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Context

With some effort we can show that Ramanujan's fastest series implies: \begin{align} \frac{8E(k_{58})K(k_{58})}{\pi^2}-\frac{aK^2(k_{58})}{\pi^2}=\frac{\sqrt{58}}{29\pi},\tag{1} \end{align} with: $$a=\left(\frac{489227532 \sqrt{58}}{{29}}- \frac{691872192\sqrt{29}}{29}+90847272\sqrt{2}-128477440\right)$$ and $$k_{58}=(13\sqrt{58}-99)(\sqrt{2}-1)^6.$$

Some related formulas

Being: \begin{align} P(q)=1-\sum_{n=1}^{\infty}\frac{24nq^{2n}}{1-q^{2n}},\tag{2} \end{align} with $q=e^{-\pi\frac{K(k')}{K(k)}}$, $K(k)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}$ and $E(k)=\int_{0}^{\pi/2}{\sqrt{1-k^2\sin^2t}}dt$ the complete elliptic integrals of the first kind and second kind respectively and $k'=\sqrt{1-k^2}$ the complementary modulus. Then we have: \begin{align} P(q)=\frac{12E(k)K(k)}{\pi^2}+\frac{(4k^2-8)K^2(k)}{\pi^2}\tag{3}. \end{align} Ramanujan dealt with $P(q)$ proving (as Paramanand has noted) that $P(q)$ would have the form: \begin{align} P(q)=\frac{K^2(k)}{\pi^2}A(n,k)+\frac{3}{\pi\sqrt{n}},\tag{4} \end{align} where $A(n,k)$ is an algebraic number. Asumming this we have for $n=58$: \begin{align} \frac{8E(k_{58})K(k_{58})}{\pi^2}+\frac{2(4k_{58}^2-8)K^2(k_{58})}{3\pi^2}=\frac{K^2(k_{58})}{\pi^2}A(58,k_{58})+\frac{\sqrt{58}}{29\pi}. \end{align} Comparing this last with $(1)$ gives $A(58,k_{58})$. This suggests that some reverse approach can be done to obtain the modular equation of degree 29 and the $f(k)$ in the following formula: \begin{align} 29P(q^{29})-P(q)=\frac{K^2(k)}{\pi^2}f(k). \end{align}

Note

  1. See this answer of mine dealing with $k_{1}$ in an elementary approach: Finding $p\cdot K\left(s\right)E\left(s\right)+q\cdot K\left(s\right)^{2}$ such that it is a known constant
  2. This answer dealing with $k_{2}$: Specific values of Elliptic integral of the second kind $E$
  3. This answer of @ParamanandSingh and the other by me dealing with $k_{3}$: How to show $\sum_{n=0}^{\infty}\frac{(6n+1)(2n)!^3}{2^{8n}n!^6}=\frac{4}{\pi}$?.
  4. This answer of @ParamanandSingh dealing with $k_{5}$ using Ramanujan ideas: A closed form for a triple integral involving Heron's formula
  5. And this answer dealing with $k_{7}$: How to show that $\frac{K(k')}{K(k)}=\sqrt{7} \implies k=k_{7}=\frac{\sqrt{2}}{8}(3-\sqrt{7})$?

Question

Can we prove $(1)$ by hand like Ramanujan did?

