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enter image description hereDespite all my efforts trying to crack these, i haven't been able to do so. An approach that i've tried gives me somewhat of an asymptotic approximation, but still fails to produce the values near x=0.

The method i've tried is to express the sum as an integral and then integrate it to obtain it's value but apparently there is something that I can't do or something wrong with it.

If anybody can suggest any ideas or (ideally) solve the problem I'd be very grateful. Thanks in advance.

I'll try to show here what I think i did wrong, maybe then you'll be able to point out where i fail at.

Since the two sums are related, i'll only try to find one of them, the second one. Since: $$ \int_a^b f(x)dx = \lim_{n\to \infty} \sum_{k=1}^n f(a+\frac{b-a}{n}k)\cdot \frac{b-a}{n}$$

I let each term of the sum be $\frac{c^k}{1+c^{2k}}$ so, if $x=a+ \frac{b-a}{n}k$ then $f(x)=\frac{n}{b-a}\cdot \frac{c^{\frac{x-a}{b-a}n}}{1+c^{\frac{x-a}{b-a}2n}}$ therefore: $$\sum_{k=1}^\infty \frac{c^k}{1+c^{2k}} = \lim_{n\to \infty} \int_a^b \frac{n}{b-a}\cdot \frac{c^{\frac{x-a}{b-a}n}}{1+c^{\frac{x-a}{b-a}2n}}dx$$ and integrating between a and b I obtain (since c<1 and n tends to infinity): $$\sum_{k=1}^\infty \frac{c^k}{1+c^{2k}}=\frac{\pi}{4log(\frac{1}{c})}$$ Therefore $$\sum_{n=1}^\infty x^{n^2} = \frac{\sqrt{1-\frac{\pi}{log(x)}}-1}{2}$$ and as x tends to 1- : $$\sum_{n=1}^\infty x^{n^2} = \frac{\sqrt{1+\frac{\pi}{1-x}}-1}{2}$$

But STILL this only works as an asymptotic function. Obviously there is something wrong with this reasoning. Hope you guys can help me. Thanks a lot again.

*Edit: In the photo i've posted you can see the difference between the graph of the functions.

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  • $\begingroup$ For the first, see the Jacobi theta functions :mathworld.wolfram.com/JacobiThetaFunctions.html . The second is probably related to the polygamma functions . $\endgroup$ – JJacquelin Feb 8 '16 at 15:57
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    $\begingroup$ These two are related between them: $$ \sum_{n=1}^\infty x^{n^2} = \frac{\sqrt{1+4 \sum_{n=1}^\infty \frac{x^n}{1+x^{2n}}} -1}{2} $$ $\endgroup$ – Rafa Feb 8 '16 at 16:03
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    $\begingroup$ What is your actual question? Seeking a closed form? Proving the above relation? How to compute values for small $|x|$ or for $|x|$ near $1$? Something else? $\endgroup$ – ccorn Feb 8 '16 at 16:16
  • $\begingroup$ My question is computing values for small x, since the ones near 1 are given by the above reasoning, but there is a gap between the graph of the sum and the formula that i've obtained near 0. $\endgroup$ – Rafa Feb 8 '16 at 16:23
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    $\begingroup$ In you methodology, how can $f$ depend on anything other than $x$? In particular, $f$ cannot depend on the same index, $n$, as used to partition the $x$ axis in the formation of a Riemann sum. Thus, we don't expect the development to lead to an equality. $\endgroup$ – Mark Viola Feb 8 '16 at 17:26
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Let me first summarize some essential comments before proceeding to an answer.

