254634506784359948830848123115730318946565006112364011333198128898650653253095268614023332692667654215384168407952132332888276981790.523004380926941615808779611316129609882583345358716090780820
Another improvement to the program would be for it to generate many
more columns in the table with a variety of computed values and have
more information to review at a time, and to compute in the
background, or to use a large table of precalculated values.
Before I thought to try and use the definition of the Gamma function I
tried a variety of expressions, here again is where a large variety of
expressions calculated for n would be helpful.
(n/4)! = (5n/4)! (n/4-1)! / ( n! n^(n/4) )
(n/4) (n/4)! = (5n/4)! (n/4)! / (n! n^(n/4))
(n/4) = (5n/4)! / (n! n^(n/4))
lim n! n^(n/4) (n/4) / (5n/4)! = 1
About the case with k=n/2, here are some calculated values
f(n) = (n n!) / ( (n/2)!(n/2)! 2^n 2^(log_2(n/2-1)) )
f(4) = 1.500000000000000000000000000000000000000000000000
f(16) = 1.571044921875000000000000000000000000000000000000
f(64) = 1.589548059967470344452933339596256701042875647544
f(256) = 1.594211517956484839624950947759399703028478483734
f(1024) = 1.595379577150690797583016460816556042629587428727
f(4096) = 1.595671726561313225912210533040091324648758050191
f(16384) = 1.595744772287097827906598649182402930052438187131
f(65536) = 1.595763034241236923518120275255359526357923810327
It appears that lim (n n!) / ( (n/2)!(n/2)! 2^n 2sqrt(n) ) =
sqrt(8/Pi). I'll call sqrt(8/Pi) a constant c.
This has n! / (n/2)!(n/2)! 2^n = ( 2c ) / ( sqrt(n) ) , besides
having
sqrt(n) = ( (n/2)!^2 2^n 2c ) / (n!)
n = ( (n/2)!^4 2^2n (2c)^2) / (n!)^2
n! = 2c (n/2)!^2 2^n / sqrt(n)
From above
n! = (5n/4)! / n^(n/4) (n/4)
That gives
(5n/4)! / ( n^(n/4) (n/4) ) = 2c (n/2)!^2 2^n / sqrt(n)
lim (5n/4)! sqrt(n) / ( n^(n/4) (n/4) (n/2)^2 2^n) = 2c
Where sqrt(n) n! / (n/2)!(n/2)! 2^n = 2c
(5n/4)! sqrt(n) / ( n^(n/4) (n/4) (n/2)^2 2^n) = sqrt(n) n! /
(n/2)!(n/2)! 2^n
(5n/4)! = n! (n^(n/4) (n/4))
leading back to
n! = (5n/4)! / (n^(n/4) (n/4))
or also (5n/4)! = n! (n/4) n^(n/4)
Let's see, n^(n/4) * n^(3n/4) = n^n, n^(n/4) = n^n / n^(3n/4)
Stirling has that
n! = sqrt(2pi) n^(n+1/2)e^(-n)
n! / n^n = sqrt(2Pi) sqrt(n) / e^n
n! / n^n = (5n/4)! / ( (n^n)^2 n^(3n/4) n/4)
n! / n^n = (5n/4)! / (n^(11n/4) (n/4) )
(5n/4)! e^n = sqrt(2Pi) sqrt(n) (n/4) (n^(11/4))
(5n/4)! e^n / (n/4) (n^(11n/4)) sqrt(n) = sqrt(2Pi)
n! e^n / (n^(5n/2) sqrt(n) = sqrt(2Pi)
n! e^n / (n^(5n/2 - 1/2) = sqrt(2Pi)
sqrt(2Pi) sqrt(n) / sqrt(n) = sqrt(2Pi)
1=1
Thus it appears I determined a correct way to replace the variables in
the Gamma function to get a different expression but it doesn't appear
to converge, thus I have made a cognitive error or not made a
cognitive jump, or perhaps I replaced variables incorrectly to get a
known true expression, or there is a flaw in my program to calculate
values for finite inputs and observe the results, or I have not used
large enough finite inputs.
How many of the 2^n sequences have equal numbers of ones and zeros?
n!/(n/2)!^2, just like n of the sequences have one zero, n!/(n-1)! 1!.
It's simple to say that there are n!/ (n/4)!(3n/4)! sequences with
one quarter zeros, the idea is to find an expression f(n) that
multiplied by that gives 2^n, 1/f(n) of the sequences have a one for
every three zeros.
Anyways I want someone, preferably myself, to show an expression or
form for 1/4, 1/8, etcetera, and also for 1/3, 1/5, etcetera. Please
do.
Ross
Luis A. Rodriguez
>
> lim n->oo n! / (2 (n/2)! 2^n) = 1 = (2 (n/2)! 2^n) / n!
>
> > I'm interested in your opinion about this, probably. Greetings.
>
> Ross F.
It is false because by Stirling :
lim n-->inf. = n!/[n^n.e^-n.sqr(2Pi.n)] =1
Your expression = [(n/2e)^(n/2]sqr(2). Its limit = inf.
L.Rodriguez
Ross A. Finlayson
> r...@tiki-lounge.com (Ross A. Finlayson) wrote in message news:<3c6b9c1e.03080...@posting.google.com>...
> > Starting with sequences with equal numbers of zeros and ones, I think
> > a new thing to consider are those sequences having 1/4, 1/8, 1/16,
> > etcetera ones, and then also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8,
> > 3/8, 1/9, 2/9, 4/9, etcetera many ones, or zeros.
> >
> Well, I got into the case where a quarter of the sequence elements are
> ones or zeros.
>
> n! / ( (n/4)!(3n/4)! 2^n )
>
> I problem here with getting an identity that converges to a value is
> that (n/4)!(3n/4)! >> (n/2)!(n/2)!. I thought I might be able to use
> Euler's formula for Gamma.
I want to determine a relation between (n/4)!(3n/4)! and (n/2)!(n/2)!,
I think I went about it in an overly complicated way.
(n/4)!(3n/4)! / (n/2)!(n/2)! = (n/4)!(3n/4)! / (n/2)!(n/2)!
So then, when I use (n/4)!(3n/4)! in an expression where there was
instead (n/2)!(n/2)! that appears to converge, then I should be able
to multiply it be (n/2)!(n/2)! / (n/4)! (3n/4)! to get the value of
the expression when it has (n/2)!^2 instead of (n/4)!(3n/4)!.
The same should hold true for all the cases like (n/8)!(7n/8)! (which
is greater than (n/4)!(3n/4)! and also (n/2)!^2) and for
(n/3)!(2n/3)!. That is to say, if f(n) is a simple ratio of
expressions and that so is g(n) where each is h(n) divided by the
expression (n/2)!(n/2)! for f(n) and (n/4)!(3n/4)! for g(n) that f(n)
= g(n) * (n/2)!(n/2)! / ((n/4)!(3n/4)!), and that that would be true
for any expression for rational a/b < 1 that f(n) = g(a, b, n) *
(n/2)!(n/2)! / ( ((a/b)n)! ((1-a/b)n)! ) where f(n) = n! /
((n/2)!(n/2)! 2^n) = g(n,1,2) and g(n,a,b) = n! / ( (a/b)n! (1-a/b)n!
2^n ). Thus I was surprised when that did not appear to be the case
for the small finite values of n that I calculated.
I guess it does, I may have had some other error in my program, I
think I already tried it.
f(4)= 1.500000000000000000000000000000000000000000000000000000000000
f(16)= 1.571044921875000000000000000000000000000000000000000000000000
f(64)= 1.589548059967470344452933339596256701042875647544860839843750
f(256)= 1.594211517956484839624950947759399703028478483734198325900710
f(1024)= 1.595379577150690797583016460816556042629587428727203086922211
f(4096)= 1.595671726561313225912210533040091324648758050191856710414061
f(16384)= 1.595744772287097827906598649182402930052438187131273212971257
I still need to understand the proof of why that value in the limit as
n->oo is sqrt(8/Pi), as I haven't seen one yet, but it appears that it
is very simple to generate the expressions for the expected number of
sequences with a given fraction of ones or zeros.
n! / ((n/2)!^2 2^n) ~= sqrt(8)/ sqrt(Pi*n)
n! / ((a/b)n! (1-a/b)n! ! 2^n) ~= ( sqrt(8) (a/b)n! (1-a/b)n! ) /
(n/2)!^2 sqrt(Pi * n)
With this and Stirling's formula it is possible to determine n^n, e^n,
(a/b)n!, 2^n, and n in terms of each other and Pi.
Another thread today on sci.math is discussing "probabilities that
equal zero". Pick a number between zero and one, I'll probably bet
it's one half, with probability at least infinitesimally greater than
one half. After all, the average value of all of the reals between
zero and one is one half. Of any characteristic of the amount of ones
and zeros of any infinite string of coin tosses, it is more probable
that it has nearly 1/2 ones and 1/2 zeros than any other fraction.
Then again, on "pick a number between zero and n for unknown n", I
pick four.
Could infinite monkeys each typing "at random" infinitely generate the
works of Shakespeare? Easily, the work of Shakespeare is a finite
string. How many cigarettes would they smoke?
The probability of selecting a given sequence "at random" from the 2^n
possible sequences is 1/ 2^n. That evaluates to zero for infinite n
in terms of x, y, or z unrelated to n, but 2^n / 2^n evaluates to
equal to 1 for any n.
Ross
Ross A. Finlayson
terms of (n!), (n/4), and n^(n/4).
(5n/4)! = n! (n/4) n^(n/4)
In calculating the value of the expression, espcially as n gets large
it may well be that it would cost less to calculate n!, (n/4),
n^(n/4), and their product than (5n/4)!. Well, maybe not. It would
take n extended precision multiplications to calculate n!, one to
calculate n/4, and n/4 to calculate n^(n/4), and two to multiply the
three values together, yielding 5n/4 + 3. To calculate (5n/4)! it
would take 5n/4 calculations, three less. Yet, it might take less
time and resources. So where x is divisible by five, then 4x/5 could
be calculated as n and then that used to generate the exact value of
(5n/4)! without multiplying (5n/4)(5n/4-1)(5n/4-2)...(n+1).
Is it so that (5n/4)! / n! (n/4) n^(n/4) = 1?
One thing here is that n is divisible by 4 or conversely x is
divisible by 5. If that is known then it takes a division or
right-shift by two bits to get n/4 or a division and two
multiplications or one and a left-shift to get x/5 and 4x/5.
Let's see, n=4, n/4=1, n!=24, n^(n/4)=4. The product n! * (n/4) *
(n^(n/4) = 96. That is not equal to 120, 5!, 120/96= 5/4, that is not
equal to one.
Let's see , n^(n/4) = 2^(n/4) 2^(n/4) (n/4)^(n/4).
I guess I should reevaluate the expression I derived from the
factorial definition of the gamma function.
Gamma(z) = lim n->oo n! n^z / ((z+1)(z+2)...(z+n))
If x = n/2 and z = n/4 =x/2, then:
Gamma(x/2) = limit n->oo (2x)! (2x)^(x/2) / ((5x/2)! /(x/2)!)
(x/2-1)! = (2x)! (2x)^(x/2) (x/2)! / (5x/2)!
(5x/2)! (x/2-1)! = (2x)! (2x)^(x/2) (x/2)!
(5x/2)! = (2x)! (2x)^(x/2) (x/2)
(5n/4)! = n! n^(n/4) (n/4)
One issue is that the Gamma function has two variables, n and z, I
relate them by saying z = n/4. In the definition of the function, it
is as the limit is taken as n diverges to infinity. Both sides of the
above relation would diverge, but due to their expression of equality
in the limit then their ratio would equal one.
lim n->oo (5n/4)! / ( n! n^(n/4) (n/4) ) = 1
Is this true or false? Why or why not?
Say y is n/8 instead of n/4.
Gamma(n/8)= limit n->oo n! n^(n/8) / ( (9n/8)! / (n/8)! )
(9n/8)! = n! n^(n/8) (n/8)
For integer x, this expression would show a similar relation:
((1+1/x)n)! = n! n^(n/x) (n/x)
lim n->oo n! n^(n/x) (n/x) / ((1+1/x)n)! = 1
This is just a reformulation of the Gamma function and is not directly
related to the Factorial/Exponential Identity.
If that is so then it isn't immediately apparent from the calculation
of f(n) = (5n/4)! / (n! n^(n/4) n/4) ):
f(4)= 1.250000000000000000000000000000000000000000000000000000000000
f(8)= 0.703125000000000000000000000000000000000000000000000000000000
f(16)= 0.443572998046875000000000000000000000000000000000000000000000
f(32)= 0.352519793887040577828884124755859375000000000000000000000000
f(64)= 0.444950048827587832031906868766203160007403471370990288846769
f(128)= 1.417186935509602084562734755564139602885006156205776690581632
f(256)= 28.747764530480100745279297021710678182560318273161584010488280
f(512)= 23656.230392256541850405423894512179641023942805494295940168978839
f(1024)= 32035924072.135932556176618190547111645083296343519866651520060543116994
f(2048)= 117500598137002400541538.031470022915735801375789783708714641436319783240693433287440
f(4096)= 3161334239219732890548900496905568901855436766817.559143447753479442386466660331677310351829378702512827084827
f(8192)= 4576763120832850605581105273558091926041601241117842406990095200363942901967894137560256891968781924.571344603817963917473527495885667696395205831275810273990727
f(16384)= 19185059304741736873902378347025151130820749241812820572375186361782239983313438378190905194904438405630125326542726807509450428857025444551240483198088691281269116834341160857218280248285816531957071631.912144049687149936446060645956473757627640835673986634036602
f(32768)= 674220418048043976976300676749250732148049671918036997984568986017553900991381380976079646552393627560560186471464571827818588215162467318553876941240140365396944153166340551921022276088532714382633367874965138049465703759289950257911889185477409102946662759777043295472466283207676469068185468353315608987821633625155599692797764514013299915413216836552466585753874803628320414103746751372
72043545644920554.290075136553133247138361866911422321755093685091910989718665
f(65536)= 166536362244043885954716941074705526807663435293445248718163478230597021413650682009573329453019670487723891160162637011951065630230022456269164535382693127273125655024933910882417710414164269205021803030077296607801504113833645740752390331750949185147956719741896320444104917689939300843596433477810966394524829245400332224383562662870929985227152025378707797729317864065080043694469977930
8528886651450983357429357473610067842411399042703987251059543373336962984726826686393395308356936594995013221857407406543218291232052632715332441786548674112052071988437170060733768825242052829784177111954813929658347565760247146239494279865386418460569084426556419933407643048434366554573622025627905522569723767693087846914161049243717063400710054982467929795910901517032032411434348954361152475576
6609412751467435655225345918.908626512801798705056045495171847632554849010334776618238475
Please explain your thoughts on this matter.
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> I think it's interesting that there is an expression for (5n/4)! in
> terms of (n!), (n/4), and n^(n/4).
>
> (5n/4)! = n! (n/4) n^(n/4)
This is trivially false for n = 4,
Left hand side = (5n/4)! = (5*4/4)! = 5! = 120,
Right hand side = n! (n/4) n^(n/4) = 4! (1) 4^1 = 24 * 4 = 96
It is less trivial, but still false for n = 8
Is there any n for which it is true?
Richard Carr
:Date: Fri, 08 Aug 2003 00:25:36 -0600
:From: Virgil <vmh...@comcast.net>
:Newsgroups: sci.math
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:In article <3c6b9c1e.03080...@posting.google.com>,
:
Fairly trivial for n=8 too. Here's a sort of strategy:
for some N, for any n>=N there is a prime p with n<p<=5n/4.
Then if 4 divides m and m>=n p will divide (5m/4)! but not m!.
So there can be finitely many n for which it is true.
You can probably attack the other side (especially if you can figure out
N) as you can raise the lower bound fairly quickly.
If n=4r, then 5 must divide r, r=5t. Then you get 77 divides t etc.
Robin Chapman
> I think it's interesting that there is an expression for (5n/4)! in
> terms of (n!), (n/4), and n^(n/4).
>
> (5n/4)! = n! (n/4) n^(n/4)
>
What's the point? Have you ever posted anything anywhere anytime
that wasn't total bollocks?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen
Virgil
<Pine.LNX.4.56.03...@cpw.math.columbia.edu>,
Richard Carr <ca...@cpw.math.columbia.edu> wrote:
It is also trivially false for n being any integer multiple of 4 and
having a prime between n and 5n/4, since that prime must divide the
left hand side but not the right hand side of the equation.
For example, when n is 100, there are 4 primes betweem 100 and 125,
they are 101.103.107.109 and 113.
Thus there are a large number of n, possibly even infinitely many,
for which it is trivial to prove that the equation is false.
And, as yet, no known values for which it is true.
Not a vary good batting average.
Richard Carr
:Date: Fri, 08 Aug 2003 11:29:39 -0600
:From: Virgil <vmh...@comcast.net>
:Newsgroups: sci.math
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:In article
:<Pine.LNX.4.56.03...@cpw.math.columbia.edu>,
That was the point I made just after the word "strategy" above.
:
:For example, when n is 100, there are 4 primes betweem 100 and 125,
:they are 101.103.107.109 and 113.
:
:Thus there are a large number of n, possibly even infinitely many,
:for which it is trivial to prove that the equation is false.
I also commented about this (rather more in fact that the set of
counterexamples is not only infinite but cofinite and, quite likely,
empty).
:
:And, as yet, no known values for which it is true.
:
Ross A. Finlayson
I know (5/4)! / (n! (n/4) n^(n/4)) does not equal one for any finite
integer n. What I have surmised is that the limit as n diverges of
that expression is equal to one. It is implicit in the context of the
discussion about the equation there that the limit case is that that
expression may be true where I already pointed out that 120 does not
equal 96, and provided values of the result of the expression for
various larger values of n.
You might notice that that expression can be expressed in terms of the
Gamma function instead of factorial, with a domain of r in R\Z-
instead of n in Z+, and that between 4 and 8 the value of the
expression goes from higher to one to less than one, and between 64
and 128 it goes back, with it being continuous, thus that for some
finite, probably irrational, value r that it is true for the Gamma
expression instead of the factorial expression.
The point is that the factorial expression is true in the limit case,
or it is not, that is the point of discussion here. I presented a
rationale why to consider it to be true, and want to clarify here that
the "prime in [n, (1+1/x)n]" argument does not affect the rationale of
the limit case argument.
This is different than lim n->oo sqrt(n Pi/2) n! / ( (n/2)!^2 2^n ) =
1. I am told that Maple evaluates that expression, yet I'm unclear as
to how, although I have an idea. I think it uses the doubling formula
for Gamma, and Gamma(1/2), and other than that I can't say how it's
derived, only that I expressed it, and observed it to correspond with
truth.
Given a "random uniform distribution over all infinite binary
sequences" the probability of a random sequence having equal numbers
of ones and zeros is sqrt(n Pi/2), which equals sqrt(n Pi/2) 2^n /
2^n, in terms of n, an "infinite variable".
So, the factorial expression doesn't hold true for a finite integer
value n, I understand, although it does hold true for various real
values in the gamma expression. I say that the expression is true in
the limit case, do you understand?
Ross
Ross A. Finlayson
> Ross A. Finlayson wrote:
>
> > I think it's interesting that there is an expression for (5n/4)! in
> > terms of (n!), (n/4), and n^(n/4).
> >
> > (5n/4)! = n! (n/4) n^(n/4)
> >
>
> What's the point? Have you ever posted anything anywhere anytime
> that wasn't total bollocks?
Hi Robin,
Why do you care?
Ross
Ross A. Finlayson
Watch as x goes to n.
((1+1/n)n)! / ( n! n^(n/n) (n/n))
((1+1/n)n)! / ( n! n^1 1)
(n+1)! / (n! n)
lim ((n+1) / n) = 1
How about ((1+1/x)n)! = n! n^(n/x) (n/x) for x =1?
lim (2n)! / (n! n^n n) = 1
The Handbook of Mathematical Functions has an identity for Gamma(2z).
gamma(2z) = sqrt(2Pi) 2^(2z-1/2) gamma(z) gamma(z + 1/2)
.
(2n)! = gamma(2n+1) = 2n gamma(2n)
.
2n gamma(2n) / ( n! n^n n ) = 1
.
2 gamma(2n) / (n! n^n ) = 1
.
gamma(2n) = (n! n^n ) / 2
.
sqrt(2Pi) 2^(2n-1/2) gamma(n) gamma(n+1/2) = (n! n^n ) / 2
.
2sqrt(2Pi) 2^(2n -1/2) gamma(n) gamma(n+1/2) = gamma(n+1) n^n
.
sqrt(Pi) 2^(2n+1) gamma(n) gamma(n+1/2) / (gamma(n+1) n^n ) = 1
.
sqrt(Pi) 2^(2n+1) gamma(n+1/2) / n^(n+1) = 1
The function gamma(n+ 1/2) is sqrt(Pi) 1*3*5*7*...*(2n-1) / 2^n.
lim ( 2Pi 2^n) (1*3*5*7*...*(2n-1)) / n^(n+1)) = 1
That looks funny. Do you know any expressions g(n) such that the
infinite product of the expression 2n-1, that is f(n) = II_n=1^oo
(2n-1), besides f(n)=g(n), that their quotient lim ( f(n) / g(n) ) =
1?
Also for x=1:
lim n->oo (2n)! / (n! n^n n) = 1
The expression has the term n^n which finds a place in Stirling's
equation:
lim n->oo n! e^n / n^n sqrt(2Pi) sqrt(n) = 1
n^n = n! e^n / sqrt(2Pi) sqrt(n)
(2n)! sqrt(2Pi) sqrt(n) / (n! n! e^n) = 1
(2n)! sqrt(2Pi) sqrt(n) / ( n!^2 e^n) = 1
We have that;
lim y->oo sqrt(y Pi/2) y! / (y/2)!^2 2^y = 1
y!/ (y/2)!^2 = 2^y / sqrt(y Pi/2)
Setting n=y/2, y = 2n
y! sqrt(2Pi) sqrt(y/2) / (y/2)!^2) e^(y/2) = 1
y! / (y/2)!^2 = e^(y/2) / sqrt(y/2)
2^y / sqrt(y Pi/2) = e^(y/2) / sqrt(y/2)
2^2n / sqrt(2n Pi/2) = e^n / sqrt(n)
2^2n / sqrt(n) sqrt(Pi) = e^n / sqrt(n)
2^2n / sqrt(Pi) = e^n
lim n-> oo 2^2n / e^n = sqrt(Pi)
lim n->oo 2^4n / e^2n = Pi
Consider again:
lim n->oo (1+1/x)n! / (n! n^(n/x) (n/x)) = 1
Let x be 1/n, approaching zero from the positive side. That isn't an
integer.
((1+n)n)! / (n! n^(n^2) n^2) = 1
(n^2 + n)! / (n! n^(n^2) n^2) = 1
Ross
Robin Chapman
Shouldn't you be glad that someone does?
Richard Carr
:Date: 8 Aug 2003 15:53:16 -0700
:From: Ross A. Finlayson <r...@tiki-lounge.com>
:Newsgroups: sci.math
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:Hi,
:
:I know (5/4)! / (n! (n/4) n^(n/4)) does not equal one for any finite
:integer n.
You had (5n/4)! on top before - and you then in fact did say they were
equal.
:What I have surmised is that the limit as n diverges of
:that expression is equal to one.
That's not what you said, so it was hardly to be inferred - but anyway the
ratio does not converge to 1. Indeed, if you do mean (5/4)! on the top (by
which one presumes you mean Gamma(9/4) then clearly the limit is 0. On the
other hand, if you really mean (5n/4)! on the top then the expression
diverges - it grows exponentially.
I might as well snip the rest of the discussion to preserve my sanity.
Ross A. Finlayson
lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
is based upon a reformulation of the formula for the Gamma function:
1: Gamma(z) = lim n->oo (n! n^z z!) / (n+z)!
Reverse the equation.
2: lim n->oo (n! n^z z!) / (n+z)! = Gamma(z)
Gamma(z) = (z-1)!.
3: lim n->oo (n! n^z z!) / (n+z)! = (z-1)!
Divide each side by (z-1)!
4: lim n->oo (n! n^z z! ) / ((n+z)! (z-1)!)= 1
Set z = n/x.
5: lim n->oo (n! n^(n/x) (n/x)! ) / ( (n+n/x)! (n/x-1)! = 1
Divide the expression by (n/x-1)! / (n/x-1)!.
6: lim n->oo (n! n^(n/x) (n/x) ) / (n+n/x)! = 1
Invert both sides of the expression.
7: lim n->oo (n+n/x)! / (n! n^(n/x) (n/x) ) = 1
Extract the term n from (n+n/x) to be (1+1/x)n.
8: lim n->oo ((1+1/x)n)! / (n! n^(n/x) (n/x) ) = 1
Then, I set x to various values including 1/n, 1, n, and observe the
terms thus generated in the expression, and relate them to each other
and to other known expressions, for example the Duplication Formula
for Gamma, Stirling's Identity, and lim n->oo ( sqrt(n Pi / 2) n! ) /
( (n/2)!^2 2^n ) = 1
I got a result that says:
lim 2^2n / e^n = sqrt(Pi).
That doesn't seem right. The constant e is around 2.7182.... For
f(n)= 2^2n / e^n, f(2) = 16 / 7.3890... ~= 2.165, f(3)= 64/
20.08554... ~= 3.186..., f(n+1)/f(n) = 4/e = 1.471518..., f(n) =
(4/e)^n.
I got that equation from the equation
8: lim n->oo ((1+1/x)n!) / ( n! n^(n/x) (n/x) ) = 1
I set x equal to one.
9: lim n->oo ((1+1/1)n)! / ( n! n^(n/1) (n/1) ) = 1
Reduce.
10: lim n->oo (2n)! / ( n! n^n n ) = 1
I notice that that identity has expressions found in other identities,
namely the Duplication Formula which has an expression for Gamma(2z),
and Stirling's Identity which has an expression with n! and n^n.
Stirling's equation, though, has the n! and n^n terms on opposite
sides of the divisor bar.
Replace (2n)! with 2n gamma(2n).
11: lim n->oo 2n gamma(2n) / ( n! n^n n) = 1
Divide left side by n/n, multiply both sides by 1/2.
12: lim n-> oo gamma(2n) / (n! n^n) = 1/2
Replace the expression gamma(2n) with the r.h.s. of the Duplication
Formula gamma(2n) = sqrt(2Pi) 2^(2n-1/2) gamma(n) gamma(n+1/2).
13: lim n->oo sqrt(2Pi) 2^(2n-1/2) gamma(n) gamma(n+1/2) / (n! n^n) =
1/2
Replace n! with n gamma(n)
14: lim n->oo sqrt(2Pi) 2^(2n-1/2) gamma(n) gamma(n+1/2) / (n gamma(n)
n^n) = 1/2
Divide expression by gamma(n)/gamma(n)
15: lim n->oo sqrt(2Pi) 2^(2n-1/2) gamma(n+1/2) / (n n^n) = 1/2
Multiply both sides by two.
16: lim n->oo 2 sqrt(2Pi) 2^(2n-1/2) gamma(n+1/2) / (n n^n) = 1
Group twos in the exponential function.
17: lim n->oo sqrt(Pi) 2^(2n+1) gamma(n+1/2) / (n n^n) = 1
The above formula is thus a result of setting x=1 and exchanging
sqrt(Pi) 2^n gamma(n) gamma(n+1/2) for gamma(2n).
About Stirling, in the case where x=1, then there is again:
10: lim n->oo (2n)! / (n! n^n n) = 1
Stirling has that
18: lim n->oo n! e^n / n^n sqrt(n) sqrt(2pi) = 1
This is equivalent to:
19: lim n->oo n^n / ( n! e^n / sqrt(n) sqrt(2Pi) ) = 1
The reformulated Gamma function with x=1 is multiplied by the Stirling
expression and the n^n terms cancel.
20: lim n->oo (2n)! sqrt(n) sqrt(2Pi) / ( n! n! e^n ) = 1
Multiply the expression by 2^2n / 2^2n
21: lim n->oo (2n)! sqrt(n) sqrt(2Pi) 2^2n / (n! n! e^n 2^2n)) = 1
There is that:
22: lim y->oo ( sqrt(y Pi/2) y! ) / (y/2)!^2 2^y = 1
This is that:
23: lim y->oo (1/ sqrt(y Pi/2)) /(y!/ ((y/2)!^2 2^y) = 1
Set y=2n, n = 2/y, and replace n in equation 22 with y/2.
24: lim y/2->oo ( y! sqrt(y/2) sqrt(2Pi) 2^y ) / ( (y/2)!^2 e^(y/2)
2^y) = 1
This is that:
25: lim y/2->oo ( y! / ((y/2)!^2 2^y)) / ( e^(y/2) /
(sqrt(y/2)sqrt(2Pi)2^y) ) = 1
Then, here I want to multiply 23 by 25, to take advantage of the
common term of the binomial coefficient (y! / ((y/2)!^2 2^y)), but
there is the problem that in 24 that y->oo and in 26 that y/2->oo.
26?: lim y->? (1/ sqrt(y Pi/2) ) / ( e^(y/2) / (sqrt(y/2)sqrt(2Pi)2^y)
) = 1
Reorder.
27?: lim y->? (sqrt(y/2) sqrt(2Pi) 2^y) / (sqrt(y Pi/2) e^(y/2)) = 1
Reduce.
28?: lim y->? sqrt(2) 2^y / e^(y/2) = 1
Multiply both sides by 1/sqrt(2).
29?: lim y->? 2^y/ e^(y/2) = 1/sqrt(2)
Square both sides.
30?: lim y->? 2^2y / e^y = 1/2
This is similar to the expression that was derived for Pi for 2^2y/e^y
in my previous post, and neither of those is equal to (4/e)^y, thus
each of the equations with question marks, 26-30, is false, where each
of 1-25 is true.
About Stirling, in the case where x=1, then there is again:
10: lim n->oo (2n)! / (n! n^n n) = 1
Stirling has that
18: lim n->oo n! e^n / n^n sqrt(n) sqrt(2pi) = 1
This is equivalent to
31: lim n->oo n! / (n^n sqrt(n) sqrt(2pi) e^(-n) ) = 1
Multiply 10 by 31.
32: lim n->oo (2n)! n! / ( n! n^n n^n n sqrt(n) sqrt(2pi) e^(-n) ) =
1
Reduce.
33: lim n->oo (2n)! / ( (n^n)^2 n sqrt(n) sqrt(2pi) e^(-n) ) = 1
Group n terms.
34: lim n->oo (2n)! / ( n^(2n+3/2) sqrt(2pi) e^(-n) ) = 1
Multiply by e^n/e^n.
35: lim n->oo ( (2n)! e^n ) / ( n^(2n+3/2) sqrt(2pi) ) = 1
Replace (2n)! with 2n gamma(2n).
36: lim n->oo ( 2n gamma(2n) e^n) / (n^(2n+3/2) sqrt(2pi) ) = 1
Reduce.
37: lim n->oo sqrt(2) gamma(2n) e^n / (n^(2n+1/2) sqrt(pi) ) = 1
Replace gamma(2n) by duplication formula.
38: lim n->oo (sqrt(2) sqrt(2Pi) 2^(2n-1/2) gamma(n) gamma(n+1/2)
e^n) / (n^(2n+1/2) sqrt(pi) ) = 1
Reduce.
39: lim n->oo ( 2^(2n+1/2) gamma(n) gamma(n+1/2) e^n ) / (n^(2n+1/2)
= 1
I want to remove the expression n^2n from formula 35. Get an
expression for n^n.
19: lim n->oo n^n / ( n! e^n / sqrt(n) sqrt(2Pi) ) = 1
Square both sides.
40: lim n->oo (n^2n) / (n!^2 e^2n / n 2pi) = 1
Get an expression for n^2n in 35.
41: lim n->oo ( (2n)! e^n ) / ( n^(2n) n^(3/2) sqrt(2pi) ) = 1
Multiply 40 by 41.
42: lim n->oo ( (2n)! e^n n 2pi ) / ( n!^2 e^2n n^(3/2) sqrt(2Pi) ) =
1
Reduce.
43: lim n->oo ( (2n)! sqrt(2pi) ) / ( n!^2 e^n sqrt(n) ) = 1
Well, that's an expression with variables n!, (2n)!, e^n, and sqrt(n),
where Stirling's formula has as variables n^n, n!, e^n, and sqrt(n).
Reorder 19:
44: lim n->oo n! e^n / ( n^n sqrt(n) sqrt(2pi) ) = 1
Multiply 43 by 44.
45: lim n->oo ((2n)! sqrt(2pi) n! e^n ) / ( n^n n sqrt(2pi) n!^2 e^n)
= 1
Reduce.