User-Refolio
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    Unfortunately no one has any idea how Ramanujan got modular equations of higher degrees (ie greater than 5). My own belief regarding your question is that Ramanujan had some simpler form of a modular equation of degree 29 in the theory of alternative bases and he used it to get his fastest series for $1/\pi$. – Paramanand Singh Mar 23 '24 at 13:31
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    I am also fairly certain that Ramanujan did the calculations for both $n=58,163$ but somehow did not use the case $n=163$ to get Chudnovsky series. See https://mathoverflow.net/q/384027/15540 for more details. – Paramanand Singh Mar 23 '24 at 13:34
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    Also note that most of the formulas Ramanujan gave for $nP(q^n) - P(q) $ have not been proved using modular equations but rather using modular forms and computer algebra packages. – Paramanand Singh Mar 23 '24 at 13:44
  • @ParamanandSingh I believe that both can be proven by hand. It seems that still we don't understand what Ramanujan developed. He proved a lot of relations like $(1)$ without knowing the implicit values. – User-Refolio Mar 23 '24 at 17:45
  • Also a proof using Selberg-Chowla seems to be doable. This will give $K(k_{58})$ and $E(k_{58})$ I have not seen this proof yet. – User-Refolio Mar 23 '24 at 17:50
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    @User: The Chowla Selberg formula works nicely for $K$, but to find $E$ one requires a relation of type $(1)$ and the value of $a$ as well. So it is not really possible to escape the evaluation of $P(q) $ or equivalently $A(n, k) $. Also Ramanujan did not assume the form $(4)$ for $P(q) $ but proved it for the first time in his paper "Modular equations and approximations to $\pi$". Also have a look at https://math.stackexchange.com/a/4733525/72031 where I find the relation between $K, E$ using Ramanujan's technique for $n=5$. – Paramanand Singh Mar 24 '24 at 01:41
  • I wasn't aware of the calculations for $n=163$ that Ramanujan did. Thanks for that reference. – User-Refolio Mar 24 '24 at 06:36
  • One is lucky in case of $k_1$ as $k_1=k'_1$ and then Legendre's identity is sufficient to give a direct link between $E(k_1)$ and $K(k_1)$. But for other singular moduli getting the corresponding relation between these elliptic integrals is difficult. – Paramanand Singh May 02 '24 at 11:30
  • @ParamanandSingh For n=1,2 and n=4 we are lucky because Landen's transformation and Legendre's relation are enough. For n=3 we can use Legendre third singular moduli and respective values of K and E. The difficulty begins with n=5. I know that the the link between K and E in general is a very hard problem to solve. – User-Refolio May 03 '24 at 14:35
  • @ParamanandSingh I have to note that we have not use the implicit values of K and E for n=1. Follows directly from Clausen's formula and Legendre's relation. For n=2 the same occurs but we need also Landen's transformation. – User-Refolio May 03 '24 at 14:48
  • I checked the paper referenced in Dec 23,2024 update and I find that their entire paper contains nothing new. The evaluation of $g_{58}$ using algebraic number theory is well known (Weber got this first if I am not wrong). The lattice sum and its proof are covered in my answer here. The final ingredient $\alpha(58)$ is taken from Borweins Pi and the AGM and I have described their approach in my answer. – Paramanand Singh Dec 29 '24 at 07:41
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    To sum up the paper is more a consolidation of existing results in one place and I have nothing against publishing such stuff, but I tend to assume that research papers usually have some new idea not readily available in prior art. A key misunderstanding on part of authors is that they think the most difficult part of Ramanujan series is evaluation of $g_{58}$ but in reality that is easy/well-known part and the hardest part is evaluation of $\alpha(58)$ and so far no one has evaluated it by hand. – Paramanand Singh Dec 29 '24 at 07:45
  • @ParamanandSingh For me are hard both parts. I have also a question. To determine Ramanuan $g_{n}$ don't we need modular equations? So the modular equation of degree 58 is related to $\alpha(58)$. – User-Refolio Dec 29 '24 at 13:32
  • @ParamanandSingh Borwein's brothers were so clever defining $\alpha(r)$ and then attacking the problem with modular forms and computers (as you said me is the modern approach). However I prefer to understand what Jacobi and Ramanujan developed. – User-Refolio Dec 29 '24 at 13:49
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    My best bet is that Ramanujan used a modular equation of degree 29 to evaluate both $g_{58}$ and an expression for $29P(q^{58})-P(q^2)$ and he did it for the theory related to alternative bases. – Paramanand Singh Dec 29 '24 at 17:48

1 Answers1

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The short answer to your question is that there is no available proof for the identity $(1)$ of your question which involves only hand calculation. In the following I present reasonable details of work by Ramanujan as well Borwein brothers.

