Closed expression for the sum $\sum _{k=1}^{\infty } \sin ^2\left(\frac{x}{3^k}\right)$?

Question

Studying and proving the interesting formula 1.439.2 of Gradshteyn/Ryzhik,

$$\frac{\sin (x)}{x}=\prod _{k=1}^{\infty } \left(1-\frac{4}{3} \sin^2 \left(\frac{x}{3^k}\right)\right)\tag{GR 1.439.2}$$

I came up with the following related question:

Define

$$f(x) = \sum _{k=1}^{\infty } \sin ^2\left(\frac{x}{3^k}\right)\tag{1}$$

Question 1: is there a closed expression for $f$ in terms of known functions?

Question 2: what are the properties of $f$ (singularities, periodicity, self similarity, asymptotic behavour)?

Here is a plot of the function

For the range up to $x=1000$ the plot is

It shows heavy oscillation about some weakly exhibited trend. This trend should be specified in the answer to question 2.

Generalizations

One obvious generalization is to take other powers $p$ of $\sin$ under the sum

$$f_p(x) = \sum _{k=1}^{\infty } \sin^p \left(\frac{x}{3^k}\right)\tag{2}$$

Another one replaces $3$ by a parameter $a$.

$$f_p(x,a) = \sum _{k=1}^{\infty } \sin^p \left(\frac{x}{a^k}\right)\tag{3}$$

Answer 1

Introduction

While question (1) most probably has a negative answer, i.e. the sum is neither itself a known function not is it expressible as a (finite) combination of known functions, there are some interesting results related to question (2) which I am going to describe in this solution.

We shall discuss the infinite sum

$$f(z,a,p) = \sum _{k=1}^{\infty } \sin ^p\left(\frac{z}{a^k}\right)\tag{1}$$

where $a \gt 1$, $p = 1, 2, 3, ...$ and $z$ is the argument which can be a complex number.

Basic properties

Unless stated otherwise we consider the analytic properties of $f$ as a function of the variable $x$ for given parameters $a$ (real, $\gt 1$) and $p$ (positive integer). $x$ can be a complex number.

1. The radius of convergence of the sum $f$ is infinite. Hence $f$ is an entire transcendental function, i.e. is has no singularities in the complex x-plane except for $x=\infty$. It admits a power series representation which, about $x=0$, is given in the section Taylor series.

2. Zeroes of f as a function of x
   For even $p$ there is only the trivial zero of $f$ at $x=0$; for odd $p$ there are real zeroes, even infinitely many (otherwise $f$ would be a polynomial). I conjecture that there are no zeroes beyond the real x-axis.

3. Periodicity $f$ is not periodic, i.e. there is no $d>0$ for which $f(x) = f(x+d)$ for all $x$.

4. Monotonity, boundedness $f$ is not monotonous. It is finite for any finite $x$; $f$ is not bounded for even $p \gt 0$ (see section patterns, approximations and asymptotic behaviour); it remains open if $f$ is bounded for odd $p$.

5. Functional equation
   As $f(x,a,p)$ is of the general form $f(x) = \sum_{k=1}^\infty g(\frac{x}{a^k})$ (with $g(z)=\sin(z)^p$) it obeys the functional equation

$$f(a x) = f(x) + g(x)\tag{2}$$

Patterns, approximations and asymptotic behaviour

Let us first study the patterns exhibited in the plots of our function.

We shall start with the cases $p=1$ ($f(x)$ antisymmetric) and $p=2$ ($f(x)$ symmetric).

p = 1

Plot 1a:

Plot 1b:

Plot 1c:

p = 2

Plot 2a:

Plot 2b:

Plot 2c:

The overall pattern of the graphs of the function can be described as follows

1. "ups" and "downs" follow each other regularly, i.e. there are no "shoulders".

2. there is a "basic pattern" which for "p=1" looks like an inverted "w" follwed by a smaller inverted "v", and for "p=2" an inverted "w" with the central "v" flipped upwards. This basic pattern is repeated indefinitely at different "altitudes".

3. on a larger scale we see heavy fluctuations about a trend

In attempting to describe the trend I found it useful to consider the arithmetic mean of the function between $0$ and $x$ instead of the function itself.

