What's 'X !' derivative?

[villaf@feedme]

I haven't received an Advanced Calculus Curse yet, so I can't tell if 'X!'
derivative exists or not. When I try to get it with my HP, I get 'der@!(X,1)'
where @ is a dot.
Could anybody please explain it to me ?

Thanks.

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[Christian Meland]

villaf@feedME wrote in article <3514c...@lightning.ica.net>...

> I haven't received an Advanced Calculus Curse yet, so I can't tell if
'X!'
> derivative exists or not. When I try to get it with my HP, I get
'der@!(X,1)'
> where @ is a dot.
> Could anybody please explain it to me ?
>

I'll try. X! is only defined for integers, and has no derivative. The hp
uses the fact that X!=Gamma(X+1). The gamma function is continious for
X+1>0. I have never seen an expression for it's derivative, but the
derivative of ln(gamma(x)) is called the trigamma-function, which can be
evaluated with a function in the library stat03 found in stat03.zip (search
on http://ftpsearch.ntnu.no)

Christian

[David Starr]

Minor correction. d/dx(ln(gamma(x)) is the Digamma function.

-David-

[JEEjohn]

In article <351555...@TheRamp.net>, David Starr <bu...@TheRamp.net> writes:

>
>Christian Meland wrote:
>>
>> villaf@feedME wrote in article <3514c...@lightning.ica.net>...
>> > I haven't received an Advanced Calculus Curse yet, so I can't tell if
>> 'X!'
>> > derivative exists or not. When I try to get it with my HP, I get
>> 'der@!(X,1)'
>> > where @ is a dot.
>> > Could anybody please explain it to me ?
>> >
>> I'll try. X! is only defined for integers, and has no derivative. The hp
>> uses the fact that X!=Gamma(X+1). The gamma function is continious for
>> X+1>0. I have never seen an expression for it's derivative, but the
>> derivative of ln(gamma(x)) is called the trigamma-function, which can be
>> evaluated with a function in the library stat03 found in stat03.zip (search
>> on http://ftpsearch.ntnu.no)
>>
>> Christian
>>

(snip)

>Minor correction. d/dx(ln(gamma(x)) is the Digamma function.
>
>-David-

When dealing with special functions, there are all sorts of names
and definitions. In this case, the logarithmic derivative of the gamma
function (what is given above), is called the Digamma function or
also the Psi function. The derivative of the log of the gamma is
equal to the derivative of the gamma function divided by the gamma
function. So that the derivative of the gamma function is the
Psi or digamma function times the gamma function.

John Edry
JEE...@aol.com

[Søren Bækhøj Kjær]

villaf@feedME wrote in message <3514c...@lightning.ica.net>...

>I haven't received an Advanced Calculus Curse yet, so I can't tell if 'X!'
>derivative exists or not. When I try to get it with my HP, I get
'der@!(X,1)'
>where @ is a dot.
>Could anybody please explain it to me ?
>

>Thanks.
>
>------------------------------------------------------------------
> Hungry for NEWSGROUPS??? USE feedME.ORG
> Read and Post to 30,000 groups through feedME.ORG
> FREE FREE FREE http://www.feedME.ORG FREE FREE FREE
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>

Hi, I will try to help you.

First the difenition of X!

X!= X*(x-1)*(X-2)****3*2*1. If X = 10, then we have:
10(10-1)(10-2)(10-3)(10-4)(10-5)(10-6)(10-7)(10-8)(10-9) = 3.628.800

If you want to find the derivate of 10! then find the derivate of
x(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) and insert x = 10.
The derivate of 10! is then equal to 10.628.640

I hope that this can help you..

[francisco guzman]

Your only problem is that the derivative of a constant is always 0.
--
francisco guzman
fgu...@cyberg8t.com
http://www.cyberg8t.com/fguzman

[Christian Meland]

David Starr <bu...@TheRamp.net> wrote in article
<351555...@TheRamp.net>...

> > Christian

> >
> > > Thanks.
> > >
> > > ------------------------------------------------------------------
> > > Hungry for NEWSGROUPS??? USE feedME.ORG
> > > Read and Post to 30,000 groups through feedME.ORG
> > > FREE FREE FREE http://www.feedME.ORG FREE FREE FREE
> > > ------------------------------------------------------------------
> > >
> > >
>

> Minor correction. d/dx(ln(gamma(x)) is the Digamma function.
>

Sorry, you're right. The Trigamma is the derivative of the digamma.

Christian

> -David-
>

[dave y]

On Mon, 23 Mar 1998 17:06:27 +0100, "Søren Bækhøj Kjær" wrote
...