In his famous paper Modular equations and approximations to $\pi$ Ramanujan discussed about evaluation of the function $P(q^2) $ where $$P(q) =1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}\tag{1}$$ for certain specific values of $q$ of the form $q=\exp(-\pi\sqrt{r}) $ where $r\in\mathbb{Q},r>0$. The function $P(q) $ is related to Dedekind's eta function $$\eta(q) =q^{1/24}\prod_{n\geq 1}(1-q^n),\tag{2}$$ via $$P(q) = 24q\frac{d}{dq}\log\eta(q)\tag{3}$$

The topic is highly interesting and is related to theory of elliptic integrals which we discuss in brief. Let us start with definitions of elliptic integrals first. Let $k\in(0,1)$ be the elliptic modulus and $k'=\sqrt{1-k^2}$ be the complementary modulus and we define $$K(k)=\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}},E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx\tag{4}$$ If values of $k, k'$ are available from context we write $K, K', E, E'$ for $K(k), K(k'), E(k), E(k') $ respectively. These integrals satisfy a relation $$KE'+K'E-KK'=\frac{\pi} {2}\tag{6}$$ which is famous by the name Legendre's identity. The identity however fails to give a direct link between $K, E$ and this makes the evaluation of $E$ considerably difficult compared to that of $K$. Ramanujan gave a very ingenious technique of evaluation of $E$ in his aforementioned paper via the use of function $P(q^2)$. If the values of $K, K'$ are known then the values of $k, k'$ can be obtained using theta functions of Jacobi as $$k=\frac{\vartheta_2^2(q)}{\vartheta_3^2(q)},k'=\frac{\vartheta_4^2(q)}{\vartheta_3^2(q)}\tag{7}$$ where \begin{align} q&=\exp(-\pi K'/K) \tag{8a}\\ \vartheta_2(q)&=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}\tag{8b}\\ \vartheta_3(q)&=\sum_{n\in\mathbb{Z}}q^{n^2}\tag{8c}\\ \vartheta_4(q)&=\vartheta_3(-q)\tag{8d} \end{align} The variable $q$ is called the nome corresponding to modulus $k$ and we also have another relation $$\vartheta_3^2(q)=\frac{2K}{\pi}\tag{9}$$ Now we talk about the relation between $k, q$ given by $q=e^{-\pi K'/K} $. It can be proved that $K'/K$ is a decreasing function of $k$ and maps $(0,1)$ to $(0,\infty)$ and hence $q$ is an increasing function of $k$ and maps $(0,1)$ to itself. More precisely we have $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}\tag{10}$$ Using product representations of theta functions it can be proved that $$\eta(q^2)=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{11}$$ and using $(3),(10),(11)$ we can prove that $$P(q^2) =\left(\frac{2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)\tag{12}$$ so that the value $E$ can be computed if $K, P(q^2)$ are known.

We now describe major theorems of Jacobi, Ramanujan and Selberg-Chowla which help us to perform these evaluations (at least in theory). Let $r$ be a positive real number and let $k$ be a given modulus then the value $rK'/K$ is positive and hence there is a unique modulus $l\in(0,1)$ such that $K(l') /K(l) =rK'/K$ so that $l$ is a function of $k$ if $r$ is a positive constant. The integrals $K(l), K(l') $ are traditionally denoted by $L, L'$ and Jacobi proved

Theorem 1 (Jacobi): If $r$ is a positive rational number and $L'/L=rK/K$ then the relation between $k, l$ is algebraic ie there is a polynomial $P(x, y) $ with integer coefficients such that $P(k, l) =0$. This relation between $k, l$ is said to be a modular equation of degree $r$.

Ramanujan added a further constraint in this equation namely $l=k'$ so that $P(k, k') =0$ and then both $k, k'$ are algebraic numbers and $L'=K, L=K'$ so that $K'/K=1/\sqrt{r}$ and $L'/L=\sqrt {r} $. And then we have

Theorem 2: If $r$ is a positive rational number and $q=\exp(-\pi\sqrt{r}) $ then the values $k, k'$ given in $(7)$ are algebraic and are called singular moduli and denoted by $k_r, k'_r$.

Ramanujan was an expert in finding modular equations and evaluating singular moduli in explicit radical form. Ramanujan further proved

Theorem 3 (Ramanujan): If $r$ is a positive rational number and moduli $k, l$ are connected by modular equation of degree $r$ so that $L'/L=rK'/K$ then $$rP(q^{2r})-P(q^2)=\frac{4KL}{\pi^2}A_r(l,k)\tag{13}$$ where $A_r(l, k) $ is an algebraic function of $l, k$.