  1. As JJacquelin has hinted, $$\sum_{n=1}^\infty x^{n^2} = \frac{\theta_3(0,x)-1}{2}$$ where $\theta_3(0,x)$ is a Jacobi thetanull $$\theta_3(0,x) = \sum_{n=-\infty}^\infty x^{n^2}\qquad(|x| < 1)$$ Cf. DLMF ch. 20 Theta functions and OEIS A010052.
  2. As you have noted, $$\sum_{n=1}^\infty x^{n^2} = \frac{\sqrt{1+4 \sum_{n=1}^\infty \frac{x^n}{1+x^{2n}}}-1}{2}$$ Consequently, $$\begin{align} \theta_3(0,x)^2 &= 1 + 4\sum_{n=1}^\infty \frac{x^n}{1+x^{2n}} \\ \sum_{n=1}^\infty \frac{x^n}{1+x^{2n}} &= \frac{\theta_3(0,x)^2-1}{4} \end{align}$$ Cf. OEIS A002654 and OEIS A004018 where you can find other representations as well, e.g. $$\sum_{n=1}^\infty \frac{x^n}{1+x^{2n}} = \sum_{k=0}^\infty (-1)^k\frac{x^{2k-1}}{1-x^{2k-1}}$$ which can be obtained by expanding the summand to a power series, changing the nesting order of the summation, and unexpanding the new inner series.
  3. As Dr. MV has pointed out, your $f$ is not allowed to depend on the step count $n$ whilst approximating the integral.

I'd like to add that the modular properties of $\theta_3$ allow you to convert between values at small $|x|$ and values at $|x|$ near $1$: $$\theta_3\!\left(0,\exp(-\pi/y)\right) = \sqrt{y}\,\theta_3\!\left(0,\exp(-\pi y)\right) \qquad(\Re y > 0)\tag{*}$$ Note that the associated lattice parameter (period ratio) is $\tau=\mathrm{i}y$ here.

To answer your question: For computation of $\theta_3(0,x)$ at nonnegative $x$, the DLMF notes on computation work quite well. If $|x|\leq\exp(-\pi)$, the series for $\theta_3(0,x)$ converges very quickly. Otherwise, you can use the transform $(*)$ which recurs to a new value of $x$ less than $\exp(-\pi)$.

For negative $x$, compute $\theta_3(0,x)$ as $\theta_4(0,-x)$ in a similar manner, but note that the modular transformation then swaps $\theta_4$ with $\theta_2$: $$\theta_4\!\left(0,\exp(-\pi/y)\right) = \sqrt{y}\,\theta_2\!\left(0,\exp(-\pi y)\right) \qquad(\Re y > 0)$$ For complex nonzero $x$, the above approach must be slightly refined. Set the lattice parameter $\tau=\frac{\log x}{\pi\mathrm{i}}$. Since $|x|<1$, we have $\Im\tau>0$. Now either recursively compute the true two-variable version $\theta_3(z\mid\tau)$ (initially for $z=0$) or the triple $$T(\tau) = \left(\theta_2(0\mid\tau),\theta_3(0\mid\tau),\theta_4(0\mid\tau)\right)$$ I'll describe the latter approach. We have the modular symmetries $$\begin{align} T(\tau+1) &= \left(\sqrt{\mathrm{i}}\,\theta_2(0\mid\tau), \theta_4(0\mid\tau),\theta_3(0\mid\tau)\right) \tag{T}\\ T(-1/\tau) &= \sqrt{-\mathrm{i}\tau} \left(\theta_4(0\mid\tau),\theta_3(0\mid\tau),\theta_2(0\mid\tau)\right) \tag{J} \end{align}$$ If $\Im\tau$ is greater than some threshold, say $\frac{1}{2}$, then the associated $|x|$ is very small. In that case, use the series representations for the entries of $T(\tau)$ with suitable truncation.

Otherwise, use (T) to reduce the real part of $\tau$ such that $|\Re\tau|\leq\frac{1}{2}$. Since $\Im\tau$ is bounded here, we also have $|\tau|<1$, in fact, with the above threshold $|\tau|\leq\frac{1}{\sqrt{2}}$. Now use (J) to turn to some $|\tau|>1$; the above-proposed threshold gets us $|\tau|\geq\sqrt{2}$; in particular, $\Im\tau$ at least doubles in that step. Repeating those recursions gets you quickly to a state where $\Im\tau$ is large enough that you can use the truncated series.

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