>First the difenition of X!
>
>X!= X*(x-1)*(X-2)****3*2*1. If X = 10, then we have:
>10(10-1)(10-2)(10-3)(10-4)(10-5)(10-6)(10-7)(10-8)(10-9) = 3.628.800
>
>If you want to find the derivate of 10! then find the derivate of
>x(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) and insert x = 10.
>The derivate of 10! is then equal to 10.628.640

...

But you forget X! is not a continuous function and has no derivatives.
It is defined only at points 1, 2, 3, 4, ... ,X. This makes it a
discrete function which has a 1st difference, 2nd difference, etc. as
the stand-in for the derivatives. There are various ways to define a
reasonable answer. e.g.,

1. trailing difference

D = X! - (X-1)!

2. bi-lateral difference

D = [ (X+1)! - (X-1)! ] / 2

3. leading difference

D = (X+1)! - X!

So the answer can be chosen to be one of:

1. 3.265.920
2. 19.776.960
3. 36.288.000

with the second one the better choice.

dave y.

[sag...@usa.net]

> If you want to find the derivate of 10! then find the derivate of
> x(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) and insert x = 10.
> The derivate of 10! is then equal to 10.628.640

Sorry, but the author of this post is incorrect. Even though
x(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) does return 10! when x=10,
this function never returns the factorial of any number other than 10 (which
should be clear). This function is clearly not a "factorial" function,
because such a function should return x! for all input values, x (without
needing to alter the function for each x value!). The only smooth function
(for positive x) which returns x! and is differentiable is the gamma function
(as previously mentioned) -- x!=gamma(x+1). The derivative of the function
x(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)
will not agree with the derivative of gamma(x+1) (= x!) since these two
functions agree only at x=10.

Scott Guth
Mathematics Dept.
Mt. San Antonio College

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[John H Meyers]

Although the discrete factorial function is defined for integers,
it can be made a continuous function by defining it as

'F(X)=\.S(1,0,LN(Z)^X,Z)' where \.S represents the Integral symbol.

This is not a very practical means for calculation, but it is correct
in theory (you can verify it using integration by parts, recursively);
the HP48 delivers the correct numerical results for any real argument
to the built-in factorial function, but via entirely different means.

The derivative is then 'derF(X)=\.S(1,0,LN(LN(Z))*LN(Z)^X,Z)'

If you want to calculate this derivative expression numerically,
then better use 'derF(X)=\.S(1,0,RE(LN(LN(Z)))*LN(Z)^X,Z)'
(to avoid complex integrand values).

It is a bit easier to estimate the derivative numerically
by more direct means:

\<< SWAP OVER DUP2 + FACT ROT ROT - FACT - SWAP / 2 / \>>

Where the arguments are X and a small interval,
e.g. 3 1E-6 ==> 7.537

Note that FACT is a synonym for the command !,
and also works as a "prefix" style algebraic function name.

If you evaluate the former integral expression for X=3 in 3 FIX mode,
you will obtain 7.535, which is fairly close, given that each of these
calculations is approximate.

If we consider Stirling's approximation for the factorial function:
'S(X)=X^X*EXP(-X)*\v/(2*\pi*X)' [omitting the final adjusting series]

Then 'derS(X)=S(X)*(LN(X)+INV(2*X))' by this approximation; note that
this is asymptotic to F(X)*LN(X) for large X, as you may confirm
using the above numerical derivative program.

Curiously, the derivative of 'P(X)=X^X' is 'derP(X)=P(X)*(LN(X)+1)'
(note the resemblance?)

Chances are, that no one actually wanted all this info, right? ;-)

-----------------------------------------------------------
With best wishes from: John H Meyers <jhme...@mum.edu>

[dave y]

On 26 Mar 1998 05:24:51 GMT, jhme...@miu.edu (John H Meyers) wrote:

>Although the discrete factorial function is defined for integers,
>it can be made a continuous function by defining it as
>
>'F(X)=\.S(1,0,LN(Z)^X,Z)' where \.S represents the Integral symbol.

But what is the point of forcing a derivative on a function which
doesn't have one. There is an infinite set of differentiable
functions which satisfy X! at X=INT(X). In what sense is your one
equation the 'right' one?

dave y.

[John H Meyers]

I wrote:

> Although the discrete factorial function is defined for integers,
> it can be made a continuous function by defining it as
> 'F(X)=\.S(1,0,LN(Z)^X,Z)' where \.S represents the Integral symbol.

In article <351c26aa...@news.concentric.net>,
sag...@cris.com (dave y) writes:

> But what is the point of forcing a derivative on a function which
> doesn't have one. There is an infinite set of differentiable
> functions which satisfy X! at X=INT(X). In what sense is your
> one equation the 'right' one?

This is a very astute inquiry, which ought to be answered.

