Archive for the ‘complex analysis’ Category

Hypergeometric formulas for Ramanujan’s continued fractions 2

(continued from yesterday’s post)

III. Icosahedral group

\begin{aligned} G(q) &= \sum_{n=0}^\infty \frac{q^{n^2}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\H(q) &= \sum_{n=0}^\infty \frac{q^{n^2+n}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\end{aligned}

I observed that,

\begin{aligned}&q^{-1/60}G(q) = j^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j}\big) = (j-1728)^{1/60}\,_2F_1\big(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{1728-j}\big)\\[2.5mm]&q^{11/60}H(q) = j^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j}\big) = (j-1728)^{-11/60}\,_2F_1\big(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{1728-j}\big)\end{aligned}

where, as in the previous post, $j=j(\tau)$ is the j-function, $q = e^{2\pi i \tau} = \exp(2\pi i \tau)$, $\tau = \sqrt{-N}$, and $N>1$.  Since it is known that,

\begin{aligned}&r(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \cfrac{q^3}{1 + \ddots}}}} = \frac{q^{11/60}H(q)}{q^{-1/60}G(q)} = \frac{q^{11/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}}{q^{-1/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(q^{5n-4})}}\end{aligned}

this implies that,

\begin{aligned}r(q) &=\frac{j^{-11/60}\,_2F_1\big(\frac{31}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{j}\big) }{j^{1/60}\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{j}\big)}\\[3mm]&=\frac{(j-1728)^{-11/60}\,_2F_1\big(\frac{41}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{1728-j}\big) }{(j-1728)^{1/60}\,_2F_1\big(\frac{29}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{1728-j}\big)}\end{aligned}

Example. Let $\tau = \sqrt{-4}$, hence $j = j(\sqrt{-4}) = 66^3$. Then,

$1/r(q) -r(q) = \left(\frac{1+\sqrt{5}}{2}\right)^4+\left(\frac{1+\sqrt{5}}{2}\right)5^{3/4} = 12.2643\dots$

Furthermore, since Ramanujan established that,

$G(q^{11})H(q)-q^2G(q)H(q^{11}) = 1$

if we define the two functions,

\begin{aligned}U(\tau) &= \big(j(\tau)\big)^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j(\tau)}\big)\\V(\tau) &= \big(j(\tau)\big)^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j(\tau)}\big) \end{aligned}

then the counterpart hypergeometric identity is also beautifully simple and given by,

$U(11\tau)V(\tau)-U(\tau)V(11\tau)=1$

In the next post, we will use one of the hypergeometric formulas to solve the general quintic.

Hypergeometric formulas for Ramanujan’s continued fractions 1

There are five Platonic solids, two are duals to another two, while the tetrahedron is self-dual. As such, this gives rise to 3 polyhedral groups: the tetrahedral group of order 12, the octahedral group of order 24, and the icosahedral group of order 60.

Amazingly, Ramanujan found 3 continued fractions that can be associated with each group. See this article for more details.  It turns out there are also corresponding hypergeometric formulas, and the numbers 12, 24, and 60 naturally appears.

First though, define the j-function as,

$j = j(\tau) = \frac{1}{q} + 744 + 196884q + 21493760q^2 + \dots$

where,

$q = e^{2\pi i \tau} = \exp(2\pi i \tau)$

This can be conveniently calculated in Mathematica as,

$j(\tau) = 1728\text{KleinInvariantJ}(\tau)$

NOTE:  In the formulas below, it will be assumed that,

$\tau = \sqrt{-N},\;\; N > 1$

I. Tetrahedral group

Given,

\begin{aligned}&c=c(q)=\cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \cfrac{q^3+q^6}{1 + \ddots}}}} = q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})^2}\end{aligned}

and,

\begin{aligned}&d = j^{1/3}\,\frac{\,_2F_1\big(\frac{1}{4},\frac{-1}{12},\frac{2}{3},\frac{1728}{j}\big)}{\,_2F_1\big(\frac{1}{4},\frac{7}{12},\frac{4}{3},\frac{1728}{j}\big)}\end{aligned}

then we have the simple relationship,

$d = 4c^2+c^{-1}$

Example.  Let $\tau = \sqrt{-2}$, hence $j=j(\sqrt{-2}) = 20^3$, then,

$d = 3\sqrt{2(11+4\sqrt{6})} = 19.3484\dots$

and c(q) can then be easily solved for as a cubic equation.

