## Posts Tagged ‘Riemann zeta function’

### 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.

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

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.

### 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 $a \not=0$.

### A missing binomial sum identity?

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?

### Apery-like formulas for zeta(2n)

It is well-known that,

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

D. Bailey, J. Borwein, D. Bradley gave a generalization. First define,

\begin{aligned}&A(a_0) = \sum_{k=1}^\infty \frac{1}{k^{a_0}\,\binom{2k}k}\\ &A(a_0, a_1, a_2,\dots) = \sum_{k=1}^\infty \frac{1}{k^{a_0}\,\binom{2k}k} \sum_{p=1}^{k-1}\frac{1}{p^{a_1}} \sum_{q=1}^{k-1} \frac{1}{q^{a_2}}\dots\end{aligned}

Obviously,

$\zeta(2) = 3A(2)$

However, a little experiment with Mathematica’s LatticeReduce command will show there are two solutions for $\zeta(4)$,

\begin{aligned} &a\big(\zeta(4)-3A(4)+9A(2,2)\big) =0\\ &b\big(5\zeta(4)-10A(4)-6A(2,2)\big) = 0\end{aligned}

where {a, b} are scaling variables.  Adding the two together,

$(a-5b)\zeta(4)-(3a-10b)A(4)+3(3a+2b)A(2,2) = 0$

hence there are an infinite number of solutions.  For appropriately chosen {a,b}, we can also eliminate one term. Thus,

\begin{aligned} \zeta(4) &= \frac{36}{17}A(4)\\ &=\frac{108}{5}A(2,2)\end{aligned}

For $\zeta(6)$, there are now three solutions. Given the five terms,

$\zeta(6),\, A(6),\, A(4,2),\, A(2,4),\, A(2,2,2)$

then the coefficients such that their sum is equal to zero are,

\begin{aligned} &\text{1st sol:}\; (5, -9, 1, -15, 3)\\ &\text{2nd sol:}\; (2, -7, 23, 9, 15)\\ &\text{3rd sol:}\;\, (10,-17,-3,6,-36)\end{aligned}

Using the same approach above, we can eliminate two of the terms.  One solution has an interesting number pop up,

$163\zeta(6) = 288A(6)+432A(2,4)$

though the appearance of 163 is probably only a coincidence.   See Bailey, Borwein, Bradley’s paper for more details.

### Borwein and Bradley’s Apery-like formulas for zeta(4n+3)

Apery gave,

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

J. Borwein and D. Bradley found this can be generalized to $\zeta(4n+3)$. Define the functions,

\begin{aligned} &B(a_0)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{a_0}\,\binom {2k}k}\\ &B(a_0,a_1,a_2,\dots)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{a_0}\,\binom {2k}k}\; \sum_{p=1}^{k-1} \frac{1}{p^{a_1}}\;\sum_{q=1}^{k-1} \frac{1}{q^{a_2}}\;\dots \end{aligned}

then,

\begin{aligned} \frac{2}{5}\,\zeta(3) &= B(3)\\ \frac{2}{5}\,\zeta(7) &= B(7)+5B(3,4)\\ \frac{2}{5}\,\zeta(11) &= B(11)+5B(7,4)-\frac{15}{2}B(3,8)+\frac{25}{2}B(3,4,4)\\ \frac{2}{5}\,\zeta(15) &= B(15)+5B(11,4)-\frac{15}{2}B(7,8) +\frac{25}{2}B(7,4,4)+\frac{130}{6}B(3,12)\\&-\frac{225}{6}B(3,8,4)+ \frac{125}{6}B(3,4,4,4) \end{aligned}

and so on.  Beautiful, aren’t they? Notice that all the $a_i$ (excepting $a_0$) are all divisible by 4. This infinite family has a generating function. Let z $\not=$ non-zero integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^3(1-z^4/k^4)}&=\frac{5}{2}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^3\;\binom {2k}k} \frac{1}{1-z^4/k^4}\prod_{j=1}^{k-1}\frac{1+4z^4/j^4}{1-z^4/j^4}\end{aligned}

On the other hand, for s = 4n+1,

\begin{aligned} 2\,\zeta(5) &= 4B(5)-5B(3,2)\\[2.5mm] 4\,\zeta(9) &= 9B(9)-5B(7,2)+20B(5,4)+45B(3,6)-25B(3,4,2)\\[2.5mm] 12\,\zeta(13) &= 28B(13)-10B(11,2)+90B(9,4)\\&+90B(7,6)-50B(7,4,2)-60B(5,8)+100B(5,4,4)\\&-310B(3,10)+75B(3,8,2)+450B(3,6,4)-125B(3,4,4,2)\end{aligned}

with this version for $\zeta(13)$ found by Jim Cullen.  There are various versions for both s = 4n+1 and 4n+3.  For example, for $\zeta(7)$, we have the relations,

\begin{aligned} 4\zeta(7) &= 8B(7,0)\,-\,5B(3,4)\,-\,8B(5,2)+5B(3,2,2)\\ 0 &= -2B(7,0) - 55B(3,4)-8B(5,2)+5B(3,2,2)\end{aligned}

Eliminating the last two terms will yield the shorter relation given by Borwein and Bradley. There is a generating function for all s = 2n+1, but none is known that is only for s = 4n+1. See Apery-Like Formulae for $\zeta(4n+3)$ for more details.