We also have a related formula (which is a consequence of the transformation formula for Dedekind's eta function) $$rP(e^{-2\pi\sqrt{r}})+P(e^{-2\pi/\sqrt{r}})=\frac{6\sqrt{r}}{\pi}\tag{14}$$ which holds for all positive real numbers $r$. Putting $q=e^{-\pi/\sqrt{r}} $ in $(13)$ and using $(14)$ Ramanujan proved

Throrem 4 (Ramanujan): If $r$ is a positive rational number and $q=e^{-\pi\sqrt{r}} $ then $$P(q^2)=\frac{3}{\pi\sqrt{r}}+A_r\cdot\left(\frac{2K}{\pi}\right)^2\tag{15}$$ where $A_r$ is an algebraic number dependent on $r$.

Ramanujan gave formulas (ie explicit radical expressions) for $A_r(l, k) $ in terms of $k, l$ for many positive integer values of $l$. He mentioned that such a formula can be obtained by differentiating the modular equation $P(k,l)=0$ twice with respect to $k$. The technique works fine for small values of $r=2,3$ but becomes unwieldy for larger values and no one really knows how Ramanujan got those not so unwieldy expressions for $A_r(l, k) $. Perhaps his great and unmatched skill in the manipulation of radicals helped him here. Using these expressions for $A_r(l, k) $ and the values of singular moduli $k_r, k'_r$ one can get the values of the constant $A_r$ in $(15)$.

Using $(12)$ and $(15)$ we can prove that $$K(k_r) E(k_r) + B_rK(k_r) ^2=\frac{\pi}{4\sqrt{r}}\tag{16}$$ where $r$ is a positive rational number and $B_r$ is some algebraic number dependent on $r$. This is exactly what your equation $(1)$ in question says for $r=58$. But to evaluate the constants $A_r, B_r$ it is essential to know the explicit form of function $A_r(l, k) $ and the values of related singular moduli. These constants $A_r, B_r$ are also needed to find certain series for $1/\pi$ which were first brought into prominence by Ramanujan. Before discussing this we also mention a theorem by Selberg-Chowla which deals with evaluation of $K(k_r) $.

Theorem 5 (Selberg-Chowla): If $r$ is a positive rational number then the value $K(k_r) $ can be expressed in a closed form involving $\pi$ and values of Euler's Gamma function at rational points. There is a direct and nice formula available if $r$ is a square free positive integer and $\mathbb{Q} (\sqrt{-r}) $ has class number $1$.

Using these aforementioned theorems it is possible in principle to evaluate $K(k_r), E(k_r) $ in closed form consisting of values of Gamma function at rational points. But in practice the biggest challenge is to find the constants $A_r, B_r$.

Borwein brothers approach this problem in a slightly different manner and they define a function $\alpha:(0, \infty) \to\mathbb{R} $ via $$\alpha(r) =\frac{E'} {K} - \frac {\pi} {4K^2}\tag{17}$$ with modulus $k$ corresponding to nome $q=e^{-\pi\sqrt{r}} $. Using Legendre's identity and the fact that $K'=\sqrt{r} K$ we can rewrite $(17)$ as $$\frac{\pi} {4}=\sqrt{r} KE-(\sqrt{r} - \alpha(r)) K^2\tag{18}$$ and from $(16)$ we see that $$B_r=\frac{\alpha(r)} {\sqrt{r}} - 1$$ so that $\alpha(r) $ is algebraic when $r$ is rational.

Borwein brothers also give an expression for $\alpha(r) $ in terms of theta functions which can be used for numerical evaluation $$\alpha(r) =\frac{1}{\vartheta_3^4(q)}\left(\frac{1}{\pi}-4q\sqrt{r}\frac{d}{dq}\log\vartheta_4(q)\right), q=e^{-\pi\sqrt{r}} \tag{19}$$ and show that $\alpha (r) \to1/\pi$ as $r\to\infty$.

