Re: Rare Event...The Full Moon Strikes Again!

[Charles Brenner]

On Feb 11, 4:55 am, Thomas Dehn <thomas-use...@arcor.de> wrote:
> Warning: off-topic mathematics below.
> No bridge content in this posting.
>
> On 02/11/2012 10:27 AM, Charles Brenner wrote:
>
> > On Feb 10, 9:24 pm, Thomas Dehn<thomas-use...@arcor.de>  wrote:
> >> On 02/11/2012 02:41 AM, Charles Brenner wrote:
>
> >>> On Feb 10, 1:40 pm, Thomas Dehn<thomas-use...@arcor.de>    wrote:
> >>>> On 02/10/2012 08:34 PM, Charles Brenner wrote:
>
> >>>>> On Feb 9, 6:17 pm, "Fred."<ghrno-goo...@yahoo.com>      wrote:
> >>>>>> Sometimes the improbable happens without intervention,
> >>>>>> but not often.
>
> >>>>>> Someone once gave me a thought experiment.  You write
> >>>>>> a program to simulate random coin flips.
>
> >>>>>> Case 1:  You start it up and it goes heads the 1st 6 times.
>
> >>>>>> Case 2:  You start it up and it goes 1000 flips
> >>>>>>             without 6 six heads in a row.
>
> >>>> p(n+1)=p(n)-1/128*p(n-6).
>
> >>> Nice -- that's how to do it. Assuming that
> >>> p(n+1)=p(n)-1/128*p(n-6)
> >>> is correct, there's a mechanistic way to write an equation for p(n)
> >>> from this. The process is tedious -- it requires writing down the
> >>> roots of the characteristic polynomial
> >>> x^7 - x^6 + 1/128, then p(n) is a linear combination of the n-th
> >>> powers of those roots (finding the coefficients more tedium)
> >>> notwithstanding that (I guess) six of the seven roots are complex.
> >>> Maybe Mathematica. Anyway my point is that conversely the formula that
> >>> this procedure produces, with all it's irrational complex numbers, is
> >>> as concise an answer as there is.
>
> >> I found a functional online polynomial root calculator here:http://xrjunque.nom.es/precis/rootfinder.aspx
>
> > Wow. That is one cool web tool.
>
> >> That gives:
> >> x^7 -x^6 +(1/128) = 0 =>
> >> =>  x^7 -x^6 +0.0078125 = 0
> >> =>  (x +0.420154549170266)*(x -0.5)*(x -0.991791421712163)*(x +(
> >> -0.195146016987811 -i*0.408930828152264))*(x +(0.23096445325876
> >> -i*0.359572188979042))*(x +(0.23096445325876 +i*0.359572188979042))*(x
> >> +( -0.195146016987811 +i*0.408930828152264)) = 0
>
> >> Real solutions:
> >> Root 1: -0.420154549170266
> >> Root 2: 0.5
> >> Root 3: 0.991791421712163
> >> Complex roots:
> >> Root 4: -0.23096445325876+0.359572188979042 * i
> >> Root 5: -0.23096445325876-0.359572188979042 * i
> >> Root 6: 0.195146016987811-0.408930828152264 * i
> >> Root 7: 0.195146016987811+0.408930828152264 * i
>
> >> Yes, 1/2 is one of the roots.
>
> > Euler and Ramanujan would have noticed that. I didn't.
>
> It is pretty easy to find the 1/2 root
> because of the simple structure
> of the characteristic polynomial.

I had in mind the famous conversation between Ramanujan and Hardy, in
which Hardy happened to mention the number 1729 and Ramanujan
instantly noted that it was the sum of cubes in two different ways.
That fact was obvious to Hardy as well, once alerted to it, for when
Hardy ("naturally") asked for a corresponding number for 4th powers
Ramanujan, knowing his friend, replied that there was no similarly
*obvious* answer. (Euler also explored the question.)

With this polynomial, maybe no manipulation is necessary to guess 1/2
as a root once you think to look. I was just admitting that I didn't
look. I only blindly attacked the polynomial with numeric tools.