II. Octahedral group

Let,

\begin{aligned}&u = u(q) = \cfrac{\sqrt{2}\,q^{1/8}}{1 + \cfrac{q}{1+q + \cfrac{q^2}{1+q^2 + \cfrac{q^3}{1+q^3 + \ddots}}}} = \sqrt{2}\,q^{1/8}\prod_{n=1}^\infty\frac{1-q^{2n-1}}{(1-q^{4n-2})^2} \end{aligned}

then,

\begin{aligned}&u = \frac{\sqrt{2}}{j^{1/8}}\left(\frac{\,_2F_1\big(\frac{5}{24},\frac{13}{24},\frac{5}{4},\frac{1728}{j}\big)}{\,_2F_1\big(\frac{7}{24},\frac{-1}{24},\frac{3}{4},\frac{1728}{j}\big)}\right)^{1/2}\end{aligned}

Example. Still using $\tau = \sqrt{-2}$, then,

$u = \sqrt{1+\sqrt{2}-\sqrt{2(1+\sqrt{2})}} = 0.4656\dots$

III. Icosahedral group

(To be discussed in the next post.)

On Riemann-like zeta functions

Given the Riemann zeta function $\zeta(s)$, there is the nice equality,

\begin{aligned}&\sum_{m=1}^\infty\big[\zeta(2m)-1] = \frac{3}{4}\end{aligned}

It can be shown that,

\begin{aligned}&\sum_{m=1}^\infty\big[\zeta(pm)-1] = \sum_{k=2}^\infty \frac{1}{k^p-1}\end{aligned}

Consider the following evaluations,

\begin{aligned}&\sum_{k=2}^\infty \frac{1}{k^2-1} = \frac{3}{4} = 0.75\\ &\sum_{k=2}^\infty \frac{1}{k^2+1} = -1+\frac{\pi\text{coth}(\pi)}{2} = 0.5766\dots\end{aligned}

In general, given a root of unity, $\omega_p = e^{2\pi i/p}$, then,

\begin{aligned}&\sum_{k=2}^\infty \frac{1}{k^p-a} = -\frac{a^{1/p}}{ap}\sum_{j=1}^p \omega_p^j\, \psi(2-a^{1/p} \omega_p^j)\end{aligned}

for integer p > 1, any non-zero real or complex a, and where $\psi(z)$ is the digamma function. Thus, since roots of unity are involved, the formula uses complex terms even though, as the two examples show, the sum may be real.  But it turns out for real a and even powers p, it can be expressed using only real terms.  First,

\begin{aligned}&\sum_{k=2}^\infty \frac{1}{k^p-a} = \frac{1-3a}{2a(1-a)} -\frac{a^{1/p}\,\pi}{ap}\sum_{j=1}^{p/2} \omega_p^j\, \cot(\pi a^{1/p} \omega_p^j)\end{aligned}

for even p and any non-zero a except a = 1,  which is given by the special case,

\begin{aligned}&\sum_{k=2}^\infty\frac{1}{k^p-1} = \frac{2p-1}{2p}-\frac{\pi}{p}\sum_{j=1}^{p/2-1}\omega_p^j\,\cot(\pi\omega_p^j)\end{aligned}

But one can split the cotangent function into its real and imaginary parts as,

\begin{aligned}&\cot(\pi u\, e^{2\pi i n}) = \frac{-\sin(2\pi u\cos(2\pi n))+i \text{sinh}(2\pi u\sin(2\pi n)) }{\cos(2\pi u\cos(2\pi n))-\text{cosh}(2\pi u \sin(2\pi n))}\end{aligned}

hence cancel out the conjugate terms and leave only the real parts.  For example, we have,

\begin{aligned}&\sum_{k=2}^\infty \frac{1}{k^4-1} = \frac{1}{8}\big(7-2\pi\text{coth}(\pi)\big) = 0.0866\dots\\ &\sum_{k=2}^\infty \frac{1}{k^6-1} = \frac{1}{12}\big(11-2\pi\sqrt{3}\text{tanh}(\tfrac{\pi\sqrt{3}}{2})\big)= 0.0175\dots\end{aligned}

and so on. It is reminiscent of the situation with the zeta function,

\begin{aligned}&\sum_{k=1}^\infty \frac{1}{k^p} = \zeta(p)\end{aligned}

which has a closed-form solution only for even p, and is expressed by the real $\pi^p$  and Bernoulli numbers.  It makes me wonder if there is  a closed-form formula for $\zeta(p)$  involving the roots of unity.

