Lambert series of $\theta^8$

Question

I need your help on proving the following identity

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Theorem

Let $q=e^{-\pi \frac{K'}{K}}$ where $K$ denotes the complete elliptic integral of the first kind,

Then $$\left(\frac{2 K}{\pi}\right)^4(1-k^2+k^4)=1+240\sum_{n=1}^{\infty}\frac{n^3 q^{2n}}{1-q^{2n}}$$

This is stated on the book Pi and the AGM equation $(3.2.17)$. It is said that the identity follows from $$\theta^8_4(q)=\left(\sum_{n=-{\infty}}^{\infty}(-1)^n q^{n^2}\right)^8=1+16\sum_{n=1}^{\infty}\frac{(-1)^n n^3 q^n}{1-q^n}$$ equation $(3.2.25)$, Which I am unable to prove.

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Possible approaches:

Famously, the theta functions have Lambert series like $$\theta^4_3(q)=1+8\sum_{n=1}^{\infty}\frac{nq^n}{1+(-q)^n}$$ $$\theta_2^4(q)=16 \sum_{n= 1}^{\infty} \frac{(2n-1)q^{2n-1}}{1-q^{4n-2}}$$ $$\theta_4^4(q)=1+8\sum_{n= 1}^{\infty} \frac{n(-1)^n q^n}{1+q^n}$$ Which could be proved using the identities $$\theta_3^4(q)=-4q\left(\frac{\theta'_4}{\theta_4}-\frac{\theta'_2}{\theta_2}\right)$$ $$\theta_4^4(q)=-4q\left(\frac{\theta'_3}{\theta_3}-\frac{\theta'_2}{\theta_2}\right)$$ $$\theta_2^4(q)=-4q\left(\frac{\theta'_4}{\theta_4}-\frac{\theta'_3}{\theta_3}\right)$$ See theorem 2.3, which it utilizes the fact that $\frac{dk}{dq}=\frac{2kk'^2K^2}{q\pi^2}$. This approach doesn't seem like that it would work since the summand has $n^3$ on the Lambert series of $\theta_4^8$ instead of $n$. Which suggests that it doesn't have a simple formula with $\log$ (I'd need to differentiate it three times). Paramanand's blog posts suggests that proving the identity algebraically feels quite tedious (Even for $\theta_3^4$). The only relevant formula I could think of using is $$\theta_3^4=\theta_2^4+\theta_4^4$$

Answer 1

I am presenting here the proof by Jacobi as given in Fundamenta Nova. Ramanujan uses almost similar technique to obtain differential equations for his functions $P, Q, R$ and then uses them to prove the theorem in your question. He however does not prove the Lambert series for $\vartheta_4^8(q)$. There is no evidence available that Ramanujan studied Jacobi's Fundamenta Nova, but apparently he had similar (or even more) skills in algebraic manipulation. Ramanujan's approach is available in my blog posts here and here.

Jacobi uses the following infinite product and series expansions of elliptic function $\text{sn} (u, k) $ as a starting point of the proof: \begin{align} \text{sn} (u, k) &= \frac{\vartheta_{3}(q) }{\vartheta_{2}(q) }\frac{\vartheta_{1}(z, q)}{\vartheta_{4}(z, q)} = \frac{2q^{1/4}}{\sqrt{k}}\sin z\prod_{n = 1}^{\infty}\frac{1 - 2q^{2n}\cos 2z + q^{4n}}{1 - 2q^{2n - 1}\cos 2z + q^{4n - 2}}\tag{1}\\ \frac{2kK}{\pi}\text{sn}(u, k) &= 4\sum_{n = 0}^{\infty}\frac{q^{n + 1 / 2}\sin(2n + 1)z}{1 - q^{2n + 1}}\tag{2} \end{align} where $z=\pi u/2K$ and $q=\exp(-\pi K'/K) $ is the nome corresponding to modulus $k$.

Next he takes logarithm of the infinite product $(1)$ above and says that after obvious reductions we get the following $$\log\text{sn} (u, k) =\log\frac{2q^{1/4}}{\sqrt{k}}+\log\sin z+2\sum_{n=1}^{\infty}\frac{q^n\cos 2nz}{n(1+q^{n})}\tag{3}$$ However those reductions are not at all obvious (at least to me) and I present them below.