The expression I wrote is one of the ways to define Gamma(X+1),
which is the same function that others have been mentioning.
Let's see whether we can justify identifying it with the
factorial function, as the HP48 certainly does (and I believe
all HP scientific calcs since the HP34C have done):

If we rest upon a definition of factorial which has no meaning for
non-integers, then that definition will not admit of having a
derivative, inasmuch as the function will not be continuous,
but will be defined only at discrete points.

Much the same is true of a good many other mathematical functions;
for example, the function A^X is first introduced to us in terms of
the result of multiplying together exactly X identical factors A;
when defined as such, then A^.5 or A^0 or A^-1 are all meaningless,
as are all exponents which are not positive integers. Therefore,
we would have to say at this point that A^X has no derivative,
because A^X does not exist for non-integer (or non-positive) X.

However, when we look at the logical rules for dividing powers, we find
that we can make these rules more universally true by considering A^0
to be equal to 1, and A^-N to be equal to 1/A^N. By further considering
the subject of taking roots, we come to an even more universal definition
of rational exponents as a combination of integer roots and powers;
finally, by some leap of faith (justifiable after some fancy limit
arguments), we can even extend this notion to irrational exponents,
giving us a universal definition, plus a derivative: A^X*LN(A)

How do we know that this definition of "exponents" is the "right" one?
Well, it fills in the original gaps in a perfectly consistent and very
useful way, which also seems to be the simplest way, and a great many
scientific foundations seem to be laid upon the bias of our minds
that the simplest and most beautiful way is the "best" and most
righteous way of things. "The eternal mystery of the universe
is its comprehensibility," said Einstein, who always seemed to
want to avoid having to add more terms to his relativity equations,
even though he also said "Everything should be as simple as possible,
but no simpler."

If I show you the sequence 1, 4, 9, 16, 25, 36, 49, 64, 81, ...
and ask you to fit a polynomial to this set of data points,
you might well say that there are infinitely many polynomials
which pass through all of the points thus far listed, yet even if we
try to fit a higher-degree polynomial, it is hard to define a "best fit"
in any way which does not lead powers higher than 2 to tend to have
small coefficients; if I further say that the polynomial must pass
through the square of every integer whatsoever, then Y=X^2 seems
to be forced upon us. Other functions exist which also pass through
all these points, e.g. Y=X^2+NI(X), where NI(X) ["not an integer"]
is zero for integers X and non-zero for non-integers X; however,
if we are emotionally attached to the beauty of continuous and infinitely
differentiable functions, I think we are stuck with only one possibility.

I don't know how the definition of the Gamma function materialized,
but considering that it happens to pass through all of the
infinitely many integer-valued factorials, it might be quite difficult
to duplicate this feat with any other definition of a continuous and
infinitely differentiable function (does Taylor's theorem cover this?)

By the way, suppose we define the "factorial polynomial X^N" as
X*(X-1)*...*(X-N+1) [temporarily redefining the ^ symbol]; note that
this product has the same number of factor terms as what we usually call
X^N, except that the factors progressively decrease, rather than all being
exactly the same. This is called a "factorial polynomial," because it is
certainly a polynomial, and also because when X=N it equals FACT(N).

Now, one curious fact about factorial polynomials, which was presented to
be proved as a math contest question in my high school, is that if you
write down the usual definition of the "binomial theorem," which is

'(a+b)^n=\GS(r=0,n,COMB(n,r)*a^(n-r)*b^r)' [Greek Sigma]

(Display this in the Equation Writer for a nicer view)

Then if you re-interpret each ^ symbol in the above definition as now
meaning a "factorial polynomial," rather than the old-fashioned meaning,
it so happens that the above formula remains true (prove it!)

Note that the COMB(n,r) function itself is basically defined and
calculated by means of factorial polynomials, as you may have noticed.

For extra credit, define a factorial polynomial in terms of
Gamma functions (or X! functions on the HP48).

Does this suggest that Nature has some highly consistent
patterns of intelligence lurking all throughout its existence,
as Einstein seemed to feel?

---------------------------

"Imagination is more important than knowledge." - Einstein

"I know truth never triumphs, but its opponents eventually grow extinct."
- Einstein

"Mathematics possesses not only truth, but supreme beauty
-- a beauty cold and austere, like that of sculpture,
without appeal to any part of our weaker nature...
sublimely pure, and capable of a stern perfection
such as only the greatest art can show." - Bertrand Russell

"A mathematician, like a painter or a poet, is a maker of patterns.
If his patterns are more permanent than theirs,
it is because they are made with ideas.

The mathematician's patterns, like the painter's or the poet's,
must be beautiful; the ideas, like the colours or the words,
must fit together in a harmonious way. Beauty is the first test:
there is no permanent place in the world for ugly mathematics."

- G. H. Hardy ("A Mathematician's Apology")