tpiezas

23 Jul

Hypergeometric formulas for Ramanujan’s continued fractions 2

Posted by tpiezas in complex analysis, geometry, identities. Tagged: continued fractions, hypergeometric function, quintics, Ramanujan. 2 Comments

(continued from yesterday’s post)

III. Icosahedral group

Given the Rogers-Ramanujan identities (see also here),

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

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.

22 Jul

Hypergeometric formulas for Ramanujan’s continued fractions 1

Posted by tpiezas in complex analysis, geometry, mathematics. Tagged: continued fractions, hypergeometric function, j-function, Ramanujan. 1 Comment

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.)

8 Jul

Algebraic values of the Hypergeometric function

Posted by tpiezas in identities, mathematics. Tagged: hypergeometric function, icosahedron, identities. Leave a Comment

The hypergeometric function,

$\begin{aligned} &\,_2F_1(a,b,c,z) = \sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!}\end{aligned}$

where $(a)_n$ is a Pochhammer symbol is, for rational {a, b, c, z}, generally not a rational value. In the link given above, equations (18) and (19) are,

$\,_2F_1\big(\frac{1}{3},\frac{2}{3},\frac{5}{6},\frac{27}{32}\big) =\frac{8}{5}$

$\,_2F_1\big(\frac{1}{4},\frac{1}{2},\frac{3}{4},\frac{80}{81}\big) =\frac{9}{5}$

There are in fact an infinite number of such equalities. One given by M. Glasser is, let $0 < x < \frac{1}{\sqrt{3}}$ , then,

$\,_2F_1\big(\frac{1}{3},\frac{2}{3},\frac{3}{2},\frac{27x^2(1-x^2)^2}{4}\big) =\frac{1}{1-x^2}$

Another, based on eq. (42) of Vidunas’ “Transformations of algebraic Gauss hypergeometric functions” is, let $0 < y < 1$ , then,

$\,_2F_1\big(\frac{1}{4},\frac{-1}{12},\frac{2}{3},\frac{(9-y^4)^3(-1+y^4)}{64y^{12}}\big) =\frac{1}{y}$

A third which yields not a rational but an algebraic number is remarkable for its connection to the Rogers-Ramanujan continued fraction. Let $0 < z < v_1$ where,

$v_1 = \text{Root}[z^4+228z^3+494z^2-228z+1=0] = 0.004428\dots$

then,

$\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728z(z^2-11z-1)^5}{(z^4+228z^3+494z^2-228z+1)^3}\big) =\frac{1}{(z^4+228z^3+494z^2-228z+1)^{1/20}}$

If the polynomials are familiar, it is because they are invariants of the icosahedron. They also appear in the j-function formula,

$j(\tau) = \frac{-(r^{20}-228r^{15}+494r^{10}+228r^5+1)^3}{r^5(r^{10}+11r^5-1)^5}$

where,

$r = r(\tau) = \cfrac{q^{1/5}}{1+ \cfrac{q}{1 + \cfrac{q^2}{1+ \cfrac{q^3}{1 + \ddots}}}}$

and

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

8 Jul

On Riemann-like zeta functions

Posted by tpiezas in complex analysis, mathematics. Tagged: Riemann zeta function, roots of unity, trigonometry. Leave a Comment

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.

11 Jun

A new formula for Apery’s constant and other zeta(2n+1)?

Posted by tpiezas in equations, mathematics, number theory. Tagged: constants, Ramanujan, Riemann zeta function. 3 Comments

I. Introduction

In Identities Inspired from Ramanujan’s Notebooks, Simon Plouffe recounts how, based on Ramanujan’s,

$\begin{aligned}\sum_{k=1}^\infty \frac{\coth(\pi k)}{k^3} = \frac{7}{180}\pi^3\end{aligned}$

he found,

$\begin{aligned}\zeta(3) &= \frac{7\pi^3}{180}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{2\pi k}-1)}\\ \zeta(7) &= \frac{19\pi^7}{56700}-2\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k}-1)} \end{aligned}$

and similar ones for other s = 4m+3. On a hunch, and using Mathematica’s LatticeReduce function, I found that,

$\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}\\ \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(e^{\pi k\sqrt{2}}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(e^{2\pi k\sqrt{2}}-1)}\\ \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k\sqrt{2}}-1)} \end{aligned}$

etc.