The infinite product part on right side of equation $(1)$ can be written as $$\prod_{n\geq 1}\frac{(1-q^{2n}e^{2iz})(1-q^{2n}e^{-2iz})}{(1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})}\tag{4}$$ Taking logarithm converts this into an infinite series and one such particular series is $$\sum_{n\geq 1}\log(1-q^{2n}e^{2iz})=-\sum_{n\geq 1}\sum_{m\geq 1}\frac{q^{2mn}e^{2imz}}{m}$$ Interchanging the order of summation the above expression equals $$-\sum_{m\geq 1}\frac{e^{2imz}}{m}\sum_{n\geq 1}q^{2mn}=-\sum_{m\geq 1}\frac{q^{2m}e^{2imz}}{m(1-q^{2m})}$$ Similarly considering logarithm of other factors of expression $(4)$ we can see that the logarithm of this expression equals $$-2\sum_{m\geq 1}\frac{q^{2m}}{m(1-q^{2m})}\cos 2mz+2\sum_{m\geq 1}\frac{q^{m}}{m(1-q^{2m})}\cos 2mz=2\sum_{m\geq 1}\frac{q^m}{m(1+q^m)}\cos 2mz$$ and thus equation $(3)$ is established.

Next Jacobi squares the series $(2)$ directly and this approach is representative of his great powers of algebraic manipulation. He says that squaring will generate expression of type $\sin mz\sin nz$ with $m, n$ being odd positive integers and hence it is better to use the formula $$2\sin mz\sin nz=\cos(m-n)z-\cos(m+n)z$$ to represent square of series in $(2)$ in the form $$a_0+a_1\cos 2z+a_2\cos 4z+\dots\tag{5}$$ He then proceeds to find the coefficients $a_i$ using smart algebraic manipulation. Handing $a_0$ is simpler and it corresponds to the term $\cos(m-n)z$ with $m=n$ and we have $$a_0=8\sum_{n\geq 0}\frac{q^{2n+1}}{(1-q^{2n+1})^2}$$ To find the coefficient $a_i$ one just needs to notice that term $\cos 2iz$ originates in two ways: first when $m-n=\pm 2i$ and second when $m+n=2i$. In the first case to account for the $\pm $ sign we need to multiply the term by $2$ and in second case the terms are with negative sign. With this in mind we have $$a_i=16b_i-8c_i$$ where $$b_i=\frac{q^{i+1}}{(1-q)(1-q^{2i+1})}+\frac{q^{i+3}}{(1-q^3)(1-q^{2i+3})}+\frac{q^{i+5}}{(1-q^5)(1-q^{2i+5})}+\dots$$ and $$c_i=\frac{q^i}{(1-q)(1-q^{2i-1})}+\frac{q^i}{(1-q^3)(1-q^{2i-3})}+\dots+ \frac{q^i}{(1-q^{2i-1})(1-q)}$$ Next Jacobi says that $$\frac{q^{i+j}} {(1-q^j)(1-q^{2i+j})}=\frac{q^i} {1-q^{2i}}\left(\frac{q^j} {1-q^j}-\frac{q^{2i+j}}{1-q^{2i+j}}\right)$$ and uses this split to rewrite $b_i$ as $$b_i=\frac{q^i} {1-q^{2i}}\left(\frac{q}{1-q}+\frac{q^3}{1-q^3}+\dots\right)-\frac{q^i}{1-q^{2i}}\left(\frac{q^{2i+1}}{1-q^{2i+1}}+\frac{q^{2i+3}}{1-q^{2i+3}}+\dots\right)$$ and thus $$b_i=\frac{q^i} {1-q^{2i}}\left(\frac{q}{1-q}+\frac{q^3}{1-q^3}+\dots+\frac{q^{2i-1}}{1-q^{2i-1}}\right)$$ To evaluate $c_i$ Jacobi uses another split $$\frac{q^i} {(1-q^j)(1-q^{2i-j})}=\frac{q^i} {1-q^{2i}}\left(1+\frac{q^j} {1-q^j}+\frac{q^{2i-j}}{1-q^{2i-j}}\right)$$ and rewrites $c_i$ as $$c_i=\frac{iq^i} {1-q^{2i}}+\frac{2q^i}{1-q^{2i}}\left(\frac{q}{1-q}+\frac{q^3}{1-q^3}+\dots+\frac{q^{2i-1}}{1-q^{2i-1}}\right)$$ Finally we get $$a_i=8(2b_i-c_i)=-\frac{8iq^i}{1-q^{2i}}$$ and thus $$\left(\frac{2kK}{\pi}\right) ^2\text{sn}^2(u,k)=a_0-8\sum_{n\geq 1}\frac{nq^n}{1-q^{2n}}\cos 2nz\tag{6}$$ Integrating the above with respect to $z$ on interval $[0,\pi/2]$ we get another expression for $a_0$ as $$a_0=\frac{2}{\pi}\left(\frac{2K}{\pi}\right)^2\int_0^{\pi/2}k^2\text{sn}^2(2Kz/\pi,k)\,dz$$ Putting $z=\pi u/2K$ we get $$a_0=\frac{4K}{\pi^2}\int_0^Kk^2\text{sn}^2(u,k)\,du=\frac{4K}{\pi^2}\int_0^K(1-\text{dn}^2(u,k))\,du =\left(\frac{2K}{\pi}\right)^2-\frac{4KE}{\pi^2}$$ and thus we have finally $$\left(\frac{2kK}{\pi}\right) ^2\text{sn}^2(u,k)=\left(\frac{2K}{\pi}\right)^2-\frac{4KE}{\pi^2}-8\sum_{n\geq 1}\frac{nq^n}{1-q^{2n}}\cos 2nz\tag{7}$$ Putting $z=0$ in above equation gives another expression for $a_0$ namely $$a_0=8\sum_{n\geq 1}\frac{nq^n}{1-q^{2n}}$$ so that equation $(7)$ can be rewritten as $$\left(\frac{2kK}{\pi}\right) ^2\text{sn}^2(u,k)=8\sum_{n\geq 1}\frac{nq^n}{1-q^{2n}}(1-\cos 2nz)\tag{8}$$ And dividing above equation by $z^2$ and taking limits as $z\to 0$ we get $$k^2\left(\frac{2K}{\pi}\right)^4=16\sum_{n\geq 1}\frac{n^3q^n}{1-q^{2n}}\tag{9}$$ Our job will be done if we find some similar series for $k'^2(2K/\pi)^4$ and add it to above equation and get a series for $\vartheta_3^8(q)=(2K/\pi)^4$. To do so we need to derive a series for $(2K/\pi)^2\text{ns}^2(u,k)$, Jacobi does this by using the identity $$\frac{d^2}{dz^2}\log\text{sn}(u,k)=\left(\frac{2K}{\pi}\right)^2\left(k^2\text{sn}^2(u,k)-\text{ns}^2(u,k)\right)\tag{10}$$ We thus just have to differentiate equation $(3)$ two times and use $(7)$ together with $(10)$. Doing these calculations leads us to $$\left(\frac{2K}{\pi}\right)^2\text{ns}^2(u,k)= \left(\frac{2K}{\pi}\right)^2-\frac{4KE}{\pi^2}+\text{cosec} ^2\,z-8\sum_{n\geq 1}\frac{nq^{2n}}{1-q^{2n}}\cos 2nz\tag{11}$$ We next replace $q$ with $-q$ and $z$ with $(\pi/2)-z$ in above equation thereby replacing $K$ with $k'K$, $E$ with $E/k'$ and elliptic function $\text{ns} $ with $\text{nc} $ to get $$\left(\frac{2k'K}{\pi}\right)^2\text{nc}^2(u,k)=\left(\frac{2k'K}{\pi}\right)^2-\frac{4KE}{\pi^2}+\sec^2z+8\sum_{n\geq 1}(-1)^{n-1}\frac{nq^{2n}}{1-q^{2n}}\cos 2nz\tag{12}$$ Putting $z=0$ in above equation we get $$\frac{4KE}{\pi^2}=1+8\sum_{n\geq 1}(-1)^{n-1}\frac{nq^{2n}} {1-q^{2n}}\tag{13}$$ From the last two equations we get $$\left(\frac{2k'K}{\pi}\right)^2(\text{nc}^2(u,k)-1)=\tan^2z+8\sum_{n\geq 1}(-1)^{n}\frac{nq^{2n}}{1-q^{2n}}(1-\cos 2nz)\tag{14} $$ Dividing by $z^2$ and taking limits as $z\to 0$ we get $$k'^2\left(\frac{2K}{\pi}\right)^4=1-16\sum_{n\geq 1}(-1)^{n-1}\frac{n^3q^{2n}}{1-q^{2n}}\tag{15}$$ Adding this to equation to $(9)$ we get desired series for $\vartheta_3^8(q)$ as $$\left(\frac{2K}{\pi}\right)^4=\vartheta_3^8(q)=1+16\sum_{n\geq 1}\frac{n^3q^n}{(1-(-q)^n)}\tag{16}$$ Replacing $q$ with $-q$ we get $$\left(\frac{2k'K}{\pi}\right)^4=\vartheta_4^8(q)=1-16\sum_{n\geq 1}(-1)^{n-1}\frac{n^3q^n}{1-q^n}\tag{17}$$ Let us rewrite the above series as $$ 1-16\sum_{n\geq 1}\frac{n^3q^n}{1-q^n} + 2\cdot 16\sum_{n\geq 1}\frac{(2n)^3q^{2n}}{1-q^{2n}}= 1-16\sum_{n\geq 1}\frac{n^3q^n}{1-q^n} + 16\cdot 16\sum_{n\geq 1}\frac{n^3q^{2n}}{1-q^{2n}}$$ Adding this to equation $(9)$ we get $$\left(\frac{2K}{\pi}\right)^4(1-k^2+k^4)=1+16\cdot16\sum_{n\geq 1}\frac{n^3q^{2n}}{1-q^{2n}}- 16\sum_{n\geq 1}\frac{n^3q^{2n}}{1-q^{2n}}$$ or $$\left(\frac{2K}{\pi}\right)^4(1-k^2+k^4) =1+240\sum_{n\geq 1}\frac{n^3q^{2n}}{1-q^{2n}}\tag{18}$$ which is $Q(q^2)$ in Ramanujan's notation.