10: lim n->oo (2n)! / (n! n^n n) = 1
So where I am tryng to figure out a resolution is in the variable
replacement for the limit variable. When I replaced n with n/2 it
resulted in an untrue expression. Otherwise, the expression lim
((1+1/x)n)! / (n! n^(n/x) (n/x) ), a restatement of Euler's formula
for Gamma(z) for z=f(n)=n/C, seems not yet to contradict Stirling's
formula, or lim n->oo (sqrt(n pi/2) n! ) / ( (n/2)!^2 2^n) = 1,
another known result.
Ross
Robin Chapman
> The expression
>
> lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
>
> is based upon a reformulation of the formula for the Gamma function:
>
The equation
lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
is false.
Ross A. Finlayson
> Ross A. Finlayson wrote:
>
> > The expression
> >
> > lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
> >
> > is based upon a reformulation of the formula for the Gamma function:
> >
>
> The equation
>
> lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
>
> is false.
Hi Robin,
That's concise but not very helpful. It might perhaps also be wrong.
Please explain your reasoning. Did you derive a simple or complex
re-expression that is definitely invalid? If so, what is it? Can you
show an issue with setting z of the Gamma function to n/x? If not,
why not? Why else would it be false?
lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
The above expression is derived from the formula for Gamma:
Gamma(z) = lim n->oo n! n^z (z-1)! / (n+z)!
I got that form from the "Handbook of Mathematical Functions", from
Chapter 6, "Gamma Function and Related Functions", equation 6.1.2,
"Euler's Formula".
Gamma(z) = lim n->oo n! n^z / ( z (z+1) ... (z+n) )
There ( z (z+1) ... (z+n) ) is (n+z)! / (z-1)!.
Then I say that z is a function of n, z=f(n)= n/x for some constant,
after grouping the variables, to then take the limit. Then I express
the Gamma function term in terms of factorial and reduce the
expression to get the reformulated form of the Gamma function.
lim n->oo n! n^z (z-1)! / (n+z)! Gamma(z) = 1
lim n->oo n! n^(n/x) (n/x-1)! / (n+n/x)! Gamma(n/x) = 1
lim n->oo n! n^(n/x) (n/x-1)! / (n+n/x)! (n/x-1)! = 1
lim n->oo n! n^(n/x) / (n+n/x)! = 1
Hey, wait a second, where did the (n/x) term go? What I have done
wrong is that Euler's formula has z(z+1)...(z+n) in the denominator
where I had treated it as if it were (z+1)...(z+n), it is
(z+n)!/(z-1)!, not (z+n)!/z!. Daang. Now all the expressions from it
would have to be re-expressed.
lim n->oo n! n^(n/x) / (n+n/x)! = 1
lim n->oo n! n^(n/x) / ((1+1/x)n)! = lim n->oo ((1+1/x)n)! / ( n!
n^(n/x) ) = 1
Hmm..., I wonder what that will do there in the equations. I should
run my program here again for x=4 without multiplying by the term n/4.
f(n) = (5n/4)! / ( n! n^(n/4) )
f(4)= 1.250000000000000000000000000000000000000000000000000000000000
f(8)= 1.406250000000000000000000000000000000000000000000000000000000
f(16)= 1.774291992187500000000000000000000000000000000000000000000000
f(32)= 2.820158351096324622631072998046875000000000000000000000000000
Hmm, that appears to be increasing. When n is 32, 4 times as much as
8, the result of the function has more than doubled in value.
f(64)= 7.119200781241405312510509900259250560118455541935844621548312
f(128)= 45.349981936307266706007512178052467292320196998584854098612225
It looks like a strictly increasing function. For n=128, f(n) is more
than six times what it is for 64, 64 less than 128.
f(256)= 1839.856929950726447697875009389483403683860369482341376671249969
f(512)= 3027997.490208837356851894258497558994051064679103269880341629291473
f(1024)= 8201196562466.798734381214256780060581141323863941085862789135499037950541
f(2048)= 60160306246145229077267472.112651732856730304404369258861896415395729019235037843169745
f(4096)= 3237206260961006479922074108831302555499967249221180.562890499562949003741860179637565800273283791373134934862938
f(8192)= 9373210871465678040230103600246972264533199341809341249515714970345355063230247193723406114752065381522.113748619190102985784311573847442217381542452859441133009973
f(16384)= 78582002912222154235504141709415019031841788894465313064448763337860054971651843597069947678328579709460993337519009003558708956598376220881881019179371279488078302553461394871166075896978704514896165404312.142027518566139683064405837716511242816862920649253013922253
f(32768)= 552321366464957625938985514392986199775682291235255908748958913345580155692139627295604446455720859697610904757423777241348987465861093227359335990263922987333176650273866180133701448571725999622253254963171441090122304519610327251281419620743093537133906132809353867651044379203728563460657535675036146882823482265727467268339928689879695290706507232503780627049574239132320083233789338724
290980725923189180744.295518643267560557460413738371659817727468272934827775303839
The integer portion of f(32768) has 413 digits.
f(4)/f(8)=0.888888888888888888888888888888888888888888888...
f(8)/f(16)=0.792569659442724458204334365325077399380804953...
f(16)/f(32)=0.629146229146229146229146229146229146229146229...
f(32)/f(64)=0.396134122038985620386617272321950791538559490...
f(64)/f(128)=0.156983541718718127975159211852108972389891667...
f(128)/f(256)=0.024648645880047745368958988798680123931357667...
f(256)/f(512)=0.000607615077588401079193415935020829919401559...
f(512)/f(1024)=0.000000369214110056406263898772465900333544234...
f(1024)/f(2048)=0.000000000000136322387205139772849761246892001...
f(2048)/f(4096)=0.000000000000000000000000018584020107599157222...
f(4096)/f(8192)=0.000000000000000000000000000000000000000000000...
f(8192)/f(16384)=0.000000000000000000000000000000000000000000000...
f(16384)/f(32768)=0.000000000000000000000000000000000000000000000...
That would be expected to go to one if f(n) was converging by the time
n was 32768.
For f(n) = (17n/16)! / (n! n^(n/16)):
f(16)/f(32)=0.242424242424242424242424242424242424242424242...
f(32)/f(64)=0.058783008036739380022962112514351320321469575...
f(64)/f(128)=0.003455838971487670725557738492169367867400451...
f(128)/f(256)=0.000011943509035745408927553693457815160966844...
f(256)/f(512)=0.000000000142651505273914914619002408821749099...
f(512)/f(1024)=0.000000000000000000020349744200911823781329770...
f(1024)/f(2048)=0.000000000000000000000000000000000000000414115...
f(2048)/f(4096)=0.000000000000000000000000000000000000000000000...
f(4096)/f(8192)=0.000000000000000000000000000000000000000000000...
f(8192)/f(16384)=0.000000000000000000000000000000000000000000000...
f(16384)/f(32768)=0.000000000000000000000000000000000000000000000...
The function appears to diverge.
For f(n)= (2n)! / ( n! n^n):
f(4)/f(8)=0.212173012173012173012173012173012173012173012...
f(8)/f(16)=0.045367698030203290107044411855453852681544638...
f(16)/f(32)=0.002066267699763389104078467561421227730755355...
f(32)/f(64)=0.000004277804990452485511398532458199199293497...
f(64)/f(128)=0.000000000018317492755107689949374501812414088...
f(128)/f(256)=0.000000000000000000000335694409004190925579058...
f(256)/f(512)=0.000000000000000000000000000000000000000000112...
f(512)/f(1024)=0.000000000000000000000000000000000000000000000...
f(1024)/f(2048)=0.000000000000000000000000000000000000000000000...
f(2048)/f(4096)=0.000000000000000000000000000000000000000000000...
f(4096)/f(8192)=0.000000000000000000000000000000000000000000000...
That appears to diverge more quickly than for x=4 or x=16.
I say "appears to diverge" instead of the perhaps more credulous
"obviously diverges" because the behavior for small finite values of n
does not reconcile with the algebraic transformation of the Gamma
function formula.
What I need here is some explanation of why I can't validly replace z
with n/x. Otherwise I need to prove that I can replace z with n/x,
besides the fact that z is defined on all positive integers and R\{Z-
U {0}}, Gamma(z) is defined for all reals except 0, -1, -2, ...., and
n is an element of N. The function is defined for variable x on the
non-negative reals. Otherwise the re-expression of the function would
belie the behavior of the function for pretty much any finite value of
n calculable on a computer.
Ross
Robin Chapman
> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> wrote in message
> news:<bh4tkj$q5u$1...@news8.svr.pol.co.uk>...
>> Ross A. Finlayson wrote:
>>
>> > The expression
>> >
>> > lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
>> >
>> > is based upon a reformulation of the formula for the Gamma function:
>> >
>>
>> The equation
>>
>> lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
>>
>> is false.
>
> Hi Robin,
>
> That's concise but not very helpful. It might perhaps also be wrong.
What is "it"? Your statement or my statement?
There are no "perhaps"s about the truth-values of either.
> Please explain your reasoning.
Stirling's formula.
<snip>
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> wrote in message
> news:<bh4tkj$q5u$1...@news8.svr.pol.co.uk>...
> > Ross A. Finlayson wrote:
> >
> > > The expression
> > >
> > > lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
> > >
> > > is based upon a reformulation of the formula for the Gamma function:
> > >
> >
> > The equation
> >
> > lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
> >
> > is false.
>
> Hi Robin,
>
> That's concise but not very helpful. It might perhaps also be wrong.
If the choice of bets is between Robin being correct and Ross being
correct here, the wise bettor will bet on Robin.
>
> Please explain your reasoning. Did you derive a simple or complex
> re-expression that is definitely invalid? If so, what is it? Can you
> show an issue with setting z of the Gamma function to n/x? If not,
> why not? Why else would it be false?
You miss the point, Ross.
When one makes an assertion, such as your assertion of the value of
a certain limit, the burden of proof is on the person making that
assertion, and never on the ones expressing doubt.
The derivation below involves enough hand waving tbetween one step
to the next that it is quitelikely to contain errors. It is your
reposibility to fill in all those gaps if you want acceptance.
Ross A. Finlayson
> Ross A. Finlayson wrote:
>
> > Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> wrote in message
> > news:<bh4tkj$q5u$1...@news8.svr.pol.co.uk>...
> >> Ross A. Finlayson wrote:
> >>
> >> > The expression
> >> >
> >> > lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
> >> >
> >> > is based upon a reformulation of the formula for the Gamma function:
> >> >
> >>
> >> The equation
> >>
> >> lim n->oo ((1+1/x)n)! / ( n! n^(n/x) (n/x) ) = 1
> >>
> >> is false.
> >
> > Hi Robin,
> >
> > That's concise but not very helpful. It might perhaps also be wrong.
>
> What is "it"? Your statement or my statement?
> There are no "perhaps"s about the truth-values of either.
>
> > Please explain your reasoning.
>
> Stirling's formula.
>
> <snip>
Hi,
"It" was that expression, I showed how I had mis-derived it.
You say Stirling's formula explains your reasoning. That "statement"
implies to me that some impossible equation would be derivable from
assuming both Stirling and the reformulated Gamma function.
The reformulated Gamma function is:
lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1
For a given value of x that equation has some common terms with
Stirling's formula. That gives us a foothold to compare the formulae.
Stirling's formula:
1: lim n->oo ( n! e^n ) / ( n^n sqrt(n) sqrt(2P)i ) = 1
The expression f(x) = lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1 for
x=1:
2: lim n->oo (2n!) / ( n! n^n) = 1
The common terms of those two expressions are n! and n^n. Put n!
alone in the numerator for 1 and 2:
3: lim n->oo n! / ( n^n e^(-n) sqrt(n) sqrt (2Pi) ) = 1
4: lim n->oo n! / ((2n)! n^(-n)) = 1
Divide 3 by 4:
5: lim n->oo ( (2n)! e^n ) / ( n^(2n) sqrt(n) sqrt(2Pi) ) = 1
Put n^n alone in the numerator for 1 and 2:
6: lim n->oo n^n / (( n! e^n)/(sqrt(n) sqrt(2Pi))
7: lim n->oo n^n / ( (2n)! / n! ) = 1
Divide 6 by 7:
8: lim n->oo ( (2n)! sqrt(n) sqrt(2Pi) ) / ( n! n! e^n ) = 1
So we have two equations, 5 and 8, each is the result of reducing
common expressions from a product of the two equations 1 and 2.
That's about how the l.h.s of each equation is equal to the same
value, one.
How about gathering the (2n)! terms of 5 and 8 and attempting to
divide those? That's probably not very informative.
9: lim n->oo (2n)! / ( n^(2n) sqrt(n) sqrt(2Pi) e^(-n) ) = 1
10: lim n->oo (2n)! / (( n! n! e^n )/(sqrt(n) sqrt(2Pi))) = 1
Divide 9 by 10.
11: lim n->oo n! n! e^n e^n / ( n^(2n) n 2Pi) = 1
That is equivalent to:
12: lim n->oo n! ^2 e^2n / (n^(2n) n 2Pi) = 1
Take the square root of both sides.
1: lim n->oo n! e^n / (n^n sqrt(n) sqrt(2Pi)) = 1
Well, that's just the same thing as Stirling's formula. I think it
was derived from Stirling's formula and f(x) for x=1 without simply
multiplying Stirling by f(x)/f(x), except in a way it is as f(x)/f(x)
= 1 / 1 = 1.
Please explain your reasoning. I'm not trying to be obtuse here, I'm
trying to be not obtuse here, if it's invalid I want to know and why.
Thanks,
Ross
Ross A. Finlayson
Here'a another way to look at
lim (2n)! / ( n! n^n ) = 1
Consider (x^2-1) / (x^2).
In the limit, lim x->oo (x^2-1) / (x^2) = 1, just like lim x->oo
(x+1)/x = lim x->oo (x-1)/x=1. Partially that's because (x^2-1)/x^2 =
(x+1)(x-1)/x^2 = (x+1)/x * (x-1)/x.
Now, for x^x, what you do is put that in the denominator and put
(x+1)(x+2)...(x+x/2) in the numerator, and (x-1)(x-2)...(x-x/2) in the
numerator. That puts x many terms in the numerator and x many terms
in the denominator.
(x-x/2)(x-x/2+1)... (x-2)(x-1) (x+1)(x+2)...(x+x/2-1)(x+x/2)
------------------------------------------------------------
x^x
Now it should be pretty obvious that the limit of that expression
evaluates to one. There is (x+n)(x-n) in the numerator and x^2 in the
denominator for n=1 to x/2.
Now what that is there is ( ( (x-1)! / (x-x/2-1)!) * ( (x+x/2)!/x!) )
/ x^x.
lim x->oo ((x+x/2)! (x-1)!) / ( (x-x/2-1)! x! x^x ) = 1
lim x->oo (x+x/2)! / ( x-x/2-1)! x x^x ) = 1
lim x->oo (3x/2)! / ((x/2 -1)! x x^x) = 1
Hmm, I have that lim x->oo (3x/2)! / (x! x^(x/2) ) = 1.
That would lead me to think that:
lim x->oo ( ((x/2)-1)! x x^x ) / ( x! x^(x/2) ) = 1
lim x->oo ( ((x/2)-1)! x x^(x/2) ) / x! = 1
lim x->oo ( (x/2)! 2 x^(x/2) / x! = 1
lim x->oo x! / ( 2 (x/2)! x^(x/2) ) = 1
Ross
Robin Chapman
> lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1
False.
Robin Chapman
> lim (2n)! / ( n! n^n ) = 1
False.
Starblade Darksquall
> From: David C. Ullrich <ull...@math.okstate.edu>
> Newsgroups: sci.math
> Subject: Re: Factorial/Exponential Identity, Infinity
>
> >>On Sun, 27 Jul 2003, Ross A. Finlayson wrote:
>
> >>> lim n->oo n! / (2 (n/2)! 2^n) = 1
>
> >>large. Thus by my reckoning lim n->oo n! / (2 (n/2)! 2^n) = oo
>
> >Um, of course the formula he gives is wrong, but you haven't shown
> >that. (_If_ I have Stirling's formula right then
> > sqrt(n) (2n)!/(4^n (n!)^2)
>
> >has a finite limit, although your "reckoning" would show it also
> >tends to infinity.)
>
> Sterling's approximation: n! = (n/e)^n sqr(2n.pi)
> 2n! 2^2n = 2(n/e)^n 2^2n sqr(2n.pi)
> (2n)! = (2n/e)^2n sqr(4n.pi) = (n/e)^2n 2^2n sqr(4n.pi)
>
> (2n)! / (2n! 2^2n) = (sqr 1/2)(n/e)^n
> I reckon so.
>
> ----
Does this have anything to do with the fact that if you multiply a
bunch of numbers together starting with just above 0 to a number, and
raise the resulting number to the power of the interval, you get about
(x/e)^x?
Imagine a kind of geometric differentiation function, g(x) = lim
(dx->0) (f(x+dx)/f(x))^(1/dx). Taking the natural log of both sides
gives us ln(g(x)) = lim (dx->0) (ln(f(x+dx))-ln(f(x)))/dx which is
simply d/dx ln(f(x)), or f'(x)/f(x) making g(x)=exp(f'(x)/f(x)), and
the reverse of which is f(x) = exp((0-x)integral(ln(g(t))dt))... or
multiply the last bit by a number c, preferably nonzero and probably
better as a positive number... but better left the way it was if we're
starting multiplying from 1.
Now plug g(x) = x into the function, and you get f(x) = exp(xln(x)-x),
or f(x) = x^x/e^x, or (x/e)^x.
That's similar to multiplying all the integers from 1 to x, and
getting x!
Is this what is meant by saying x! is similar to x^x?
(...Starblade Riven Darksquall...)
Virgil
> Hi,
>
> "It" was that expression, I showed how I had mis-derived it.
>
> You say Stirling's formula explains your reasoning. That "statement"
> implies to me that some impossible equation would be derivable from
> assuming both Stirling and the reformulated Gamma function.
The phrase "to me" explains it.
Virgil
> Hi,
>
> Here'a another way to look at
>
> lim (2n)! / ( n! n^n ) = 1
Robin Chapman's suggestion that you use Stirlings formula to
approximate factorials, will readily allow you to see that the
indicated limiting value in the above equation is wrong.
Ross A. Finlayson
> (x-x/2)(x-x/2+1)... (x-2)(x-1) (x+1)(x+2)...(x+x/2-1)(x+x/2)
> ------------------------------------------------------------
> x^x
>
> Now it should be pretty obvious that the limit of that expression
> evaluates to one. There is (x+n)(x-n) in the numerator and x^2 in the
> denominator for n=1 to x/2.
>
To get an expression that equals one there what is necessary is
II ( (x+n) (x-n) + n^2 )
That actually equals II x^2. The expression (x+n)(x-n) is greater
than (x+n+1)(x-n-1) for positive x and n.
Ross
Ross A. Finlayson
> Ross A. Finlayson wrote:
>
> > lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1
>
> False.
Stirling's "approximation" for n! is actually a transformation of a
formula that's only true in the limit. That's why some use Gosper's
approximation for n!, it's more accurate for some values of n, I have
also seen mentioned Burnside's formula, and a Lanczos approximation,
and Godfrey's improved Lanczos approximation.
Here are some links with information about factorial approximations:
http://mathforum.org/library/drmath/view/60692.html
http://www.google.com/search?q=factorial+approximation
http://www.google.com/search?q=%22Gamma+function%22+approximation
Stirling's identity is:
lim n->oo (n! e^n) / (n^n sqrt(n 2Pi)) = 1
An approximation is:
n! ~= n^n sqrt(n2Pi) / e^(n-(1/12)n), n >> 1
"The value of n factorial is approximately equal to n to the n'th
power times the square root of n times 2Pi divided by e to the n'th
power, for n much greater than one."
What that says is that for given large finite values of n, that the
result of the approximation of n! is "near" n!. The benefit of using
the other approximations is that they are "nearer" n!.
This thread isn't concerned about four trillion factorial or eight
hundred and thirty seven thousand factorial, it's about the result of
expressions as the variable diverges to infinity, n >> 4000000000000.
Particularly, at this point it's concerned with setting the value of
one variable of the gamma function to an expression of the other
variable which diverges.
lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1
Here I'll call that the Reformulated Gamma expression, RG(x). RG(1)
is:
lim n->oo ((2n)!) / (n! n^n) = 1
If that is not true, then it it is perhaps possible to validly compare
it with known results of the value of n! or the other terms in the
variable expresion and to get a known untrue result. That would
definitely help establish the truth value of that identity.
As we have previously discussed in this thread, there is a known
result for an expression with n!, (n/2)!, 2^n, and sqrt(n).
Stirling's formula has terms of n!, n^n, e^n, and sqrt(n). The
Duplication Formula for the Gamma function gives an expression for
Gamma(2z) in terms of 2^(2z-1/2), Gamma(z), and Gamma(z+1/2).
Stirling's formula has n! and n^n on opposite sides of the divisor,
where RG(1) has them on the same side of the divisor.
Stirling's formula:
Divide 3 by 4:
Divide 6 by 7:
Divide 9 by 10.
That is equivalent to:
You might notice that the preceding is exactly what I wrote before.
That's because it has the same, true meaning.
Why do I say lim n->oo ( (1+1/x)n! ) / ( n! n(n/x) ) = 1? That's not
asking why it is so, it is asking why I say it is so. I've explained
why I think it is so. I think you should address that.
Do you have some reason why you say it is not so? You said "False"
and "Stirling's formula". That's not an explanation. I use "False"
sometimes, normally in protracted conversations where an explanation
is already evident, it's fun as hell, but it's not an explanation,
just the assertion of the existence of an explanation. It is so that
2 n! =/= (2n)!, for n =/=1.
So anyways, what's your explanation? That is to say, for the benefit
of people who don't know what your explanation is, present it.
I think it would be great if RG(1) was true. I think it is, for the
reasons that I have provided, namely that I think it is a valid use of
Euler's Gamma function and that it appears as shown here to not
contradict Stirling's formula. While that may be true, I'm not going
to ignore valid objections.
So, if you say it's false, then explain for our benefit why. I don't
mind if you reiterate your explanation and answer the questions of
others about it, I think it would be good.
Thanks,
Ross
Ross A. Finlayson
Set x=1/n:
lim n->oo ((1+n)n)! / (n! n^(n^2))
lim n->oo (n^2+n)! / (n! n^(n^2) )
Set x=n:
lim n->oo ((1+1/n)n)! / (n! n^(1) )
lim n->oo (n+1)! / (n! n)
lim n->oo (n+1) / (n)
The result is equal to one.
Set x=n/2:
lim ((1+2/n)n!) / (n! n^(2))
lim (n+2)! / (n! n^2)
lim (n+1)(n+2) / n^2
lim (n^2 + 3n + 2) / n^2
lim (n^2/n^2) + (3n/n^2) + (2/x^2)
That appears to start diverging, and also appears to be equal to one.
Set x = 2n:
lim ((1+ 1/2n)n)! / (n! n^(1/2))
lim (n+1/2)! / (n! n^(1/2))
lim Gamma(n+3/2) / (Gamma(n+1) n^(1/2))
That appears to diverge.
Set x=1:
lim (2n)! / (n! n^n)
lim (n+1)(n+2)...(2n) / ( n^n)
lim (n^n/n^n) + (an^(n-1)/n^n) + (bn^(n-2)/n^n) + ... + n!/n^n
That might also appear to evaluate to one as all but the first term
which evaluates to one evaluate to equal zero, depending on what a, b,
... equal. The value of a might be sum n, which is greater than n,
with b being the value of the sum of the product of each pair of N, I
wonder if that is n^2. If that were so on and so forth for the other
terms then the value of the expression would be n. If a, b, ... were
each less than n, n^2, etcetera, that would not be so.
lim (n^n/n^n) + (an^(n-1)/n^n) + (bn^(n-2)/n^n) + ... + n! / n^n
If a is sum n, then a is (n+1)(n/2) = n^2/2 + n/2, and an^(n-1) is
(n^(n+1) + n^n)/2, and that divided by n^n is n/2+1/2.
The limit of the n'th term, n! / n^n, is zero.
Let's see, consider a finite case of (x+1)(x+2)...(x+4).
(x^2 + 3x + 2)(x+3)(x+4)
(x^3 + 6x^2+11x + 6)(x+4)
(x^4 + 10x^3 + 35x^2 + 50x + 24)
Yep, it appears a = sum n for n= 1 to 4, 5*2, (n+1)/(n/2). The value
of b is the sum of each product of each pair of {1, 2, 3, 4), the
value of c would be the sum of each product of each triple of 1, 2, 3,
4, etcetera, d is the product 1, 2, 3, 4. Normally in the polynomial
the coefficient a is on the first term, this starts with a on the
second.
1*2 + 1*3 + 1*4 +2*3 + 2*4 + 3*4 = 35
1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50
1*2*3*4 = 24
Let's see, 35 = 1*(2+3+4) + 2*(3+4) + 3*4, the sum for i from 1 to n-1
of the product of i and the sum of n from i+1 to n.
The third term c is 1*(2*3 + 2*4 + 3*4) + 2*(3*4) ..., a slightly more
convoluted expression where more terms would be necessary to display
it by example.
So, it appears that the expression diverges. Now I want to figure out
what it diverges to, symbolically. Then you could divide the
divergent expression by that and get a convergent expression.
Set x=1/2:
lim ((1+2)n)! / (n! n^(2n))
lim (3n)! / (n! n^(2n))
More of the same.
lim (n+1)(n+2)...(3n) / (n^(2n)
lim n^2n/n^2n + a n^(2n-1) / n^(2n) + b n^(2n-2) / n^(2n) + ...
If a is (2n+1)n = 2n^n + n, then the second term would be (n^(2n+1)+
n^2n) / n^2n = n+1, where in the above it was (n+1)/2.
What are the closed forms for the coefficients of the polynomial of
(x+1)(x+2)...(x+n)? They are {1, sum n, ..., n!), what is the form of
each of the intervening coefficients?
Going back, set x=1/n:
lim n->oo ((1+n)n)! / (n! n^(n^2))
lim n->oo (n^2+n)! / (n! n^(n^2) )
lim (n+1)(n+2)...(n^2+n) / n^(n^2)
lim (n^(n^2))/(n^(n^2)) + ((n^2+n+1)(n^2+n)/2) (n^(n^2-1)) / (n^(n^2))
+ ...
lim 1 + (n^4 + 2n^3 + n^2 + n) (n^(n^2-1)) / 2(n^(n^2)) + ...
lim 1 + (n^(n^2+3) + 2n^(n^2+2) + n^(n^2+1) + n^(n^2))/2(n^(n^2)) +
...
lim 1 + (n^3)/2 + n^2 + n/2 + n/2 + 1/2 + ...
So there appear to be five or seven cases, where x is a constant
f(x)=c, x is a function of a constant times n, f(x)=cn, for c=1,
0<c<1, and c>1, and x is a function of a constant times the reciprocal
of n, f(x)=c/n, for c=1, 0<c<1, and c>1.
For f(x)= lim n->oo ((1+1/x)n)! / (n! n^(n/x)) = 1, that is true when
x=cn, 0<c<=1.
Bollocks, indeed.
Ross
Robin Chapman
> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> wrote in message
> news:<bh7h6b$8g7a7$1...@athena.ex.ac.uk>...
>> Ross A. Finlayson wrote:
>>
>> > lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1
>>
>> False.
> lim n->oo (n! e^n) / (n^n sqrt(n 2Pi)) = 1
So, use Stirling's formula to obtain the asymptotics
(at least the dominant term) of ((1+1/x)n)!/(n! n^(n/x)).
Richard Carr
:
:That appears to start diverging, and also appears to be equal to one.
:
and then later.
:
:Bollocks, indeed.
:
:Ross
:
Speaks for itself.
Ross A. Finlayson
(n+1)(n/2), and basically for each an expression of the sum of the
product of each triple, quadruple, quintuple, etcetera, of {1, 2, 3,
4, ..., n}, which would be the same as for {0, 1, 2, ..., n}.
HMF: "An infinite series whose general term is a rational function of
the index may always be reduced to a finite series of psi and
polygamma functions."
Let's see, f(x) = II_i=1^n (n+x)
I go to MathWorld and read about polynomial.
http://mathworld.wolfram.com/Polynomial.html
The function is a univariate polynomial with constant coefficients.
There are n many monomial summands. The order or degree of f is n.
"Exchanging the coefficients of a univariate polynomial end-to-end
produces a polynomial whose roots are the reciprocals of the 1/x_i of
the original roots x_i."
The roots x_i of f are -1, -2, -3, ..., their reciprocals 1/x_i are
-1, -1/2, -1/3, ....
Consider Vieta's Formulas.
http://mathworld.wolfram.com/VietasFormulas.html
I noticed the s1 = r1 + r2 + r3 + r4, it is similar to the 1 + 2 + 3 +
4, and so on for the following values of s.
Consider Stirling Numbers of the First Kind.
The Stirling Number generating functions are defined in terms of a
falling factorial, where f is defined in terms of a rising factorial,
almost. The rising factorial or Pochhammer is x(x+1)(x+2)...(x+n-1),
f is Pochhammer(x,n) * 2.
The function f is easily defined as a falling factorial
(2n)(2n-1)...(n+1), that is (2n)_n.
What is to be done here is to determine the symbolic form of the
coefficients of the n_1 monomials of the polynomial,
(n+1)(x+2)...(n+n), to sum them together to get another polynomial.
The first coefficient is 1, the second (n+1)(n/2), and the last n!.
There are thus the n-2 other coefficients that require determination.
The coefficients form another polynomial with each i'th coefficient
multiplied by n^(n+1-i). Their sum is a univariate polynomial of n.
I'm trying to figure out what its degree would be, as the number of
monomials is a function of the variable, and the coefficients are each
a function of the variable. Call the polynomial g(n).
This polynomial is for the case of RG(1), the expression
lim n->oo (2n!) / (n! n^n)
is under consideration.
lim n->oo (2n)! / (n! n^n g(n)) = 1
Basically g(n) = (n+1)(n+2)...(n+n) / n^n = RG(1).
In that sense g(n) for RG(x) is RG(x), and when x=cn, 0<c<=1,
RG(cn)=1.
Heh, the limit of any expression divided by its reciprocal is equal to
one.
This leads to saying "well, big deal" and "so what, 97x! x^2.1 / 97x!
x^2.1 = 1, big deal". The idea though is that there are some special
characteristics of lim n->oo ((1+1/x)n!)/(n! n^(n/x)), it equals one
for certain values of x without explicitly multiplying it by its
reciprocal. Also, it should be possible to determine forms for g(n).
Ross
Virgil
> lim n->oo ((1+1/x)n!) / (n! n^(n/x))
In this expression x is a free variable and n is a bound (dummy)
variable. Thus this expression represents a function of x defined
whichever x's the limit exists for, if any.
>
> Set x=1/n:
>
> lim n->oo ((1+n)n)! / (n! n^(n^2))
This is no longer a functional expression because it does not
contain a free variable.
Thus its meaning and the meaning of your original functional
expression are unrelated.
The remainder of this derivation is useless , so [snip].
Virgil
> I want
[snipped the bit that didn' make sense]
Ross A. Finlayson
denominator.
lim n->oo ((1+1/n^2)n! / ( n! n^(1/n) )
lim (n + 1/n)! / ( n! n^(1/n))
That appears to converge to one.
It looks like that for any expression of x that is greater than n that
RG(x) is true.
How about x= -1?
lim ((1-1)n)! / (n! n^(-n) )
lim 0! / (n! n^-n)
lim n^n / n!
That diverges.
Then, for x = -n:
lim ((1-1/n)n)! / (n! n^(-1))
lim (n-1)! n / n!
RG(-n) is true.
For x = -1/n:
lim ((1-n)n)! / (n! n^(-n^2))
lim (n-n^2)! n^(n^2) / n!
The Gamma function is not defined for non-positive integers.
Going back to (n+1)(n+2)...(n+n), the numerator of the fraction of
RG(1) ((2n)! 1/n!) / n^n, it has n+1 monomial summands, not n many.
There are n+1 terms, from n^n to n^0.
What is the name for the characteristic of a univariate polynomial
being that it has n+1 terms for order n? I guess any univariate
polynomial can have n+1 terms, with having a coefficient of zero for
the terms that are not otherwise in the polynomial, e.g. x^4-x^2-1 =
x^4 +0x^3 -x^2 +0x -1. Anyways I wonder what is the characteristic of
a univariate polynomial where each of the n+1 coefficients is nonzero.
As the coefficients are functions of the variable, then each
coefficient a_i is a polynomial, I want to figure out what the degree
of each is, d(a_i). The i'th term has degree n+1-i. Then a_i
n^(n+1-i) has a degree of d(a_i) + n+1-i. Basically I want to figure
out how much more than n is the highest degree of the polynomial, to
get a grip on what the degree of a reciprocal would be.
There are n+1 many summands of the numerator, the degree of the first
is n, as it is n^n. The degree of the second is that of (n+1)(n/2)
n^(n-1) or (n^2 +n)/2 n^(n-1) or (1/2) n^(n+1) + (1/2) n^n, the degree
of that is n+1.