II. Functions

If we define,

$\begin{aligned} &U_a(s) = \sum_{k=1}^\infty \frac{1}{k^s(e^{a\pi k}-1)}\end{aligned}$

then Plouffe discovered integer relations between,

$\zeta(s), \pi^s, U_1(s),\, U_2(s),\, U_4(s)$

for odd s, with s = 3 being,

$\begin{aligned} &\zeta(3) = 28U_1(3)-37U_2(3)+7U_4(3)\\ &\text{and,}\\ &\pi^3 =720U_1(3)-900U_2(3)+180U_4(3)\end{aligned}$

Eliminating $U_1(s),\, U_4(s)$ leads to the 3-term equalities in the Introduction. See Chamberland’s and Lopatto’s Formulas for Odd Zeta Values. On the other hand, by defining the function,

$\begin{aligned} &V_b(s) = \sum_{k=1}^\infty \frac{1}{k^s(e^{b\pi k\sqrt{2}}-1)}\end{aligned}$

I observed integer relations between,

$\zeta(s), \pi^s \sqrt{2},\, V_1(s),\, V_2(s),\, V_3(s),\, V_6(s)$

also for odd s, with s = 3 as,

$\begin{aligned} &\zeta(3)-102V_1(3)+99V_2(3)+10V_3(3)-5V_6(3) = 0\\ &\text{and,}\\ &\pi^3\,\sqrt{2}-3720V_1(3)+3540V_2(3)+360V_3(3)-180V_6(3) = 0\end{aligned}$

and so on. Eliminating $V_3(s),\, V_6(s)$ leads to the 4-term equalities in the Introduction.

III. Conjecture

The 4-term equalities have coefficients that are simple except for one term. Recall that,

$\begin{aligned} V_1(s) &= \sum_{k=1}^\infty \frac{1}{k^s(e^{\pi k \sqrt{2}}-1)}\\ V_2(s) &= \sum_{k=1}^\infty \frac{1}{k^s(e^{2\pi k \sqrt{2}}-1)}\end{aligned}$

Conjecture:

“Using the positive case of $\pm 1$ for s = 4m+3, and the negative for s = 4m+5, then in the equation,

$\begin{aligned}&F(s)\zeta(s-1)\pi\sqrt{2} = \big(2^{(s-1)/2}\pm 1\big)\frac{\zeta(s)}{2}+2^{(s-1)/2}\,V_1(s) \pm V_2(s) \end{aligned}$

$F(s)$ is a rational number.”

The first few for s = {3, 7, 11,…} are $F(s) = \frac{1}{4}, \frac{41}{40}, \frac{319}{84}, \dots$ while for s = {5, 9, 13,…} are $F(s) = \frac{1}{3}, \frac{19}{11}, \frac{5017}{691},\dots$ These rationals may have a closed-form expression in terms of Bernoulli numbers, but I do not yet know the exact formulation.

11 Jun

On Bailey and Crandall’s unusual sum

Posted by tpiezas in equations, identities, mathematics. Tagged: identities, logarithm, roots of unity, trigonometry. Leave a Comment

In page 20 of Bailey and Crandall’s On the Random Character of Constant Expansions, they give the wonderfully unusual sum,

$\begin{aligned} x_1 = \sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5}{5n+2}+\frac{1}{5n+3}\right)&=\frac{5^2}{2}\ln\left(\frac{781}{256}\left(\frac{57-5\sqrt{5}}{57+5\sqrt{5}}\right)^{\sqrt{5}}\right)\\ &= 2.833601\dots\end{aligned}$

I didn’t think this was an isolated result so set about to find a generalization. I found its counterpart,

$\begin{aligned} x_2 = \sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5^3}{5n+1}+\frac{1}{5n+4}\right)&=\frac{5^3}{2}\ln\left(\frac{781}{256}\left(\frac{57+5\sqrt{5}}{57-5\sqrt{5}}\right)^{\sqrt{5}}\right)\\ &= 125.256703\dots\end{aligned}$

Note that,

$(57+5\sqrt{5})(57-5\sqrt{5}) = 4(781) = 5^5-1$

We can demystify the sum a bit by splitting the log function into parts. After some algebraic manipulation, we find that the first one becomes,

$\begin{aligned}&x_1 = 25\Big(-5\ln(2)+\tfrac{1-\sqrt{5}}{2}\,\ln(57+5\sqrt{5})+\tfrac{1+\sqrt{5}}{2}\,\ln(57-5\sqrt{5})\Big)\end{aligned}$

Thus it can be expressed in the form,

$x_1 = r\ln\big({p_1}^{q_1}\,{p_2}^{q_2}\big)$

where { $p_1, p_2$ } are roots of the same equation, { $q_1, q_2$ } are roots of another, and r is a rational. The fact that,