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To be honest, Jacobi uses a slightly different approach and first proves equation $(18)$ and then deduces equation $(16)$ with help of equation $(9)$. His proof of equation $(18)$ is a bit complicated and uses the Taylor series for $\text{sn} (u, k) $ given by $$\text{sn} (u, k) =u-\frac{1+k^2}{3!}u^3+\frac{1+14k^2+k^4}{5!}u^5-\dots\tag{19}$$ The above series is arrived by laboriously calculating the derivatives of $\text{sn} (u, k) $ upto fifth order. Next the above series is squared to get $$\frac{\text{sn} ^2(u,k)}{u^2}=1-\frac{1+k^2}{3}u^2+\frac{2+13k^2+2}{45}u^4+\dots $$ and next we find its reciprocal as $$\frac{u^2}{\text{sn}^2(u,k)}=1+\frac{1+k^2}{3}u^2+\frac{1-k^2+k^4}{15}u^4+\dots$$ or in terms of $z=\pi u/2K$ $$\frac{(2K/\pi)^2}{\text{sn}^2(u,k)}=\frac{1}{z^2}+\frac{1+k^2}{3}\left(\frac{2K}{\pi}\right)^2+\frac{1-k^2+k^4}{15}\left(\frac{2K}{\pi}\right)^4z^2 +\dots\tag{20}$$ Noting that $$\text {cosec} ^2z=\frac{1}{z^2}+\frac{1}{3}+\frac{z^2}{15}+\dots$$ and comparing coefficients of $z^2$ in equations $(11)$ and $(20)$ we get equation $(18)$.

