Posts Tagged ‘trigonometry’

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.

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On Bailey and Crandall’s unusual sum

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.

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.