Charles


> I'll transform it a bit to make it more obvious:
>
> Let y=1/x:
>
> y^(-7) - y^(-6) + (1/128) = 0 | multiply by 128*y^7
>
> 128 - 128y + y^7 = 0.
>
> This is a polynomial where all coefficients are integer,
> and furthermore the coefficient at the highest power of
> y is 1 (or -1).
> It is pretty easy to see that any rational solution is
> integer (*), and must furthermore be a divisor of 128.
>
> So all you have to do is try all divisors of 128:
> 1, -1, 2, -2, 4, -4, 8, -8, 16, -16, ..., 128, -128.
>
> (*) To see that any rational solution is integer, consider
> y^n + a_(n-1)*y^(n-1) + ... + a_0 = 0,
> where all a_i are integer.
>
> Let y=p/q be any non-integer rational solution, where p and
> q are integers, have no common divisors, and q>1.
> Insert y=y/q, multiply the equation by q^n:
>
> p^n + a_(n-1)*p^(n-1)*q + ... + a_0*q^n = 0.
>
> If you look at this modulo q, p^n mod q is not 0, as
> p and q have no common divisors, and all other parts
> are zero modulo q. This is a contradiction, and thus
> such a non-integer rational solution y=p/q cannot exist.
>
> Thomas

[Gerben Dirksen]

> I had in mind the famous conversation between Ramanujan and Hardy, in
> which Hardy happened to mention the number 1729 and Ramanujan
> instantly noted that it was the sum of cubes in two different ways.
> That fact was obvious to Hardy as well, once alerted to it, for when
> Hardy ("naturally") asked for a corresponding number for 4th powers
> Ramanujan, knowing his friend, replied that there was no similarly
> *obvious* answer. (Euler also explored the question.)
>
> With this polynomial, maybe no manipulation is necessary to guess 1/2
> as a root once you think to look. I was just admitting that I didn't
> look. I only blindly attacked the polynomial with numeric tools.
>
> Charles

Although correct, apparently Euler discovered that:

158^4 + 59^4 = 134^4 + 133^4 = 635,318,657

Gerben

[Charles Brenner]

Yes, and needless to say he discovered it by brainpower and not by
sheer calculating. In fact he may (my memory is vague) have had an
algebraic identity of the form, hence an infinite family of solutions.
He also (of course) explored the corresponding question for fifth
powers as did I with a 1960's computer (the 7090, "-ty" standing for
Transistorized), and as have others with much more powerful computers
and more ingenious programs. Discuss: Which is the stronger evidence
that there is no solution for fifth powers, that computers haven't
found one, or that Euler didn't?

Charles

[Michael Stemper]

In article <804945be-ec83-47a0...@l14g2000vbe.googlegroups.com>, Gerben Dirksen <gerb...@hotmail.com> writes:

>> With this polynomial, maybe no manipulation is necessary to guess 1/2
>> as a root once you think to look. I was just admitting that I didn't
>> look. I only blindly attacked the polynomial with numeric tools.
>

>Although correct, apparently Euler discovered that:
>
>158^4 + 59^4 = 134^4 + 133^4 = 635,318,657

Presumably, he used this to factor 635,318,657.

When I learned of his method to factor numbers that can be expressed as
the sum of two positive squares in two ways, I coded it up. Unfortunately,
the only way that I knew to find decompositions into the sum of two squares
was brute force, so the program was painfully slow.

Are there any good ways to perform this decomposition? (I wouldn't
consider "factor it and reverse Euler's method" to be very good.)

--
Michael F. Stemper
#include <Standard_Disclaimer>
This message contains at least 95% recycled bytes.

[Thomas Dehn]

On 02/15/2012 02:46 PM, Michael Stemper wrote:
> In article<804945be-ec83-47a0...@l14g2000vbe.googlegroups.com>, Gerben Dirksen<gerb...@hotmail.com> writes:
>
>>> With this polynomial, maybe no manipulation is necessary to guess 1/2
>>> as a root once you think to look. I was just admitting that I didn't
>>> look. I only blindly attacked the polynomial with numeric tools.
>>
>> Although correct, apparently Euler discovered that:
>>
>> 158^4 + 59^4 = 134^4 + 133^4 = 635,318,657
>
> Presumably, he used this to factor 635,318,657.
>
> When I learned of his method to factor numbers that can be expressed as
> the sum of two positive squares in two ways, I coded it up. Unfortunately,
> the only way that I knew to find decompositions into the sum of two squares
> was brute force, so the program was painfully slow.
>
> Are there any good ways to perform this decomposition? (I wouldn't
> consider "factor it and reverse Euler's method" to be very good.)

Depends on the size of numbers you are thinking about.
I'd guess that computing a rainbow table once isn't
that expensive for, say, all numbers up to 10^12.


Thomas

[micro...@hotmail.com]

Is the Moon Mooning you?