### Sequences 1, Tribonacci numbers

In this 3-part series of posts, we’ll discuss well-known sequences with the recurrence,

$aP_{n-3} + bP_{n-2} + cP_{n-1} = P_n$

where {a, b, ccan only be zero or unity.  Aside from the Fibonacci and Lucas numbers which is a = 0, there is the Narayana sequence with b = 0, the Padovan and Perrin with c = 0, and the tribonacci has a = b = c = 1.  All four cases may then share similar properties and one of which, interestingly enough, is that their limiting ratios, a root of the following equations,

\begin{aligned} x^2 &=x+1\\ y^3 &= y^2+1\\ z^3 &= z+1\\ t^3 &= t^2+t+1\end{aligned}

can also be used to express $\zeta(3)$,  or Apery’s constant.

I. Fibonacci and Lucas numbers

Given the two roots of,

$x^2=x+1$

with $x_1 > x_2$, the larger root being the golden ratio, we get the Lucas numbers L(n) and Fibonacci numbers F(n),

\begin{aligned} L_n &= x_1^n+x_2^n = 2,1,3,4,7,11,18,29,\dots\\[2mm] F_n &= \frac{x_1^n-x_2^n}{\sqrt{5}} = 0,1,1,2,\,3,\,5,\,8,\,13,\dots\end{aligned}

(The starting index is n = 0.)  Expanding powers of the golden ratio, then for n > 0,

\begin{aligned} & {x_1}^n = \Big(\frac{1+\sqrt{5}}{2}\Big)^n = \frac{L_n+F_n\sqrt{5}}{2}\end{aligned}

We’ll see this can be generalized to powers of the tribonacci constant.

II. Tribonacci numbers

These are a generalization of the Fibonacci numbers, being,

$t_n = t_{n-1}+t_{n-2}+t_{n-3}$

Pin-Yen Lin has a nice paper involving these numbers.  First, define the following three sequences with this recurrence, but with different initial values,

\begin{aligned}S_n &=0,0,1,1,2,4,7,13,24,\dots\\U_n &=0,3,2,5,10,17,32,49,\dots\\V_n &=3,1,3,7,11,21,39,71,\dots \end{aligned}

(The starting index as usual is n = 0.)  The first and the third are recognized by the OEIS, with the first being the tribonacci numbers.  The limiting ratio for all three is the tribonacci constant, T, the real root of,

$x^3=x^2+x+1$

or,

$T = \frac{1}{3}+\frac{1}{3}(19+3\sqrt{33})^{1/3}+\frac{1}{3}(19-3\sqrt{33})^{1/3}$

I’ve already written about the tribonacci constant before.  But I want to include how Lin found that powers of x can be expressed in terms of those three sequences. Define,

$a =\sqrt[3]{19+3\sqrt{33}}$

$b =\sqrt[3]{19-3\sqrt{33}}$

then, similar to the golden ratio,

$T^n = \frac{1}{9}(a^2+b^2)\,S_n+\frac{1}{9}(a+b)\,U_n+\frac{1}{3}\,V_n$

Hence, starting with = 1,

$T = \frac{0}{9}(a^2+b^2)+\frac{3}{9}(a+b)+\frac{1}{3}$

$T^2 = \frac{1}{9}(a^2+b^2)+\frac{2}{9}(a+b)+\frac{3}{3}$

$T^3 = \frac{1}{9}(a^2+b^2)+\frac{5}{9}(a+b)+\frac{7}{3}$

and so on.  Interesting, isn’t it, that powers of the tribonacci constant can be expressed in this manner.

There is a primality test regarding Lucas numbers: if n is a prime then $L_n-1$ is divisible by n.  For example $L_5 = 11$, minus 1, is divisible by 5.  However there are Lucas pseudoprimes, composite numbers that pass this test, with the smallest being n = 705.
The third tribonacci sequence can be formed analogously to the Lucas numbers.  Given the three roots $x_1, x_2, x_3$ of,
$x^3=x^2+x+1$
$V_n = x_1^n+x_2^n+x_3^n = 3,1,3,7,11,21,39,71,\dots$
I notice that likewise, if n is prime, then $V_n-1$ is divisible by n.  But there are also tribonacci-like pseudoprimes.  The smallest is n = 182.  Steven Stadnicki was nice enough to compute the first 36.  It turns out they are relatively rarer, as there are only 21 less than $10^8$, while there are  852 Lucas pseudoprimes in the same range.