first euler's identity

[lg]

hi,

could anybody help me to prove the first euler's identity?
the equation is:

(1+x)(1+x^3)(1+x^5).... = sum(from k=1 to infinity) ( ( x^
( k^2 ) ) ) / ( (1-x^2)(1-x^4)(1-x^6)... )

the second equation is similar:

(1+x^2)(1+x^4)(1+x^6).... = sum(from k=1 to infinity) ( ( x^( k(k
+1) ) ) ) / ( (1-x^2)(1-x^4)(1-x^6)... )

thanks for any kind of help:)

[Robert Israel]

> hi,
>
> could anybody help me to prove the first euler's identity?
> the equation is:
>
> (1+x)(1+x^3)(1+x^5).... = sum(from k=1 to infinity) ( ( x^
> ( k^2 ) ) ) / ( (1-x^2)(1-x^4)(1-x^6)... )

This is wrong. In fact

(product_{k=0}^infty (1+x^(2k+1))) (product_{k=1}^infty (1 - x^(2k)))
= product_{k=0}^infty (1-(-x)^k)
= sum_{k=-infty}^infty (-1)^k (-x)^(k(3k-1)/2)
(for |x| < 1) by Euler's pentagonal number theorem (see e.g.
<http://en.wikipedia.org/wiki/Pentagonal_number_theorem>

> the second equation is similar:
>
> (1+x^2)(1+x^4)(1+x^6).... = sum(from k=1 to infinity) ( ( x^( k(k
> +1) ) ) ) / ( (1-x^2)(1-x^4)(1-x^6)... )

Again, wrong.

(product_{k=1}^infty (1+x^(2k))) (product_{k=1}^infty (1 - x^(2k)))
= product_{k=1}^infty (1 - x^(4k))
= sum_{k=-infty}^infty (-1)^k x^(2k(3k-1))

for |x| < 1.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[lg]

strange, this equation gave me my discrete mathematic teacher so I
suppose it's correct