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Tough inequality challenge

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Opalg

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Feb 7, 2012
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In a recent challenge thread, anemone asked for a proof that $1-x + x^4 - x^9 + x^{16} - x^{25} + x^{36} > 0$. When I graphed that function, I noticed that in fact it is never less than $\frac12$. If you add more terms to the series, this becomes even more apparent:

Screenshot 2020-06-04 at 14.40.26.png

So the challenge is to prove that\(\displaystyle \sum_{r=0}^{2n}(-1)^rx^{r^2} = 1-x + x^4 - x^9 + x^{16} -\ldots + x^{(2n)^2} > \frac12\) for all $x$ such that $0<x<1$ (outside that interval the result is obvious anyway).
 
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Opalg

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Feb 7, 2012
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Opalg

MHB Oldtimer
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Feb 7, 2012
2,680
For $0<x<1$ the infinite series \(\displaystyle \sum_{r=0}^\infty (-1)^rx^{r^2}\) is convergent. It is an alternating series, where the terms alternate in sign and decrease in absolute value. It has the property that the partial sums are alternately greater than, and less than, the sum of the whole series.

Let \(\displaystyle S = \sum_{r=0}^\infty (-1)^rx^{r^2}\), and for $n\geqslant 1$ let \(\displaystyle S_n = \sum_{r=0}^n (-1)^rx^{r^2}\). Then $S_{2n-1}<S<S_{2n}$. So to prove that $S_{2n} > \frac12$ it will be sufficient to show that $S\geqslant\frac12$.

To investigate $S$, use the Jacobi triple product, which says that $$\sum_{n=-\infty}^\infty x^{n^2}y^{2n} = \prod_{m=1}^\infty (1-x^{2m})(1+x^{2m-1}y^2)\left(1+\frac{x^{2m-1}}{y^2}\right)$$ for complex numbers $x,y$ with $|x|<1$ and $y\ne0$. For $0<x<1$ and $y=i$ that gives $$\sum_{n=-\infty}^\infty(-1)^n x^{n^2} = \prod_{m=1}^\infty (1-x^{2m})\bigl(1-(-1)^nx^{2m-1}\bigr)^2.$$ Each factor in that product is positive, so the infinite product must be positive or zero. Therefore \(\displaystyle \sum_{n=-\infty}^\infty(-1)^n x^{n^2} \geqslant 0.\) But $$\sum_{n=-\infty}^\infty(-1)^n x^{n^2} = \sum_{n=-\infty}^0(-1)^n x^{n^2} + \sum_{n=0}^\infty(-1)^n x^{n^2} - 1$$ (the $-1$ coming from the fact that otherwise the $n=0$ term would be counted twice). Also, $(-1)^n x^{n^2}$ is the same for $-n$ as it is for $+n$. Therefore $2S-1\geqslant0$, so that $S\geqslant\frac12$, as required.