Answer 2

I managed to prove the following identity

For $x=\frac{\pi u}{2K}$, $$\left(\frac{2K}{\pi \text{sn}(u,k)}\right)^2=\frac{1}{\sin^2 x}+\text{cnst.}-8\sum_{k=1}^{\infty}\frac{k q^{2k}}{1-q^{2k}}\cos(2kx)\tag{1}$$

and was able to solve the question.

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Preliminary remark: I started with the Fourier series $$\frac{2K}{\pi \text{sn}(u,k)}=\frac{1}{\sin x}+4\sum_{n=0}^{\infty}\frac{q^{2n+1}\sin((2n+1)x)}{1-q^{2n+1}}\tag{2}$$ This can be shown by recognizing the poles and residues to obtain the Mittag-Leffler pole expansion of $$\frac{1}{\text{sn}(u,k)}=\sum_{(n,m)\in\mathbb Z^2}\frac{(-1)^n}{u+2Kn+2iK'm}$$ Summing through $n$ gives $(2)$. With this, I squared both sides in $(2)$. For simplicity, I will use the symbol $$\eta_k=\sum_{n=0}^{k}\frac{q^{2n+1}}{1-q^{2n+1}}\quad , \quad \eta=\eta_\infty$$

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Lemma 1 $$\sum_{m\geq 0}\frac{q^{4m+2+2k}}{(1-q^{2m+1})(1-q^{2m+2k+1})}=\frac{\eta_{k-1}}{1-q^{2k}}-\eta$$

Proof Applying partial fraction decomposition gives $$\frac{q^{4m+2+2k}}{(1-q^{2m+1})(1-q^{2m+2k+1})}=\frac{q^{2k}}{1-q^{2k}}\left(\frac{q^{2n+1}}{1-q^{2n+1}}-\frac{q^{2n+1}}{1-q^{2n+2k+1}}\right)$$ Summing both sides gives $$\sum_{m\geq 0}\frac{q^{4m+2+2k}}{(1-q^{2m+1})(1-q^{2m+2k+1})}=\frac{q^{2k}}{1-q^{2k}}(\eta-q^{-2k}(\eta-\eta_{k-1}))=\frac{\eta_{k-1}}{1-q^{2k}}-\eta$$ This completes the proof.

Lemma 2 $$\sum_{n=0}^{\infty}\frac{1}{(1-q^{2n+1})(1-q^{2k-2n-1})}=\frac{k+2\eta_{k-1}}{1-q^{2k}}$$

Proof The proof is similar to lemma 1 but this time the partial fraction is $$\frac{1}{(1-q^{2n+1})(1-q^{2k-2n-1})}=\frac{1}{1-q^{2k}}\left(\frac{1}{1-q^{2n+1}}+\frac{q^{2k-2n-1}}{1-q^{2k-2n-1}}\right)$$

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We can now start the derivation, squaring $(2)$ gives

$$\left(\frac{2K}{\pi \text{sn}(u,k)}\right)^2=\frac{1}{\sin^2 x}+8\sum_{n=0}^{\infty}\frac{q^{2n+1}}{1-q^{2n+1}}\frac{\sin((2n+1)x)}{\sin x}+16\sum_{m,n=0}^{\infty}\frac{q^{2n+1}}{1-q^{2n+1}}\frac{q^{2m+1}}{1-q^{2m+1}}\sin((2n+1)x)\sin((2m+1)x)$$ $$=\frac{1}{\sin^2 x}+8S_1+16 S_2$$ We start with computing $S_1$, $$\begin{split}S_1=& \sum_{n=0}\frac{q^{2n+1}}{1-q^{2n+1}}\frac{\sin((2n+1)x)}{\sin x} \\ = & \sum_{n=0}\frac{q^{2n+1}}{1-q^{2n+1}}\{1+2\cos(2x)+2\cos(4x)+\cdots+2\cos(2nx)\} \\ S_1= & \eta+2\cos(2x)(\eta-\eta_0)+2\cos(4x)(\eta-\eta_1)+\cdots\end{split}$$

