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        riddles >> putnam exam (pure math) >> lim (1-x)(1+x+x^4+x^9+...)^2
        
(Message started by: Eigenray on Aug 15^th, 2005, 1:36pm)

Title: lim (1-x)(1+x+x^4+x^9+...)^2
Post by Eigenray on Aug 15^th, 2005, 1:36pm

Determine the limit, as x approaches 1 from below, of
(1-x)(1+x+x⁴+x⁹+...+x^{n^2}+...)².

Title: Re: lim (1-x)(1+x+x^4+x^9+...)^2
Post by Raghavan on Aug 19^th, 2005, 1:37pm

(1+n)^2

Title: Re: lim (1-x)(1+x+x^4+x^9+...)^2
Post by Eigenray on Aug 19^th, 2005, 2:47pm

I think you misunderstood... n is just to denote the general term x^{n^2}. I'm looking for
\lim_{x\to 1^-} (1-x)(\sum_{n=0}^\infty x^{n^2})^2

Title: Re: lim (1-x)(1+x+x^4+x^9+...)^2
Post by Michael_Dagg on Aug 23^rd, 2005, 12:40pm

Without the square on the series, the limit is 0. But with the series squared, it is much more complicated. The answer is pi/4.

Title: Re: lim (1-x)(1+x+x^4+x^9+...)^2
Post by Barukh on Aug 27^th, 2005, 11:25pm

As nobody seems to give a solution, I decided to describe one I’ve found really beautiful.

Square the second term, and write it as a power series:

P(x) = [sum] s_nxⁿ,
where s_n is the number of ways to represent n a sum of 2 squares of non-negative integers. Now use the known identity P(x) = (1-x)P’(x), where
P’(x) = [sum] S_nxⁿ,
and S_n = s₀ + .. + s_n (why?). S_n then is the number of ways to represent all the numbers <= n as a sum of 2 non-negative squares. To put it differently, S_n is the number of lattice points in a quadrant of the circle with radius [sqrt]n. Referring to Gauss’s circle problem (http://mathworld.wolfram.com/GausssCircleProblem.html), we get that S_n ~ [pi]n/4 (this maybe stated in a completely formal manner to justify the answer). Therefore, the sought limit equals:

lim (1-x)P(x) = lim (1-x)²P’(x) = lim (1-x)² [sum] [pi]/4 nxⁿ = [pi]/4.

Title: Re: lim (1-x)(1+x+x^4+x^9+...)^2
Post by Eigenray on Aug 29^th, 2005, 6:07pm

Yep. One can also proceed as follows:
For simplicity replace the second factor with
A(x) = (1+2x+2x⁴+2x⁹+...)² = [sum] r_nxⁿ,
where r_n is the number of representations of n as the sum of 2 squares of integers. This has the effect of multiplying the answer by 4. Now,
r_n = 4[sum]_d|n X(d),
where X(d) is the non-principal character mod 4:
X(2n)=0, X(4n+1)=1, X(4n+3)=-1.
It follows that
A(x) = [sum] r_n xⁿ = 4[sum] X(n)xⁿ/(1-xⁿ),
and therefore
(1-x)A(x) = 4[sum] X(n)xⁿ/(1+x+...+x^n-1).
At this point I mumble something about
x⁴ⁿ⁺¹/(1+x+...+x⁴ⁿ) - x⁴ⁿ⁺³/(1+x+...+x⁴ⁿ⁺²)
maybe being monotonic for 0<x<1, so by the MCT,
lim (1-x)A(x) = 4(1 - 1/3 + 1/5 - 1/7 + ....) = pi.

Maybe a bit more direct, but I don't know a nice way of finishing it.