So now for the first two terms of:
(n+1)(n+2)...(n+n) / n^n
There is:
n^n + (1/2)n^(n+1) + (1/2)n^n + ... / n^n
or:
(3/2) + (1/2)n + ...
There is a convenient expression for the second coefficient sum n,
that being (n+1)(n/2). What would be good here would be similar
convenient expressions for the i'th coefficient. The n+1'th
coefficient is n!, 1*2*3*...*n.
Would the coefficient a_2 equal (n/2)(n+1)^2? It does for the case
when n=4, but not n=2. Here n is considered to be composite. Does it
work for n=6? The coefficient a_2 is the sum of the product of each
pair of {1, 2, ..., n}.
1*2 + 1*3 + 1*4 + 1*5 + 1*6 + 2*3 + 2*4 + 2*5 + 2*6 + 3*4 + 3*5 + 3*6
+ 4*5 + 4*6 + 5*6
(5+4+3+2+1) * 6 + (4+3+2+1) * 5 + (3+2+1) * 4 + (2+1) * 3 + 1*2
(6+5+4+3+2) * 1 + (6+5+4+3) * 2 + (6+5+4) * 3 + (6+5) * 4 + 6*5
15 * 6 + 10 * 5 + 6 * 4 + 3 * 3 + 2
90 + 50 + 24 + 9 + 2
175
Is that equal to (6/2)(6+1)^2 = 3*49? No. What I need here is a
function in terms of n that has a result the sum of the product of
each pair of postive integers less than or equal to n.
f(2)= 1*2 = 2
f(3)= 1*2 + 1*3 + 2*3 = 11
f(4)= 50
f(5)= 74
f(6)= 175
It appears that f(n) is a function of n and f(n-1). f(n) = f(n-1) +
(sum i<n) * n. f(1)=1. For example, f(7) = 175 + 21*7 = 175 + 147 =
322. Maybe there is a sequence with these values that has a form, (1,
2, 11, 50, 74, 175, 322), that is not a recursive function.
http://www.research.att.com/~njas/sequences/
No match is found.
Then I would need forms to find the 3'rd, 4'th, 5'th, etcetera
coefficients in terms of n. These are the sums of the product of each
triple, quadruple, quintuple, etctera. Then I want to multiply each
coefficient a_i by the term x^(x+1-i) and sum those to divide the
whole thing by n^n to get (2n)!/ (n! n^n).
So (2n)!/n! is the sum of the product of each pair, triple, etc., of
n. There are simple functions describing how many subsets of size i
that there are of a set, it's the sum of the numbers one less than the
size of the set for i=2.
Summing the elements of an even sized set {1, 2, ..., n} is simple,
(n+1)(n/2). So there are (n+1-1)(n-1)/2 pairs of a set of size n.
For example, there are 6 pairs in {1, 2, 3, 4}, and 15 in {1, 2, 3, 4,
5, 6}. The idea then is to get the product of each pair, and sum
those products. The set {1, 2, ..., n} would have (n^2-n)/2 pairs.
For n=6, here are five pairs with one, their products are 2, 3, 4, 5,
6. There are four other pairs with two, with products 6, 8, 10, 12,
then three other pairs with three, 12, 15, 18, then two other pairs
with four, 20 and 24, and then the pair with five and six, 30.
2, 3, 4, 5, 6
6, 8, 10, 12
12, 15, 18
20, 24
30
2, 4, 6, 8, 10, 12
3, 6, 12, 18, 24
5, 10, 15, 20, 30
That's a different way to reorder those products. What is there is
that each list starts with a prime, and then the first list increments
by two, the second list increments by three once and then six, and the
third list increments by five four times and then ten, for six
multiples of two, five multiples of three, and five multiples of five.
For the case n=4 there are these products:
2, 3, 4, 6, 8, 12
2, 4, 8
3, 6, 12
For the case n=8 there are (n^2-n)/2 = 28 pairs. Their products are
2, 3, 4, 5, 6, 7, 8
6, 8, 10, 12, 14, 16
12, 15, 18, 21, 24
20, 24, 28, 32
30, 35, 40
42, 48
56
2, 4, 6, 8, 10, 12
3, 6, 12, 18, 24
5, 10, 15, 20, 30, 40
7, 14, 21, 28, 35, 42, 56
8, 16, 24, 32
Not each of those lists starts with a prime.
The idea here is to get a function that has as a result the sum of
those products as a function of n.
2*(1+2+3+4+5+6)
3*(1+2 +4 +6+8)
5*(1+2+3+4 +6+8)
7*(1+2+3+4+5+6+8)
8*(1+2+3+4)
For n=10 there are (n^2-n)/2= 45 pairs. Their products are:
2, 3, 4, 5, 6, 7, 8, 9, 10
6, 8, 10, 12, 14, 16, 18, 20
12, 15, 18, 21, 24, 27, 30
20, 24, 28, 32, 36, 40
30, 35, 40, 45, 50
42, 48, 54, 60
56, 63, 70
72, 80
90
2, 4, 6, 8, 10, 12
3, 6, 12, 18, 24, 30
5, 10, 15, 20, 30, 40, 50
7, 14, 21, 28, 35, 42, 56, 70
8, 16, 24, 32, 48
9, 18, 27, 36, 45, 54, 63, 72, 90
20, 40, 60, 80
2*(1+2+3+4+5+6)
3*(1+2 +4 +6+8+10)
5*(1+2+3+4 +6+8+10)
7*(1+2+3+4+5+6+8+10)
8*(1+2+3+4+5+6)
9*(1+2+3+4+5+6+7+8+10)
20*(1+2+3+4)
So, for n=4, 6, 8, 10:
2*(1+2+3)
3*(1+2 +4)
2*(1+2+3+4+5+6)
3*(1+2 +4 +6 +8)
5*(1+2+3+4 +6)
2*(1+2+3+4+5+6)
3*(1+2 +4 +6 +8)
5*(1+2+3+4 +6 +8)
7*(1+2+3+4+5+6 +8)
8*(1+2+3+4)
2*(1+2+3+4+5+6)
3*(1+2 +4 +6 +8 +10)
5*(1+2+3+4 +6 +8 +10)
7*(1+2+3+4+5+6 +8 +10)
8*(1+2+3+4+5+6)
9*(1+2+3+4+5+6+7+8 +10)
20*(1+2+3+4)
The polynomial (2n)!/n! has a simple expression, (2n)!/n!.
Ross
Ross A. Finlayson
Virgil,
The variable x is assigned the value n.
What the issue is, assigning the variable to a function of n, occurs
earlier in the setting of the variable z in the Gamma function to a
function of n. Setting x to n/z basically converts the variable of
the Gamma function back to a finite z.
Consider the Gamma function around {0, -1, -2, ...}. The function is
asymptotic at those points. Does that have anything to do with the
roots of (n+1)...(n+n) being {0, -1, -2, ...}? If only it does. What
is 1/Gamma(z)?
1/Gamma(z) = lim n->oo (n+z)! / ( n! n^z (z-1)! )
http://mathworld.wolfram.com/GammaFunction.html
The function 1/Gamma(z) would equal zero at 0, -1, -2, etcetera.
So, the "no free variable" comment is applicable not to what you say
it is but to something earlier. I disagree. Sometimes it's unclear
how we write so much in so little, or not.
What I'm doing is picking points halfway, one half, out to infinity
and calling them a function of one scalar infinity n: n/2, by
preserving the variable across each equation and derivation.
There's this branch of this thread, then I need to get back to the
probability distributions of infinite sequences. The probability of
selecting a given binary sequence of length n from among all of the
2^n many binary sequences of length n is 1/2^n, the selection of each
is unique, and the sum of their probabilities is one.
I'm not quite sure where to proceed with the consideration of the
factorial. Mostly as much of this is new material to me I want to
integrate it into my understanding so that I can use it later. I do
see some directions for research.
Ross
Virgil
> Virgil <vmh...@comcast.net> wrote in message
> news:<vmhjr2-24E6E4.21111012082003@[63.218.45.211]>...
> > In article <3c6b9c1e.03081...@posting.google.com>,
> > r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> >
> > > lim n->oo ((1+1/x)n!) / (n! n^(n/x))
> >
> > In this expression x is a free variable and n is a bound (dummy)
> > variable. Thus this expression represents a function of x defined
> > whichever x's the limit exists for, if any.
> > >
> > > Set x=1/n:
> >
> > >
> > > lim n->oo ((1+n)n)! / (n! n^(n^2))
> >
> > This is no longer a functional expression because it does not
> > contain a free variable.
> >
> > Thus its meaning and the meaning of your original functional
> > expression are unrelated.
> >
> > The remainder of this derivation is useless , so [snip].
>
> Virgil,
>
> The variable x is assigned the value n.
Which value of n?
In 'lim n->oo ((1+1/x)n!) / (n! n^(n/x))' there are, implicitly,
infinitely many values of n involved in the indicated limiting
process.
But x can only assume ONE value of n in any evaluation of the limit
lim n->oo ((1+1/x)n!) / (n! n^(n/x)).
Which one?
Ross A. Finlayson
If x is a function of n, and n is the only variable, and the
expression is considered for n as n in the limit as n goes to
infinity, relating f(n) to f(n+1), that's an expression.
lim n->oo n! / n! n^0
That expression is equal to one, considering it equal to one is true.
lim n->oo (2n)! / (n! n^n)
That expression is not equal to one. It's a reformulation of the
Gamma function with an infinite or infinitesimal value for the
variable z. it should be equal to one for that, but it's not.
Instead it's equal to (n+1)...(n+n) / n^n, and the limit of its
reciprocal is equal to zero, as the expression diverges.
Gamma(z) = lim n->oo n! n^z / (n+z)!
(z-1)! = lim n->oo n! n^z (z-1)! / (n+z)!
1 = lim n->oo n! n^z / (n+z)!
This is true for z not equal to any of 0, -1, -2, .... The variable n
is always positive, the function is defined and true for any value of
n. I simply set z to be an expression of n, z = n/x. As z diverges,
the equation should still be true for any value of z. Is not the
expression true for any defined value of z?
Consider lim n->oo z/n, for positive z. It evaluates to zero, for z <
n. Consider lim n->oo n/n, it evaluates to one. If z is an
expression of n, f(z) = Cn, lim z/n = C.
You might say z = f(n) is not a defined expression for z except for
similarly to f(n)=1/n, z=1.
So here I have z = f(z, n).
Consider the limit process. Evaluate f(n)/f(n+1), it has to be a
decreasing function, or increasing function, where |f(n+1)|-|f(n)| <
|f(n)|-|f(n-1)|. For the limit l, l-f(n+1) < l-f(n). Then again I
say that the limit of sinc(x), that being sin(x)/x, is zero as x
diverges, it's the average value of that function, more or less,
between f(x) and f(x+2Pi) the maximum value of the absolute value is
greater than the maximum value of the absolute value of any value
between f(x+2Pi) and f(x+4Pi).
Anyways the functions we consider here are not periodic and are
increasing or decreasing, or constant.
About taking the limit when z is an expression of n, it's saying to
compare each f(n) as if the variable z was the expression of n, for
all intents and purposes eliminating the variable z from the
expression and replacing it with an expression of the other variable,
n. The expression should be true for all z, in this notation n is
always an element of N and a defined value for z, it should be true,
but it appears to not be true except where n >> z.
So I figure I'll determine what it is when n approximates z, and z >>
n.
What I think is key to bring into focus here is the concept of n and
the limit of f(n) as n being infinite, and of n representing a unit
scalar infinity as the additive identity in a ring and field of scalar
infinities. Where there is the Gamma function there, it doesn't say
oo! oo^z / (oo+z)! . Yet, it's only true in the limit, in this case a
limit ordinal-like scalar infinity, that represents one unit. The
expression appears to be only true when z is infinitesimal relative to
n. When z is on the same infinite scale as n, the expression is
itself representative of a huge infinite quantity.When z is
infinitesimal relative to one, the regular, finite, unit, the
expression also happens to be true, as n is infinite relative to one.
lim n! n^0 / n! = 1
There are reasons why the limit is used instead of claiming n to be a
plain old unit scalar infinity. Perhaps they should be reexamined.
About (2n)! / (n! n^n), it's a polynomial, which can be considered as
having ever more finite terms, it has some simple properties. Its
roots are the non-positive intgers, its a monic polynomial, and there
are expressions to derive each of its n+1 many coefficients alpha_i.
I think there is a case for forgoing the limit expression and dealing
with plain infinite values. That's not really necessary to explore
the properties of the Gamma function, its more a broader consideration
in numbers and number theory.
Ross
Virgil
> I dispute that.
As Ross has left nothing to visible as the target of his
disagreement, one can only conclude that he is disputing nothing.
Ross A. Finlayson
The unit scalar infinity is in no way the additive identity. The
additive identity is zero.
What are implications of using an infinity as a number in the same
equations as finite numbers and infinitesimals?
Say there's a scalar infinity, I, capital Latin I, or capital Greek
iota. What is 1 compared to I? Does 1/I equal zero? No, it equals
1/I. Then there's infinitesimal, little i. 1/I = i/1.
For any non-zero n in N or r in R, i < |n| < I and i < |r| < I.
Associativity, commutativity, distributivity, and symmetry hold in
addition and multiplication of like-infinitely-valued operands as they
do for finitely valued-operands and like-infinitesimally-valued
operands.
For a, b, c, d, e, f in R, infinitesimal should have a different
symbol so it doesn't conflict with i, a square root of negative one.
Anyways hereafter in this post i means infinitesimal unit:
(ai + b + cI) + (di + e + FI) = (a+d)i + (b+e)(1) + (c+f)I
a * (bi + c + dI) = (ab)i + (ac)(1) + (ad)I
(ai + b + cI) * (di + e + fI) = (ad)i_2 + (ae+bd)i + (af + be + cd)(1)
+ (bf+ce)I + (cf)I_2
There, i_2 is an infinitesimal of i, its fluxion, where I_2 is an
infinity of I, its fluent.
1/I_1 = i_1/1
i_n * i_1 = i_(n+1)
I_n * I_1 = i_(n+1)
It takes more record keeping to keep track of what may be
infinitesimals and infinities. I'm still trying to figure out why and
how and whether I_2 is I^2.
Does it have any use? Probably, otherwise it wouldn't seem
intuitively appealing.
What's the length of the infinite binary sequence? It's I. Here's
how many permutations there are of the sequence: 2^I.
Here's how many positive integers there are, arbitrarily enough: I.
Here's how many of them are even: I/2, and the probability of one at
random being even is I/2 / I: 1/2. Here's how many integers are
non-negative: I + 1, and here's how many are integers: 2I + 1.
Ross
Richard Carr
:Date: 14 Aug 2003 01:03:26 -0700
:From: Ross A. Finlayson <r...@tiki-lounge.com>
:Newsgroups: sci.math
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:The above is riddled with error.
The content of the subject or Ross A. Finlayson.
(Not necessarily mutually exclusive - in fact, one might say equivalent.)
Ross A. Finlayson
> The above is riddled with error.
>
>
> What are implications of using an infinity as a number in the same
> equations as finite numbers and infinitesimals?
>
This thread started because I think that half of the sequences have
equal numbers of ones and zeros. Yet, I am unable to prove that.
What was shown, and it was known, or at least a variation of it was an
exercise in a book, is that:
lim n->oo n! sqrt(n) / ( (n/2)!^2 2^n) = sqrt(2 / (Pi n ))
I wanted to get an expression for how many of the sequences (the
infinitely many binary sequences of infinite length) had as a fraction
of the population / probability of random occurrence xn or (1-x)n for
0<x<1. It's only crude, though, multiplying to remove the terms of x
and replace the terms of n/2.
lim n->oo n! / ( ((x)n)! ((1-x)n)! 2^n) = ( ((x)n)! ((1-x)n)! /
(n/2)!^2 ) (sqrt(2/ (Pi n)))
Here's a thought: it goes back to the binomial distribution where
there is probability p of success and q of failure, each of p and q in
the above is 1/2, what if p and q multiply to be 1/e, where their sum
equals one? Then instead of 2^n it would be e^(n/2).
p + q = 1
p * q = 1/e
Is there more than one pair of possible values for p and q? Is there
a solution? There isn't. The value of 1/e is 1/2.7182... or
.36789.... , and (1/2)(1/2) = 1/4 = 0.25 is about the maximum value
of pq for p+q=1 and 0<p<1, 0<q<1. There are values of p and q thus
that p+q=1 and p*q = 1/xe for x > 4/e =. Heh, 4/e = 1.472.... For
example p*q can equal 1/(2e). Then instead of 2^n it would be
(2e)^(n/2). This is only when k=n/2. Otherwise it would be p^k
q^(n-k), where all of the consideration in this thread has been with
p=q=1/2, as half of the sequence elements are one and the other half
zero.
p+q = 1
q = 1-p
p*q = 1/2e
p(1-p) = 1/2e
p-p^2 = 1/2e
p^2 - p + 1/2e = 0
p= -(-1 +- sqrt(1-2/e))/2
p = (1-sqrt(1-2/e))/2
n! / (( n/2)!^2 2^(n/2) e^(n/2) )
That's the expression for the probability of half successes and half
failures of a sequence of length n where p+q=1 and p*q = 1/(2e).
Anyways, from that a variety of different expressions for n! were
shown vis-a-vis Stirling's formula.
I looked to Euler's formula for Gamma, and got the bright idea of
replacing the variable z with an expression of the variable n, z =
n/x, to find more identities fo n! for basically infinite n, the limit
of n! as n diverges. I soon found that it was only true for z being
small finite values instead of presumably finite values that were
expressions of n, assumed to be an arbitrarily large and then ever
larger finite variable.
That led into consideration of polynomials of the form (n+1)...(n+n),
for the specific case of the expression (2n)! / ( n! n^n).
Polynomials for other values of x would be about (3n/2)! / (n!
n^(n/2)), etcetera, (n+1)...(n+ n/2), for ((1+1/x)n)! / (n! n^(n/x)).
Thus I wanted to get an idea of what expressions would reducedly
symbolize the polynomial coefficients of (n+1)...(n+n) as a way to get
started. I figured out that a recurrence relation would give the
coefficient in terms of n, and have yet to find a form for it. The
idea of determing what those coefficients are is that they would form
another polynomial as divided by n^n that equals (2n)! / (n! n^n).
At that point I digressed about the variable as an infinite value and
discussed scalar infinities and scalar infinitesimals. I think those
might actually be visible in boundary value problems. They might
certainly be useful in a variety of ways. I'll talk about that more
later.
Anyways, back to the polynomials, I note in a recent post that Ullrich
says polynomials have to have a finite number of terms. Here I'm
talking about a case where a polynomial would have n+1 many summands
for infinite n. For example, n! is (n)(n-1)...(n-n+1), I am here more
concerned with (n+1)...(n+n). So, as I hope to discover the
recurrence relations for the coefficienct a_i for i=1 to n+1, it would
have to be in the form where n is assumed to diverge, and I will call
(2n)!/n! a polynomial.
The univariate polynomial (n+1)...(n+n) has as its first coefficient
at least a simple result: 1. The second coefficient is the sum of
each value of n, from 1 to n. That value has a simple expression:
(n+1)(n/2). In the form of a recurrence relation, f(1)=1, f(n) =
f(n-1)+n. The coefficient a_3 is not so simple to compute. It is the
sum of the product of each pair of {1, ..., n}. From i = two onwards,
the coefficient a_i is the sum of the product of each of the elements
of each subset of length i-1 of N.
One way to approach that is to sum all the n-1 pairs with 1's in them,
then all the n-2 remaining pairs with 2's, then all the remaining
pairs with n-3 3's, for each of the (n^2-n)/2 pairs. Then for the
first set of pairs their products' sum is (n+1)(n/2)-1. The pairs
with 2 have the product ((n+1)(n/2)-1-2)*2. The pairs with 3 have the
product ((n+1)(n/2)-1-2-3)*3.
f(x) = ((n+1)(n/2)-sum(x))*x
sum x=1^n f(x) = a_3
(n+1)(n/2)(n+1)(n/2) - sum x=1^n sum(x)*x
(n^2 + n)^2 / 4 - sum x=1^n sum(x)*x
(n^2 + n)^2 / 4 - sum x=1^n (x+1)(x/2) * x
(n^2 + n)^2 / 4 - sum x=1^n (x^3 + x^2)/2
I think the above is an expression for a_3. Then trying to find an
expression for a_4, the sum of the product of each triple of N, is
quickly more complicated, where in trying to find a recurrence
relation to a_3, it is to start with a_3 minus something and call it
the sum of each triple with 1. Then there are all of the triples with
2, etcetera.
Here are a couple things I have come across here that I want to
understand better: it's the plain old sum function that is the the
sum of integers from 1 to n, much like the factorial is the product of
integers from 1 to n. Then also I have the cumulative sum function,
which is another recurrence relation which is the sum of the integers
1 to n plus the sum of the integers 1 to n-1 plus the sum of the
integers 1 to n-2, etcetera. These are parts of the recurrence
relations about summing the the products of pairs, triples, etcetera
of sets of integers with contiguous values starting at one.
There's a lot of available information about the recursive product of
numbers, factorial, where's the information about recursive sums? I
read in the algorithm books about Theta space and big-O, with the
expression in few variables of more complicated recurrence relations,
I'm talking about more exactitude.
The sum of n integers 1 to n is f(n)=1, f(n+1)= f(n) + n+1, it's also
(n+1)(n/2). What's a function for f(n)=1,
f(n+1)=f(n)+f(n-1)+f(n-2)... + n+1? How about a recurrence relation
for the products? Is there a compendium or dictionary of recurrence
or difference equations?
So, we're discussing factorial/exponential identities.
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> This thread started because I think that half of the sequences have
> equal numbers of ones and zeros. Yet, I am unable to prove that.
If Ross means that of the collection of all infinite sequences all
of whose terms are zeros or ones only half have "equal" numbers of
zeros and ones, he underestimates.
Any such sequence which has infinitely many zeros and infinitely
many ones has equally (denumerably) many of each digit, and it may
be shown that there are uncountably many of such sequences, but the
set of sequences containing only finitely many zeros or finitely
many ones is countable.
Thus "more" sequences have equally many zeros and ones than have
unequally many.
Ronald Bruck
<vmh...@comcast.net> wrote:
But another, and perhaps more reasonable, way of measuring is to use
asymptotic density instead of cardinality. And only infinitesimally
many sequences have asymptotic density 1/2.
--Ron Bruck
Virgil
Ronald Bruck <br...@math.usc.edu> wrote:
But does that say that the number having asymptotic density 1/2 is
or is not countable?
Randy Poe
> r...@tiki-lounge.com (Ross A. Finlayson) wrote in message news:<3c6b9c1e.03081...@posting.google.com>...
> > The above is riddled with error.
> >
> > ...
> >
> > What are implications of using an infinity as a number in the same
> > equations as finite numbers and infinitesimals?
> >
>
>
> This thread started because I think that half of the sequences have
> equal numbers of ones and zeros. Yet, I am unable to prove that.
It's not easy giving meaning to "half of the sequences". Virgil
offers one interpretation and proves that under that interpretation,
almost all sequences have equal numbers of ones and zeros (the
fraction that don't is zero).
Here's another interpretation that perhaps has already occurred
in this thread: In even-length random binary sequences of length
2N, what is the asymptotic probability as N->oo that the number
of 0's equals the number of 1's? Note that I am asking a
question about large finite N, not infinite N.
In your roundabout way you show that you're aware of the
binomial distribution. This probability is quite easy to
calculate. It's
(2N)!/[2^(2N)*(N!)^2]
> What was shown, and it was known, or at least a variation of it was an
> exercise in a book,
The distinction between "what was shown" and "a variation
of what was shown" is important. I have no idea whether or
not you have taken a valid result and made a "variation"
of it that is no longer valid. I suspect that to be the
case since your statement as written is nonsense.
>
> lim n->oo n! sqrt(n) / ( (n/2)!^2 2^n) = sqrt(2 / (Pi n ))
>
The limit as n->oo can't have a dependence on n in it.
However, the large-n asymptotic behavior might. What was
the original result that you "varied"?
Using the Stirling approximation (I *strongly* suspect I'm
repeating ground that was already covered in this thread)
I get
(2N)!/(N!)^2 = (approx)(1/sqrt(2*pi)) * 2^(2N+1)/sqrt(2N)
So
(2N)!/[2^(2N)*(N!)^2] is asymptotically approximated by
sqrt(2/N)/sqrt(2*pi) = 1/sqrt(pi*N)
That appears to be equivalent to what you wrote if you
delete the factor of sqrt(n) from the numerator on the
left-hand side. I guess that's your "variation".
Now, this probability decreases as 1/sqrt(N). That means
that the fraction becomes smaller and smaller as N increases.
For N = 10^6, less than 1/3000 of the sequences have
equal ones and zeros.
For N = 10^12, less than 1 in 3 million have equal numbers
of ones and zeros.
Where do you think this trend goes as N increases without
bound?
Does it really surprise you that you're having trouble
proving that the limit of an expression proportional to
1/sqrt(N) is 1/2?
- Randy
Ross A. Finlayson
One thing about this discussion is that where the binomial choice
function is normally (n k), here we have been discussing (n n/2). The
point to observe there is that instead of two variables it is
considered with one variable n and an expression of n as the second
variable, and the expression is considered for n in the limit. Yet,
in Euler's formula for Gamma, replacing z with an expression of n
yields various results depending on the expression of z in terms of n.
It's valid for some expressions. Why isn't it for all?
When I put a variable on both sides of a limiting equation:
lim n->oo f(n) = g(n)
that means
lim n->oo f(n) = lim n->oo g(n)
and
lim n->oo f(n) / g(n) = 1
Is it a countable number of ones and zeros? I don't care. It doesn't
matter. It's irrelevant. It's also useless and counterproductive.
If there are half ones and half zeros in a sequence, then removing a
pair of one one and one zero, repeated ad infinitum, would never
change that set from having half ones and half zeros. I realize that
any infinite sequence with infinite ones and infinite zeros could have
a pair of each removed any number of times.
An infinite sequence of the form .0101(01)... has half ones and half
zeros. You can interchange any two sequence elements as many times as
you want, it still has half ones and half zeros, and would never be
the sequence .001001(001)... or .011011(011)..., for 1/3 and 2/3, yet
it can easily be .00000000000000001111111111111111(01)..., for 1/2.
The same holds true for other fractions greater than zero and less
than one, there is a canonical sequence describing it and it is a
different sequence than for any other fraction.
What about those sequences with one or n more more or less than 1/2?
They differ from the canonical sequence after canonicalization at
finitely n many points. If it differs by more than finitely many
points then it's a finite difference of a different canonical
sequence.
Why is it not possible to switch some elements of a canonical sequence
and get a different canonical sequence?
Here I've used the terms "canonical sequence" and "canonicalization".
What are they? What is their meaning and context?
There's a quantity that defines how many of the infinite length
sequences have equal numbers of ones and zeros. It's infinitesimal.
When summed with the quantities of there being all the other
expressions of n between 0n and 1n many zeros, and their finite
differences, that is equal to one.
Ross Finlayson
Virgil
> When I put a variable on both sides of a limiting equation:
>
> lim n->oo f(n) = g(n)
Ross is breaking one of the customs, if not rules, of logic in using
the same symbol, "n", as both a bound variable and a free variable
in the same statement. The custom has the desirable goal of avoiding
ambiguity, which Ross seems rather to relish than avoid.
As this "equation" has no free variable on its left hand side but
does have one on its right hand side, the only sensible
interpretation is that the right hand side "function", denoted by
g(n), must be a constant.
If this is the case, and the constant is denoted by c, and the
constant is not zero, then one trivially gets what Ross seems to
want, namely
if lim n->oo f(n) = c and c <> 0,
then lim n->oo f(n)/c = 1
>
> that means
>
> lim n->oo f(n) = lim n->oo g(n)
>
> and
>
> lim n->oo f(n) / g(n) = 1
Ross' assumption that if lim n->oo f(n) = lim n->oo g(n) then
lim n->oo f(n) / g(n) = 1 is flawed without the additional
assumption that neither lim n->oo f(n) nor lim n->oo g(n) is zero.
Consider f(n) = 1/n^2 and g(n) = 1/n.
Then lim n->oo f(n) = lim n->oo g(n) = 0
but lim f(n)/g(n) = 0, not 1 and g(n)/f(n) diverges.
Ross A. Finlayson
sequence to a sequence with repeating terminating sequences and some
finite beginning sequence.
Can any infinite binary sequence be converted to a sequence with a
repeating terminating sequence and a finite beginning sequence?
There are the sequences similar to Champernowne's that can not,
perhaps.
.01001000100001000001...
These sequences have infinitesimally many ones compared to zeros, but
infinitely many ones.
Thus canonicalization might result in a sequence similar to those, or
to the repeating sequence.
There is a function of n to approximate the density of primes in
[1,n], it is similar to that.
The above sequence has x many ones and sum x many zeros for a sequence
length of x + sum x. The function sum x for x={1, 2, ...} is equal to
(x+1)x/2. Thus the number of ones compared to sequence elements is x
/ (x+(x+1)(x/2)). Thus for n = (x + (x+1)(x/2)), the number of ones
is x. Solving for x in terms of n:
n = (x + (x+1)(x/2))
n = (x+ x^2/2 + x/2)
n = (x^2/2 + 3/2 x)
2n = x^2 + 3x
2n+ (3/2)^2 = (x+3/2)^2
(2n + 9/4)^(1/2) - 3/2 = x
The idea here is to say that (n x) = (n f(n)), so instead of (n n/2) /
2^n it's along the lines of (n ((2n + 9/4)^(1/2) - 3/2)) ).
n! / ( Gamma(2n + 9/4)^(1/2) - 1/2) Gamma( n - ((2n + 9/4)^(1/2) -
1/2) ) 2^n )
Then, besides the sequences canonicalizable to
.01001000100001...
With a zero then a one then two zeros then a one then three zeros then
a one then four zeros then a one, etcetera, there would be sequences
with various numbers of zeros and various number of ones, where in the
limit there are infinitely many more zeros than ones, with the density
of the infinitely many ones being infinitesimal. For example instead
of the number of contiguous zeros incrementing, it might increment by
two or double.
n = x + (x+1)x
n = x + sum 2^x
These would be sequences representing irrational numbers, and they
could not be canonicalized into a rational sequence. That is
different than sequences with about equal or fractional numbers of
ones and zeros which are canonicalizable into sequences with repeating
sequences. These are sequences with infinitely many ones, and
relatively infinitely many more zeros, and vice versa.
a) sequences with finite ones or zeros
b) sequences with infinitely many ones and relatively infinitely many
more or less zeros
c) sequences with infinitely many ones and relatively finitely many
more or less zeros
It is not the simple case that all the irrationals are of type b, some
are of type c, but all the sequences of type b are irrational. The
canonical sequences of type c are rational.
It's easy enough to characterize the canonical sequences with
repeating terminal sequences, particularly the cases with terminating
1... or 0.... Those are basically the rationals, with the sequences
terminating with 1 or 0 having finitely many zeros or ones,
respectively.
Type a might be considered a subset of type c, as the finite number of
ones or zeros goes to an expression of n, one is 1/n, two is 2/n,
etcetera, each is less than n/x for finite x, e.g n/2, n/3, n/4, 2n/5,
n/5, n/6, etc., 1/n of n is 0n +1.
What I'm interested in determining now is how many of the sequences of
type c are irrational, and how many sequences there are of type b.
There are 2^n possible sequences, I want to see if there are 2^(n-1)
each of rational and irrational sequences. The problem there is
determining how many sequences of type c are irrational or rational,
besides determining how many sequences of type b and c there are.
It is apparent that there are more than n many rational sequences:
there are n many sequences with a one and infinitely many zeros, each
is rational.
I guess I can start with considering the canonical sequence with half
ones and half zeros.
.101010(10)...
It's representative of a rational. So are
.000111(10)...
.001011(10)...
.010011(10)...
.100011(10)...
.100101(10)...
.101001(10)...
.110001(10)...
.110010(10)...
.110100(10)...
.111000(10)...
.001101(10)...
.011001(10)...
.011010(10)...
.011100(10)...
.001110(10)...
.010110(10)...
.100110(10)...
.101100(10)...
.010101(10)...
.101010(10)...
Each there still has "equal numbers of ones and zeros", they
canonicalize to .(10).... For the beginning sequence of x many
elements there are (x x/2) or in the above example for x=6 there are
6! / (3! 3!) = 6*5*4/3*2*1 = 20. That accounts for about x of the n
many elements of the sequences, the terminating n-x many elements have
obviously (n-x (n-x)/2) -1 permutations besides (10)....
We can just consider sequences in terms of how many ones there are,
from 1 to n many ones. There are sequences with n/2, n/3, 2n/3, n/4,
3n/4, etcetera, +-x, many ones, those are all the sequences
canonicalizable into sequences with repeating endings. The sequences
can be rational or irrational, their canonical form is rational.