The zeta function and roots of unity

In Mathworld’s entry on the Riemann zeta function, one finds in eq. 119-121 the curious evaluations,

\begin{aligned}\sum_{n=1}^\infty [\zeta(2n)-1] &= \frac{3}{4}\\ \sum_{n=1}^\infty [\zeta(3n)-1] &= \frac{1}{3}\left[-(-1)^{2/3}H_{(3-\sqrt{-3})/2}+(-1)^{1/3}H_{(3+\sqrt{-3})/2} \right]\\ \sum_{n=1}^\infty [\zeta(4n)-1] &= \frac{1}{8}\,(7-2\pi\coth(\pi))\end{aligned}

However, using the Inverse Symbolic Calculator, the first and the third, plus another one, can also be expressed as,

\begin{aligned}\sum_{n=1}^\infty [\zeta(2n)-1] &= \frac{5}{4}-\sum_{n=1}^\infty \frac{1}{2n^2+2n} = \frac{3}{4}\\ \sum_{n=1}^\infty [\zeta(4n)-1] &= \frac{5}{8}-\sum_{n=1}^\infty \frac{1}{2n^2+2}=\frac{7}{8}-\frac{1}{4}\,\pi i \cot(\pi w_4)\\ \sum_{n=1}^\infty [\zeta(6n)-1] &= \frac{5}{12}-\sum_{n=1}^\infty \frac{1}{2n^2+2n+2}=\frac{11}{12}-\frac{1}{6}\sqrt{3}\pi i\cot(\pi w_6)\end{aligned}

where $w_p = e^{2\pi i/p}$.  Interesting similar forms, isn’t it?

Unfortunately, it doesn’t seem to generalize to $\zeta(pn)$ for p = 8.  However, there is still p = 3 and, based on the even case, I assumed perhaps roots of unity are also involved.  First, given the Euler-Mascheroni constant $\gamma$, and the digamma function,

$\psi_0(z) = \psi[z]$

where we suppress the subscript for ease of notation.  Define,

$u_p = e^{\pi i/p } = (-1)^{1/p}$

and the pth root chosen such that $(-1)^{1/p} \not = -1$, then I found that p = 3 generalizes as,

\begin{aligned} 3\sum_{n=1}^\infty [\zeta(3n)-1] &= 3 + \gamma + u_3^{-1}\, \psi[u_3^{-1}]+u_3\,\psi[u_3]\\&= 0.66506\dots\\ 5\sum_{n=1}^\infty [\zeta(5n)-1] &= 6 + \gamma + \sum_{k=0}^1 \Big(u_5^{-(2k+1)}\, \psi[u_5^{-(2k+1)}]+u_5^{(2k+1)}\,\psi[u_5^{(2k+1)}]\Big)\\&=0.18976\dots\\ 7\sum_{n=1}^\infty [\zeta(7n)-1] &= 9 + \gamma + \sum_{k=0}^2 \Big(u_7^{-(2k+1)}\, \psi[u_7^{-(2k+1)}]+u_7^{(2k+1)}\,\psi[u_7^{(2k+1)}]\Big)\\&=0.05887\dots\end{aligned}

and so on, though a rigorous proof is needed that it holds true for all odd numbers p.

P.S. Going back to even p, note that p = 2, 4, 6 can also be expressed by the digamma function since,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{an^2+bn+c} = \frac{1}{\sqrt{b^2-4ac}}\Big(\psi[\tfrac{2a+b+\sqrt{b^2-4ac}}{2a}]-\psi[\tfrac{2a+b-\sqrt{b^2-4ac}}{2a}]\Big)\end{aligned}

for $a \not=0$.

Roots of unity and binomial sums

Thanks to Robert Israel who answered my question in mathstackexchange, we have a generalization of the binomial sums of the previous post.  Interestingly, it turns out roots of unity are involved.  Given,

$w = e^{2\pi\, {\rm i}/k}$

where k is an even integer then,

\begin{aligned}&\sum_{n=0}^\infty \frac{(2z)^{kn}}{\binom{kn}{kn/2}} = \frac{1}{1-z^k}+\frac{1}{k}\sum_{j=0}^{k-1}\frac{w^j z \arcsin(w^j z)}{(1-w^{2j}z^2)^{3/2}}\end{aligned}

for appropriate z such that the sum converges.  For the special case when,

$2z =w^{1/2}=e^{\pi\,{\rm i}/k}$

then,

\begin{aligned}&\sum_{n=0}^\infty\frac{(-1)^n}{\binom{kn}{kn/2}}=\frac{2^k}{2^k+1}+\frac{1}{k}\sum_{j=0}^{k-1}\frac{w^{j+1/2} \arcsin(\tfrac{1}{2}\,w^{j+1/2})}{2(1-\tfrac{1}{4}\,w^{2j+1})^{3/2}}\end{aligned}

Note that the terms are complex, but the sum is a real number so they must come in conjugate pairs. The arcsin of a complex root of unity can be given as,

$\arcsin(\frac{1}{2}\,e^{\pi\,{\rm i}/k}) = -\arcsin\big(\frac{-a+b}{4}\big)+{\rm i}\ln\Big(\frac{a+b+\sqrt{-6+2ab}}{4}\Big)$

where,

\begin{aligned}a &= \sqrt{5+4\cos(\pi/k)}\\ b &= \sqrt{5-4\cos(\pi/k)}\end{aligned}