Integrating the summand for $p=1$ and $p=2$ gives, respectively,

$$f_{m}(1)=\frac{1}{x}\int_0^x \sin \left(\frac{t}{3^k}\right) \, dt = \frac{3^k}{x} \left(1-\cos \left(3^{-k} x\right)\right)$$

$$f_{m}(2)=\frac{1}{x}\int_0^x \sin ^2\left(\frac{t}{3^k}\right) \, dt=\frac{1}{4 x} \left(2 x-3^k \sin \left(2\ 3^{-k} x\right)\right)$$

Now, in addition, we consider another "smooting" method, we replace the k-sum by an integral, and consider two different expressions depending on the starting value of $k$ in the integral.

This leads to expresions designated $f_{m}(p,k_{0})$ where $m$ stands for "mean", "p" is the power $p$ and $k_{0}$ is the lower limit of the k-integral:

$$f_{m}(1,0)=\int_0^{\infty } f_{m}(1) \, dk=\frac{x \text{Si}(x)+\cos (x)-1}{x\log (3)}$$

$$f_{m}(1,1)=\int_1^{\infty } f_{m}(1) \, dk=\frac{x \text{Si}\left(\frac{x}{3}\right)+3 \cos \left(\frac{x}{3}\right)-3}{x \log (3)}$$

$$f_{m}(2,0)=\int_0^{\infty } f_{m}(2) \, dk=\frac{2 x (-\text{Ci}(2 x)+\log (2 x)+\gamma -1)+\sin (2 x)}{x \log (81)}$$

$$f_{m}(2,1)=\int_1^{\infty } f_{m}(2) \, dk=\frac{2 x \left(-\text{Ci}\left(\frac{2 x}{3}\right)+\log \left(\frac{2 x}{3}\right)+\gamma -1\right)+3 \sin \left(\frac{2 x}{3}\right)}{x \log (81)}$$

We show now the plots for $p=1$ of the arithmetic mean of $f$ and the approximation replacing the k-sum by the integral for different ranges of $x$:

Plot 1ma:

Plot 1mb:

Plot 1mc:

Conclusions for $p=1$

1. The mean oscillates in a peculiar "sawtooth" manner about $f_{m}(1,1)$ which for large $x$ approaches $\pi / \log(9)$

2. There is a kind of self-similarity in the mean: the pattern roughly repeats in different scales. This can be traced back to the functionl euqation (2).

For $p=2$ we also show the two versions of the k-sums

Conclusions for $p=2$

1. the most prominent feature is that $f_{m}$ is enclosed between the two approximations differing in the lower limit of the k-integral

2. the self similarity of $p=1$ is also observed here, but less pronounced

Taylor series

Inspired by the idea of Yuriy S. I have calculated (with a different method) the Taylor series of $f$ for general powers $p$.

After inserting the series expansion of $\sin$, and doing the geometric series in $k$ the main task left is to calculate the multinomial sums of the type

$$\sum_{n_i\ge 0,\;r=\sum _{i=1}^p n_i} \frac{1}{\prod _{i=1}^p (2 n_i+1)!}\tag{t1}$$

and the result can be written in the form

$$f_T(z,a,p) = \sum _{r=0}^{\infty } \frac{(-1)^r z^{p+2 r} d(p,r)}{(p+2 r)! \left(a^{p+2 r}-1\right)}\tag{t2}$$

where the first 9 coefficients $d(p,r)$ in the format $\{p,d(p,r)\}$ are

$$ \begin{array}{l} \{1,1\} \\ \left\{2,2^{2 r+1}\right\} \\ \left\{3,\frac{3}{4} \left(9^{r+1}-1\right)\right\} \\ \left\{4,2^{2 r+3} \left(4^{r+1}-1\right)\right\} \\ \left\{5,\frac{5}{16} \left(-3^{2 r+5}+25^{r+2}+2\right)\right\} \\ \left\{6,-3 2^{2 r+1} \left(2^{2 r+7}-3^{2 r+5}-5\right)\right\} \\ \left\{7,\frac{7}{64} \left(-5^{2 r+7}+9^{r+4}+49^{r+3}-5\right)\right\} \\ \left\{8,4^{r+2} \left(7\ 2^{2 r+7}+2^{4 r+13}-9^{r+4}-7\right)\right\} \\ \left\{9,\frac{9}{256} \left(4\ 5^{2 r+9}-7^{2 r+9}-28\ 9^{r+4}+81^{r+4}+14\right)\right\} \\ \end{array}\tag{t3} $$

This result provokes the desire to have the general formula for $d(p,r)$.