$2\cos(\pi/5) = \frac{1+\sqrt{5}}{2}$

was my clue that trigonometric functions may be involved. Define,

$c_k = -2\cos(2k\pi/p)$

then for p = 5,

$\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{5^{5n}}\left(\tfrac{5}{5n+2}+\tfrac{1}{5n+3}\right) &= 5^2\ln\big((5c_1+26)^{c_2}(5c_2+26)^{c_1}/4^2\big)\\&=2.833601\dots\end{aligned}$

p = 7

$\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{7^{7n}}\left(\tfrac{7}{7n+3}+\tfrac{1}{7n+4}\right) &= 7^3\ln\big((7c_1+50)^{c_3}(7c_2+50)^{c_1}(7c_3+50)^{c_2}/6^2\big)\\&=2.583334\dots\end{aligned}$

p = 9

$\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{9^{9n}}\left(\tfrac{9}{9n+4}+\tfrac{1}{9n+5}\right) &= 9^4\ln\big((9c_1+82)^{c_4}(9c_2+82)^{c_1}(9c_3+82)^{c_3}(9c_4+82)^{c_2}/8^2\big)\\&=2.450000\dots\end{aligned}$

with the constants { $26, 50, 82$ } easily ascertained as { $5^2+1, 7^2+1, 9^2+1$ }, and so on. On the other hand, their counterparts are easier as the exponent $c_k$ has the same subscript as the base. Still defining,

$c_k = -2\cos(2k\pi/p)$

then for p = 5,

$\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{5^{5n}}\left(\tfrac{5^3}{5n+1}+\tfrac{1}{5n+4}\right) &= 5^3\ln\big((5c_1+26)^{c_1}(5c_2+26)^{c_2}/4^2\big)\\&=125.256703\dots\end{aligned}$

p = 7

$\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{7^{7n}}\left(\tfrac{7^5}{7n+1}+\tfrac{1}{7n+6}\right) &= 7^5\ln\big((7c_1+50)^{c_1}(7c_2+50)^{c_2}(7c_3+50)^{c_3}/6^2\big)\\&=16807.169\dots\end{aligned}$

p = 9

$\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{9^{9n}}\left(\tfrac{9^7}{9n+1}+\tfrac{1}{9n+8}\right) &= 9^7\ln\big((9c_1+82)^{c_1}(9c_2+82)^{c_2}(9c_3+82)^{c_3}(9c_4+82)^{c_4}/8^2\big)\\&\approx 4.7829\, {\rm x}\, 10^6\end{aligned}$

etc.

11 Jun

The zeta function and roots of unity

Posted by tpiezas in complex analysis, identities, mathematics. Tagged: constants, gamma function, Riemann zeta function, roots of unity. Leave a Comment

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 $a \not=0$ .

10 Jun

Roots of unity and binomial sums

Posted by tpiezas in complex analysis, identities, mathematics. Tagged: binomial, logarithm, roots of unity, trigonometry. Leave a Comment

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.

3 Jun

Fermat primes and Binomial sums

Posted by tpiezas in equations, identities, mathematics. Tagged: binomial, Dedekind eta function, logarithm, primes. 3 Comments

We have,

$\begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n}{\binom n{n/2}} &= \frac{4}{3}-\frac{4\pi\sqrt{3}}{27}\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} &= \frac{4}{5} - \frac{4\sqrt{5}}{25}\ln\left(\frac{1+\sqrt{5}}{2}\right) \end{aligned}$

For the next step, Renzo Sprugnoli gave the Ramanujan-like identity,

$\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}+\frac{4\sqrt{34}\,(-2+\sqrt{17}\,)}{17^2\,\sqrt{-1+\sqrt{17}}}\arctan\left(\frac{\sqrt{2}}{\sqrt{-1+\sqrt{17}}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{17^2\,\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right) = 0.846609\dots \end{aligned}$

(The sign of the third term has been changed by this author.) However, to make it more symmetrical, we can express the arctan in terms of the log function. Since,

$\begin{aligned}&\arctan(z) = \frac{i}{2}\ln\left(\frac{1-i z}{1+i z}\right)\end{aligned}$

then,

$\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}-\frac{2\sqrt{-34}\,(-2+\sqrt{17}\,)}{17^2\,\sqrt{-1+\sqrt{17}}}\ln\left(\frac{\sqrt{-1+\sqrt{17}}+\sqrt{-2}}{\sqrt{-1+\sqrt{17}}-\sqrt{-2}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{17^2\,\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right) =0.846609\dots \end{aligned}$

In this manner, it reduces to the concise,

$\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}-\frac{x_1}{17}\ln(y_1)-\frac{x_2}{17}\ln(y_2)\end{aligned}$

where, $x_1,x_2$ and $y_1,y_2$ are the appropriate roots of,

$\begin{aligned} &289x^4-799x^2-676 = 0\\ &y^4-5y^3+4y^2-5y+1 = 0\end{aligned}$

I found that, curiously, the argument of the log can be expressed in terms of the Dedekind eta function, $\eta(z)$ . Let,

