S.O.S. Mathematics CyberBoard

[Metru]

It is known (from Euler) that:

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Respectfully, Metru.

[jinydu]

Doesn't anyone have anything to say about Zeta(3)? Its such a tantalizing problem, simple enough for a middle school student to understand.

1 + (1/8) + (1/27) + (1/64) + (1/125) + (1/216) + (1/343) + (1/512) + (1/1000) + ...

I realize this is an unsolved problem, but has anyone thought about it, just for the sake of fun?

[CHAVS]

zeta(3) is irrational (proved by Aperey).

[jinydu]

[shmoe]

[jinydu]

Now, I guess I'll try to (finally) post my thoughts on Zeta(3), and hopefully not lose it all to an invalid session.

As I hinted at in an earlier post, my idea was to use Euler's method in reverse.

I start out with a function written as an infinite product:

f(x) = (1-(x^3)/(pi^3))*(1-(x^3)/(8pi^3))*(1-(x^3)/(27pi^3))*...

Unfortunately, my textbook only tells me how to prove that an infinite sum converges, I don't know how to prove the convergence of an infinite product. But I will assume that this infinite product converges, otherwise this whole method collapses.

Working backwards, I try to factorize the infinite product into linear terms. This is a bit too complicated to type into a text box, but suffice to say that each factor splits up into three factors, each one involving the three (complex) cube roots of unity. Of course, I'm now assuming that the infinite product will converge for complex values of x.

Working backwards again, I now have to find the function that satisfies the infinite product. I can state the conditions more precisely:

f(0) = 1
f(n*pi) = 0
f(n*(-0.5+(i*sqrt(3)/2))*pi) = 0
f(n*(-0.5-(i*sqrt(3)/2))*pi) = 0

where n is a natural number greater than or equal to 1.

Once I have found this function, I have to prove that it is equal to its Taylor Series.

Finally, I calculate the third derivative at 0, multiply by pi^3, divide by 6, and I would have the exact value of zeta (3)!

[Theo Moore]

[jinydu]

[jinydu]

I may have celebrated too soon. I took another look at the proof, and there's actually a much simpler way to do it. Here's the simpler way.

Suppose r is a root of f(x). Then, by definition of a root:

k*(e^(ar) + e^(br) + e^(cr)) = 0

Now, in order for this function to work, nr (where n is any positive integer) must also be a solution. Therefore:

k*(e^(anr) + e^(bnr) + e^(cnr)) = 0

Since n is a natural number, the property:

e^(yn) = (e^y)^n must hold

k*((e^(ar))^n + (e^(br))^n + (e^(cr))^n) = 0

Obviously, k cannot be zero (otherwise, the function would just be f(x) = 0, which can't be right). Thus, the expression in the parenthesis must be 0 for all positive integer n.

Now, let A = e^(ar), B = e^(br), C = e^(cr)

The equation becomes

A^n + B^n + C^n = 0

For the first three values of n, this generates the equations:

A + B + C = 0
A^2 + B^2 + C^2 = 0
A^3 + B^3 + C^3 = 0

In the first equation, I solved for C. Then, I plugged that into the second equation and used the quadratic formula to solve for B (of course, I got two possible values of B). Finally, I plugged both values into the third equation. To my surprise, just about everything cancelled or simplified out, and I got A = B = C = 0.

But that is impossible, since e^(ar), for instance, can't possibly be 0. Thus, since the equations don't work for n = 1, 2 and 3, they can't possibly work for all positive integer n. Thus, the original assumption that f(x) had the form k*(e^(ar) + e^(br) + e^(cr)) must be false.

I've been able to generalize a bit and also rule out:

k*(+-e^(ar) +- e^(br) +- e^(cr))
k*(+-e^(ar) +- e^(br) +- e-^(cr))/(x^n), n a positive integer

This also brings up the question, what are the possible solutions to:

A + B + C + D = 0
A^2 + B^2 + C^2 + D^2 = 0
A^3 + B^3 + C^3 + D^3 = 0
A^4 + B^4 + C^4 + D^4 = 0

Can I generalize further to n variables and n such equations?

Also, I haven't tried:

f(x) = a*e^(tx) + b*e^(ux) + c*e^(vx)

But that equation has 6 arbitrary constants, so it would probably be harder to deal with.

[coolguy]

14 different evaulations of zeta(2):

http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf

[jazzmaster]

Very nice, coolguy!

[PiDeltaPhi]

Haha, I just wanted to post a link and I notice that someone has already posted it 6 months ago...
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"The paradise of the moment is greater than any invented eternity."-

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