I think it is safe to say that there are relatively infinitely many
more sequences of type c than type b. Yet, it is much easier to
specify the relative quantities of zeros and ones for all sequences of
type c than type b.
In a way, that's about comparing the density in the naturals of
integer multiples, parallel to type c, to squares, cubes, primes,
sums, factorials, etcetera, infinite sets with infinitesimal density,
type b, except sequences of type b are exclusive from sequences of
type c.
MathWorld notes that abundant numbers have a finite, positive density
in the naturals.
Then, there's the problem of separating the sequences of type c into
rationals and irrationals.
For the sequences S_(1/2) that canonicalize to a sequence with n/2 of
n many ones, how many are irrational and thus the rest rational? We
have an expression for how many sequences there are with n/2 many
ones, in terms of n. What would answer that question then is how many
of those are rational. The problem I confront here in that
consideration is that the sequence is of infinite length. One way to
consider it is to determine how many possible repeating sequences
there could be, in terms of having some amount left over to be the
finite beginning sequence with other than the repeating sequence yet
still with half ones. I'll start with considering repeating sequences
of length l 2, 4, 6, etcetera, the repeating sequence has to have a
length of a multiple of 2 just as for the case of S_(1/3) it woud have
to be a multiple of three. Let's say it is two:
(01)...
(10)...
Each of those is one of the cases for l=4, 6, 8, ...,
(0101)...
(010101)...
(01010101)...
Where l=2, I will assume then that the n many elements of the sequence
are divided into n/2 many subsequences. Then, of those subsequences,
another variable's as many of those, a finite number, are subsequences
for the beginning sequence, and the remaining subsquences each have
the repeating sequence.
The length of the repeating subsequence can be any multiple of two,
although it should be arbitrarily less than n. Again, the idea here
is to determine how many sequences end with repeating subsequences and
are thus rational.
How is it gone about quantifying the number of rationals compared to
the number of irrationals among all the sequences with half ones and
half zeros?
I think it helps that I think 2^(n-1) of the sequences are rational
and the other 2^(n-1) are irrational. Determining how many sequences
of type c are rational would actually show that, or not.
That leads me back to thinking about how the rationals and irrationals
are each dense in the reals and complementary where their union is the
reals.
Here's a non sequitur, for integer n, n! +-1 is prime.
Go back to the posts about functions for the recurrence relations.
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> I guess
We've noticed.
Ross A. Finlayson
I examine the products of pairs of a set of integers {1, ..., n}. For
example, for a set {1, 2, 3}, the pairs are {1, 2}, {1, 3}, and {2,
3}. The number of pairs is (n^2-n)/2.
I put the products in a list. Each list element is a number, prime or
composite, multiplied by the sum of a series of numbers. For example,
the sum of these two list elements:
2*(1+2 +4)
3*(1+2 +4)
is the sum of the products of the pairs of {1, 2, 3, 4}.
There seems to be a rational way to enumerate the list elements as I
start with small values of n. I wrote a small program to generate the
products and sort them to compare to how an algorithm would enumerate
them on a list. Here are the working lists for various values of n:
2:
2*(1)
3:
2*(1)
3*(1+2)
4:
2*(1+2 +4)
3*(1+2 +4)
5:
2*(1+2 +4)
3*(1+2 +4)
5*(1+2+3+4)
6:
2*(1+2+3+4 +6)
3*(1+2 +4 +6 +8)
5*(1+2+3+4 +6)
Notice that the multipliers increment by one to an even number and
then sometimes increment by two.
7:
2*(1+2+3+4 +6)
3*(1+2 +4 +6 +8)
5*(1+2+3+4 +6)
7*(1+2+3+4+5+6)
8:
2*(1+2+3+4 +6)
3*(1+2 +4 +6 +8)
5*(1+2+3+4 +6 +8)
7*(1+2+3+4+5+6 +8)
8*(1+2+3+4 +6)
The number 8 is not prime. It is the cube of a prime, and product of
a prime and a composite.
9:
2*(1+2+3+4 +6)
3*(1+2 +4 +6 +8)
5*(1+2+3+4 +6 +8)
7*(1+2+3+4+5+6 +8)
8*(1+2+3+4 +6)
9*(1+2+3+4+5+6+7+8)
The number 9 is not prime. It is the square of a prime.
10:
2*(1+2+3+4+5+6)
3*(1+2 +4 +6 +8 +10)
5*(1+2+3+4 +6 +8 +10)
7*(1+2+3+4+5+6 +8 +10)
8*(1+2+3+4 +6)
9*(1+2+3+4+5+6+7+8 +10)
20*(1+2+3+4)
The number 20 is greater than n. It's used instead of 10, the product
of two primes or a semiprime, to have the multipliers follow the
sequence starting with 1.
Notice that the multipliers, (1+2+...), increment by one and then
increment by two.
11:
2*(1+2+3+4+5+6)
3*(1+2 +4 +6 +8 +10)
5*(1+2+3+4 +6 +8 +10)
7*(1+2+3+4+5+6 +8 +10)
8*(1+2+3+4 +6)
9*(1+2+3+4+5+6+7+8 +10)
20*(1+2+3+4)
11*(1+2+3+4+5+6+7+8+9+10)
Eleven is a prime, its list entry is each multiple of 11 less than 11.
12:
2*(1+2+3+4+5+6)
3*(1+2 +4 +6 +8 +10 +12)
5*(1+2+3+4 +6 +8 +10 +12)
7*(1+2+3+4+5+6 +8 +10 +12)
8*(1+2+3+4 +6)
9*(1+2+3+4+5+6+7+8 +10 +12)
20*(1+2+3+4 +6)
11*(1+2+3+4+5+6+7+8+9+10 +12)
12*(1+2 +4 +6 +8)
Twelve has as factors 1, 2, 3, 4, 6, and 12.
13:
2*(1+2+3+4+5+6)
3*(1+2 +4 +6 +8 +10 +12)
5*(1+2+3+4 +6 +8 +10 +12)
7*(1+2+3+4+5+6 +8 +10 +12)
8*(1+2+3+4 +6)
9*(1+2+3+4+5+6+7+8 +10 +12)
20*(1+2+3+4 +6)
11*(1+2+3+4+5+6+7+8+9+10 +12)
12*(1+2 +4 +6 +8)
13*(1+2+3+4+5+6+7+8+9+10+11+12)
Thirteen is a prime number, its list entry has multipliers of the sum
from 1 to n-1. Calculating the list entry for thirteen is evaluating
13*(12+1) 12/2 = 13^2 *6. Compare that to the list entry for three.
Most of the incrementations in the multipliers of three are increments
by two. An expression of the list entry for three is the sum of the
multipliers times three, and as well a partial sum of the mutipliers
tmes three plus the remainder sum of the multipliers times three. This
is about how (1+2+4+6+8+10+12) is 1+ 2*(1+2+3+4+5+6) = 1+2*(6+1)*6/2 =
43. That's different than the list entry for 7, most of its
multiplier increments are by one. (1+2+3+4+5+6+7+8+9+10+11+12) -
(7+9+11), or (12+1)*12/2 - 3*9 = 51. To calculate the list entry for
3 takes four multiplications/divisions and two additions/subtractions,
for list entry 7 there are three multiplications and two additions,
besides knowing the sequence of multiples and when they increment.
14:
2*(1+2+3+4+5+6)
3*(1+2 +4 +6 +8 +10 +12 +14)
5*(1+2+3+4 +6 +8 +10 +12 +14)
7*(1+2+3+4+5+6 +8 +10 +12 +14)
8*(1+2+3+4 +6)
9*(1+2+3+4+5+6+7+8 +10 +12 +14)
20*(1+2+3+4 +6)
11*(1+2+3+4+5+6+7+8+9+10 +12 +14)
12*(1+2 +4 +6 +8)
13*(1+2+3+4+5+6+7+8+9+10+11+12 +14)
14*(1+2 +4 +6 +8 +10 +12)
Fourteen is semiprime, its factors besides itself and one are primes.
As I get into higher numbers it gets a bit more complicated. The
prime is the simple case, but as I add composites I want to preserve
aspects of the multiplier progression. For example, with fifteen, it
is in some ways like a prime in its multiplier series changes, yet it
is a semiprime, it is also odd. Sixteen, a 4'th power of a prime and
a square of a square of a prime and a composite of a prime and a cube
of a prime, might lead to increments by four.
This is about recurrence relations, or recurrence equations or
difference equations. Calculating the sum of the elements of a set
{1, ..., n} given n is a matter of evaluating (n+1)n/2, the product of
all elements of the set is n!. This strategy explores the sum of the
products of the pairs (triples/trios, quadruples, etc.) of a set
{1,..., n} or equivalently {0, ..., n}.
The outer product of a pair of vectors is a matrix with the entries of
the matrix being the products of each i,j of the two vectors. That's
somewhat different from the above where the diagonal and the upper
lower elements are zero, eg, for (1, 2, 3, 4):
1 2 3 4
1 1 2 3 4
2 2 4 6 8
3 3 6 9 12
4 4 8 12 16
1 2 3 4
1
2 2
3 3 6
4 4 8 12
The sum of the elements of the outer product of {1,2,3,4} and itself
is
1*(1+2+3+4)
2*(1+2+3+4)
3*(1+2+3+4)
4*(1+2+3+4)
That is sum n squared, or (n+1)n/2 *(n+1)n/2, or 10*10=100. Each row
and column sums to a multiple of ten, the sum of 1, 2, 3, and 4.
The sum of the products of the pairs of {1, 2, 3, 4} is
(2+3+4)+(6+8)+12 = 9+14+12 = 35.
It would be possible to calculate the values of the diagonal of the
outer product and then subtract that from the sum of the outer
product, and then divide that result by two. The diagonal has the
value sum for i from 1 to 4 of x^2, that being 1+4+9+16, 30. 100,
calculated from (sum n) squared, minus 30, calculated from sum (n
squared), equals 70, dividing that by two would give the result of 35.
Now then, there is the question of deriving the sum of the squares,
heh, sum of squares, in a way besides adding them together.
Products of pairs for (1, ..., 48)
Number of pair products: (48^2 - 48)/2 = 1128
Sum of pair products: 672476
Generated products:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37,
38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 6, 8, 10, 12, 14, 16, 18,
20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52,
54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86,
88, 90, 92, 94, 96, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45,
48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96,
99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138,
141, 144, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76,
80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136,
140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192,
30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110,
115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180,
185, 190, 195, 200, 205, 210, 215, 220, 225, 230, 235, 240, 42, 48,
54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138,
144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 222,
228, 234, 240, 246, 252, 258, 264, 270, 276, 282, 288, 56, 63, 70, 77,
84, 91, 98, 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175,
182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273,
280, 287, 294, 301, 308, 315, 322, 329, 336, 72, 80, 88, 96, 104, 112,
120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224,
232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336,
344, 352, 360, 368, 376, 384, 90, 99, 108, 117, 126, 135, 144, 153,
162, 171, 180, 189, 198, 207, 216, 225, 234, 243, 252, 261, 270, 279,
288, 297, 306, 315, 324, 333, 342, 351, 360, 369, 378, 387, 396, 405,
414, 423, 432, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210,
220, 230, 240, 250, 260, 270, 280, 290, 300, 310, 320, 330, 340, 350,
360, 370, 380, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 132,
143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286,
297, 308, 319, 330, 341, 352, 363, 374, 385, 396, 407, 418, 429, 440,
451, 462, 473, 484, 495, 506, 517, 528, 156, 168, 180, 192, 204, 216,
228, 240, 252, 264, 276, 288, 300, 312, 324, 336, 348, 360, 372, 384,
396, 408, 420, 432, 444, 456, 468, 480, 492, 504, 516, 528, 540, 552,
564, 576, 182, 195, 208, 221, 234, 247, 260, 273, 286, 299, 312, 325,
338, 351, 364, 377, 390, 403, 416, 429, 442, 455, 468, 481, 494, 507,
520, 533, 546, 559, 572, 585, 598, 611, 624, 210, 224, 238, 252, 266,
280, 294, 308, 322, 336, 350, 364, 378, 392, 406, 420, 434, 448, 462,
476, 490, 504, 518, 532, 546, 560, 574, 588, 602, 616, 630, 644, 658,
672, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375, 390, 405, 420,
435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630,
645, 660, 675, 690, 705, 720, 272, 288, 304, 320, 336, 352, 368, 384,
400, 416, 432, 448, 464, 480, 496, 512, 528, 544, 560, 576, 592, 608,
624, 640, 656, 672, 688, 704, 720, 736, 752, 768, 306, 323, 340, 357,
374, 391, 408, 425, 442, 459, 476, 493, 510, 527, 544, 561, 578, 595,
612, 629, 646, 663, 680, 697, 714, 731, 748, 765, 782, 799, 816, 342,
360, 378, 396, 414, 432, 450, 468, 486, 504, 522, 540, 558, 576, 594,
612, 630, 648, 666, 684, 702, 720, 738, 756, 774, 792, 810, 828, 846,
864, 380, 399, 418, 437, 456, 475, 494, 513, 532, 551, 570, 589, 608,
627, 646, 665, 684, 703, 722, 741, 760, 779, 798, 817, 836, 855, 874,
893, 912, 420, 440, 460, 480, 500, 520, 540, 560, 580, 600, 620, 640,
660, 680, 700, 720, 740, 760, 780, 800, 820, 840, 860, 880, 900, 920,
940, 960, 462, 483, 504, 525, 546, 567, 588, 609, 630, 651, 672, 693,
714, 735, 756, 777, 798, 819, 840, 861, 882, 903, 924, 945, 966, 987,
1008, 506, 528, 550, 572, 594, 616, 638, 660, 682, 704, 726, 748, 770,
792, 814, 836, 858, 880, 902, 924, 946, 968, 990, 1012, 1034, 1056,
552, 575, 598, 621, 644, 667, 690, 713, 736, 759, 782, 805, 828, 851,
874, 897, 920, 943, 966, 989, 1012, 1035, 1058, 1081, 1104, 600, 624,
648, 672, 696, 720, 744, 768, 792, 816, 840, 864, 888, 912, 936, 960,
984, 1008, 1032, 1056, 1080, 1104, 1128, 1152, 650, 675, 700, 725,
750, 775, 800, 825, 850, 875, 900, 925, 950, 975, 1000, 1025, 1050,
1075, 1100, 1125, 1150, 1175, 1200, 702, 728, 754, 780, 806, 832, 858,
884, 910, 936, 962, 988, 1014, 1040, 1066, 1092, 1118, 1144, 1170,
1196, 1222, 1248, 756, 783, 810, 837, 864, 891, 918, 945, 972, 999,
1026, 1053, 1080, 1107, 1134, 1161, 1188, 1215, 1242, 1269, 1296, 812,
840, 868, 896, 924, 952, 980, 1008, 1036, 1064, 1092, 1120, 1148,
1176, 1204, 1232, 1260, 1288, 1316, 1344, 870, 899, 928, 957, 986,
1015, 1044, 1073, 1102, 1131, 1160, 1189, 1218, 1247, 1276, 1305,
1334, 1363, 1392, 930, 960, 990, 1020, 1050, 1080, 1110, 1140, 1170,
1200, 1230, 1260, 1290, 1320, 1350, 1380, 1410, 1440, 992, 1023, 1054,
1085, 1116, 1147, 1178, 1209, 1240, 1271, 1302, 1333, 1364, 1395,
1426, 1457, 1488, 1056, 1088, 1120, 1152, 1184, 1216, 1248, 1280,
1312, 1344, 1376, 1408, 1440, 1472, 1504, 1536, 1122, 1155, 1188,
1221, 1254, 1287, 1320, 1353, 1386, 1419, 1452, 1485, 1518, 1551,
1584, 1190, 1224, 1258, 1292, 1326, 1360, 1394, 1428, 1462, 1496,
1530, 1564, 1598, 1632, 1260, 1295, 1330, 1365, 1400, 1435, 1470,
1505, 1540, 1575, 1610, 1645, 1680, 1332, 1368, 1404, 1440, 1476,
1512, 1548, 1584, 1620, 1656, 1692, 1728, 1406, 1443, 1480, 1517,
1554, 1591, 1628, 1665, 1702, 1739, 1776, 1482, 1520, 1558, 1596,
1634, 1672, 1710, 1748, 1786, 1824, 1560, 1599, 1638, 1677, 1716,
1755, 1794, 1833, 1872, 1640, 1680, 1720, 1760, 1800, 1840, 1880,
1920, 1722, 1763, 1804, 1845, 1886, 1927, 1968, 1806, 1848, 1890,
1932, 1974, 2016, 1892, 1935, 1978, 2021, 2064, 1980, 2024, 2068,
2112, 2070, 2115, 2160, 2162, 2208, 2256,
Sorted products:
2, 3, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, 12, 12, 12, 13, 14, 14, 15,
15, 16, 16, 17, 18, 18, 18, 19, 20, 20, 20, 21, 21, 22, 22, 23, 24,
24, 24, 24, 25, 26, 26, 27, 27, 28, 28, 28, 29, 30, 30, 30, 30, 31,
32, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 36, 36, 37, 38, 38, 39,
39, 40, 40, 40, 40, 41, 42, 42, 42, 42, 43, 44, 44, 44, 45, 45, 45,
46, 46, 47, 48, 48, 48, 48, 48, 50, 50, 51, 52, 52, 54, 54, 54, 55,
56, 56, 56, 57, 58, 60, 60, 60, 60, 60, 62, 63, 63, 64, 64, 65, 66,
66, 66, 68, 68, 69, 70, 70, 70, 72, 72, 72, 72, 72, 74, 75, 75, 76,
76, 77, 78, 78, 78, 80, 80, 80, 80, 81, 82, 84, 84, 84, 84, 84, 85,
86, 87, 88, 88, 88, 90, 90, 90, 90, 90, 91, 92, 92, 93, 94, 95, 96,
96, 96, 96, 96, 98, 99, 99, 100, 100, 102, 102, 104, 104, 105, 105,
105, 108, 108, 108, 108, 110, 110, 111, 112, 112, 112, 114, 114, 115,
116, 117, 117, 119, 120, 120, 120, 120, 120, 120, 123, 124, 125, 126,
126, 126, 126, 128, 128, 129, 130, 130, 132, 132, 132, 132, 133, 135,
135, 135, 136, 136, 138, 138, 140, 140, 140, 140, 141, 143, 144, 144,
144, 144, 144, 145, 147, 148, 150, 150, 150, 152, 152, 153, 154, 154,
155, 156, 156, 156, 160, 160, 160, 160, 161, 162, 162, 164, 165, 165,
168, 168, 168, 168, 168, 170, 170, 171, 172, 174, 175, 175, 176, 176,
176, 180, 180, 180, 180, 180, 180, 182, 182, 184, 184, 185, 186, 187,
188, 189, 189, 190, 190, 192, 192, 192, 192, 195, 195, 196, 198, 198,
198, 200, 200, 200, 203, 204, 204, 205, 207, 208, 208, 209, 210, 210,
210, 210, 210, 215, 216, 216, 216, 216, 217, 220, 220, 220, 221, 222,
224, 224, 224, 225, 225, 228, 228, 230, 230, 231, 231, 232, 234, 234,
234, 235, 238, 238, 240, 240, 240, 240, 240, 240, 242, 243, 245, 246,
247, 248, 250, 252, 252, 252, 252, 252, 253, 255, 256, 258, 259, 260,
260, 261, 264, 264, 264, 264, 266, 266, 270, 270, 270, 270, 272, 272,
273, 273, 275, 276, 276, 279, 280, 280, 280, 280, 282, 285, 286, 286,
287, 288, 288, 288, 288, 288, 290, 294, 294, 296, 297, 297, 299, 300,
300, 300, 301, 304, 304, 306, 306, 308, 308, 308, 310, 312, 312, 312,
315, 315, 315, 319, 320, 320, 320, 322, 322, 323, 324, 324, 325, 328,
329, 330, 330, 330, 333, 336, 336, 336, 336, 336, 338, 340, 340, 341,
342, 342, 344, 345, 348, 350, 350, 351, 351, 352, 352, 352, 357, 360,
360, 360, 360, 360, 360, 363, 364, 364, 368, 368, 369, 370, 372, 374,
374, 375, 376, 377, 378, 378, 378, 380, 380, 384, 384, 384, 385, 387,
390, 390, 390, 391, 392, 396, 396, 396, 396, 399, 400, 400, 403, 405,
405, 406, 407, 408, 408, 410, 414, 414, 416, 416, 418, 418, 420, 420,
420, 420, 420, 423, 425, 429, 429, 430, 432, 432, 432, 432, 434, 435,
437, 440, 440, 440, 442, 442, 444, 448, 448, 450, 450, 450, 451, 455,
456, 456, 459, 460, 460, 462, 462, 462, 464, 465, 468, 468, 468, 470,
473, 475, 476, 476, 480, 480, 480, 480, 480, 481, 483, 484, 486, 490,
492, 493, 494, 494, 495, 495, 496, 500, 504, 504, 504, 504, 506, 506,
507, 510, 510, 512, 513, 516, 517, 518, 520, 520, 522, 525, 525, 527,
528, 528, 528, 528, 532, 532, 533, 540, 540, 540, 540, 544, 544, 546,
546, 546, 550, 551, 552, 552, 555, 558, 559, 560, 560, 560, 561, 564,
567, 570, 570, 572, 572, 574, 575, 576, 576, 576, 578, 580, 585, 585,
588, 588, 589, 592, 594, 594, 595, 598, 598, 600, 600, 600, 602, 608,
608, 609, 611, 612, 612, 615, 616, 616, 620, 621, 624, 624, 624, 627,
629, 630, 630, 630, 630, 638, 640, 640, 644, 644, 645, 646, 646, 648,
648, 650, 651, 656, 658, 660, 660, 660, 663, 665, 666, 667, 672, 672,
672, 672, 675, 675, 680, 680, 682, 684, 684, 688, 690, 690, 693, 696,
697, 700, 700, 702, 702, 703, 704, 704, 705, 713, 714, 714, 720, 720,
720, 720, 720, 722, 725, 726, 728, 731, 735, 736, 736, 738, 740, 741,
744, 748, 748, 750, 752, 754, 756, 756, 756, 759, 760, 760, 765, 768,
768, 770, 774, 775, 777, 779, 780, 780, 782, 782, 783, 792, 792, 792,
798, 798, 799, 800, 800, 805, 806, 810, 810, 812, 814, 816, 816, 817,
819, 820, 825, 828, 828, 832, 836, 836, 837, 840, 840, 840, 840, 846,
850, 851, 855, 858, 858, 860, 861, 864, 864, 864, 868, 870, 874, 874,
875, 880, 880, 882, 884, 888, 891, 893, 896, 897, 899, 900, 900, 902,
903, 910, 912, 912, 918, 920, 920, 924, 924, 924, 925, 928, 930, 936,
936, 940, 943, 945, 945, 946, 950, 952, 957, 960, 960, 960, 962, 966,
966, 968, 972, 975, 980, 984, 986, 987, 988, 989, 990, 990, 992, 999,
1000, 1008, 1008, 1008, 1012, 1012, 1014, 1015, 1020, 1023, 1025,
1026, 1032, 1034, 1035, 1036, 1040, 1044, 1050, 1050, 1053, 1054,
1056, 1056, 1056, 1058, 1064, 1066, 1073, 1075, 1080, 1080, 1080,
1081, 1085, 1088, 1092, 1092, 1100, 1102, 1104, 1104, 1107, 1110,
1116, 1118, 1120, 1120, 1122, 1125, 1128, 1131, 1134, 1140, 1144,
1147, 1148, 1150, 1152, 1152, 1155, 1160, 1161, 1170, 1170, 1175,
1176, 1178, 1184, 1188, 1188, 1189, 1190, 1196, 1200, 1200, 1204,
1209, 1215, 1216, 1218, 1221, 1222, 1224, 1230, 1232, 1240, 1242,
1247, 1248, 1248, 1254, 1258, 1260, 1260, 1260, 1269, 1271, 1276,
1280, 1287, 1288, 1290, 1292, 1295, 1296, 1302, 1305, 1312, 1316,
1320, 1320, 1326, 1330, 1332, 1333, 1334, 1344, 1344, 1350, 1353,
1360, 1363, 1364, 1365, 1368, 1376, 1380, 1386, 1392, 1394, 1395,
1400, 1404, 1406, 1408, 1410, 1419, 1426, 1428, 1435, 1440, 1440,
1440, 1443, 1452, 1457, 1462, 1470, 1472, 1476, 1480, 1482, 1485,
1488, 1496, 1504, 1505, 1512, 1517, 1518, 1520, 1530, 1536, 1540,
1548, 1551, 1554, 1558, 1560, 1564, 1575, 1584, 1584, 1591, 1596,
1598, 1599, 1610, 1620, 1628, 1632, 1634, 1638, 1640, 1645, 1656,
1665, 1672, 1677, 1680, 1680, 1692, 1702, 1710, 1716, 1720, 1722,
1728, 1739, 1748, 1755, 1760, 1763, 1776, 1786, 1794, 1800, 1804,
1806, 1824, 1833, 1840, 1845, 1848, 1872, 1880, 1886, 1890, 1892,
1920, 1927, 1932, 1935, 1968, 1974, 1978, 1980, 2016, 2021, 2024,
2064, 2068, 2070, 2112, 2115, 2160, 2162, 2208, 2256
There's a simpler form for the product of the pairs, it's ((sum n)^2 -
sum (n^2))/2, but there might be use of an algorithm for mechanistic
computation of that value.
I consider then the next function, that to generate the product of
each triple. Consider a three-dimensional cubic array, and the values
under the diagonal.
Ross
Ross A. Finlayson
each element of N or Z+ less than n: {1, ..., n}? It's the set of
counting numbers less than or equal to n, it's the set of contiguous
elements of N from 1 to n, etcetera.
How is done summing the products of each triple of this set?
Summing the products of pairs of the set is as an expression ((sum
n)^2 - sum(n squared))/2. Is the sum of products of triples in the
form ((sum n)^3 - sum(n cubed))/8?
Consider the case with 1, 2, 3. There is only one triple, {1, 2, 3},
and the sum of the product of itself is 6.
1 1 2 3
1 1 2 3
2 2 4 6
3 3 6 9
2 1 2 3
1 2 4 6
2 4 8 12
3 6 12 18
3 1 2 3
1 3 6 9
2 6 12 18
3 9 18 27
The diagonal of that, in a way, would be 1, 8, 27, for (1, 1, 1), (2,
2, 2), (3, 3, 3). The value of (sum n)^3 is 6^3 = 216. sum (n^3) is
36. (sum n)^3 - sum (n^3) = 180. 180 / 30 = 6, the result. The
result is also the sum and the product of (1, 1, 1), (1, 1, 2), and
(1, 1, 3).
For n=4, there are these triples:
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
The value of (sum n)^3 is 1000, the value of sum n^3 is 100. The sum
of the products of the triples is 6+8+12+24=50. 900/18 = 50.
For n=3, the divisor x is 30, for n=4, 18, to get a correct expression
for the sum of the products of triples being ((sum n)^3 - sum n^3) /
x.
The idea here is to determine forms for the coefficients of the
polynomial (n+1)...(n+n)/n^n. Determining a function of n for the sum
of the product of each triple of {1, ...n} would give an expression
for the fourth coefficient, a_4. The third coefficient, a_3, has as
an expression ((sum n)^2-sum n^2)/2, the second coefficient has the
form sum n = (x+1)x/2, the first coefficient is equal to 1. It would
be acceptable/preferable to have expressions of the coefficients that
only apply to an infinite value of n, or fractional expressions of n.
For n=5, sum n is 15, (sum n)^3 is 3375, and sum (n^3) is 225, their
difference is 3150. The triples of {1,...5} are
{1, 2, 3}
{1, 2, 4}
{1, 2, 5}
{1, 3, 4}
{1, 3, 5}
{1, 4, 5}
{2, 3, 4}
{2, 3, 5}
{2, 4, 5}
{3, 4, 5}
2*(4)
3*(2+4+8+12)
5*(2+3+4+6+8)
The sum of their products is 6+8+10+12+15+20+24+30+40+60 = 225. The
value of x for 3150/x=225 is 14. Thus for some f(x), f(3)=30,
f(4)=18, f(5)=14.
I guess what I should make here is an algorithm to list each subset of
a given number of elements, here 3, of {1, ..., n}. Also I should
research the function that gives the number of subsets of a given
length of a set of a given length. What is the analysis of a bit
sequence of length n with x "on" bits, and each of those sequences?
I look to MathWorld and it notes that subsets of length k of a set are
called k-subsets.
http://mathworld.wolfram.com/k-Subset.html
There are n choice k many k-subsets of a set of length n: n! / (n-k)!
k!. What I want to determine here are what their products are for {1,
..., n}. In the case of k=3, for n=3, 4, 5 there are 1, 4, and 10
permutations, the sum of their products is 6, 50, 225.
Earlier, I noted that
((sum n)^2 - sum(n^2)) / 2 = (n^2+n)^2 / 4 - sum(n^3+n^2)/2.
(sum n)^2 - sum(n^2) = ((n^2+n)^2) / 2 - sum(n^3+n^2)
(sum n)^2 = ((n^2+n)^2) / 2 - sum(n^3+n^2) + sum(n^2)
(sum n)^2 = ((n^2+n)^2) / 2 - sum(n^3)
Here when I have been writing sum (f(n)) it means sum for i=1 to n of
f(n).
(sum n)^2 = (2 (sum n)^2 - sum(n^3)
sum(n^3) = (sum n)^2
Hmm, that's kind of interesting. I wasn't aware of that relation.
The sum of the cubes of the numbers from 1 to n is equal to the square
of the sum of the numbers from 1 to n.
That's saying that taking the sum of the cubes of the numbers from one
to a zillion, that that sum is equal to the square of the sum of the
numbers from one to a zillion.
What's up with that? Is that well-known or trivial, and if so, in
what context? Here a zillion is an arbitrarily large integer, and a
power of ten, and is less than a gazillion, another abitrarily large
power of ten.
Ross
Ross A. Finlayson
..., n} and it's taking a little longer than I thought it would.
I notice that a subset of a set of n elements can be represented as an
n-bit word, with an on bit for that the element at that index is in
the subset, and an off bit for that it is not. For example, for an
eight bit word, that can represent a subset of a set of eight
elements, in this case {1, ..., 8}. The set {1, ..., 8} is 11111111
or in hexadecimal notation 0xFF. The set {1} would be 10000000, {8}
would be 00000001, with the empty set {} being 00000000.
Anyways, I have started writing a recursive method to enumerate the
subsets, and it's a hassle to write and has quite a few expressions of
n and k. In frustration I started typing patterns and noticed
something about how the subsets may be generated with a seemingly
simpler method, at least computationally.
Consider a set of say, six elements, {1, 2, 3, 4, 5, 6}, and the
3-subsets of that, each subset with three elements. Start with {1, 2,
3} in the binary notation, and then shift right, carrying over the
right bit back to the beginning of the sequence.
1, 1, 1, 0, 0, 0
0, 1, 1, 1, 0, 0
0, 0, 1, 1, 1, 0
0, 0, 0, 1, 1, 1
1, 0, 0, 0, 1, 1
1, 1, 0, 0, 0, 1
Then, I start with {1, 3, 4}, having shifted the bits after the first
one bit right. Then I repeat the right-shift with wraparound n=6
times again.
1, 0, 1, 1, 0, 0
0, 1, 0, 1, 1, 0
0, 0, 1, 0, 1, 1
1, 0, 0, 1, 0, 1
1, 1, 0, 0, 1, 0
0, 1, 1, 0, 0, 1
Next is {1, 4, 5}, shifted six times.
1, 0, 0, 1, 1, 0
0, 1, 0, 0, 1, 1
1, 0, 1, 0, 0, 1
1, 1, 0, 1, 0, 0
0, 1, 1, 0, 1, 0
0, 0, 1, 1, 0, 1
Finally there is {1, 3, 5}, there have been eighteen sequences
representing different subsets thus far, two more will make the
remaining 3-subsets.
1, 0, 1, 0, 1, 0
0, 1, 0, 1, 0, 1
I thought I had stumbled upon some incredibly simple method to
enumerate the k-subsets of a set. Yet, it wasn't immediately clear
how it would work with other values of n and k, besides the fact that
it is very easy to implement as a machine.