In their book Pi and the AGM Borwein brothers give a host of formulas to evaluate $\alpha(r) $ and also indicate its role in deriving Ramanujan type series for $1 /\pi$. I will mention here a few such formulas below. Let us use the symbol $k_r$ not just for singular modulus (for rational $r$) but for the modulus corresponding to nome $q=e^{-\pi\sqrt{r}} $ and then we have $$\alpha(4r)=(1+k_{4r})^2\alpha(r)-2\sqrt{r}k_{4r}\tag{20}$$ and $$\alpha(9r)=s(r)^2\alpha(r)-\frac{\sqrt{r}(s(r)^2+2s(r)-3)} {2} \tag{21}$$ where $$s(r) =\sqrt{1+4\cdot\frac{(k_{9r}k'_{9r})^{3/4}}{(k_rk'_r)^{1/4}}}\tag{22}$$ There are formulas available for $\alpha(2r),\alpha(3r)$ as well but they are a bit complicated. The basis of these formulas is the expression $A_r(l, k) $ which was given explicitly as radicals by Ramanujan for many integer values of $r$ and Borweins say that these radical expressions can be verified for small values of $r=2,3,4$ and for higher values we rely on Ramanujan. These expressions were later verified using theory of modular forms and mathematical software by Bruce C. Berndt and his collaborators.

Let us now discuss how Borweins established the Ramanujan type series for $1/\pi$. They start with a formula of the form $$\left(\frac{2K}{\pi}\right)^2=m(k) \sum_{n\geq 0}a_n\phi(k)^n=m(k) F(\phi(k)) \tag{23}$$ where $m(k), \phi(k) $ are algebraic functions of $k$ and the above series is a hypergeometric type series. Differentiating the series with respect to $k$ we get $$\frac{8K}{\pi^2}\frac{dK}{dk}=m'(k)\sum_{n\geq 0}a_n\phi(k) ^n+m(k)\sum_{n\geq 1}na_n\phi(k) ^{n} \frac{\phi'(k)} {\phi(k)}\tag{24} $$ Now we can note that $$\frac{dK} {dk} =\frac{E-k'^2K}{kk'^2}$$ and using $(18)$ to express $E$ terms of $K$ we get $$\frac{1}{\pi}=\sqrt{r}k_r{k'}_r^2\cdot\frac{4K}{\pi}\frac{dK}{dk}+(\alpha(r) - \sqrt{r} k_r^2)\frac{4K^2}{\pi^2}$$ Using $(23),(24)$ in above equation we get the following series for $1/\pi$ $$\frac{1}{\pi}=\sum_{n\geq 0}a_n\left(\frac{\sqrt{r}} {2}\cdot kk'^2m'(k)+(\alpha(r) - \sqrt{r} k^2)m(k) +\frac{n\sqrt{r}} {2}\cdot kk'^2m(k)\cdot\frac{\phi'(k)}{\phi(k)}\right) \phi(k) ^n\tag{25}$$ with $k=k_r$.

Ramanujan's fastest series is based on the formula $$\frac{4K^2}{\pi^2}=\frac{1}{1+k^2}{}_3F_2\left(\frac{1}{4},\frac{3}{4},\frac{1}{2};1,1;\left(\frac{g(k)^{12}+g(k)^{-12}}{2}\right)^{-2}\right)$$ where $g(k) =(2k/k'^2)^{-1/12}$. Using $m(k) =(1+k^2)^{-1}$ and $\phi(k) =4/(g(k)^{12}+g(k)^{-12})^2$ in $(25)$ one gets the following series $$\frac{1}{\pi}=\sum_{n\geq 0}\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}d_n(r)x_r^{2n+1}\tag{26}$$ where $$x_r=\frac{4k_r{k'}_r^2}{(1+k_r^2)^2}\tag{27}$$ and $$d_n(r) =\left(\frac{\alpha(r) /x_r} {1+k_r^2}-\frac{\sqrt{r}}{4}g(k_r)^{-12}\right) +n\sqrt{r} \left(\frac{g(k_r) ^{12}-g(k_r)^{-12}}{2}\right)\tag{28} $$ Ramanujan's series in question is obtained by putting $r=58$ in above formula. The stumbling block however is the calculation of $\alpha(r)$ for $r=58$ and Borweins mention that they first evaluated $d_0(58)$ given in $(28)$ numerically and then compared the first term of their series with that given by Ramanujan (containing the constant $1103$) and argued that if their difference were beyond a certain margin of error the series would differ from $1/\pi$ by a different margin. That the series sums to $1/\pi$ within a certain margin of error confirms that $d_0(58)$ must match corresponding value of the first term as given by Ramanujan. And then they evaluated $\alpha(58)$ in symbolic form. A similar procedure was used to evaluate $\alpha(37)$ based on a different series given by Ramanujan.