With this transformation, it is now possible to have an expression all in real terms.  The case k = 2, 4 was given in the previous post.  For k = 6, we have the counterpart to Sprugnoli’s equality as,

\begin{aligned}\sum_{n=0}^\infty \frac{(-1)^n}{\binom{6n}{3n}}&=\tfrac{64}{65}-\tfrac{2\sqrt{26}\,(1+7\sqrt{13})}{3\cdot13^2\,\sqrt{1+\sqrt{13}}}\arcsin\left(\tfrac{-\sqrt{6}}{2\sqrt{5+\sqrt{13}}}\right)\\&-\tfrac{\sqrt{26}\,(-1+7\sqrt{13})}{3\cdot13^2\,\sqrt{-1+\sqrt{13}}}\ln\left(\tfrac{\sqrt{7+2\sqrt{13}}\,+1}{\sqrt{7+2\sqrt{13}}\,-1}\right)-\tfrac{4\sqrt{5}}{3\cdot5^2}\ln\left(\tfrac{1+\sqrt{5}}{2}\right) = 0.95106\dots\end{aligned}

Note that the prime factors of 65 are 5 and 13, and the square root of both appear above. However, for k = 8, while the expression contains the fraction $\frac{256}{257}$ as expected, the argument of the log and arcsin do not factor over the quadratic extension $\sqrt{257}$, but rather only over $\sqrt{2}$.  Furthermore, the argument of the log for both k = 6, 8 are no longer simply expressible in terms of the Dedekind eta function, so observations for lower k do not generalize to higher ones.

Shanks’ approximation to pi

In “Pi Approximations“, line 58, Weisstein mentions one by Shanks (1982) that differs by a mere $10^{-82}$ as,

$\pi \approx \frac{6}{\sqrt{3502}}\, \ln(2u)$

where u is “…a product of four simple quartic units”. Frustratingly, he doesn’t give u but I eventually found the primary source online. Hence,

$u = \big(a+\sqrt{a^2-1}\big)^2 \big(b+\sqrt{b^2-1}\big)^2 (c+\sqrt{c^2-1}) (d+\sqrt{d^2-1})$

where,

\begin{aligned}a &= \tfrac{1}{2}\, (23+4\sqrt{34}\,)\\ b &= \tfrac{1}{2}\, (19\sqrt{2}+7\sqrt{17}\,)\\ c &= 429+304\sqrt{2}\\ d &= \tfrac{1}{2}\,(627+442\sqrt{2}\,) \end{aligned}

with a slight modification by this author since Shanks didn’t realize the first two quartic factors were in fact squares.  (The product of the last two factors is also a square.)

A cute thing about these numbers is that their defining polynomials are palindromic, the same read forward or backward.  For example, the first factor (unsquared) is the root of,

$x^4-46x^3-13x^2-46x+1 = 0$

Author’s note:  Daniel Shanks (1917-96) was a mathematician best known as the first to calculate pi up to 100,000 decimal places, as well as for his book, Solved and Unsolved Problems in Number Theory.

In general, Shanks’ approximation belongs to the family,

\begin{aligned} e^{\pi\sqrt{2m}} &\approx \left(\tfrac{\eta(\frac{1}{2}\sqrt{-2m})}{\eta(\sqrt{-2m})}\right)^{24}\\ &\approx 2^6 x^{k}\end{aligned}

where $\eta$ is the Dedekind eta function and, for m a positive odd integer, then x is an algebraic integer that is the root of an equation P(x) with palindromic (if unsigned) coefficients.  For appropriate k, then P(x) has degree equal to the class number h(-2m).  Furthermore, it is solvable in radicals.

For example, given prime m, with 2m = {10, 14, 26} which has class number 2, 4, 6, respectively, then k = 12 and,

\begin{aligned} x_{10}&\; \text{is a root of}\, x^2-x-1 =0\\ x_{14}&\;\text{is a root of}\, x^4-2x^3+x^2-2x+1 = 0\\ x_{26}&\;\text{is a root of}\, x^6-2x^5-2x^4+2x^2-2x-1 = 0 \end{aligned}

and so on.  It then is a simple matter to take the natural logarithm of both sides to bring down pi and have a relation of form,

$\pi \approx \frac{1}{\sqrt{2m}} \ln(2^6 x^{k})$

Shanks chose 2m = 3502 since d = 4(2m) is the largest fundamental discriminant d divisible by 4 with class number h(-d) = 16.  Here is a list of of d with small class number.  You can calculate it (among many other things) in www.wolframalpha.com simply as,

ClassNumber[Sqrt[-d]]