The 1-subsets of {1, ..., 6} are simple enough:
1, 0, 0, 0, 0, 0
0, 1, 0, 0, 0, 0
0, 0, 1, 0, 0, 0
0, 0, 0, 1, 0, 0
0, 0, 0, 0, 1, 0
0, 0, 0, 0, 0, 1
That is {1}, {2}, {3}, {4}, {5}, {6}. The 2-subsets of {1, ..., 6}
are not quite so simple. There are 6!/4!2! many of them, or
6*5/2*1=15:
1, 1, 0, 0, 0, 0
0, 1, 1, 0, 0, 0
0, 0, 1, 1, 0, 0
0, 0, 0, 1, 1, 0
0, 0, 0, 0, 1, 1
1, 0, 0, 0, 0, 1
1, 0, 1, 0, 0, 0
0, 1, 0, 1, 0, 0
0, 0, 1, 0, 1, 0
0, 0, 0, 1, 0, 1
1, 0, 0, 0, 1, 0
0, 1, 0, 0, 0, 1
1, 0, 0, 1, 0, 0
0, 1, 0, 0, 1, 0
0, 0, 1, 0, 0, 1
Very good. It seems that there is a simple pattern to generate
sequences representing the subset elements. Yet, this is so far only
for values of n and k such that n is around 2k, and that k is a factor
of n, for small n. I guess I should implement this and see if it
works.
Consider the 4-subsets of {1, ..., 6}. There are just as many
4-subsets of six as there are 2-subsets of six, 6!/2!4! = 6!/4!2!.
Those sequences above are the 2-subsets of 6, their complements are
the 4-subsets of six. Thus I can consider only the cases where k <=
n/2. The process generates the k-subsets and (n-k)-subsets at about
the same time, it is a matter of a complement operation on each.
Of course, my goal here is to sum the products of those subsets, the
4-subsets would have higher products than the 2-subsets.
Given one of these sequences, I wonder if there's a way to quickly
calculate the product of the elements that it represents. The bit
sequence up to word size is typically stored on a machine register in
a computer program.
I can see some problems with this method for higher values of n. The
size of n isn't a problem, it's when k is small compared to n,
generating the starter sequences requires its own algorithm. It's
when k is relatively prime compared to n that this is more complex,
when k is a factor of n it is more simple. For example, consider the
3-subsets of {1, ..., 8}. There would be 8!/5!3! of them, 8*7*6/3*2 =
56. These would be the starter sequences with 8 many each:
1, 1, 1, 0, 0, 0, 0, 0
1, 0, 1, 1, 0, 0, 0, 0
1, 0, 0, 1, 1, 0, 0, 0
1, 0, 0, 0, 1, 1, 0, 0
1, 0, 0, 0, 0, 1, 1, 0
That's 40 of them, then there are 16 more.
1, 0, 1, 0, 1, 0, 0, 0
1, 0, 0, 1, 0, 0, 1, 0
Hmm..., well there they are, those two sequences each generate eight
subsets. That's different from the case where the sequence 0, 1, 0,
1, ... generates only two subsets.
I wonder how it is different for odd n.
How many starter sequences are there for k many sequences apiece?
I'll call them the blocked starter sequences. Let's look at that:
for n=6, k=1 there is one blocked sequence, for n=6, k=2 there are two
blocked sequences, for n=6, k=3 there are 3. For n=8, k=1 there is
one, n=8, k=2, there could be 6 except {1, 3} is considered under the
alternating starter sequence, it would have to be not considered. It
should be a blocked starter sequence and not repeated in the
alternating starter sequence. For n=8, k=3 there are 5, and for n=8,
k=4, there is four.
Then there are the alternating starter sequences, there is a given
number of them based upon n and k and whether k is a factor of n, and
then they generate a given number of sequences based upon n!/k!(n-k)!
- k(n-k) and the number of sequences. Here I'm going to work with k <
n/2, so the function there is about the greatest integer x thus that
k/n < 1/x. For example for n=8, k=3, 1/3 < 3/8 < 1/2, x =2, there are
2 alternating starter sequences. I guess it's easier to compare n/k
to an integer, 8/3 in integer math will give the answer of x=2. Then,
to generate the alternating starter sequences there is 1, 0, 1, 0,
..., spacing the on bits by one off bit, then 1, 0, 0, 1, 0, 0,
etctera, the sequence yn+1 for postive integer y <= x, i.e. {1, 3, 5,
...}, {1, 4, 7, ...}, {1, 5, 9, ...}, each of those. For example in
the above description for k=3 and n=8, x =2 and thus for y=1 and y=2
the starter sequences are for {1, 3, 5} and {1, 4, 7}.
Why is it that there are k many subsets from alternating starter
sequences where k is not a factor of n and n/k many for k being a
factor of n? Maybe it is more apt to say that as k divides into n
that there are only n/k many subsets from the alternating starter
sequence.
In a way the k-subsets relate back to the concept of the canonical
sequence. This is where the canonical sequence for k=n/2 is
101010..., and the other k-subset's sequences canonicalize to that.
For n=8, there are more starting sequences. This is where there are
enough sequence elements to separate blocks of multiple elements.
For n=10, k=5, there are 10!/5!5! or 10*9*8*7*6/5*4*3*2*1 =
2*3*2*7*3=252 5-subsets of {1, ..., 10}.
1, 1, 1, 1, 1, 0, 0, 0, 0, 0
1, 0, 1, 1, 1, 1, 0, 0, 0, 0
1, 0, 0, 1, 1, 1, 1, 0, 0, 0
1, 0, 0, 0, 1, 1, 1, 1, 0, 0
1, 0, 0, 0, 0, 1, 1, 1, 1, 0
1, 1, 0, 1, 1, 1, 0, 0, 0, 0
1, 1, 0, 0, 1, 1, 1, 0, 0, 0
1, 1, 0, 0, 0, 1, 1, 1, 0, 0
1, 1, 0, 0, 0, 0, 1, 1, 1, 0
1, 1, 0, 1, 1, 0, 1, 0, 0, 0
1, 1, 0, 1, 1, 0, 0, 1, 0, 0
1, 1, 0, 1, 1, 0, 0, 0, 1, 0
1, 1, 0, 1, 0, 1, 1, 0, 0, 0
1, 1, 0, 1, 0, 1, 0, 1, 0, 0
1, 1, 0, 1, 0, 1, 0, 0, 1, 0
1, 1, 0, 1, 0, 0, 1, 1, 0, 0
1, 1, 0, 0, 1, 1, 0, 1, 0, 0
1, 1, 0, 0, 1, 1, 0, 0, 1, 0
1, 1, 0, 0, 0, 1, 1, 0, 1, 0
1, 0, 1, 0, 1, 1, 1, 0, 0, 0
1, 0, 1, 0, 0, 1, 1, 1, 0, 0
1, 0, 1, 0, 0, 0, 1, 1, 1, 0
1, 0, 0, 1, 0, 1, 1, 1, 0, 0
1, 0, 0, 1, 0, 0, 1, 1, 1, 0
1, 0, 0, 0, 1, 0, 1, 1, 1, 0
Those 25 sequences, if correct, rotate to each form ten sequences
forming a total of 250 sequences representing 5-subsets of 10.
With those blocked starting sequences there is the alternating
sequence, it generates 10/5 = 2 5-subsets:
1, 0, 1, 0, 1, 0, 1, 0, 1, 0
At that point again it is about as not simple as the regular recursive
method to enumerate the subsets. Yet, it is much more computationally
efficient.
Figuring this out would probably help me determine how the product
lists as discussed earlier would progress.
I thought to look at discussion of Gray codes, they are progressions
of sequences, but their use is not immediately apparent here as the
sequences here have fixed numbers of on and off bits. The Hamming
distance is the minimum number of bits to be changed to convert one
sequence to another. What's the regular way to enumerate the
sequences with a given number of on bits that are each different on
rotation?
I figure the easiest way to generate the byte arrays of each rotation
of a starter sequence will be to put two copies of the starter
sequence into a sequence of bits, and then to extract subsequences.
For example, for the starter sequence
1, 0, 0
put into memory
1, 0, 0, 1, 0, 0
and then extract the three subsequences of length three starting at
the index from one to three:
[ 1, 0, 0, ] 1, 0, 0 -> 1, 0, 0
1, [ 0, 0, 1, ] 0, 0 -> 0, 0, 1
1, 0, [ 0, 1, 0, ] 0 -> 0, 1, 0
This is similar to having the sequence being written about a circle,
and then reading off a sequence of that length from each element,
beads on a string. Some machine architectures might have instructions
to rotate a register.
I was reading MathWorld and notice that Mathematica has a KSubsets
method to list the k-subsets of a set, I'm still in the process of
implementing it. The other day I was reading the documentation of
HYPERG, a Hypergeometric function package for Maple, based on HYP.
Today I'll read the documentation of HYP.
Heh, bait.
Ross
Virgil
> I started writing a program to enumerate the k-subsets of a set {1,
> ..., n} and it's taking a little longer than I thought it would.
If Ross means subsets with k elements, he should try the binomial
coefficient, C(n,k).
If he does not mean that, he should explain up front what he does
mean.
Ross A. Finlayson
Determining which subsets of a set have a given product appears to be
known as a subset product problem.
http://www.google.com/search?q=%22Subset+Product+Problem%22
http://www.csc.liv.ac.uk/~ped/teachadmin/COMP202/annotated_np.html
Here, I am concerned with the sum of each product of each k-subset of
a set for a given k of a set {1, ..., n}, not the subset product
problem.
For one-subsets, the sum of the elements of {1, ..., n} and as well
the sum of the products of the elements is (n+1)n/2. For a set {x,
2x, ..., xn}, the sum of the elements and sum of the products of the
one-subsets is x(n+1)n/2.
For a set {1, ..., n} the sum of the products of the 2-subsets is
((sum n)^2 - sum (n^2))/2, the sum of the products of the 3-subsets
is ((sum n)^3 - sum (n^3))/f(n), for f(n) being some function of n.
It may well be that the sum of the products of the k-subsets of {1,
..., n} is ((sum n)^k - sum (n^k))/f(n,k).
So anyways I want to determine the starter sequences of n-bit words
for an algorithm to enumerate the k-subsets of a set. Here's what I
think it has to do with: each of the ways that a positive integer can
be expressed as the sum of positive integers. For example, for a few
values of k:
1: 1+0 = 1
2: 2+0, 1+1 = 2
3: 3+0, 2+1, 1+1+1 = 3
4: 4+0, 3+1, 2+2, 2+1+1, 1+1+1+1 = 5
5: 5+0, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1 = 7
For a given integer k, as k increases from one then the number of ways
to represent k as a sum of smaller positive integers rises rapidly
past k. In a way this is similar to factoring for addition,
summorization, addorization. Are there P(k) summorizations of k for
P(x) being the x'th prime? I keep in mind the law of small numbers.
There also has to be a consideration of of when a number can be the
sum of, for example, 2, 1, 2, and 1, 2+1+2+1=6, that in the sequences
that the ordering of those contiguous subsequences of that many on
bits could be 1, 2, 2, 1 or 1, 2, 1, 2, with as well then having as
many zeros as there could be between each contiguous on subsequence.
1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, ...
...
1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, ...
...
The number of k-subsets of a set of length n is (n choice k): n!/
k!(n-k)!: n(n-1)...(n-k+1) / k(k-1)...(1). Here, k <= n/2. For each
of the sequences that is generated, it rotates to generate either n or
an integer multiple less than k times n/k sequences. Say that x many
sequences rotate fully to generate n many sequences each, and then
there are, say, y many other sequences that rotate partially to
generate n/k many sequences each, then x*n + y*n/k = n!/ k!(n-k)!.
Yet, there might be some sequences that rotate partially for 2n/k
sequences, where others are for 4n/k, for integer multiples, where the
y many subsequences are divided into y_i's.
The idea here is still to generate forms for the coefficients of the
polynomial (n+1)...(n+n), and generally for (n+1)...(n+n/x). That is
about examination of Euler's Gamma function for arbitrarily large
values of z.
"Voidness is that which stands right in the middle between this and
that. The void is all-inclusive, having no opposite - there is nothing
which it excludes or opposes. It is living void, because all forms
come out of it and whoever realizes the void is filled with life and
power and the love of all being."
"The knowledge and skills you have achieved are meant to be forgotten
so you can float comfortably in emptiness, without obstruction." -
Bruce Lee
Ross
Ross A. Finlayson
numbers.
The triangular numbers numbers of the form n(n+1)/2, the tetrahedral
numbers are n(n+1)(n+2)/6. Following in this pattern would be
n(n+1)...(n+k+1)/(k+1)!.
A set {1, ..., n} can be called an n-set. The combinations with
k-subsets of elements of an n-set are its k-subsets, the enumeration
of combinations of an n-set are its k-subsets.
There are a variety of algorithms to enumerate the combinations of a
set, combinatorial algorithms, and various orders of those
combinations in the enumerated list, for example named lexicographic,
co-lexicographic, RD(recursive descent?), transpositional, other
order. Referenced are Nijenhuis and Wilf, Ruskey, Kreher, Stinson,
Skiena, combinatoric algorithms.
http://www.cs.sunysb.edu/~algorith/implement/wilf/implement.shtml
Hey, cool, the graphic on that webpage has the "summorizations" of 5,
maybe it wil explain what they are called.
For my algorithm that is to generate a list of sequences and then
rotate them to generate the rest of the sequences, I need to figure
out how to generate the sequences.
What the "summorization" is is called the number of partitions,
additive partitions.
http://www.mathpages.com/home/kmath091.htm :
"The total number of partitions A(n) is obviously just A_n(n) from
this table, so we have the sequence of values for A(n) for n=1,2,...
n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
A(n) = 1 2 3 5 7 11 15 22 30 42 56 77 101 135 ..."
So in generating the starter sequences, what is of use is something
between A(n) and a(n).
"We can also define a(n) the same as A(n) except that distinct
orderings of the terms are counted as distinct partitions."
What is of interest is not all of the distinct orderings, but those
that are distinct under rotation, for example considering 1+2+1+2
different than 1+1+2+2 but the same as 2+1+2+1.
Then, for that there k=6, and n>=2k, then there are n-k zeros. The
partition has 4 summands, four blocks of one or more zeros are
required, from n-k many zeros. For example, where n=12, there are 6
zeros, so each of the distinct additive partitions with 4 summands is
of interest.
http://mathworld.wolfram.com/PartitionFunctionP.html
So I am wondering about rotationally distinct additive partitions of k
and n-k.
1: 1+0
2: 2+0, 1+1
3: 3+0, 2+1, 1+1+1
4: 4+0, 3+1, 2+2, 2+1+1, 1+1+1+1
5: 5+0, 4+1, 3+2, 2+2+1, 3+1+1, 2+1+1+1, 1+1+1+1+1
Each of the above partitions is distinct from the others, its summands
are different. Also, reordering the summands in any way does not make
an order of the summands that is not created by rotation of the
summands. For example, any way that you switch the summands of
2+1+1+1, any permutation of (2, 1, 1, 1), it is either 2+1+1+1,
1+2+1+1, 1+1+2+1, or 1+1+1+2.
6: 6+0, 5+1, 4+2, 3+3, ..., 1+2+1+2, ...
Here of note is 1+2+1+2. It is "rotationally distinct" from 1+1+2+2.
This is where the rotations of 1+1+2+2 are 2+1+1+2, 2+2+1+1, and
1+2+2+1, none of which is 1+1+2+2. Similarly, for 3+2+1 there would
also be 3+1+2 but not 1+3+2, where 1+3+2 rotates to 3+2+1 so is not
"rotationally distinct".
Here I hope there is an existing term for "rotationally distinct",
like calling an additive partitioning a "summandorization". A search
of the term returns some results that appear to be consistent in their
usage.
Here what this is about is having k many elements of an n-set, the
combinations of elements of an n-set with k many elements. The n-set
is represented as a sequence of n bits, and then there are k on bits
and n-k off bits in each bit sequence representing a k-subset. For
each of the ways that the number k can be additively partioned that is
rotationally distinct, that generates a variety of sequences along
with how n-k may be additively partitioned into as many partitions as
there are partitions f k, to interleave the partitions.
For example, for k=2, there are two possible rotationally distinct
partitions of k: 2 and 1+1. The first partition has 1 element, 2,
the second partition 2, the first and second one. Then in an n-set
with obviously n elements, there are n-k zeros, these can be
partitioned in one way with one element in the partition: n-k, and a
variety of ways with two elements in the partition, and that is what
is relevant to the consideration of how many "rotationally distinct"
sequences there are with k many on bits and n-k many off bits.
Then when we get to a partition of k that has elements that may be
"rotationally distinct" in addition to the weaker condition
"distinct", then those have to be considered in the algorithm to
enumerate the combinations of an n-set. The idea behind using this
algorithm is that it might be computationally efficient, yet all this
business about determining additive partitions for k and n-k to
determine sequences that are unique under rotation might lead it to be
inefficient.
It might be very efficient.
Ross
Ross A. Finlayson
the sequences A_m that sum to n, with A_i >= A_(i+1). Equivalently
they are the sets the elements of which sum to n.
Another consideration is a superset of the integer partitions which
are each permutation of the sets of which the elements sum to n.
What I'm interested in are the subset of the permutations of the
integer partitions which are having the cyclic characteristic of being
distinct on rotation, cyclic permutation. For example, the sequence
(2, 1, 2, 1) is indistinct from (1, 2, 1, 2), but distinct from (1, 1,
2, 2).
http://mathworld.wolfram.com/Permutation.html
(That reference lists permutations as sets, I think they should be
sequences instead, i.e. {1, 2, 3} is the same set as {3, 2, 1} but (1,
2, 3) is a different sequence than (3, 2, 1), a notation quibble. )
The reason is that I require algorithms to generate a variety of
integer partitions. I want to generate the rotationally distinct
unordered partition sequences of a given number of partitions of for a
number A_k_p, and then generate the unordered partition sequences with
the same number of partitions for another larger number, B_(n-k)_p.
The idea then is that for A_k_p x B_(n-k)_p that there is a sequence
(a_1_p, b_1_p, ..., ..., a_i_p, b_i_p). This sequence then can
represent a sequence with a_1_p many ones, then b_1_p many zeros,
etcetera, to a_i_p many ones and b_i_p many zeros. The resulting
sequence represents a k-subset of an n-set, and can be rotated to
generate n or a fraction of n many other k-subsets of the n-set. The
idea is that it is an algorithm to enumerate k-subsets, sets of a
given size, of an n-set.
The idea behind generating k-subsets is then to multiply their
elements together to sum those proucts and get the result of the sum
of the product of each k-subset of an n-set, {1, ..., n}. There are
simpler methods to enumerate the k-subsets.
Well this is pretty simple, the sum of the products of the unordered
pairs of {1, ..., n} are the Stirling numbers of the first kind s(n+1,
n-1).
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000914
http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html
There's an error in the Encyclopedia listing, it says "Sum of product
of unordered pairs of numbers from {1..n+1}", where it is actually
"Sum of product of unordered pairs of numbers from {1..n}".
I search the Encyclopedia of Integer Sequences for "Sum of product"
and get some search results, but none of them is the Sum of product of
unordered trios of {1, ..., n}.
Would that be s(n+2, n-1)? Gee, wouldn't that be great. Perhaps not,
as they're all negative, but -s(n+1, n-2) might be closer to the mark.
Let's see, going back to (n+1)...(n+n)/n^n, the first coefficient of
the polynomial is 1, the second is then n(n+1)/2, the third is here
being s(n+1, n-1).
Yet, I want an equation in terms of n that is this value, the
"Stirling number of the first kind" does not really help here, unless
I find a way to derive a simple form.
It is so that ((sum n)^2 - sum (n^2))/2 = s(n+1, n-1), for n.
It does appear that the sum of the products of each unordered trio of
elements from {1 ,..., n}, for n>=3, is the Stirling number of the
first kind s(n+1, n-2).
It isn't so that ((sum n)^3) - sum (n^3))/6 = -s(n+1, n-2). It
appears to be some other function. Here are the first values of
-s(n+1, n-2) from a table:
n=3, -s(4, 1)=6
n=4, -s(5, 2)=50
n=5, -s(6, 3)=225
n=6, -s(7, 4)=735
n=7, -s(8, 5)=1960
n=8, -s(9, 6)=4536
n=9, -s(10, 7)=9450
n=10, -s(11, 8)=18150
The sum of i for i=1 to n is the n'th pyramidal number, n(n+1)/2. Its
cube is (sum n)^3.
n=3,(sum n)^3=6^3=216
n=4,(sum n)^3=10^3=1000
n=5,(sum n)^3=15^3=3375
n=6,(sum n)^3=21^3=9261
n=7,(sum n)^3=28^3=21952
n=8,(sum n)^3=36^3=46656
n=9,(sum n)^3=45^3=91125
n=10,(sum n)^3=55^3=166375
The sum of i^3 for i=1 to n is perhaps a numerate number. It equals
(sum n)^2.
n=3, sum (n^3)=1+8+27=36
n=4, sum (n^3)=1+8+27+64=100
n=5, sum (n^3)=1+8+27+64+125=225
n=6, sum (n^3)=1+8+27+64+125+216=441
n=7, sum (n^3)=1+8+27+64+125+216+343=784
n=8, sum (n^3)=1+8+27+64+125+216+343+512=1296
n=9, sum (n^3)=1+8+27+64+125+216+343+512+729=2025
n=10, sum (n^3)=1+8+27+64+125+216+343+512+729+1000=3025
The difference of (sum n)^3 and sum (n^3) is going to be a multiple of
-s(n+1, n-2), the sum of the products of each trio of {1, ..., n}, t
was for small values and maybe for these yet small values it will show
a pattern based upon n.
n=3, ((sum n)^3-sum(n^3))=216-36=180
n=4, ((sum n)^3-sum(n^3))=1000-100=900
n=5, ((sum n)^3-sum(n^3))=3375-225=3150
n=6, ((sum n)^3-sum(n^3))=9261-441=8820
n=7, ((sum n)^3-sum(n^3))=21952-784=21168
n=8, ((sum n)^3-sum(n^3))=46656-1296=45630
n=9, ((sum n)^3-sum(n^3))=91125-2025=89100
n=10, ((sum n)^3-sum(n^3))=166375-3025=163350
Then, I will divide ((sum n)^3-sum(n^3)) by -s(n+1, n2), to see what
is the ratio of the difference of the two constructions.
n=3, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=180/6 = 30
n=4, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=900/50 = 18
n=5, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=3150/225 = 14
n=6, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=8820/735 = 12
n=7, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=21168/1960 = 10.8
n=8, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=45630/4536 = ~10.05952
n=9, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=89100/9450 = ~9.428751
n=10, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=163350/18150 = 9
So then I'm looking for a function thus that
f(3)= 1/30
f(4)= 1/18
f(5)= 1/14
f(6)= 1/12
f(7)= 1/10.8
f(8)= 1/ ~10.05952
f(9)= 1/ ~9.428751
f(10)= 1/9
...
where the function f(n) gives -s(n+1, n-2) for ((sum n)^3 - sum(n^3))
* f(n), in a similar way to how s(n+1, n-1) = ((sum n)^2 -sum (n^2)) *
1/2.
The table of Stirling numbers has an entry for n=24, -s(25, 22) =
3795000. Let's see, sum n for n=24 is 24(25)/2= 300, (sum n)^3 is
thus 27000000, sum (n^3) is (sum n)^2 is 90000, their difference is
27000000-90000= 26910000, twenty six million nine hundred and ten
thousand. Dividing that by -s(25, 22) yields around 7.090909, 7
1/11.
Is there a function that works for all values of n that maps ((sum
n)^3 - sum (n^3)) to -s(n+1, n-2)? What is it? Where it's known, it
might be simple to calculate s(n+1, n-2). It's pretty simple to
calculate s(n+1, n-1). How about for ((sum n)^4 - sum (n^4)) and
s(n+1, n-3), and so on for ((sum n)^x - sum (n^x)) and s(n+1,
n-(x-1))?
The Stirling number of the first kind, s(n, m), is (-1)^(n-m) times
the number of permutations of n symbols which have exactly m cycles.
So, the sum of the products of trios of a set {1, ..., n} is the
number of permutations of n+1 symbols which have exactly n-2 cycles.
Well, that's about that. This is about something different, what's
the sum of the reciprocals of n!? That is, what is 1/(1!) + 1/(2!) +
1/(3!) ... + 1/(n!) + ...?
What are the coefficients of the polynomial (n+1)(n+2)...(n+n)?
Ross
Ross A. Finlayson
I figured out a function for f(n) of ( (sum n)^3 - sum (n^3) ) / f(n)
= | s(n+1, n-2) |.
It appears that f(n)= 6(n+2)/(n-2) works.
With this then there is a way to calculate the Stirling number of the
first kind s(n+1, n-2) without calculating intermediate results of a
recurrence relation.
The value of (sum n) is n(n+1)/2, (n^2+n)/2 the value of sum (n^3) is
(sum n)^2, or (n(n+1)/2)^2. The expression ((n^2+n)/2)^2 is
(n^4+2n^3+n^2)/4, ((n^2+n)/2)^3 is (n^6+2n^5+n^4 + n^5 + 2n^4 + n^3)/8
= (n^6+3n^5+3n^4+n^3)/8.
So, there is:
( (sum n)^3 - (sum n)^2 ) (n-2)/ 6(n+2)
( ((n^2+n)/2)^3 - ((n^2+n)/2)^2 ) (n-2) / 6 (n+2)
( ((n^6+3n^5+3n^4+n^3)/8) - ((n^4+2n^3+n^2)/4) ) (n-2) / 6(n+2)
( ((n^6+3n^5+3n^4+n^3)/8) - ((2n^4+4n^3+2n^2)/8) ) (n-2) / 6(n+2)
( n^6+3n^5+3n^4+n^3 - (2n^4+4n^3+2n^2) ) (n-2) / 48(n+2)
( n^6+3n^5+n^4-3n^3-2n^2) (n-2) / 48(n+2)
( n^7 + 3n^6 +n^5-3n^4-2n^3 -2n^6 -6n^5 -2n^4 +6n^3 +4n^2) / 48(n+2)
( n^7 +n^6 -5n^5 -5n^4 +4n^3 +4n^2 ) / 48(n+2)
Evaluating for n=3:
( 2187 + 729 -1215 - 405 + 108 + 36 ) / (48 * 5)
1440 / (48*5) = 6
Yep, that there is -s(4, 1).
So, it appears that the absolute value of s(n+1, n-2), for n >= 3, is:
( n^7 +n^6 -5n^5 -5n^4 +4n^3 +4n^2 ) / 48(n+2)
I wonder if there is a similar method to get a formula representing
Stirling numbers s(n+1, n-3). One thing that would not be as simple
to represent is that where sum (n^3) = (sum n)^2, sum (n^4) is a
recurrence relation instead of a function. What's the most simple way
to represent sum (n^4) as a function? For what other values of x, if
any, is sum (n^x) = (sum n)^f(x)?
Ross
Ross A. Finlayson
of the first kind | s(n, m) | ? It can be.
I search plaintively and come across a form for s(n, n-2).
http://mathforum.org/library/drmath/view/61259.html
Thanks, Doctor Math.
s(n, n-2) = n(n-1)(n-2)(3n-1)/24
s(n, n-2) = (n^2-n)(3n^2-7n+2)/24
s(n, n-2) = (3n^4 -7n^3 +2n^2 -3n^3 +7n^2-2n)/ 24
s(n, n-2) = (3n^4 -10n^3 +9n^2 -2n) / 24
Is the sum of products of pairs of {1, 2, 3} equal to |s(3+1, 3-1)|?
(3*4^4 -10*4^3 +9*4^2-2*4) / 24
(768 -640 +144 -8 ) / 24 = 11
((sum n)^2 - sum (n^2)) / 2
(n(n+1)/2)^2 - sum (n^2)) / 2
(3(4)/2)^2 - (1+9+16) / 2
(36-24) / 2
11
Yes, it is.
So now there are these equations:
s(n, n-2) = (3n^4 -10n^3 +9n^2 -2n) / 24
s(n+1, n-2) = ( n^7 +n^6 -5n^5 -5n^4 +4n^3 +4n^2 ) / 48(n+2)
Also something to consider is that:
(3n^4 -10n^3 +9n^2 -2n) / 24 = ((sum (n-1))^2 - sum ((n-1)^2)) / 2
(3n^4 -10n^3 +9n^2 -2n) / 12 = ( (n(n-1)/2)^2 - sum ((n-1)^2))
(3n^4 -10n^3 +9n^2 -2n) / 12 = ((n^2-n)/2)^2 - sum ((n-1)^2))
(3n^4 -10n^3 +9n^2 -2n) / 12 = (n^4-2n^3+n^2)/4 - sum ((n-1)^2))
(3n^4 -10n^3 +9n^2 -2n) / 12 = (3n^4-6n^3+3n^2)/12 - sum ((n-1)^2))
(3n^4-6n^3+3n^2)/12 - (3n^4-10n^3+9n^2-2n)/12 = sum ((n-1)^2))
(4n^3-6n^2+2n)/12 = sum ((n-1)^2))
Huh. I'll try that for n=2, (4*8-6*4+4)/12 = 1. how about for n=3,
(4*27-6*9+2*3)/12 = 5 = 1^2+2^2, excellent!
So now besides a function for n for sum (n^3) I have a function for
sum ((n-1)^2).
Let's see, in terms of sum (n^2) then the expression would be:
(4(n+1)^3 - 6(n+1)^2 +2(n+1)) / 12
(4(n^2+2n+1)(n+1) - 6(n^2+2n+1) + 2(n+1) ) / 12
( 4(n^3+n^2+2n^2+2n+n+1) - 6(n^2+2n+1) + 2(n+1) ) / 12
( 4n^3 +12n^n +12n +4 -6n^2 -12n -6 +2n +2 ) / 12
( 4n^3 +6n^2 +2n ) / 12
Is it so that 1+2^2 = 5 = (4*8 + 6*4 + 2*2)/12? It is!
So I have something along the lines of
sum_(i=1^n) i^2 = ( 4n^3 +6n^2 +2n ) / 12
sum_(i=1^n) i^3 = ( n^4 +2n^3 +n^2) / 4
The sum of the products of the pairs of {1, ..., n}, | s(n+1, n-1) |,
is:
(n^4 +2n^3 -3n^2 -2n) / 24
The sum of the products of the trios of {1, ..., n}, | s(n+1, n-2) |,
is:
( n^7 +n^6 -5n^5 -5n^4 +4n^3 +4n^2 ) / 48(n+2)
So now I have the forms for the first four coefficients of
(n+1)(n+2)...(n+n).
Here's something I want you to write if you know: what are forms for
sum(n^x)? That is, what is a function the result of which is the sum
of fourth, fifth, etcetera powers of one through n? Those might
enable the evaluation of Stirling numbers of the first kind
non-recursively.
I notice Leroy Quet had mentioned an integral form of the Stirling
number of the first kind but I can't make it out. Here's something,
on http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html, in
the little permutation diagrams, doesn't it look like the third
diagram of S_1(5, 4) is missing a dot?
About evaluating more Stirling numbers, here's one function that finds
ready application here:
s(n+2, n-1) = (n+1) s(n+1, n-1) + s(n+1, n-2)
There's already a form for s(n+1, n-2).
For what other values of x, if any, is sum (n^x) = f(x)?
Ross
Leroy Quet
>...
>
> I notice Leroy Quet had mentioned an integral form of the Stirling
> number of the first kind but I can't make it out. Here's something,
> on http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html, in
> the little permutation diagrams, doesn't it look like the third
> diagram of S_1(5, 4) is missing a dot?
>
I have not been following this thread too closely; for I do not follow
any thread once it gets really big...
But I decided to click on it today, and I spotted my name.
DID I mention an integral-form for Stirling numbers of any kind???
I would not be too surprised, but I can't find anything by searching
Google.
I think I READ about an integral-form someplace.
And I DID post the occasional Stirling-number equals {something}
identity, especially years ago.
Thanks,
Leroy Quet
Ross A. Finlayson
"Another way is with calculus.
s(k,m) = (1/m!) d^k{(-ln(1-x))^m}/dx^k_{x=0},
where d^k{(-ln(1-x))^m}/dx^k_{x=0}
is the k_th derivative of (-ln(1-x))^m at x = 0."
- Leroy Quet,
http://groups.google.com/groups?selm=68ryb6u79zdl%40legacy
Leroy,
I've read some of your posts on Stirling numbers, they have some
interesting points, except I don't understand some of them, so am not
qualified to comment.
I researched some posts and haven't really found anything that will
help me get a function for s(n+1, n-3) for integer n greater than 3,
or s(n+1, n+1-k) for integer k>0 and integer n>k. Please let me know
if you know the general form of a function that is not recursive or
the product or summation of a series for any Stirling number of the
first kind.
Have a good one, please explain how s(n, m) is computed with computing
s(n-1, m) or s(n, m-1). I think it might have to do with
differentiating the generating function, yet I don't know.
Thanks,
Ross
Ross A. Finlayson
http://cosmic.swau.edu/~turner/math101/materials/series/
S_2 = n(n+1)(2n+1)/6
S_4 = n(n+1)(2n+1)(3n^2+3n-1)/30
S_6 = n(n+1)(2n+1)(3n^4+6n^3-3n+1)/42
S_1 = n(n+1)/2
S_3 = n^2(n+1)^2/4
S_5 = n^2(n+1)^2(2n^2+2n-1)/12
S_7 = n^2(n+1)^2(3n^4+6n^3-2n^2-4n+2)/24
Each of those is an integer for integer n.