To the best of my knowledge (from online sources including math overflow and mathse) there is no hand calculation available for $\alpha(37),\alpha(58),\alpha(163)$ and most of the modern work in this direction is highly assisted by mathematical software.

  • (+1) Amazing Answer. Considering that Ramanujan did all of this using hand calculations, there might be some other method. (As Bruce Berndt himself stated "However, it is unwise ever to assume that one has, for certain, obtained Ramanujan's original proofs") – Miracle Invoker Mar 28 '24 at 16:15
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    @MiracleInvoker: It is very unfortunate that Ramanujan never discussed his technique to obtain modular equations. All his published work considers modular equations as a given and then further work is done starting with them. But there is nothing mysterious here, he was just way better than typical math software when doing algebraic manipulation. And he did these calculations all the time (24x7) without any regard to sleep or food. – Paramanand Singh Mar 28 '24 at 16:23
  • Sorry but this answer is not what I expected. I can't give you +1. Best. – User-Refolio Mar 30 '24 at 13:53
  • @User: no problem. Rep is the secondary aspect here. I just wanted to summarize whatever I knew in one place about this problem. – Paramanand Singh Mar 30 '24 at 14:55
  • Thanks for the effort and for the summary. I will give in the below partial answer a discussion for $k_{4}$ and obtain the respective Ramanujan's series. It only requires $(1)$, $(2)$ Landen's transformation and Legrendre's relation. No modular forms needed. Best. – User-Refolio Mar 30 '24 at 15:07
  • @MiracleInvoker The question is if Ramanujan could have had a computer, how far would he have gone? – User-Refolio Mar 30 '24 at 15:12
  • @User: Landen transform is sufficient to handle $k_{2^n}$ but the theory and computation both get difficult for $k_n$ if $n$ has large prime factors. – Paramanand Singh Mar 30 '24 at 17:14
  • @ParamanandSingh It's a Lambert series was you thinking of that ? – Barackouda Mar 30 '24 at 17:18
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    @MissandMistercassouletchar: yeah $P(q) $ is a typical example of Lambert series. – Paramanand Singh Mar 30 '24 at 17:32
  • Hi @ParamanandSingh I have been reading the Borwein's brothers Pi and the AGM. I find this book interesting but they claim to proof this series without full details. So clever defining $\alpha(r)$ and improvements are very interesting but again opaque. Best all. – User-Refolio Apr 30 '24 at 06:06
  • Ramanujan's tecnique is fairly clair once one has studied the related stuff you have posted. It's a pitty that Ramanujan didn't have more resources to get the full details. – User-Refolio Apr 30 '24 at 06:08
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    @User: let me mention another unfortunate event. Jacobi did have the resources to publish all his development in theory of theta functions and elliptic functions and his complete works are available, but again no one really wants to use his technique (which was based on algebraic manipulation and more in line with Ramanujan) and modern mathematicians and have ditched these ideas completely and rely on theory of modular forms and computers. – Paramanand Singh Apr 30 '24 at 06:29
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    I'm also reading Jacobi_Fundamenta Nova. I'm so atonishing about this matemagician. I think Euler, Jacobi and Ramanujan have been some of the best mathemagicians. They didn't have computers and some of his calculations are far beyond computers. But this is only my opinion. All credits go directly to Archimede. – User-Refolio Apr 30 '24 at 06:41
  • Finally I decided to give you +1, but I can't accept the answer. Best all. – User-Refolio Jan 12 '25 at 13:13
  • Seems a typo on formula $(13)$ should be $\frac{4KL}{\pi^2}A_{r}(l,k)$. – User-Refolio Mar 24 '25 at 07:27
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    @User: Thanks for catching typo. I have fixed it now. – Paramanand Singh Mar 24 '25 at 11:31