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Topic: lim (1-x)(1+x+x^4+x^9+...)^2 (Read 1442 times) |
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Eigenray
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lim (1-x)(1+x+x^4+x^9+...)^2
« on: Aug 15th, 2005, 1:36pm » |
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Determine the limit, as x approaches 1 from below, of
(1-x)(1+x+x4+x9+...+xn^2+...)2.
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Raghavan
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Re: lim (1-x)(1+x+x^4+x^9+...)^2
« Reply #1 on: Aug 19th, 2005, 1:37pm » |
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(1+n)^2
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Eigenray
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Re: lim (1-x)(1+x+x^4+x^9+...)^2
« Reply #2 on: Aug 19th, 2005, 2:47pm » |
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I think you misunderstood... n is just to denote the general term xn^2. I'm looking for
\lim_{x\to 1^-} (1-x)(\sum_{n=0}^\infty x^{n^2})^2
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Michael Dagg
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Re: lim (1-x)(1+x+x^4+x^9+...)^2
« Reply #3 on: Aug 23rd, 2005, 12:40pm » |
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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.
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Regards,
Michael Dagg
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Barukh
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Re: lim (1-x)(1+x+x^4+x^9+...)^2
« Reply #4 on: Aug 27th, 2005, 11:25pm » |
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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] snxn,
where sn 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] Snxn,
and Sn = s0 + .. + sn (why?). Sn then is the number of ways to represent all the numbers <= n as a sum of 2 non-negative squares. To put it differently, Sn is the number of lattice points in a quadrant of the circle with radius [sqrt]n. Referring to Gauss’s circle problem, we get that Sn ~ [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.
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Eigenray
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Re: lim (1-x)(1+x+x^4+x^9+...)^2
« Reply #5 on: Aug 29th, 2005, 6:07pm » |
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Yep. One can also proceed as follows:
For simplicity replace the second factor with
A(x) = (1+2x+2x4+2x9+...)2 = [sum] rnxn,
where rn 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,
rn = 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] rn xn = 4[sum] X(n)xn/(1-xn),
and therefore
(1-x)A(x) = 4[sum] X(n)xn/(1+x+...+xn-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.
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