There appears to be a book or paper "Simplified Computation of Sums of
Powers of Integers", by Michael Budin and Arnold Cantor.
MathWorld has here a good page:
http://mathworld.wolfram.com/PowerSum.html
It notes that (sum n)^2 = sum (n^3) can be derived from Nicomachus'
theorem.
It says there is more information in the CRC Standard Mathematical
Tables and Formulae, I look there at section 1.2.12, page 20, and see
some information. It says
S_x = (x+1)^(-1) [ B_(x+1) (n+1) - B_(x+1) 0 ]
where B_x is the Bernoulli polynomial, referencing section 1.2.8.
http://mathworld.wolfram.com/BernoulliPolynomial.html
My idea here is to determine the form of sum (n^x) and then subtract
it from (sum n)^x, and then multiply that by some f(x), and get
|s(n+1, n-x+1)|, the sum of the products of x-subsets of {1, ..., n}.
So there is a way to get the sum of the x'th powers of {1, ..., n} but
it again involves recurrence relations where I want a closed form. I
think generating function is an unintuitive nomenclature vis-a-vis the
generated function.
I think that:
|s(n+1, n-k+1)| = ((sum n)^k - sum (n^k))/ f(n, k)
where it is so for f(n, 1) = 2 and f(n, 2) = (n+2)/(n-2)6.
I guess I could plot some values for k=4.
n=4, s(5, 1)=24
n=5, s(6, 2)=274
n=6, s(7, 3)=1624
n=7, s(8, 4)=6769
n=8, s(9, 5)=22449
n=9, s(10, 6)=63273
n=10, s(11, 7)=157773
n=4, (sum n)^4 = 10^4 = 10000
n=5, (sum n)^4 = 15^4 = 50625
n=6, (sum n)^4 = 21^4 = 194481
n=7, (sum n)^4 = 28^4 = 614656
n=8, (sum n)^4 = 36^4 = 1679616
n=9, (sum n)^4 = 45^4 = 4100625
n=10, (sum n)^4 = 55^4 = 9150624
n=4, sum (n^4) = n(n+1)(2n+1)(3n^2+3n-1)/30 = 4*5*9*59/30=354
n=5, sum (n^4) = 5*6*11*89/30 = 979
n=6, sum (n^4) = 6*7*13*125/30 = 2275
n=7, sum (n^4) = 7*8*15*167/30 = 4676
n=8, sum (n^4) = 8*9*17*215/30 = 8772
n=9, sum (n^4) = 9*10*19*267/30 = 15219
n=10, sum (n^4) = 10*11*21*329/30 = 25333
n=4, (sum n)^4 - sum (n^4) = 10000-354 = 9646
n=5, (sum n)^4 - sum (n^4) = 50625-979 = 49646
n=6, (sum n)^4 - sum (n^4) = 194481-2275 = 192206
n=7, (sum n)^4 - sum (n^4) = 614656-4676 = 609980
n=8, (sum n)^4 - sum (n^4) = 1679616-8772 = 1670844
n=9, (sum n)^4 - sum (n^4) = 4100625-15219 = 4085406
n=10, (sum n)^4 - sum (n^4) = 9150624-25333 = 9125291
n=4, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 9646 / 24 ~= 401.9167
n=5, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 49646 / 274 ~= 181.1898
n=6, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 192206 / 1624 ~= 118.3534
n=7, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 609980 / 6769 ~= 90.11375
n=8, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 1670844 / 22449 ~= 74.42844
n=9, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 4085406 / 63273 ~= 64.56792
n=10, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 9125291 / 157773 ~=
57.8381
Well, that is definitely not as obvious as for 30, 18, 14, 12, 10.8,
etcetera. At least it's a decreasing function.
Let's see, for n=4, 9646/24 reduces to 4823/12. 12 could be
(n+2)(n-2). 7 divides into 4823, the result is 689. 689 is 13 times
53. 13 is 3n+1, and n^2-3. For n=5, 49646/274 reduces to 24823/137,
137 is prime. 103*241/137.
Let's see, here are the prime factorizations of those, some via
http://www.alpertron.com.ar/ECM.HTM:
n=4, f(n)=9646/24 = 2*7*13*53 / 2*2*2*3
n=5, f(n)=49646/274 = 2*103*241 / 2*137
n=6, f(n)=192206/1624 = 2*7*13729 / 2*2*2*7*29
n=7, f(n)=609980/6769 = 2*2*5*7*4537 / 7*967
n=8, f(n)=1670844/22449 = 2*2*3*7*19891 / 3*7*1069
n=9, f(n)=4085406/63273 = 2*3*3*13*13*17*79 / 3*7*23*131
n=10, f(n)=9125291/157773 = 7*130613 / 3*7*11*683
For x=2, the function f(n) is f(n)=2. For x=3, f(n)= 6(n+2)/(n-2).
This is about ( (sum n)^x - sum (n^x) ) / f(n, x) = |s(n+1, n-x+1)|.
There are no factors common to each f(4) through f(10). I see a lot
of 2's and 7's, 4^0+1 and 2*4^1-1.
I'll compare 5 to 2*137, n=5. 2n^3= 250, 2n^3+n^2-1 = 274. For n=6,
2n^3+n^2-1 = 467, n^3+n^2 = 232 = 1624/7, 7n^3+7n^2=1624=n^4+2n^3+n^2.
For n=7, n^4+2n^3+n^2= 2401+686+49=3136, 2n^4+4n^3+2n^2+71n = 6769 =
2n^4+5n^3+2n^2+22n = 2n^4+5n^3+5n^2+n=4802+1715+245+7=6769.
n=5, 2n^3 +n^2 -1 = 274 = s(6, 2)
n=6, n^4 +2n^3 +n^2 = 1624 = s(7, 3)
n=7, 2n^4 +5n^3 +5n^2 +n = 6769 = s(8, 4)
Let's see, for 8 and 22449, 2n^4 +5n^3 +5n^2 +n = 8192 + 2560 + 320 +
8 = 11080, 22249-(11080*2) = 89, 11*8+1, so I need to multiply by two,
and add 11*8 and 1, n^2+3n+1, 4n^4 +10n^3+10n^2+2n + n^2 + 3n + 1,
4n^4 +10n^3 +11n^2 +5n +1, 5n^4 + 3n^3 + 3n^2 + 5n +1, 20480 +1536
+192 +40 +1 = 22249.
n=8, 5n^4 + 3n^3 + 3n^2 + 5n +1 = s(9, 5)
For n=9, I expect to double the polynomial for n=8 and add multiples
of 9. 5*9^4+3*9^3+3*9^2+5*9+2 = 32805 + 2187 + 243 + 45 +2 = 70563 =
63273 + 7290. So, doubling the polynomial coefficients would be too
much, by 9^4. For n^5 + 5n^3 + 6n^2 + 10n + 3, 59049+3645+486 + 90 +
3 = 63273.
n=9, n^5 + 5n^3 + 7n^2 + n + 3 = s(10, 6)
For n=10, doubling the polynomial gives 28*10^5 + 10*10^3 + 14*10^2 +
2*10 + 6, or 2800000+10000 +1400 +20 + 6 = 2811426, that's somewhat
greater than 157773, 2653653, much greater, I'm beginning to think
that this line of guessing polynomials that equal s(n+1, n-3) for
values of n is worthless.
n=10, n^5 + 5n^4 + 7n^3 + 7n^3 + 7n + 3 = s(11, 7).
What I want here is to determine the function(s) f(n) so that ((sum
n)^4 - sum (n^4))/s(n+1, n-3) = f(n). More generally I want to
determine if there are forms ((sum n)^x - sum (n^x))/|s(n+1, n-x+1)| =
f(n, x) <=> ((sum n)^x - sum (n^x))/f(n, x) = |s(n+1, n-x+1)|. I have
forms for x=2 and x=3, thus a closed form for s(n+1, n-1) and |s(n+1,
n-2)|. The idea here is that |s(n+1, n-x+1)| represents the sum of
the products of each x-subset of {1, ..., n}.
n=4, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 9646 / 24 ~= 401.9167
n=5, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 49646 / 274 ~= 181.1898
n=6, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 192206 / 1624 ~= 118.3534
n=7, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 609980 / 6769 ~= 90.11375
n=8, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 1670844 / 22449 ~= 74.42844
n=9, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 4085406 / 63273 ~= 64.56792
n=10, ((sum n)^4 - sum (n^4))/s(n+1, n-3) = 9125291 / 157773 ~=
57.8381
The form (sum n)^x - sum (n^x) / f(n, x) = s(n+1, n-x+1) might only
exist for x=2, 3. For x=1 it is obvious that (sum n)-(sum n) = 0.
The idea of ((sum n)^2 - sum (n^2))/2 came from the lower half of the
n x n matrix m with m_ij=i*j as the products of pairs of {1, ..., n}.
What are other known closed forms for s(n, m), the Stirling numbers of
the first kind, besides for s(n+1, n-1), s(n+1, n-2), and those shown
explicitly on the MathWorld web page, or in Abramowitz and Stegun
(Handbook of Mathematical Functions), Zwillinger (CRC Standard
Mathematical Tables and Formulae), Gradshteyn and Rizhik (Tables of
Integrals, Series, and Products), or Knuth (Fundamental Algorithms),
with no summations or productions?
Every time I read Knuth he amazes me. He gives a description of
generating functions in section 1.2.9 of Fundamental Algorithms. He
describes some of Stirling's circa 1730 results, and how generating
functions come from de Moivre and were expanded by Euler and Laplace.
s(n+1, n-x+1) = (-1)^(n-x) Sum i=1..(n choice x) II j=1..x (a_i_j) ,
a_i_j is the j'th element of the i'th x-subset of {1, ..., n}, x E
N>=1, n E N>=x
I haven't proven that above statement, it just appears to be true.
Well, I figure I should do some more examination of the difference of
(sum n)^x and sum (n^x) to see if there is any use for it besides the
cases with x=2 or 3.
The sum of 1-subsets is (sum n), also |s(n+1, n)|. One way to
consider that is:
( (sum n)^1 - 0 (sum (n^1)) ) / 1 = |s(n+1, n+1-1)|
With x=2:
( (sum n)^2 - 1 (sum (n^2)) ) / 2 = |s(n+1, n+1-2)|
for x=3:
( (sum n)^2 - 1 (sum (n^2)) ) (n-2) / 6(n+2) = |s(n+1, n+1-3)|
I wonder if for x=4 that I could multiply the sum (n^4) term by
something other than one and get a workable result. I'll consider 2,
3, and 4.
f(n)=sum (n^4) = n(n+1)(2n+1)(3n^2+3n-1)/30
n=4, f(n) = 354, 2f(n)= 708 , 3f(n)= 1062 , 4f(n)= 1416
n=5, f(n) = 979, 2f(n)= 1958 , 3f(n)= 2937 , 4f(n)= 3916
n=6, f(n) = 2275, 2f(n)= 4550 , 3f(n)= 6825 , 4f(n)= 9100
n=7, f(n) = 4676, 2f(n)= 9352 , 3f(n)= 14028 , 4f(n)= 18704
n=8, f(n) = 8772, 2f(n)= 17544 , 3f(n)= 26316 , 4f(n)= 35088
n=9, f(n) = 15219, 2f(n)= 30438 , 3f(n)= 45657 , 4f(n)= 60876
n=10, f(n) = 25333, 2f(n)= 50666 , 3f(n)= 75999 , 4f(n)= 101322
Then, I will subtract those values from (sum n)^4
n=4, (sum n)^4 = 10^4 = 10000
n=5, (sum n)^4 = 15^4 = 50625
n=6, (sum n)^4 = 21^4 = 194481
n=7, (sum n)^4 = 28^4 = 614656
n=8, (sum n)^4 = 36^4 = 1679616
n=9, (sum n)^4 = 45^4 = 4100625
n=10, (sum n)^4 = 55^4 = 9150624
n=4: 9646, 9292, 8938, 8584
n=5: 49646, 48667, 47688, 46709
n=6: 192206, 189931, 187656, 185381
n=7: 609980, 605304, 600628, 595952
n=8: 1670844, 1662072, 1653300, 1644528
n=9: 4085406, 4070187, 4054968, 4039749
n=10: 9125291, 9099958, 9074625, 9049292
Then, if any of those is good, then I can divide the above numbers by
s(n+1, n-3) and get something recognizable.
n=4, s(5, 1)=24
n=5, s(6, 2)=274
n=6, s(7, 3)=1624
n=7, s(8, 4)=6769
n=8, s(9, 5)=22449
n=9, s(10, 6)=63273
n=10, s(11, 7)=157773
Those are generally intractable or useless values, although their
prime factorization might show some insight. None of their quotients
are integer.
10000/24 is about 416.667, 400 times 24 is 9600, difference is 400.
50625/274 is about 184.7628, 180 times 274 is 49320, difference is
1305.
194481/1624 is about 119.7543, 115 times 1624 is 191632, difference is
2849
614656/6769 is about 90.80455, 90 times 6769 is 609210, difference is
5446.
1679616/22449 is about 74.81919, ...
Here I'm trying to figure out a function to subtract from (sum n)^4 to
get a difference that divided by s(n+1, n-3) gives a function of n.
The goal here is to get s(n+1, n-x+1) without explicitly summing the
products of the x-subsets of {1, ..., n} or inductively from s(1, 1).
There are solutions for x=1, 2, 3, tell me the solutions for the other
ones.
Ross
Ross A. Finlayson
Stirling numbers of the first kind".
On page four, section 4: History and Remarks, he notes:
"The Stirling numbers of the first kind are notoriously difficult to
compute. Unlike the numbers of the second kind, the closed formulas
for [n k] whose summands are the familiar factorials, binomial
coefficients, etc., involve two summation signs. The first asymptotic
formula for them was given by Jordan *, who showed, for k fixed, that
1/(n-1)! [n k] ~ (log n + gamma)^(k-1) / (k-1)!
Moser and Wyman gave a comprehensive study of the asymptotics of [n k]
for large n. Further, their study was not restricted to fixed k, but
indeed covered the entire range of 1 <= k <= n by braking it into
three subrangse and dealing separately with each of them. The
formulas that they obtain for the low end of the range, i.e., for
fixed k, are however not phrased in terms of the usual atoms of
asymptotics, namely powers of n and of log n. Instead the formulas
are expressed as series of polynomials in the partial sums of the zeta
function, containing p(k) monomials, p(k) being the number of
partitions of k."
* C. Jordan, The calculus of finite differences, 2nd ed., Chelsea,
1947.
I find a paper by N.M. Temme, "Bernoulli Polynomials Old and New:
Generalizations and Asymptotics", 1995. The abstract notes "In the
second problem we are concerned with the asymptotic behavior of
B_n^mu(z) when the degree n tends to infinity." The introduction
remarks that "At present Bernoulli numbers are introduced through
generating functions, as we shall do below, but historically they
arose in connection with the sums of the p-th power of the first n-1
integers 1 + 2^p + ...(n-1)^p. The Greeks, Hindus, and Arabs all had
rules amounting to" the polynomials representing sum (n^x).
Equation 3.5 says s(n, m) = ((n-1) choose (m-1)) B_(n-m)^n. The
bibligraphy references other works that would appear of interest,
Temme's 1993 "Asymptotic estimates of Stirling numbers". Also
referenced is Milne-Thompson "The calculus of finite differences" ,
Macmillan,1933, and Butzer, Hauss, and Schmidt, "Factorial functions
in Stirling numbers", Results in Mathematics 16, 1989.
CiteSeer returns some more results for "Stirling first",even for
Stirling engines. I download papers by Gessel, Krattenthaler, Das and
Ghosh, Frisch and Ward, Flajolet and Prodinger, Labelle, Leroux, et
alia, Hwang, Chelluri, Richmond, and Temme.
Gessel has some interesting stuff, he describes various stuff for
generating functions. So does Krattenthaler.
Das, Ghosh, and Deo describe Stirling tables, Stirling matrices,
Stirling graphs, and Stirling networks, basically in terms of S_1(n,
k), Stirling numbers of the first kind. "It is interesting to see
that the Stirling network recursively includes complete binary trees
at each level of the spanning tree." Mentioned in passing are
"leaf-ringed, half- and full-ringed binary trees, hypertrees, X-trees,
wheel-augmented binary trees, binary hypercubes, cube-connected
cycles, generalized hypercubes, pyramids, de Bruijn and Kautz graphs,
Moore graphs, fault-tolerant graphs, group graphs, and hypernets."
Referenced are Das and Deo, "Stirling graphs and their properties",
and Graham, Knuth,and Patashnik's "Concrete Mathematics."
Frisch and Ward present "Congruences for Stirling numbers". They
reference L. Comtet. "These Stirling numbers are closely related to
symmetric functions. Let e_n denote the elementary symmetric function
or polynomial of degree n (sum of all products of n different
variables)..." The authors discuss congruences of Stirling numbers.
Referenced are Comtet's "Nombres de Stirling generaux et fonctions
symetriques" and Gessel and Stanley's "Stirling polynomials", among
others.
Flajolet and Prodinger's abstract: "Cauchy coefficient integrals and
Hankel countours provide a natural generalizations of Stirling numbers
for unrestricted complex values of their arguments. Many classical
identities survive such an extension." They begin:
"Richmond and Merlini have introduced in [*] an extension of
Stirling's subset numbers { x y } and cycle numbers [ x y ] when x-y
is an integer. They also propose a further generalization when x-y is
not an integer but most classical properties are no longer preserved.
As the authors say about their most general extension on [* p.76]:
"It seems to us that the ideas used to derive identities and
recurrences lead to complicated formulas in general. There are
significant terms resulting from the fact that the integrands are not
single valued and also from the fact that th countours change."
In this note we give an alternative and more natural extension of
Stirling numbers of complex arguments for which most classical
identities are stll satisfied."
Flajolet and Prodinger call Stirling numbers of the second kind subset
numbers, and of the first kind cycle numbers. They go on to described
complex valued subset, Bell, Bernoulli, and cycle numbers.
"One encounters once more an instance of the duality relation [ x y ]
= { -y -x } that together with (**) confirms that Stirling numbers
eventually reduce to a single family."
*: B. Richmond and D. Merlini, Stirling numbers for complex arguments
**: E.T. Whittaker and G.N. Watson, A Course of Modern Analysis
Labelle, Leroux, Pergola, and Pinzani present "Stirling Numbers
Interpolation using Permutations with Forbidden Subsequences." Woww.
In abstract: "We present a family of number sequences which
interpolates between the sequences B_n, Bell numbers, and N!. It is
defined in terms of permutations with forbidden patterns or
subsequences. The introduction, as a parameter, of the number m of
irght-to-left minima yields an interpolation between Stirling numbers
of the second kind S(n, m) and Stirling numbers of the first kind
(signless) c(n, m). Moreover, q-counting the restricted permutations
by special inversions gives an interpolation between the usual
q-analogues of these numbers."
"We refer to Guibert and West for an exhaustive survey on the results
and on the tools used to study permutations with forbidden
subsequences and to Bona for recent results."
"The classical q-analogues of the Stirling numbers of the first kind
(signless) c_q(n, m) and of the second kind S_q[n, m], as defined by
Gould in 1961, are characterized by the generating functions ..."
"The concept of generating tree was introduced by Chung, Graham,
Hoggatt, and Kleiman in "The number of Baxter permutations" for the
study of Baxter permutations. Later West applied it to the study of
various permutations with forbidden subsequences. Generating trees
and succession rules can be used in combinatorics to deduce
enumerative results about various combinatorial objects, permitting
also their random generation."
The last paragraph begins "The meaning of Stirling numbers
interpolation lits in the observation that the permutations of length
n having m right-to-left minima are counted by the second kind
Stirling numbers for j = 1 and by the first kind Stirling numbers c(n,
m) for j = oo."
This paper might direct us to a methodology for calculating s( n+1,
n-k-1), but it doesn't. The references do appear to be valuable in
enumerating combinatoric objects.
Hwang, in "Asymptotic Expansions for the Stirling Numbers of the First
Kind", provides some new asymptotics in light of the work of Temme and
Wilf.
"We may roughly interpret Theorem 1 as saying that the Stirling
numbers of the first kind are asymptotically Poisson distributed of
parameter log n. This certainly implies the asymptotic normality of
s(n, m)/n! in the sense of of convergence in distribution."
What I'm looking for here is "s(n, m) = f(n, m)". Basically Hwang
gives an asymptotic approximation of s(n, m)/n!. "This technique can
be applied ... The advantage is that if we use only the first l terms
to approximate s(n, m)/n!, the error term is to some extent
minimized."
Again the references suggest many directions for research.
I'll have to read Chelluri, Richmond, and Temme's "Asymptootic
estimates for generalized Stirling numbers" later, the PDF browser
here mangles it.
To reiterate Wilf: "The Stirling numbers of the first kind are
notoriously difficult to compute."
Ross
Ross A. Finlayson
about Iverson notation, Iverson being the inventor of the programming
language APL: A Programming Language, the second note is about
Stirling numbers.
http://www-cs-staff.stanford.edu/~uno/papers/tnn.tex.gz
http://pckoenig1.mathematik.tu-muenchen.de/lehre/comb-ws2001/erg/tnn.pdf
Knuth notes that Stirling numbers of the first kind were investigated
as being the sum of the products of the k-subsets of {1, ..., n} back
in the nineteenth century.
"During the nineteenth century, Stirling's connections with these
numbers had been almost entirely forgotten. The numbers themselves
were studied, in the role of "sums of products of combinations of the
numbers {1, 2, ..., n} taken k at a time". Let C_k(n) and Gamma_k(n)
denote these sums, when the combinations are respectively without or
with repetitions; thus, for example,
C_4(4) = 1*2*3 + 1*2*4 + 1*3*4+2*3*4 = 50;
Gamma_3(3) = 1*1*1 + 1*1*2 + 1*1*3 + 1*2*2 + 1*2*3
+1*3*3 + 2*2*2 + 2*2*3 + 2*3*3 + 3*3*3 = 90
It turns out that
C_k(n) = [ n+1 n+1-k] and Gamma_k(n) = { n+k n }."
I want to note that there appears to be an error, C_3(4) is 50, C_4(4)
is 24.
Then, he explains a rationale for calling Stirling numbers of the
first kind Stirling cycle numbers, and Stirling numbers of the second
kind Stirling subset numbers, the appelations of first and second kind
originate with work in the early twentieth century by a Niels Nielsen,
and for bracketing first kind and bracing second kind pairs, where the
binomial coefficient uses parentheses. I'm just going to write s(n,
m) and S(n, m).
I also found a series of posts made to the math-history list circa
1997 that make it clear that a known method of generating Stirling
numbers is as the sum of the products of the k-subsets of {1, ..., n}.
http://www.polaris.net/~kermit/math/calculus/stirling
The methods of calculating s(n+1, n-1), and s(n+1, n-2) were put forth
here on this list, one from a method of calculating s(n, n-2)
alternatively to calculating s(n+1, n-1) from ((sum n)^2 - sum
(n^2))/2 and another from dividing (sum n)^3 - (sum n)^2 by
6(n+2)/(n-2).
A method to calculate s(n+1, n-3) would allow the calculation of a lot
of different numbers because of the identity s(n+1, k) = n s(n, k) +
s(n, k-1). Let's see, here.
s(n+1, k) = n s(n, k) + s(n, k-1)
s(n, k-1) = s(n+1, k) - n s(n, k)
s(n, k) = (s(n+1, k) - s(n, k-1)) / n
Well, maybe it wouldn't. For example if s(n+1, n-3) was known, then
s(n+1, n-3) = n s(n, n-3) + s(n, n-4), those values are all easily
calculable already by knowing s(n+1, n-2) and s(n+1, n-3). The
identity here allows the computation of s(n, x) for n-x < n-k, but
there are already formulas for s(n+1, n), s(n+1, n-1), and s(n+1,
n-2), as a univariate polynomial of n or a ratio of univariate
polynomials of n. Thus, knowing a formula for s(n+1, n-3) wouldn't in
itself automatically enable the computation of s(n+1, n-4), but the
process of deriving it might.
Calculating s(n+1, n-1), the sum of each product of pairs of {1, ...,
n}, I was listing the prime and composite factors with their multiple
frequencies to then add them together. Then, I noticed that I could
just make a square matrix, and have each element be the product of its
row and column indices, and remove the diagonal, and remove the upper
half, and the sum of the remaining elements was s(n+1, n-1). The
sum of the elements of the original matrix was (sum n)^2, removing the
diagonal was the subtraction of sum (n^2), and then the simple
division by two removed the symmetric upper half. I thought the case
for x=3 would be similar, and it is, make a 3d matrix, sum its
elements, remove the diagonal, and then divide by something, instead
of two to halve the remaining elements it is 6(n+2)/(n-2) to divide
the remaining elements.
I hope that it will be possible in four and more dimensions. Figuring
it out would allow the rapid solution of a problem that currently
takes quite a few more calculations to compute. This wouldn't figure
out the sum of the products of k-subsets of an arbitrary set of
integers, only those of a set of integers from 1 through n.
(Calculating the products of k-subsets of an arbitrary set of integers
is about enumerating k-subsets, where enumerating k-subsets of a set
{1, ..., n} does immediately apply to enumerating k-subsets of an
arbitrary set.)
What I'm concerned with are the square, cubic, etcetera k-d Euclidean
matrices with equal rank n in each dimension thus that the element at
(a, b, c, ..., k) is (a*b*c*... *k). The "diagonal" of these matrices
with a=b=c=... from 1 to n is the sequence (1^k, 2^k, 3^k, ..., n^k).
A subset of the elements of this matrix represent the k-subsets of {1,
..., n}, that is to say for each k-subset of {1, ..., n} that the
elements of the k-subset represent coordinates and the element at
those coordinates is the product of the elements of the k-subset. The
sum of all elements of the matrix is (sum n)^k. The elements of the
diagonal do not represent k-subsets, each k-subset has a =/= b =/= c
=/= .... Also, for the k-subset {x, y, z, ...} , if a=x, b=y, c=z,
... then there are also the elements a=y, b=z, c=x, ..., etcetera that
would be repeated k-subsets. The idea is to sum all the elements of
the matrix for each of the k-subsets of {1, ..., n} but with no
duplications, i.e., only one element of the matrix per each k-subset
is part of the sum.
With a square matrix, this is easy. It's a 2-d matrix, of rank nxn,
about the 2-subsets of {1, ..., n}, each element (i, j) for i, j= 1 to
n is i * j. We can remove the diagonal and are left with the upper
and lower halves of the matrix, they are symmetric about the diagonal,
the sum of the elements of the upper half is the same as the sum of
the elements of the lower half, dividing the sum of the elements by
two yields the result of the sum of the products of the 2-subsets of
{1, ..., n}.
With higher values of k, an n_1 x n_2 x ... n_k matrix, then besides
the diagonal there are "subdiagonals" with not each coordinate being
equal, but at least two coordinates being equal. For example with n=3
for a 3x3x3 cube, there is the "diagonal" (1, 1, 1), (2, 2, 2), (3, 3,
3) and then there are also the edges of the cube, (1, 1, 2), (1, 1,
3), (2, 2, 1), (2, 2, 3), (3, 3, 1), (3, 3, 2), (1, 2, 1), (1, 3, 1),
(2, 1, 2), (2, 3, 2), (3, 1, 3), (3, 2, 3), (2, 1, 1), (3, 1, 1), (1,
2, 2), (3, 2, 2), (1, 3, 3), and (2, 3, 3). None of those points have
coordinates that represent a k-subset because at least two of their
coordinates are equal. These are about the edges and diagonals of the
square sides of the cube of the matrix, and the diagonals between
corners of the cube. Here I'm looking at a Rubik's cube. It's
jumbled, I can solve a side. Starting from a solved cube I can get the
x's, h's or boxcars. Anyways, there are plenty of elements with
unequal coordinates, a fraction of these represent unique k-subsets.
Consider the cube with 3x3x3 subcubes, the center subcube is not used
as it is at (2, 2, 2). Here I'm about to start writing on this
Rubik's cube. OK, I've solved a side. Let's see, the 3-subset of {1,
2, 3} is {1, 2, 3}. Its permutations are (1, 2, 3), (1, 3, 2), (2, 1,
3), (2, 3, 1), (3, 1, 2), and (3, 2, 1).
In the context of matrices, for an nxnxn matrix, what is 6(n+2)/(n-2)
vis-a-vis an nxn matrix and 2?
I dig those dual sum integral identities.
Ross
Ross A. Finlayson
n-k+1), for n=4 and k = 3, |s(5, 2)| = 50.
The sum of the elements of the matrix are (sum n)^3 = 1000.
Subtracting from that by sum (n^3)= 100 gives 900 and dividing by
6*(4+2)/(4-2)= 18 yields 50, |s(n+1, n-2)|.
There is the main diagonal (1, 1, 1), (2, 2, 2), (3, 3, 3), (4,4, 4).
Its elements are 1, 8, 27, 64, its sum is 100. Then, there are the
other three diagonals:
(1, 1, 4), (2, 2, 3), (3, 3, 2), (4, 4, 1)
(1, 4, 1), (2, 3, 2), (3, 2, 3), (4, 1, 4)
(4, 1, 1), (3, 2, 2), (2, 3, 3), (1, 4, 4).
Their elements are each 4, 12, 18, 16. The sum of each is 50, the sum
of those are 150.
Those 12 coordinate triples are each of the coordinate triples with a
matching pair of coordinates x where the other coordinate is 5-x.
The entire hypermatrix has 64 coordinate triples, there are thus 48
remaining coordinate triples.
There are coordinate triples with two matching coordinates x and the
third coordinate being neither x nor n+1-x.
(1, 1, 2), (1, 2, 1), (2, 1, 1) 3*2 = 6
(1, 1, 3), (1, 3, 1), (1, 1, 3) 3*3 = 9
(2, 2, 1), (2, 1, 2), (1, 2, 2) 3*4 = 12
(2, 2, 4), (2, 4, 2), (4, 2, 2) 3*16 = 48
(3, 3, 1), (3, 1, 3), (1, 3, 3) 3*9 = 27
(3, 3, 4), (3, 4, 3), (4, 3, 3) 3*36 = 108
(4, 4, 2), (4, 2, 4), (2, 4, 4) 3*32 = 96
(4, 4, 3), (4, 3, 4), (3, 4, 4) 3*48 = 144
150 150 150
The sum of their elements is 450.
Thus 4 + 12 + 12 = 28 coordinate triples represent coordinates that
are not k-subsets of {1, 2, 3, 4}. That leaves 64-28 = 36 coordinate
triples that do represent k-subsets of {1, 2, 3, 4}. Dividing that by
three factorial, as there are three coordinates, yields 12, the number
of 3-subsets of {1, 2, 3, 4}. Thus, after removing the elements with
matching coordinates, then the remaining elements could be summed and
then divided by the number of elements to get the desired result.
1000-100-150-450=300
300/6 = 50
I guess I wonder why the sum of the diagonals besides the main sum is
equal to 2 (n+2) / (n-2) times as much as the sum of the elements with
no matching coordinates, where their sum is 3 times the sum of the
products of the 3-subsets.
Also, I want to figure out how in a 4-d matrix to determine the ratio
of the elements with no matching coordinates.
In a 4-d matrix, the concept of a diagonal is in this case about how
of the four coordinates of each matrix element that a diagonal has
matching coordinates. The main diagonal has each coordinate equal.
For the 4-d case, k=4, we consider a 5x5x5x5 matrix, for s(n+1, n-k+1)
= s(6, 2) = 274.
So with the diagonal there is the main diagonal: (1, 1, 1, 1), (2, 2,
2, 2), (3, 3, 3, 3), (4, 4, 4, 4), (5, 5, 5, 5). Then, the next type
of diagonal has 3 matching elements, and the other element is 6 minus
the matching element.
(1, 1, 1, 5) (2, 2, 2, 4), (3, 3, 3, 3),
The value of n is odd in this 4-d hypermatrix, many diagonals go
through the center element (3, 3, 3, 3), just like a 3x3x3 3-d cube
and (2, 2).
(1, 1, 1, 5) (2, 2, 2, 4), (3, 3, 3, 3), (4, 4, 4, 2), (5, 5, 5, 1)
(1, 1, 5, 1) (2, 2, 4, 2), ...
(1, 5, 1, 1) ...
...
A 4-d 5x5x5x5 hypercube has 5^4=625 elements, it is quickly cumbersome
to enumerate them by hand.
Then there are "diagonals" with two pairs of matching coordinates.
Also there are those with two matching coordinates and two differing
coordinates.
Finally there are elements with no matching coordinates, 1/24 of these
sum together to 274.
It's plainly more simple to enumerate the k-subsets and sum their
products than to enumerate the elements with matching pairs and
subtract their products.