So $S_1$ is solved, we now compute $S_2$. For simplicity, let $a_n=\frac{q^{2n+1}}{1-q^{2n+1}}$.

$$\begin{split}S_2=& \sum_{m,n=0}^{\infty}a_na_m\sin((2n+1)x)\sin((2m+1)x) \\ = & \frac{1}{2}\sum_{m,n\geq 0}a_na_m\{\cos(2(m-n)x)-\cos(2(m+n+1)x)\} \\= & \frac{1}{2}\sum_{k\geq 0}c_{2k}\cos(2kx)\end{split}$$

For $c_0$, it is $c_0=\sum_{n-0}^{\infty}\left(\frac{q^{2n+1}}{1-q^{2n+1}}\right)^2$, for $c_{2k}$, $k\geq 1$, we have to sum the cases for which the expression inside cos is $2kx$. That is,

$$\begin{split}c_{2k}=&\left(\sum_{n-m=k}+\sum_{m-n=k}-\sum_{n+m+1=k}\right)a_na_m\\ = & 2\sum_{m=0}^{\infty}a_m a_{k+m}-\sum_{n=0}^{k-1}a_na_{k-n-1} \\ = &2\sum_{m=0}^{\infty}\frac{q^{4m+2+2k}}{(1-q^{2m})(1-q^{2m+2k+1})}-q^{2k}\sum_{n=0}^{k-1}\frac{1}{(1-q^{2n+1})(1-q^{2k-2n-1})} \\ = & 2\left(\frac{\eta_{k-1}}{1-q^{2k}}-\eta\right)-\frac{q^{2k}}{1-q^{2k}}(k+2\eta_{k-1})\end{split}$$ where lemma 1 and 2 are used in the last step. We therefore obtain $$c_{2k}=2(\eta_{k-1}-\eta)-\frac{kq^{2k}}{1-q^{2k}}$$ With this, we have $$S_2=\frac{c_0}{2}+\sum_{k=1}^{\infty}(\eta_{k-1}-\eta)\cos(2kx)-\frac{1}{2}\sum_{k=1}^{\infty}\frac{kq^{2k}}{1-q^{2k}}\cos(2kx)$$ Putting back in the original expression $$\left(\frac{2K}{\pi\text{sn}(u,k)}\right)^2=\frac{1}{\sin^2 x}+8\left(\eta+2\sum_{n=1}^{\infty}(\eta-\eta_{n-1})\cos(2nx)\right)+16 \left(\frac{c_0}{2}+\sum_{k=1}^{\infty}(\eta_{k-1}-\eta)\cos(2kx)-\frac{1}{2}\sum_{k=1}^{\infty}\frac{kq^{2k}}{1-q^{2k}}\cos(2kx)\right)$$ $$\left(\frac{2K}{\pi\text{sn}(u,k)}\right)^2=\frac{1}{\sin^2 x}+8(c_0+\eta)-8\sum_{k=1}^{\infty}\frac{kq^{2k}}{1-q^{2k}}\cos(2kx)$$ So $(1)$ is proved. By applying the Taylor seires of both sides we have

$$\frac{1}{x^2}+\frac{1+k^2}{3}\left(\frac{2K}{\pi}\right)^2+\frac{1-k^2+k^4}{15}\left(\frac{2K}{\pi}\right)^4 x^2+\cdots$$ $$=\frac{1}{x^2}+\frac{1}{3}+\frac{x^2}{15}+\cdots+8(c_0+\eta)-8\left(\sum_{k}\frac{kq^{2k}}{1-q^{2k}}-\frac{(2x)^2}{2!}\sum_{k}\frac{k^3q^{2k}}{1-q^{2k}}+\cdots\right)$$ The $x^2$ coeffecient is $$\frac{1-k^2+k^4}{15}\left(\frac{2K}{\pi}\right)^4=\frac{1}{15}+16\sum_{n=1}^{\infty}\frac{n^3 q^{2n}}{1-q^{2n}}$$ Or $$Q(q^2)=\left(\frac{2K}{\pi}\right)^4(1-k^2+k^4)$$

and the proposition is proved.