$\begin{aligned} t_1 &=\frac{1+\sqrt{-5}}{2}\\ t_2 &= \frac{1+\sqrt{-17}}{2}\\ \zeta_{48} &=\exp(\pi i/24)\end{aligned}$

then,

$\begin{aligned} &\frac{1}{2}\left(\frac{\zeta_{48}\, \eta(t_1)}{\eta(2t_1)}\right)^4 = \frac{1+\sqrt{5}}{2}\\ &\frac{1}{2}\left(\frac{\zeta_{48}\, \eta(t_2)}{\eta(2t_2)}\right)^4 = \frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\end{aligned}$

Is this coincidence? Furthermore, using these as the argument of the polylogarithm,

$\begin{aligned} &L_s(z) = \text{Li}_s (z) = \sum_{k=1}^\infty\frac{z^k}{k^s}\end{aligned}$

one can find a polylogarithm ladder to express Apery’s constant. For example, getting the square root and reciprocal of $y_2$ so that z < 1,

$\begin{aligned} z &= \sqrt{\frac{\sqrt{1+\sqrt{17}}-\sqrt{2}}{\sqrt{1+\sqrt{17}}+\sqrt{2}}} = 0.480533\dots\end{aligned}$

then,

$-12L_3(z)+75L_3(z^2)-68L_3(z^3)-33L_3(z^4)+43L_3(z^6)+12L_3(z^8)-7L_3(z^{12})+2\log^3(1/z) = 3\zeta(3)$

A simpler one exists for the other argument. The next step, of course, is,

$\begin{aligned}&\sum_{n=0}^\infty \frac{(-1)^n}{\binom{8n}{2n}}=\,?\end{aligned}$

Since the first three Fermat primes 3, 5, 17 have already appeared, it should be interesting to conjecture if 257 will be next.

3 Jun

A missing binomial sum identity?

Posted by tpiezas in equations, identities, mathematics, number theory. Tagged: binomial, constants, Riemann zeta function. Leave a Comment

D.Bailey, J. Borwein, and D.Bradley found the beautiful pair involving binomial sums. In Theorem 1 of this paper (2008), let x $\not=$ non-zero integer, then,

$\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^2-x^2} &= 3\,\sum_{k=1}^\infty \frac{1}{k^2 \binom{2k}k (1-x^2/k^2)} \prod_{m=1}^{k-1} \left(\frac{1-4x^2/m^2}{1-x^2/m^2}\right)\\[2.5mm] \sum_{k=1}^\infty \frac{1}{k^3(1-x^4/k^4)} &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3 \binom{2k}k (1-x^4/k^4)} \prod_{m=1}^{k-1} \left(\frac{1+4x^4/m^4}{1-x^4/m^4}\right)\end{aligned}$

When x = 0, they reduce into,

$\begin{aligned} \zeta(2) &= 3\,\sum_{k=1}^\infty \frac{1}{k^2\binom{2k}k }\\ \zeta(3) &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\binom{2k}k }\end{aligned}$

However, there is a third single-term equality,

$\begin{aligned} \frac{17}{36}\,\zeta(4) &= \sum_{k=1}^\infty \frac{1}{k^4\binom{2k}k }\end{aligned}$

so there might be a third identity that reduces to this as the special case x = 0.

To compare, there are three identities such that as $x \to 0$ , then those zeta values are the respective limit. For x $\not=$ integer, then,

$\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\,\frac{3k^2+x^2}{k^2-x^2}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{\pi x \csc(\pi x)-1}{x^2}\\[2.5mm] \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\,\binom{2k}k}\,\frac{5k^2-x^2}{2(k^2-x^2)}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{-\psi^{(0)}(1-x)-\psi^{(0)}(1+x)-2\gamma}{2x^2}\\[2.5mm] \sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\,\frac{1}{k^2-x^2}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{\pi x \csc(\pi x)+3\cos(\pi x/3)-4}{4x^4}\end{aligned}$

The first two were found by Leshchiner and Koecher, respectively, while the third is Theorem 2 in the same paper by Bailey, Borwein, and Bradley. The function $\psi^{(0)}$ is given in Mathematica as,

$\psi^{(0)}(z) = \text{PolyGamma[0,z]}$

while $\gamma$ is the Euler-Mascheroni constant. So are the Bailey-Borwein-Bradley pair of binomial sum identities in fact a triplet?

« Older Entries

tpiezas A Collection of Algebraic Identities

Hypergeometric formulas for Ramanujan’s continued fractions 2

Hypergeometric formulas for Ramanujan’s continued fractions 1

Algebraic values of the Hypergeometric function

On Riemann-like zeta functions

A new formula for Apery’s constant and other zeta(2n+1)?

On Bailey and Crandall’s unusual sum

The zeta function and roots of unity

Roots of unity and binomial sums

Fermat primes and Binomial sums

A missing binomial sum identity?

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