For a change of subject, here are a variety of terms to consider:
hypermatrix
n-d matrix
n-d Euclidean matrix
Karnaugh map representation
main diagonal
diagonal
hypercube
Now I've stumbled across a solution for k=3, and look further for what
might lead to a solution for k=4. One place to start might be to
continue in subtracting multiples of sum (n^4) from sum (n^4). Yet,
that is probably tedious if not worthless.
One thing to consider is the symmetry around the main diagonal.
Again, it's difficult to visualize n-d space. At least I have
Schaum's Tensor Calculus to keep me warm, I may light it on fire.
Anyways for the 4-d 5x5x5x5 solution, there are many elements that
have at least two matching coordinates, diagonals. Then, of the
remaining elements, with four distinct coordinates, there are 4! = 24
multiples for each k-subset, and the sum of the remaining elements is
24 times the result, s(6, 2), 274.
Let's see, 274*24 is 6576. The sum of each element in the entire
matrix is (sum n)^4 = 15^4 = 50625. 50625-6576 = 44049. The result
of 44049/979 is 44.9387. 45 times 979 is 44055. 44055-44049=6.
Hmm... 44049-979= 43070. This method of meandering is inefficient.
Here's something to consider, it's about breaking up the 5^4 elements
of the hypercube matrix into how many there are of various numbers of
matching elements. There are (n choice k) times k! many elements that
have distinct coordinates. The rest have at least one matching pair
of coordinates.
That the number of of k-subsets is represented 24=4! times implies
that the 4-d hypercube has 24-fold symmetry, and that would imply that
the 3-d cube has some form of 6-fold symmetry, where the square has
2-fold symmetry about the main diagonal. This is where the cube has 8
corners, I would think the hypercube has 26. There are the corners
(1, 1, 1) and (3, 3, 3), and then the other six corners
(1, 1, 3), (1, 3, 1), (3, 1, 1)
(1, 3, 3), (3, 1, 3), (3, 3, 1)
Then the 5x5x5x5 4-cube would have these 24 other corners:
(1, 1, 1, 5), (1, 1, 5, 1), (1, 5, 1, 1), (5, 1, 1, 1)
(1, 1, 5, 5), (1, 5, 1, 5), (1, 5, 5, 1), (5, 1, 5, 1), (5, 5, 1, 1),
(5, 1, 1, 5)
(1, 5, 5, 5), (5, 1, 5, 5), (5, 5, 1, 5), (5, 1, 5, 5)
Hey that's only 14. Which ones are opposite each other?
(1, 1, 1) < (2, 2, 2) > (3, 3, 3)
(1, 1, 3) < (2, 2, 2) > (3, 3, 1)
(1, 3, 1) < (2, 2, 2) > (3, 1, 3)
(3, 1, 1) < (2, 2, 2) > (1, 3, 3)
(1, 1, 1, 1) <-> (5, 5, 5, 5)
(1, 1, 1, 5) <-> (1, 5, 5, 5)
(1, 1, 5, 1) <-> (5, 1, 5, 5)
(1, 5, 1, 1) <-> (5, 5, 1, 5)
(5, 1, 1, 1) <-> (5, 5, 5, 1)
(1, 1, 5, 5) <-> (5, 5, 1, 1)
(1, 5, 5, 1) <-> (5, 1, 1, 5)
(1, 5, 1, 5) <-> (5, 1, 5, 1)
Connecting those corners are diagonals.
(1, 1, 1, 1) < (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4) > (5, 5, 5, 5)
(1, 1, 1, 5) < (2, 2, 2, 4), (3, 3, 3, 3), (4, 4, 4, 2) > (1, 5, 5, 5)
(1, 1, 5, 1) < (2, 2, 4, 2), (3, 3, 3, 3), (4, 4, 2, 4) > (5, 1, 5, 5)
(1, 5, 1, 1) < (2, 4, 2, 2), (3, 3, 3, 3), (4, 2, 4, 4) > (5, 5, 1, 5)
(5, 1, 1, 1) < (4, 2, 2, 2), (3, 3, 3, 3), (2, 4, 4, 4) > (5, 5, 5, 1)
(1, 1, 5, 5) < (2, 2, 4, 4), (3, 3, 3, 3), (4, 4, 2, 2) > (5, 5, 1, 1)
(1, 5, 5, 1) < (2, 4, 4, 2), (3, 3, 3, 3), (4, 2, 2, 4) > (5, 1, 1, 5)
(1, 5, 1, 5) < (2, 4, 2, 4), (3, 3, 3, 3), (4, 2, 4, 2) > (5, 1, 5, 1)
Each of those "diagonals" shares the center element (3, 3, 3, 3),
there is no center element in a 6x6x6x6 hypercube, where the diagonals
would share eight center elements. Otherwise there are no shared
coordinates. Thus these "diagonals" represent 29 of the 625 elements
of the matrix, in the 3x3x3 cube 9 of 27 elements are on the main or
"pseudo-main" diagonals.
Besides the main and pseudo-main diagonals, there are a variety of
other elements with a matching pair or triple of coordinates that are
part of a sequence that interpolates from one extreme of the sequence
to another. For example, there is the element at (1, 1, 1, 2), its
opposite would be at (4, 5, 5, 5), and its shortest path connecting
sequence would interpolate over intervening elements, each coordinate
can increment or decrement.
(1, 1, 1, 2) < (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4) > (4, 5, 5, 5)
Most of those elements are shared with the main diagonal.
(1, 1, 1, 3) < (2, 2, 2, 3), (3, 3, 3, 3), (3, 4, 4, 4) > (3, 5, 5, 5)
(1, 1, 1, 4) < (2, 2, 2, 3), (3, 3, 3, 3), (3, 4, 4, 4) > (2, 5, 5, 5)
Basically the elements with three one's in the coordinates are corner
pieces if the other element is a one or a five, or else they're edge
pieces. All corner and edge pieces have at least all but one
coordinate being ones or fives.
I'm looking at a 3x3x3 Rubik's cube, although I would need a 4x4x4
cube to evaluate it in terms of s(5, 2). It reminds me of the game
Welltris.
Anyways, the idea here is to subtract out the diagonals until some
symmetrical part is left that when divided by a function of the number
of elements yields the required result.
I'll look at the main and pseudo-main diagonals for 5x5x5x5.
(1, 1, 1, 1) < (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4) > (5, 5, 5, 5)
sum (n^4) = 979
(1, 1, 1, 5) < (2, 2, 2, 4), (3, 3, 3, 3), (4, 4, 4, 2) > (1, 5, 5, 5)
5+2^5+ 0 +2^7+125
(1, 1, 5, 1) < (2, 2, 4, 2), (3, 3, 3, 3), (4, 4, 2, 4) > (5, 1, 5, 5)
5+2^5+ 0 +2^7+125
(1, 5, 1, 1) < (2, 4, 2, 2), (3, 3, 3, 3), (4, 2, 4, 4) > (5, 5, 1, 5)
5+2^5+ 0 +2^7+125
(5, 1, 1, 1) < (4, 2, 2, 2), (3, 3, 3, 3), (2, 4, 4, 4) > (5, 5, 5, 1)
5+2^5+ 0 +2^7+125
21 + 144 + 81 + 768 + 1125 = 2139
(1, 1, 5, 5) < (2, 2, 4, 4), (3, 3, 3, 3), (4, 4, 2, 2) > (5, 5, 1, 1)
25+2^6+0+2^6+25
(1, 5, 5, 1) < (2, 4, 4, 2), (3, 3, 3, 3), (4, 2, 2, 4) > (5, 1, 1, 5)
25+2^6+0+2^6+25
(1, 5, 1, 5) < (2, 4, 2, 4), (3, 3, 3, 3), (4, 2, 4, 2) > (5, 1, 5, 1)
25+2^6+0+2^6+25
75 + 192 + 0 + 192 + 75 = 534
Now, (50625 - 2139 - 534) =47952, 47952/ 274 is close to an integer,
which is no good because it has to be exactly an integer. 47950/274
is 175. A problem with working with these pseudo-diagonals is that
there isn't a simple function like sum (n^4) for them, although they
do evaluate to equal each other.
2675 = 107 * 25
175= 7*5*5
Let's see, 6(n+2)/(n-2).
Eh. I look at Google and it returns results for "pseudo-diagonal",
I'm not sure if it's being used in the same way. I look at MathWorld
about hypercube and tesseract. Heh, planet. I note that after rows
and columns are pillars and files, i-rows.
"Perfect Magic Hypercubes":
http://www.geocities.com/~harveyh/cube_perfect.htm
I guess instead of diagonal the term for use in the context of
hypercubes in n-d is the n-agonal.
So, for this tesseract, the above paths are quadragonals. Then there
would be triagonals from corner to corner of each cube of the
hypercube, four triagonals for each cube, and diagonals from corner to
corner of each square of the hypercube, two diagonals for each square.
None of the elements in an n-agonal represents a k-subset, at least
two of its coordinates are matched.
I'll look at the 4x4x4x4 hypercube, s(5, 1)=24. The sum of its
elements is 10000.
(1, 1, 1, 1) (2, 2, 2, 2) (3, 3, 3, 3) (4, 4, 4, 4)
1 + 16 + 81 + 256 = 354
(1, 1, 1, 4) (2, 2, 2, 3) (3, 3, 3, 2) (4, 4, 4, 1)
(1, 1, 4, 1) (2, 2, 3, 2) (3, 3, 2, 3) (4, 4, 1, 4)
(1, 4, 1, 1) (2, 3, 2, 2) (3, 2, 3, 3) (4, 1, 4, 4)
(4, 1, 1, 1) (3, 2, 2, 2) (2, 3, 3, 3) (1, 4, 4, 4)
16 + 96 + 216 + 256 = 584
(1, 1, 4, 4) (2, 2, 3, 3) (3, 3, 2, 2) (4, 4, 1, 1)
(1, 4, 4, 1) (2, 3, 3, 2) (3, 2, 2, 3) (4, 1, 1, 4)
(1, 4, 1, 4) (2, 3, 2, 3) (3, 2, 3, 2) (4, 1, 4, 1)
48 + 108 + 108 + 48 = 312
Sum of 4-agonals: 354+584+312 = 1250. 10000-1250 = 8750, 8750/24
~=364.5833, 8736/24 = 364, 364=2*2*7*13.
8750 = 2*5^4*7, 47952 = 2^3*3^2*661
What is the significance, in the case of the cube, of 6(n+2)/(n-2)?
That would probably be a good question to understand before trying to
apply to higher dimensions. What are those values?
n=3: 6*5/1
n=4: 6*6/2
n=5: 6*7/3
n=6: 6*8/4
Maybe I should skip over the case of k=4 as it is even and look to
k=5.
The 5x5x5x5x5 matrix, for k=5 and n=5, for s(6, 1) = 120, has as the
sum of its elements 15^5= 759375. The sum of i^5 for i from 1 to 5 is
n^2(n+1)^2(2n^2+2n-1)/12 = 25*36*59/12=4425. 759375-4425= 754950.
754950/120= 6291.25, appreciably near 6291 1/4.
n=5:759375 4425 754950 120 6291.25
n=6:4084101 12201 4071900 1764 2308.3333333333335
n=7:17210368 29008 17181360 13132 1308.358208955224
n=8:60466176 61776 60404400 67284 897.7528089887641
n=9:184528125 120825 184407300 269325 684.7017543859649
n=10:503284375 220825 503063550 902055 557.6861167002012
n=11:1252332576 381876 1251950700 2637588 474.6574142739503
Well hell, the other values than for five and six are quite a bit
farther from a fractional result. Yet, the values for five and six
are encouragingly near a fractional result, considering dividing
4071900 by 1764 and getting a result with a non-integer part of 1/3.
6291 = 3^3 *233
2308 = 2^2 *577
1308 = 2^2*3*109
897 = 3*13*23
684 = 2^2*3^2*19
557 = 557
474 = 2*3*79
6291.25 * 4 = 25165
2308.3333 * 3 = 6925 = 5^2*277
Onward, I decide I need to write the extended precision arithmetic to
generate as many values of s(n+1, n-k+1) with associated values of the
sets and stuff as is feasible, and use those to draw conclusions about
the number-theoretic behavior of these hypermatrices. vis-a-vis
Stirling cycle numbers and the infinite polynomial (n+1)(n+2)...(n+n).
I start writing my software, and to generate the Stirling cycle
number, the Stirling number of the first kind, I use the definition
that is on MathWorld, and in Abramowitz and Stegun and Zwillinger:
s(n, m) = Sum_k=0^(n-m) (-1)k C(n-1+k, n-m+k) C(2n-m, n-m-k)
s_2(n-m-k, k)
There, s(n, m) is the Stirling number of the first kind, C(n, m) is
the binomial coefficient, and s_2(n, m) is the Stirling number of the
second kind. I implement this.
One thing to note: CRC Standard Mathematical Tables and Formulae is
calling the Stirling numbers of the second kind the cycle numbers,
where everything else is calling them the subset numbers, where the
Stirling numbers of the first kind are the cycle numbers.
What I'm doing with the program is calculating (sum n)^x - sum (n^x)
and then dividing it by s(n+1, n-x+1), to get an idea of the ratio of
the difference to the Stirling number. For x=4, at n=500 the ratio is
around 24.3877. It got to 24 at n=197 and appears to be decreasing.
The only result that appears to be rational is for n=4, the first
entry with a ratio of 401.91666....
Here are some values for x=5, of ( (sum n)^x - sum (n^x) ) / |s(n+1,
n-x+1)|:
n=5 %=6291.250000000000000000000000000000000000000000000000000000000000
n=6 %=2308.333333333333333333333333333333333333333333333333333333333333
n=7 %=1308.358208955223880597014925373134328358208955223880597014925373
n=8 %=897.752808988764044943820224719101123595505617977528089887640449
n=9 %=684.701754385964912280701754385964912280701754385964912280701754
n=10 %=557.686116700201207243460764587525150905432595573440643863179074
n=11 %=474.662813102119460500963391136801541425818882466281310211946050
I notice that the first two numbers appear obviously rational. For
n=7, there is appearing to be the repeating sequence
358208955223880597014925373134328.... For n=8, and so on:
(75280898876404494382022471910112359550561797)...
(701754385964912280)...
(there doesn't appear to be a repeating sequence in 60 digits for ten)
(6628131021194605009633911368015414258188824)...
A number of the form .(abc)... is abc/999. What the hell is 474 and
6628131021194605009633911368015414258188824/9999999999999999999999999999999999999999999?
I know there are some things known about sequences of ones in terms of
their primality and whatnot, that might be helpful here, in terms of
factoring them and them. Maybe I should extract the sequences and
then mutiply them by that many nines in a row or see if they have a
factor of nine and that many ones in a row, and then get some integer
values for the ratios and consider them, in as well considering how
long the repeating sequences are for varying values. Damn.
I look back at x=4, there are repeating sequences in the possibly
rational results:
x = 4
n=4 %=401.916(6)...
n=5 %=181.(18978102)...
n=6 %=118.35(3448275862068965517241379310)...
n=7 %=90.113753877973112719751809720785935884177869700103412616339193
I don't see a repeating sequence in that one.
n=8 %=74.428437792329279700654817586529466791393826005612722170252572
Ditto.
n=9 %=64.566118249490303921103788345739889052202361196718979659570432
Ditto.
It might be that the sequences are farther out. All the ratios are
necessary rational, the divisors are integer.
Consider x=2:
For any integer n>=2:
% = 2
For x=3:
n=497 %=6.048484848484848484848484848484848484848484848484848484848484
n=498 %=6.048387096774193548387096774193548387096774193548387096774193
n=499 %=6.048289738430583501006036217303822937625754527162977867203219
n=500 %=6.048192771084337349397590361445783132530120481927710843373493
n=501 %=6.048096192384769539078156312625250501002004008016032064128256
n=502 %=6.048000000000000000000000000000000000000000000000000000000000
For any integer n>=3, % = 6(n+2)/(n-2).
I'll run a couple thousand values for x=4, and browse through the list
for something approaching an integer or otherwise having a very short
repeating sequence. I'll post a text file containing 5000 values on
http://www.tiki-lounge.com/~raf/x4_5000.txt. What I want to determine
is which polynomial expression of n, a polynomial divided by another
polynomial with no common monomial factors, this ratio represents.
Pleased with my toy, I try some higher values of x.
x = 6
n=6 %=119026.319444444444444444444444444444444444444444444444444444444444
n=7 %=36861.454239363330272421181512090602999693908784817875726966636057
n=8 %=18424.159120923775016084792252209542514645626629643425552808912668
n=9 %=11473.009092416537696219323457881936767632102586778686712359053725
x = 7
n=7 %=2676930.200000000000000000000000000000000000000000000000000000000000
n=8 %=715075.801576872536136662286465177398160315374507227332457293035479
n=9 %=318633.386799693016116653875671527244819646968534151957022256331542
n=10 %=181035.394964028776978417266187050359712230215827338129496402877697
n=11 %=118600.642569647225949017441401625759608554968037250414331939073474
x = 8
n=8 %=69967391.439583333333333333333333333333333333333333333333333333333333
n=9 %=16379749.438221816991630429700285219993453967363351568709963996820498
n=10 %=6565513.088979279223333126332567430499492848123538057090811236001573
n=11 %=3420547.266886029723743002864692301614937365946153148381697571595361
n=12 %=2084752.445592677919132295818055063566015615820526857696897115050702
x = 9
n=9 %=2085207419.18303571428571428571428571428571428571428571428571
n=10 %=433297675.87192622950819672131147540983606557377049180327868
n=11 %=157454245.75232350054840385614500952490908041332332736823875
n=12 %=75578349.79545956520367273530432587637648692140220237907424
n=13 %=42974745.07367298806795312338874123059420728563397523110648
x = 10
n=10 %=69801352294.09135251322751322751322751322751322751322751322751
n=11 %=13010510035.67688942047971924571176760255853803894085338578893
n=12 %=4315567552.56959774750967657454860599607337397796161761830059
n=13 %=1917007230.87213025767692928347680454646794039186543827627115
n=14 %=1019833612.70615209764400109554420803015308129101197500558460
x = 11
n=11 %=2593149593966.36331018518518518518518518518518518518518518518518
n=12 %=437413722662.34622762497271338135778214363676053263479589609255
n=13 %=133237747517.88422667085126733620277379244380679100908656145384
n=14 %=54996733816.95570874125272526613591058914907789529570520178644
n=15 %=27449560633.32401585895298978756446397210344831395750748067391
I guess I should figure out what is the form of a polynomial that
given n evaluates to 9999999, for a repeating sequence of length 7, or
a multiple of that.
It is apparent that the ratio, if it is the same ratio for all values
of n, increases much more quickly in the numerator than the
denominator. At this point I'm happy that there appear to be so many
rationals.
I run the program for x=256, it appears that it would take a while for
it to evaluate:
((sum n)^256 - sum(n^256)) / |s(257, 1)|
It's working on it, though. Then again it's still working on
calculating:
n=5000, ((sum n)^4 - sum(n^4)) / |s(5001, 4997)|
I guess I should make a Bernoulli polynomial generator for the
summation of powers and n(n+1)/2 for sum n at least, and/or store
intermediate results, I'm using a brute force summation algorithm.
x=256
n=256 %=2852572961688342538050353413371354570640712417122816880001
535818275751190053340797595487281712778969522102534280498409
831871220362897410225280035120379235536395991685336241863749
724858796598835255173280863452740733168633870462309896550025
564847767312928266655412458246163387139699236058941967578667
527556201094083483706501925740963148455495219725310512719867
890461682533849034363791776095136083577947059972371543287206
405242982212349450353655109131425517048434637072909504346311
605438174115806976127099962981600604707740631418406131820639
375475981069659554409505689547384586844219837985133485716715
7301931630230127981624905115703697763172821414858201.1482275
2168349610236969390803340044027251191911424
Huh. Well, this is good, but it looks like I will have to do more
work to determine the repeating sequences to determine the number that
is a multiple of the denominator polynomial of the ratio for a given
value of n.
The idea here is again to generate Stirling numbers of the first kind,
Stirling cycle numbers, alternatively coefficients of the rising or
falling factorial.
Here's something to do: I'll run the program on high enough values of
n for given values of x to see if it approaches, similarly for x=2 to
2 or x=3 to 6 or x=4 to 24, some value. For x=5, at n=504 it's been
around 123 since n=408, before that it was 124 since 328. How about
x=6 and 720, 6!? If it's so for higher factors then it would be
similar to the example for x=3 with 3! (x+2)/(x-2), the limit of that
expression as x diverges is 3!.
For x=5, at n=500 it's only down to 749. It's got a ways to go to
720. The program is still running to generate 5000 values for x=4.
The output file is at 190677 bytes.
This leads me to think that in the large, that ((sum n)^x - sum
(n^x))/x! = |s(n+1, n-x+1)|. Now, |s(n, 1)| = n! for all positive
integer n. Thus, I think that lim n->oo ((sum n)^n - sum (n^n)) /
(n!)^2 = 1. The expression (sum n)^n is (n^2+n)^n / 2^n. Yet, we see
where we've gotten with Gamma.
How are Bernoulli polynomials generated? What's a closed form to
generate the coefficients of the Bernoulli polynomial with no
summations or productions? The Handbook of Mathematical Functions
categorizes them with Euler polynomials and the Euler-MacLaurin
formula.
Ross
Ross A. Finlayson
sequences of ones.
Here's the first few integer sequences of ones:
1
11
111
1/1 = 1, 1/11=.(09)..., 1/111=.(009)...
The reciprocal of a sequence of x many ones is the repeating sequence
of x-1 many zeros and a nine.
There are sequences of multiples of ones, less than base, for us, ten.
2 1/2=.50...
22 1/22=.0(45)...
222 1/222=.0(045)...
3 1/3=.33...
33 1/33=.0(30)...
333 1/333=.0(030)...
4 1/4=.250...
44 1/44=.02(27)...
444 1/444=.00(225)...
5 1/5=.20...
55 1/55=.0(18)...
555 1/555=.00(180)...
6 1/6=.1(6)...
66 1/66=.0(15)...
666 1/666=.0(015)...
7 1/7=.(142857)...
7 1/77=.(012987)...
777 1/777=.(001287)...
8 1/8=.1250...
88 1/88=.011(36)...
888 1/888=.001(126)...
9 1/9=.(1)...
99 1/99=.(01)...
999 1/999=.(001)...
I grep x4_5000.txt for the reciprocal sequences of sevens.
[space:~/math/Stirling] space% grep 001287 x4_5000.txt
[space:~/math/Stirling] space% grep 142857 x4_5000.txt
[space:~/math/Stirling] space% grep 012987 x4_5000.txt
n=3167 %=24.06071783274848487397_012987_254459190883538850373922493977380496558042975948178954
Notice that f(n, x) >= 24.
Hmmm..., I guess I should convert the ratio into integer pairs to see
what I can derive about their existence. Now, I already know, or
rather firmly believe, that 24, 4!, is part of the ratio that I
should discard and then concentrate on the non-integer part of the
ratio. Here I encounter an implementation limit of the computer
programming language type that I use, I need to get more than 80
digits out of the decimal past the radix. I don't know if the type
stores more than 80 digits, that is to say, the last digit in the
generated listings might be rounded or truncated.
The point is to discover the two polynomials which are assumed to be
the same for all values of n. It appears that the evaluation of the
polynomial/polynomial ratio is rational, that is to say the arguments
that result in the same ratio are integer.
Here's an idea, to reduce the fraction of ((sum n)^x - sum (n^x)) /
|s(n+1, n-x+1)|. For that I would need a factoring algorithm,
factorization algorithm. One type, perhaps the simplest, is a sieve:
for each number test it against each number less than half of it to
see if the remainder of division is zero. Let's see, the numbers are
stored as "java.math.BigInteger"'s. I can test the right side for
zeros to see if it is divisible by two or powers of two, and shift
left until an on bit is at the far right. That would quickly factor
2^64 and skip over 2^64-1. Then, another idea is to use the "casting
out nines" concept as we have discussed on sci.math, for base 4, 8,
16, and perhaps other powers of two, simply. I think that this was
where if the number was a mutiple of b-1 then the digits in base b
would add to a multiple of b-1. I'll review that thread, it might be
fast to check for factors of 3, 7, 15, etcetera, and perhaps some of
their factors. A sequence of ones has some factors, etcetera. There
are a wide variety of number-, factor-, and base-specific
number-theoretic tests for factors.
This is the use of the sieve: apply it to each integer and run it
forever, and then store the numbers that are not trivially factorable
into relational stores, exploiting the high capacity, random, rapid,
access digital storage peripherals of our modern age.
n=4482 %=24.04288425421220569888446616223798192202161043032416981807079866557299249136371500
Yep, that's not less than 24. I definitely should subtract 24 from
that and multiply the difference by |s(n+1, n-x+1)|. Then, if I
determine a pair of polynomials for that, it might be easier to
generate the polynomials that evaluated to that plus 24 times |s(n+1,
n-x+1)| times that.
The web page I used for one-off factorizations of integers, Dario
Alpern's elliptic curve factorization is good, I might look at its
source code. I'll search for elliptic curve factorization, the
problem with searching for factorization is that it's full of
cryptography links.
Hmm, factorization: the process of determining the prime factors of an
integer.
In the abscence of a function that generates the n'th prime, it would
probably be good to make tables of primes. Partially this is about
entering numbers from a table or getting lists of the first n primes
from network symbolic number theory data archives, were they to exist.
The elliptic curve method appears to be called Lenstra elliptic curve
factorization method. The initial page of search results also denote
the Pollard method, probably Pollard rho.
Anyways the integer factorizations of the divisor and what, divisand?,
should show some avenues of expression.
http://mathworld.wolfram.com/search/?config=scienceworld_wolfram_com&words=divisor
Some crypto is just, so bad. Then again I'm having a difficult enough
time "unfactoring", determining polynomials from values, fitting data.
I think elliptic curve cryptography is about P1363.
http://www.crypto-world.com/FactorPapers.html
http://modular.fas.harvard.edu/edu/Fall2001/124/
http://research.microsoft.com/~pleyland/factorization/main.htm
http://www.loria.fr/~zimmerma/records/factor.html
It looks like there is research into n^x +-1, and 2^n +-1, and then
there is 6n+-1, n!+-1, n(n+1)/2 +-1, n(n+1)(n+2)/3 +-1, other likely
points for primality as they differ by one from a definite composite,
and perhaps an abundant number.That's where to look for primes an+-1
for a an abundant number, with the sum of its factors more than it.
Rep-units! That's what they're called: repunits, repeating sequences
of ones.
http://mathworld.wolfram.com/Repunit.html
I've leafed through the Handbook of Mathematical Functions, CRC
Standard Tables and Mathematical Formulae, Tables of Integrals,
Series, and Products, Tensor Calculus, Tensor Analysis on Manifolds,
Vector and Tensor Analysis with Applications, Tensors, Differential
Forms, and Variational Principles, and Tensor Analysis for Physicists.
x=4
n=4527 %=24.04245751396553152984000120359584908522047256812845316562530409286489276926495933
x=5
n=3346 %=120.47930442608446264774143718865417931237489587229033381243534130409435106914623261
x=6
n=3184 %=724.53899289873525576141166585182898805954901673203676723741458241921980771559225951
Ross
Ross A. Finlayson
I read about how there are a variety of methods, many influenced by
the work of Fermat, these are efficient methods, but I thought I would
try and start from scratch to see what foundation the research into
the question area would give me in understanding the contemporary
methods.
One thing I saw recently that I thought was really great was the
quadratic sieve. Basically it plots x=y^2, then for it connected with
a line each point with integer coordinates. Then, where those lines
cross the integers on the x-axis there are only composites, and and
all composites, and all primes are missed.
So anyways I go to implement a factorization method. I'm working with
numbers stored using the "java.math.BigInteger" arbitrary precision
integers. From them I can get an array of the bytes (octets) that
represent the integer in signed 2's complement form, binary. I call
the input variable n and the result of the method is to be an array
(multiset) of the prime factors of n, besides n (if prime) and not
including one.
I figure first I should factor any twos of the integer. I use the
method that gets the index of the rightmost set bit of the binary
representation of n, and it returns the result of how many twos are in
the prime factorization of n. I then shift n right that many places
and proceed, having added that many twos to the list of prime factors.
Next, I think it would be efficient to try and remove small prime
factors three, five, etcetera. I think I can use the method where the
digits of a number add to be a somewhat smaller number, and that sum
is a multiple of the factor when n is a multiple of the factor, and
trial division is faster on the smaller number. This is about casting
nines in decimal, where decimal is base b=10 and nine is is b-1. For
example, sum the digits of a decimal number, if the sum is a multiple
of nine, then the number is a multiple of nine.
That method appears to work for b-1 and roots of b-1. Anyways I can
group the bits of the binary representation into digits of higher
bases that are powers of two, and then sum those digits to see if the
sum is a multiple of b-1. For example, here is a short list of powers
of two and one less than them:
2^2 = 4 4-1=3
2^3 = 8 8-1=7
2^4 = 16 16-1=15
The numbers 2^n-1 are called Mersenne numbers. They are binary
repunits. They are composite unless n is prime, in which case they
might be prime.
2^5 = 32 32-1=31
2^6 = 64 64-1=63
2^7 = 128 128-1=127
2^8 = 256 256-1=255
2^9 = 512 512-1=511
2^10 = 1024 1024-1=1023
Anyways for 2^2 and 2^3, groups of 2 and 3 bits, it is possible to sum
those digits into an integer result and then check the remainder of
the sum's division by 3 or 7 respectively for congruence of n modulo 3
or 7. For small numbers, its usefulness is trivial, but for
3^10000000000 * 7^100000000000, it might save quite a bit of time
compared to trial division dividing the entire number in base two by 3
or 7 until the remainder is non-zero, from knowing that the remainder
is zero. The method says that the Mersenne number is a factor of the
integer, then division can be done to get the product of the remaining
multiplicative partitions.
Of the numbers on that short list of Mersenne numbers, none are
perfect squares or cubes which would allow checking for the square or
cube root of that number, and some of them are prime and some
composite, for example 31 and 127 are primes, as is 8191, in using the
composite values then their use would only work when the composite is
a factor of the number, and it might be worthwhile to try to factor
the composites from n first, in that way leaving, for example, 3, in
the number, where 3 is a factor of those composites, eg, 63 and 255.
I don't know the least Mersenne number that is a square or cube.
I was hoping that the digit summation congruence method would have a
method for 5, then with the binary power of two then all primes less
than ten would be factored by those methods. Instead, I still have to
test divide 5, 11, and 13, etcetera, the digit summation congruence
method only works for 3 and 7, small primes.
15 = 5 * 3
31 = 31
63 = 7 * 3 * 3
127 = 127
255 = 17 * 5 * 3
511 = 73 * 7
1023 = 31 * 11 * 3
2047 = 89 * 23
4095 = 13 * 7 * 5 * 3 * 3
The digit summation congruence method for 2^12, 4095 will factor out a
13, a 7, a 5, and two 3's if they're there. So in using the digit
summation congruence method, if n is greater than (2^x)-1 for x from
twelve to two then I'll try it.
Still, the computer is fast at that binary arithmetic, it's good to
get as much as can be accomplished with its most efficient atomic
operations. I figure I can go to digits of about twelve bits, base
b=4096, where the sum would fit in a 32-bit signed integer a minimum
of 524288 bytes worth of integer, 2^4194304. It's probably fastest for
summing the base 2^8 or 2^16 digits.
I wonder if there's a good way to consider overflow, without actually
doing the work.
After that so far I only have the notion of using trial division by
primes, testing each prime by division into the number until a
remainder is returned, and then going on to the next prime until the
number left to factorize is equal to one and the multiset contains a
copy of each of the prime factors.
To that end I should develop a list of prime numbers. One way to do
this is to use the trial division sieve on integers from one towards
infinity, and as primes are discovered store them in the list of
primes, and as composites are factored store their factors in the list
of composites and their factors. What concerns me about that is that
it would take some storage. The good thing about it would be that in
factoring any number, if a composite in a range of numbers that was
pre-factorized was the result then it would probably take less
computation to load the precomputed result than to recalculate the
result.
Here's something about perfect squares, obviously it factors into its
square roots, but one less than a perfect square x^2-1 factors into
x+1and x-1, I guess that's obvious enough, yet it's enough of a
problem to factor a number in the first place without determining it a
square. On the other hand, for populating a list of prime
factorizations, where the idea is to get the prime factorizations for
a given range, that would allow ready factorizations of perfect
squares.
Fermat's method and other more modern methods use the square root of n
as a jumping-off point for trial division. They get to better than
%50 on something and bet on it.
The cryptography prime factorization problem is about factoring a
large composite into exactly two primes that are each nearer the
square root than one. Most numbers have more than two prime factors,
a very few only have one.
I cobble away on the digit summation congruence implementation.
[space:~/math/Stirling] space% java Factorize 17080198121677824
2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7* 7*
7* 7* 7* 1*
[space:~/math/Stirling] space% java Pow 63 60
63^60 = 912920516330797989897501319100671163424552283060748314636667478807055142893152629668193590354000850926342401
[space:~/math/Stirling] space% java Factorize
912920516330797989897501319100671163424552283060748314636667478807055142893152629668193590354000850926342401
7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3*
7*3*3*
7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3*
7*3*3*
7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3*
7*3*3*
7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3*
7*3*3*
7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3*
7*3*3*
7*3*3* 7*3*3* 7*3*3* 7*3*3* 7*3*3* 1*
Well, it seems to be a good computational method for determining prime
factors of numbers that are powers of 3, 7, 15, 31, 63, 127, 255, 511,
1023, or products thereof. It's still necessary to divide the
determined factor out of the number to get the quotient to recursively
factorize.
I'm considering something along the lines of binary coded decimal as I
want a method to determine if 5, 11, or 13 are factors besides where
3*5, 3*11*31, and 3*3*5*7*13 are factors.
Here's one method to consider, say the number is not a multiple of 3.
I want to check it for being a multiple of 5. I multiply the number
by 3 and then check using digit summation congruence whether it is a
multiple of 15. For eleven is is similar, I use the DSC method to see
if the number is a multiple of 3 or 31. Where it isn't, it might
still be a multiple of 11. I multiply the number by 3*31 and can then
use DSC(31*11*3) on that to see if it is a multiple of 11. It might
be faster to simply test divide the number by 5 and check the
remainder for zero, but multiplication might be much faster than
division.
So, after attempting factorization of 15 and 3, I can multiply the
number to factorize by 3 and then check that for a factor of 15 using
digit summation congruence to see if the original number was a
multiple of 5.
[space:~/math/Stirling] space% java Pow 5 20
5^20 = 95367431640625
[space:~/math/Stirling] space% java Factorize 95367431640625
5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 1*
Then for each resulting factor of the number, it's multiplied by
three, determined that product had a factor of fifteen, divided by
five, and then repeated. Using plain test division, it would divide
the number by five and check the remainder for equality with zero.
[space:~/math/Stirling] space% java Factorize
100000000000000000000000000
2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2* 2*
2* 2* 2*
5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5* 5*
5* 5* 5*
1*
For 11 it is similar. The factors 31 and 3 would have already been
factorized from the number by DSC(31) and DSC(3), before that
DSC(1023) would have determined a factor of 1023 = 31*11*3. So where
it is known that the number has no factors of 31 or 3, it might still
have factors of 11. Thus, the number is multiplied by 93, and that
product has applied to it the DSC(1023) method, which if returning
true on the product means that the number is a multiple of eleven.
This is useful where the cost of performing a multiplication (by 3 or
93) and the DSC test is less than performing division and test of the
remainder. Multiplication by three can be accomplished with a one-bit
leftshift and an add. I think a digit summation congruence is an
effective method to factor small primes from random integers, after
all, a third of the integers are multiples of three, and a fifth are
multiples of five. After that then the contemporary methods are very
good.
I wonder about tests for factors that could make use of the relative
numbers of ones and zeros in the base two number, for example a test
that would be more or less likely to get a factor given a number with
3/4 ones, nearer in Hamming distance to a Mersenne number, than 1/2
ones, or 1/4 ones, nearer in Hamming distance to a power of two.
I haven't yet implemented trial division by primes. Partially that's
because I don't yet have a method to return the n'th prime.
Returning to (sum n)^x - sum (n^x) and s(n+1, n-x+1), I begin to
factorize some of those values. Now, for x=2 the difference always
has a factor of 2, and the ratio in the limit appears to be equl to 2.
For x=3, the difference always has factors 2 and 3, and the ratio in
the limit appears to be equal to 2*3. For x=4, the difference does
not always have factors of 2, 3, and 2 and 2, yet the ratio in the
limit does appear to be 24. The first value of n where the difference
is a multiple of 24 is n=17.
n=17 %= (2*2*2*3*13*17*79*1307*)/(2*7*3*17*19463*)
n=17 %= (2*2*13*79*1307) / (7*19463)
My prime factorization implementation is very slow.
n=17 %=39.40925272128067175079454789674180312828003317650340205958558730484949464551786907
I look for patterns in the factorizations.
n=32 %= (2*2*2*2*31*7*5*11*19*21419*)/(2*2*31*11*23*29*2753*)
n=32 %= (2*2*7*5*19*21419)/(23*29*2753)
Here are the values of n that contain a 7 in the factorization of the
numerator:
4
6, 7, 8
11
13, 14, 15
18
20, 21, 22
25
27, 28, 29
32
34, 35, 36
39
41, 42, 43
46(x2)
...
Here are values of n that contain an 11 in the factorization of the
numerator:
10, 11, 12
21, 22, 23
32, 33, 34
43, 44, 45
...
For prime factor 13:
4
12, 13, 14
17
25, 26, 27
30
39, 40
43...
For prime factor 17:
13
16, 17, 18
30
33, 34, 35
...
Those patterns would seem to predict some of the factors of the
difference, but it appears that the simple pattern of groups of three
in a row does not hold.
For |s(n+1, n+2)|, here are the values of n where 7 is a prime factor
of |s(n+1, n+2)|:
6, 7, 8, 9, 10
13, 14, 15, 16, 17
20, 21, 22, 23, 24
27, 28, 29, 30, 31
34, 35, 36, 37, 38
41, 42, 43, 44, 45
...
Initially it appears that 7 is a prime factor of |s(n+1, n-3)| for n =
7z-1, 7z, 7z+1, 7x+2, and 7z+3 for positive integer z. For prime
factor 11:
10, 11, 12, 13, 14
21, 22, 23, 24, 25
32, 33, 34, 35, 36
43, 44, 45, 46, 47
...
Initially it appears that 11 is a prime factor of |s(n+1, n-3)| for n
= 11z-1, 11z, 11z+1, 11x+2, and 11z+3 for positive integer z. For
prime factor 13:
12, 13, 14, 15, 16
25, 26, 27, 28, 29
38, 39, 40, 41, 42
...
Initially it appears that 13 is a prime factor of |s(n+1, n-3)| for n
= 13z-1, 13z, 13z+1, 13x+2, and 13z+3 for positive integer z.
The same appears to hold for 17. The same appears to hold for 19.
For n=21=1*19+2 and n=40 = 2*19+2 the denominator has two prime
factors of 19.
The number 7 is the fourth prime, 11 the fifth, etcetera. A short
list of primes from 7 onwards is 7, 11, 13, 17, 19, 23, 29, 31, 37.
The prime factor 29 is a factor of the Stirling cycle number for n=28,
29, 30, 31, 32, and also for 35. The prime factor 31 is a factor of
the Stirling cycle number for n=30, 31, 32, 33, 34, and also for 28.
My factorization method is only thus far to n=47.
Five is not a prime factor of |s(n+1, n-3)| until n=24.
24, 25, 26, 27, 28
...
For prime factor 3:
4,
8, 9, 10, 11, 12, 13
17, 18, 19, 20, 21, 22x2
26x2, 27x2, 28x2, 29x2, 30x2, 31
35, 36, 37, 38, 39, 40
44, 45, 46, 47, ...
Well, I suppose I can conjecture that for prime p greater than or
equal to 7 that for n=pz-1, pz, pz+1, pz+2, pz+3 that p divides
|s(n+1, n-3)|, but that does not do much good in determining the form
of the polynomials to directly calculate |(s(n+1, n-3)| from (sum n)^4
- sum (n^4).
For x=5, the prime factor 11, the fifth prime number, is present as a
factor in n=11z-1, 11z, 11z+1, 11z+2, 11z+3, and 11z+4, 6 consecutive
Stirling numbers |s(n+1, n-4)|, with two occurrences in 11z-1 and 11z.
Those are the only occurrences of 11 in the prime factorizations for
n < 45. The same holds true for p=17. For p=19, there are those
prime factors and then also occurrences in n=9, and an extra
occurrence in each 19z+4.
The prime factorization for x=5 seems to be much faster than that for
x=4. The largest prime factors for x=5 are smaller than those for
x=4. The prime factorization for x=5 for a hundred values completed
rapidly, that for x=4 didn't after hours.
Here are reduced factorizations in ratio for x=5, ((sum n)^5 - sum
(n^5)) / |s(n+1, n-4)|:
n=5 %= (7*5*719)/(2*2)
n=6 %= (5*5*277)/(3)
n=7 %= (2*2*3*3*5*487)/(67)
n=8 %= (2*2*17*5*5*47)/(89)
n=9 %= (2*2*11*887)/(3*19)
n=10 %= (2*5*3*9239)/(7*71)
n=11 %= (2*5*5*13*379)/(3*173)
n=12 %= (2*5*7*3697)/(3*3*3*23)
n=13 %= (2*5*5*5023)/(11*61)
n=14 %= (2*2*2*173*193)/(11*71)
n=15 %= (2*2*17*5*43*1013)/(3*11*13*109)
n=16 %= (2*2*5*3*5*3*1597)/(13*373)
n=17 %= (2*2*5*19*67)/(7*13)
n=18 %= (5*5*2521)/(3*79)
n=19 %= (7*3*108869)/(17*23*23)
n=20 %= (5*11*132929)/(17*3*587)
n=21 %= (5*5*5*23*59*109)/(3*3*3*3*3*17*19)
n=22 %= (2*5*38557)/(19*89)
n=23 %= (2*2*5*5*5*6553)/(19*19*41)
...
n=100 %= (17*5*3*10931807)/(2*7*97*14947)
n=101 %= (5*5*43*103*353*1049)/(2*7*7*3*11*97*953)
n=102 %= (13*47*50333)/(2*2*7*3*3*3*3*3*3*11)
n=103 %= (2*2*7*5*5*3443051)/(11*101*15859)
n=104 %= (2*2*53*89451179)/(17*5*19*23*37*101)
...
So, in assuming that those ratios represent two polynomials g(n) and
h(n), that are the same for each value n, then I am looking for
something, for example multiplying the ratio by any x/x to get a
polynomial g(n) that matches that value for each n, and also the
polynomial h(n).
I thought I might discover some orthogonal polynomials, but I don't
know.
The evaluation of g(n) starts much larger than h(n) and as n grows
g(n)/h(n) goes to approximate 120. This implies that the univariate
polynomials g(n) and h(n) have the same degree and the coefficient a_0
of g(n) is 120 and of h(n) is 1, or rather, coefficient g_0 = 120 h_0.
I assume a variety of things, among them are that the coefficients are
integer, that a pair of polynomials with the required properties
exists, etcetera. Given a few prospective values of g_0, then here
are values of h_0.
30 1/4
60 1/2
120 1
180 3/2
240 2
360 3
480 4
600 5
720 6
n=5, % =
25165 / 4
50330 / 8
75495 / 12
100660 / 16
n^3 = 125
120 n^3 = 15000
600 n^3 = 75000
n=6,
n^3 = 216
600 n^3 = 129600
% = 6925/3
131575 / 57
1975
495 19...
1975 54.86
Here I am just trying to determine a polynomial g(n) with its first
coefficient g_0 being a multiple of 120, and, also, I don't know the
order of the first term. This is for x=5, for x=3 the order of the
polynomials is 1, it might be the case that it is 2 for x=4, but that
is not yet determined, and it might be 3 for x=5, but it could be 4.
n=5
720 * n^3 = 90000
720 n^3 + 420 n^2 = 100500
720 n^3 + 420 n^2 + 30 n = 100660
n=6
720 n^3 + 420 n^2 + 30 n = 155520 + 15120 + 180 = 170820
170820/6925 = 24.66715, bzzt.
n=5
720 n^3 + 400 n^2 + 130 n = 100660
n=6
720 n^3 + 400 n^2 + 130 n = 155520 + 14400 + 780 = 170700
170700/6925 = 24.64982...
n=5
720 n^3 + 360 n^2 + 330 n = 100660
n=6
720 n^3 + 360 n^2 + 330 n = 155520 + 12960 + 1980 = 170460
24.61516...
I guess I can enumerate many possible values of the coefficients,
assuming that the order of the polynomial is three, for a first term
of g_0 n^3, and determine how to get various values multiples of the
numerator of the ratio, and then see if the same equation works for
n=6 and even integer multiples of that numerator.
I write a program for x=4 to try and determine a polynomial g(n) that
is valid for n=4, 5,6. I assume that it is of the form g_0 n^2 + g_1n
+ g_3 where g_0 is a multiple of 24. I get quite a few polynomials
that work with n=4 and 5, now I'm trying for n=6.
I run the program for a reasonable amount of time, it doesn't get any
results. I try for multiple of 4 and 6 instead of 24, still no luck,
luck in mathematics is arbitrary, luck in mathematical pursuit is part
of inspiration.
Maybe I should divide the numerator by n^2, n^3, etcetera, to see
which one is proportional to the value.
I try to find a polynomial of rank 4, g_0 n^4 + g_1 n^3 + g_2 n^2 +
g_3 n + g_4. Say, how does one determine the rank and coefficients of
a polynomial given values of the expression for variables?
I guess it is pretty simple:
(n(n+1)/2)^4 -n(n+1)(2n+1)(3n^2-3n+1)/30
(n^8 +4n^7 +6n^6 +4n^5 +n^4)/8
-(6n^5 + 15n^4 +10n^3 -n)/30
(15n^8 + 60n^7 + 90n^6 +60n^5 +15n^4)/120
-(24n^5 +60n^4 +40n^3 -4n)/120
(15n^8 +60n^7 +90n^6 +36n^5 -45n^4 -40n^3 +4n)/120
Assuming I correctly transformed the algebraic expression, the above
is the polynomial that is (sum n)^4 - sum(n^4).
Then, I plan to divide that by a ratio of polynomials that results in
the unsigned Stirling cycle number |s(n+1, n-3)|, the ratio evaluates
to 4! = 24 in the limit. I'm going to just write s(n+1, n-x+1) for
unsigned numbers. These polynomials exist for s(n+1, n-1) and s(n+1,
n-2), that is s(n+1, n-x+1) for x=2 and 3, but I don't know if the
polynomials with real, integer coefficients g(n) and h(n) exist for
higher values of x, which would allow direct computation of the
Stirling cycle number as a ratio of polynomials without summing the
products of x-subsets of an n-set, summing over indexes, or via the
recurrence relation, except where sum (n^x) is expressed as a
Bernoulli polynomial, the coefficients of which are currently
generated by a recurrence relation.
It's unclear to me whether the polynomials g(n) and h(n) exist, or
whether there might be polynomials satisfying the expression with
non-integer coefficients, or not, one of the few reaons that I would
think they do is as they exist for x=2, 3.
The polynomial ratio g(n)/h(n) for x=2 is 2. The polynomial ratio
g(n)/h(n) for x=3 is 6(n+2)/(n-2), but the polynomials might be 3(n+2)
and 1/2 (n-2).
Perhaps it is so then that the first polynomials for g(n) and h(n) are
as so:
g_2(n) = 1
g_3(n) = 3 (n+2)
h_2(n) = 1/2
h_3(n) = 1/2 (n-2)
Then I'm trying to determine if there are polynomials g_4(n) and
h_4(n), in the limit g_x(n)/h_x(n) = x!.
I guess I would think something along the lines of
g_4(n) = 6 (n+2)(n+...)
h_4(n) = 1/4 (n-2)(n-...)
or
g_4(n) = 3 (n+2)(2n+1)...
h_4(n) = 1/2 (n-2) 1/2 (n-2)...
Yet, the polynomial ratio g_4(4)/h_4(4) would have to be equal to
9646/24 = 4823/12. So, the rank of each g_4 and h_4 would have to be
higher, as they are equal, to get to 9646 as a function of 4.
Oh well, at least s(n+1, n-3) ~ n^8/192, or s(n+1, n-x+1) ~ n^(2x) /
(2^x x!).
[space:~/math/Stirling] space% java Stirling1 200 195
s(200, 195) = -24311576926544407500
[space:~/math/Stirling] space% java Pow 199 10
199^10 = 97393677359695041798001
[space:~/math/Stirling] space% java Multiply 32 120
32 * 120 = 3840
[space:~/math/Stirling] space% java Divide 97393677359695041798001
3840
97393677359695041798001 / 3840 = 25362936812420583801
They have the same number of digits, it's off by about 5%.
[space:~/math/Stirling] space% java ApproxStirling1 199 5
|s(199+1, 199-5+1)| ~ 25362936812420583801
|s(199+1, 199-5+1)| = 24311576926544407500
[space:~/math/Stirling] space% java ApproxStirling1 1000 10
|s(1000+1, 1000-10+1)| ~
269114445546737213403880070546737213403880070546737
|s(1000+1, 1000-10+1)| =
255944587934420402547409683097079125714353332944500
Not very accurate, but closer than n^(2x+1) / (2^x x!). It's worse
for higher values of x. There are easy modifications to make it more
accurate.
Back to g_x(n) and h_x(n), I'm still trying to figure out g_4(n) and
h_4(n), if they exist, and of course generally g_x(n) and h_x(n).
Do you know any polynomials in these forms?
g_2(n) = 1
g_3(n) = 3 (n+2)
h_2(n) = 1/2
h_3(n) = 1/2 (n-2)
If so, what are g_x(n) and h_x(n)?
How do I fit a polynomial to known data values? Please don't suggest
polynomial regression.
Have a nice day,
Ross
Ross A. Finlayson
the unsigned Stirling cycle numbers s(n+1, n-x+1). The signed
Stirling number is (-1)^x s(n+1, n-x+1).
For x = 0, s(n+1, n+1) = 1.
s(n+1, n+1) = 1
For x=1, s(n+1, n) = n(n+1)/2 = (n^2 + n)/2.
s(n+1, n) = (n^2 + n) / 2
For x=2, s(n+1, n-1) = ((n(n+1)/2)^2- n(n+1)(2n+1)/6)/2 = (n^2+n)^2/8
- (n^2+1)(2n+1)/12 = (n^4+2n^3+n^2) / 8 - (2n^3+n^2+2n+1)/12 =
(3n^4+2n^3+n^2-6n-3)/24
s(n+1, n-1) = (3n^4+2n^3+n^2-6n-3)/24
for x=3, s(n+1, n-2) = ((n^2+n)^3 / 8 - (n^2+n)^2 / 4 ) (n-2) /
6(n+2) = ( (n^4+2n^3+n^2)(n^2+n)/8 - (n^4+2n^3+n^2)/4 )(n-2) / 6(n+2)
= ( (n^6+3n^5 + 3n^4 + n^3)/8 - (2n^4+4n^3+2n^2)/8 )(n-2) / 6(n+2) = (
n^6+3n^5+n^4-3n^3+2n^2)(n-2) / 48 (n+2) = ( n^7+3n^6+n^5-3n^4+2n^3
-2n^6-6n^5-2n^4+6n^3-4n^2 ) / 48 (n+2) = (n^7 + n^6 -5n^5 -5n^4 +8n^3
-4n^2) / 48(n+2) = n^2(n^5 + n^4 -5n^3 -5n^2 +8n^1 -4) / 48(n+2)
s(n+1, n-2) = n^2(n^5 + n^4 -5n^3 -5n^2 +8n^1 -4) / 48(n+2)
Thus s(n+1, n-2) is represented by a polynomial with rank 7, s(n+1,
n-1) is represented by a polynomial of rank 4, s(n+1, n) by a
polynomial of rank 2, and s(n+1, n+1) by a polynomial of rank 0.
One avenue of determining s(n+1, n-x+1) is that it is equal to the sum
of the products of each x-subset of {1, ..., n}. Then it is obvious
in a 2-D nxn matrix M where m_ij = i*j that the matrix is symmetrical
about the main diagonal and that the entries either above or below the
diagonal represent the 2-subsets of {1, ..., n}, thus that the sum of
all entries is easily calculated as (sum n)^2, the sum of the diagonal
entries is obviously sum (n^2) and that their difference is twice the
sum of the 2-subsets.
In the 3-d hypermatrix case M with m_ijk = ijk it is not so obvious
how to divide the elements where the elements on the triagonal are not
3-subsets and many other elements with matching indices are also not
3-subsets, and then each 3-subset is represented 6 times in the
matrix: eg (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and
(3, 2, 1). Then, (n+2)/(n-2) represents the value of the sum of those
cells relative to the cells with matching coordinates besides the main
diagonal with all matching coordinates.
I guess I could enumerate all the cells in a 4-matrix, from (1, 1, 1,
1) to (n, n, n, n). Then, I'll remove the elements on the main
diagonal, and get metrics of those elements having two and three
matching elements. Then, of the remaining elements with four distinct
coordinates, there would be 4!=24 representations of each 4-subset of
{1, ..., n}. So what it would do is generate the n^x-many element
array with each of the coordinates, the multidimensional array, and
then start counting things about it.
One thing I do is check the sums of each of the elements of the x-d
hypermatrix with all matching elements (the main n-agonal) and also
with x-1, x-2, matching elements, and for elements with no matching
elements that represent the x-subsets of n. One thing I've noticed in
the case of x=4 is that the sum of the elements with no matching
coordinates is 24 times s(n+1, n-3). That is to say, the elements at
coordinates that represent k-subsets sum to 4! times s(n+1, n-4+1).
This is good to know, but then I am left trying to determine how to
calculate that value from n and x as a univariate polynomial of n or a
ratio of univariate polynomials of n.
I'm looking at the count and sum of elements that have at least a pair
of matching coordinates.
n = 4 x = 4
Sum of elements: 10000
Sum of elements with 3 matching elements: 2037 = 7*3*97
Sum of elements with 2 matching elements: 7033 = 13*541
Number of elements with 3 matching elements: 39 = 3*13
Number of elements with 2 matching elements: 189 = 3*3*3*7
I'm not quite sure how to calculate how many elements there are with
at least two matching coordinates and less than x many matching
coordinates, or their elements' sum.
n = 5 x = 4
Sum of elements: 50625 = (sum n)^x
Sum of elements with 4 matching elements: 979 = sum (n^x)
Sum of elements with 3 matching elements: 7412 = 2*2*17*109
Sum of elements with 2 matching elements: 35658 = 2*7*3*3*283
Sum of elements with no matching elements: 6576 = x! * s(n+1, n-x+1)
Number of elements with 4 matching elements: 5 = n
Number of elements with 3 matching elements: 64 = 2*2*2*2*2*2*2*2
Number of elements with 2 matching elements: 436 = 2*2*109
Number of elements with no matching elements: 120 = n! ~ (x+1)!
Finding simple expressions for those items above denoted as equal to
factorizations would help determine s(n+1, n-x+1) from the other known
quantities.
Once again, in examining an x-d nxnx...xn hypermatrix m where m_ijk...
= i*j*k*... in an attempt to discern a method to calculate unsigned
Stirling cycle numbers, there are simple methods for x=1, 2, 3 but I
want to determine the forms for x=4, 5, ..., and they are not so
simple.
n = 4 x = 4
Sum of elements: 10000
Sum of elements with 4 matching elements: 354
Sum of elements with 3 matching elements: 2037
Sum of elements with 2 matching elements: 7033
Sum of elements with no matching elements: 576
Number of elements with 4 matching elements: 4
Number of elements with 3 matching elements: 39
Number of elements with 2 matching elements: 189
Number of elements with no matching elements: 24
n = 6 x = 4
Sum of elements: 194481
Sum of elements with 4 matching elements: 2275
Sum of elements with 3 matching elements: 21398
Sum of elements with 2 matching elements: 131832
Sum of elements with no matching elements: 38976
Number of elements with 4 matching elements: 6
Number of elements with 3 matching elements: 95
Number of elements with 2 matching elements: 835
Number of elements with no matching elements: 360
n = 7 x = 4
Sum of elements: 614656
Sum of elements with 4 matching elements: 4676
Sum of elements with 3 matching elements: 52611
Sum of elements with 2 matching elements: 394913
Sum of elements with no matching elements: 162456
Number of elements with 4 matching elements: 7
Number of elements with 3 matching elements: 132
Number of elements with 2 matching elements: 1422
Number of elements with no matching elements: 840
What should I read to get more information about hypermatrices, what
is the definitive reference on hypermatrices?
I noticed someone asking about space being 3-D, I noticed the other
day in reading about knots that MathWorld noted that knots don't exist
in dimensions >=4. I don't quite understand that.
Besides knot theory, another thing I am trying to grasp is the concept
of the matroid. It appears to be some generalization of graphs and
matrices.
I notice some posters asking about multitudes of infinities and a
universal set. A set of all sets would be its own powerset, and would
not contain an element of all sets not containing themselves. Are
there more even than odd numbers? No, there are not. Are there more
integers than even integers? Yes, there are, in the integers (or
rationals, reals, complex, or hypercomplex numbers). There are also
more rational than integers in the rationals, etcetera, and more
positive integers than non-negative integers in the integers, for
various reasons. Are there functions that map among all those sets?
Consider a space-filling curve. An infinite set a is greater than
another infinite set b if infinitely many proper supersets of b are
proper subsets of a, |[a]| > |[b]|, an infinite set a is greater than
a proper superset b, |(a)| > |(b)|, the set or a linear translation.
Anyways, I want to know about how many permuations of x elements of
{1, ..., n} there are with multiples of the elements in the multiset.
A combination of the elements of an n-set {1, ..., n} is a subset of
that set. A permutation of a combination of an n-set is an ordered
list, a sequence, of elements of a combination of an n-set. A
multiset is a set that many contain multiple indistinguishable
instances of an otherwise unique set element. The xn-set is here
defined as the multiset with x-many instances of each element of an
n-set. The hypermatrix indices or coordinates of an x-d nxnx...xn M
= m_ijk... are the set of permutations of the x-subsets of the xn-set.
For example, for a 5-d hypermatrix of rank or degree 4, coordinates
include (1, 1, 1, 1, 1), (1, 2, 3, 3, 4), and (4, 4, 4, 4, 4),
permutations of {{1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4,
4, 4, 4, 5, 5, 5, 5, 5}}, the xn-set that is the (4)(5)-set, using two
braces to signify a multiset.
About a multiset, a set still has to have unique elements. There are
a variety of ways to represent the unique elements of a set. The von
Neumann ordinals are the most direct way to represent natural
integers, very large composites could also represent the natural
integers and as well multitudes of integers. The composites would
equate to each other as integers, a subset of the composite could be
the von Neumann ordinal, the integer operations would operate on
uniquified composite multiset element ordinals.
I've seen on Dr. Math's site on Math Forum a form for the Stirling
cycle number, the unsigned Stirling number of the first kind, for n
and n-2, as noted in this thread. I found a different one for n+1 and
n-1 before that, I assume it is already known. I found a form for
s(n+1, n-2), I have not seen any other method directly calculating
that value via a ratio of univariate polynomials. I want to know
where it was already known, because the same source might be able to
tell me all the other ones that are known. I want to determine forms
for the other values of x for s(n+1, n-x+1) without having to learn
umbral calculus, Gian-Carlo Rota's "shadow"-y exploration of
recurrence relations and perhaps linear relations, although it seems a
good thing to know.
Have I discovered a form for s(n+1, n-4+1) since my previous post?
Have you?
I'm thinking about investing in one of those CAS, computational
algebra systems. Yet, I want to actually write one. I'll look to the
noted references.
I was going to add a political spiel but "this is sci.math, humor is
irrelevant", maybe later.
Why did Isabel cross the Atlantic? To get to the other side. I hope
that Isabel doesn't do too much damage. What's an F-6 hurricane? I
heard Isabel has gusts of more than 190 miles per hour. Buckle up.
How about a war against climate change? People should start
evacuating. Hopefully Isabel will careen off towards Greenland, yet
it currently appears that it might march up the Potomac, and be the
most damaging "Act of God" to ever hit the United States. Some day a
volcano will erupt, although hopefully far enough in the future that
we can alleviate its pressure via tunneling or applied explosives, the
same goes for meteor strikes with anti-meteor tugs, rockets, and
lasers, etcetera, and of course far in the future there will be
significant control of the weather besides the crude burning of
petrochemicals to release huge amounts of greenhouse gasses,
"petrified" muck. Seriously, people on the Maryland shore should get
warnings.
Anyways, I think I found another factorial identity, as described
previously, but it's kind of strange, lim n->oo ((sum n)^n - sum(n^n))
/ n!^2 = 1, I'm still trying to understand it.
Anyways, i want to learn more about the theory of hypermatrices, n-d
arrays, "matroids", I guess I should read more of the tensor material.
Bye bye,
Ross
Erik Aronesty
did you use a good source of entropy?
Ross A. Finlayson
numbers. It isn't a polynomial ratio that is the Stirling cycle
number s(n+1, n-x+1) for arbitrary x and n but it's an approximation.
I haven't researched cycle numbers heavily but haven't seen other
explanations of these phenomena.
One thing to determine is besides the term n^2x what would be the
following terms. Basically the form is of (n(n+1)/2 )^x minus the
polynomial that is the sum of the powers of x.
That, divided by x!, is an "approximation" of s(+1, n-x+1). It
reflects its asymptotic behavior.
The approximation is accurate for n >> x.
I'm still trying to evaluates (n+1)(n+2)...(n+2) / n^n in terms of n.
That is, in a way, about infinite scalar variables, replacing z with
f(n) in Euler's factorial formula for Gamma, that kind of thing.
Other things to consider in terms of things infinite are the
asymptotic densities of sets and as well linear translations, a
constant added to each element of a set, various isometries.
Then there's also the concept from this thread of determining how many
of the infinite binary sequences that canonicalize into rational
sequences there are. We're not quite halfway there.
Here's something to consider, it's about "random probability
distributions" over the integers or reals besides "random uniform
probability distributions". Basically, it has the highest probability
of being zero. By example, start with null, zero, and then with one
binary modulus, zero and one. So there P(0) is 2/3, and P(1) is 1/3.
With the second binary modulus, then P(0)=3/7, P(1)=2/7, P(2)=1/7,
P(3)=1/7. What this is about is sum 2^x.
x=0, sum 2^x = 1
x=1, sum 2^x = 3
x=2, sum 2^x = 7
x=3, sum 2^x = 15
x=4, sum 2^x = 31
The value of sum 2^x is 2^(x+1)-1, the Mersenne number, the binary
repunit.
Then for example 2^4 = 16, representing 0 to 15, and then there are
2^5-1 or 31 values: {{0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4,
4, 5, 5, 6, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 15 }}. Then, the
probability distribution over N[0,15] is
P(0) = 5/31
P(1) = 4/31
P(2) = 3/31
P(3) = 3/31
P(4) = 2/31
P(5) = 2/31
P(6) = 2/31
P(7) = 2/31
P(8) = 1/31
P(9) = 1/31
P(10) = 1/31
P(11) = 1/31
P(12) = 1/31
P(13) = 1/31
P(14) = 1/31
P(15) = 1/31
Is this useful for anything? Is it? Anyways the idea of this kind of
distribution over the unit reals is basically the inverse as for the
integers.
P(0) = 5/31
P(1/2) = 4/31
P(1/4) = 3/31
P(3/4) = 3/31
P(1/8) = 2/31
P(3/8) = 2/31
P(5/8) = 2/31
P(7/8) = 2/31
P(1/16) = 1/31
P(3/16) = 1/31
P(5/16) = 1/31
P(7/16) = 1/31
P(9/16) = 1/31
P(11/16) = 1/31
P(13/16) = 1/31
P(15/16) = 1/31
A difference between this and a "random uniform probability
distribution" over [0,1] is that in the uniform case P(0) is 1/(2^n)
where in this case P(0) = n/(2^n). It might have some relevance to
earlier discussion on this thread about the distribution of the reals.
I still think that rationals and irrationals alternate in the reals.
I consider cases where they do not.
I'm still trying to determine a method for evaluating s(n+1, n-x+1)
for x=4 as a univariate polynomial, instead of factorial or recurrence
relation, although deriving the polynomial itself involves the
Bernoulli polynomial coefficient generating recurrence relation.
There are solutions for x={0, 1, 2, 3}, as noted here. I want to
understand it vis-a-vis generating functions, orthogonal polynomials,
hypergeometric functions, the umbral calculus of finite differences,
hypermatrices, and combinatorics.
Please explain.
Ross
Ross A. Finlayson
> When you wrote the computer program to test these random-number flips,
> did you use a good source of entropy?
Please expand.
When I wrote, the computer program, testing random numbr flips (?),
was a robust source of entropy used?
I don't understand your question. In writing prime factorization
software, thus far I have not implemented a method using random
alternative selection for factor checking. I'm only using digit
summation congruence and brute force prime divisor checking.
There are probably a variety of decent entropy generation methods,
those generated by computer algorithms are repeatible given identical
input seeds. Those generated from basically statistically random
external events, we cannot replicate the input events as time is a
direction.
I think a goal is to find the best results from a worst source of
entropy, that would be a method for determining an efficient
algorithm.
Please explain your question. Oh, uh, assume cryptography is a
facade.
Ross