Up to now I could only see the factor $\frac{p}{2^{p-1}}$ for even $p$ but I was not able to identify the terms for odd $p$ and even less the structure of the "polynomials" in powers of some small numbers.

What I did find out up to now for the "polynomials" is

1. for even $p$ there is a term $p^{2r+p-1}$
2. the number of summands is $\left\lceil \frac{p}{2}\right\rceil$

Heureka! I found simple expressions after picking up a suggestion I found browsing through OEIS: the generating functions of the coefficients.

Defining

$$g(p, t) =\frac{1}{p!} \sum_{r=0}^\infty d(p,r) t^r\tag{t4}$$

explicit formulae for these sums are found easily, and we write down the expressions for $1/g$:

$$\frac{1}{g(p_{even},t)}=\prod _{m=1}^{\frac{p}{2}} \left(1-(2 m)^2 t\right)\\=(1-4t)(1-16t)(1-36t) ...(1-p^2 t)\\=(2 i)^p t^{p/2} \left(1-\frac{1}{2 \sqrt{t}}\right)_{\frac{p}{2}} \left(1+\frac{1}{2 \sqrt{t}}\right)_{\frac{p}{2}}\tag{t5e}$$ $$\frac{1}{g(p_{odd}, t)}=\prod _{m=0}^{\frac{p-1}{2}} \left(1-(2 m+1)^2 t\right)\\= (1-t)(1-9t)(1-25t) ... (1-p^2 t)\\=(2 i)^{p-1} (1-t) t^{\frac{p-1}{2}} \left(\frac{3}{2}-\frac{1}{2 \sqrt{t}}\right)_{\frac{p-1}{2}} \left(\frac{1}{2} \left(3+\frac{1}{\sqrt{t}}\right)\right)_{\frac{p-1}{2}}\tag{t5o}$$

Here we have also expressed the finite products via the Pochhammer symbol defined as $(x)_k = x(x+1).. (x+k-1)$.

Answer 2

While I doubt a closed form exists, we can rewrite the series in another form, specifically (I consider another generalization):

$$f(x)=\sum_{k=1}^\infty \sin^2 \frac{x}{a^k}=\frac{1}{2} \sum_{n=1}^\infty \frac{(-1)^{n+1} (2x)^{2n}}{(2n)!~(a^{2n}-1)}$$

To find this form, consider the derivative:

$$f'(x)=\sum_{k=1}^\infty \frac{2}{a^k} \sin \frac{x}{a^k} \cos \frac{x}{a^k}=\sum_{k=1}^\infty \frac{1}{a^k} \sin \frac{2x}{a^k}=$$

Using the Taylor series for $\sin$ we obtain:

$$=\sum_{k=1}^\infty \sum_{n=0}^\infty \frac{1}{a^k} \frac{(-1)^n (2x)^{2n+1}}{(2n+1)! ~a^{k(2n+1)}}=\sum_{n=0}^\infty \frac{(-1)^n (2x)^{2n+1}}{(2n+1)! } \sum_{k=1}^\infty \frac{1}{a^{k(2n+2)}}=$$

Using the geometric series:

$$=\sum_{n=0}^\infty \frac{(-1)^n (2x)^{2n+1}}{(2n+1)!~(a^{2n+2}-1) }$$

We found the derivative, now we integrate by term, noting to fix the constant of integration, that:

$$f(0)=0$$

Since:

$$\int x^{2n+1}=\frac{x^{2n+2}}{2n+2}$$

We have:

$$f(x)=\frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n (2x)^{2n+2}}{(2n+2)!~(a^{2n+2}-1) }=\frac{1}{2} \sum_{n=1}^\infty \frac{(-1)^{n+1} (2x)^{2n}}{(2n)!~(a^{2n}-1)}$$

This form is more useful, because it represents the Taylor series for the function $f(x)$.

Numerically, both the series converge very fast (provided $|a|>1$ of course), so from computational standpoint one is as good as the other, though the series derived here has a rational form.

Answer 3

Another way to find the Yuriy S' series, just for fun. Let $\mathfrak{M}\left(*,s\right)$ the Mellin transform. Using the identity $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\,s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\mathfrak{M}\left(g\left(x\right),s\right) $$ we have $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\sin^{2}\left(\frac{x}{3^{k}}\right),\,s\right)=\sum_{k\geq1}3^{ks}\mathfrak{M}\left(\sin^{2}\left(x\right),s\right)$$ and since $$\mathfrak{M}\left(\sin^{2}\left(x\right),s\right)=-\Gamma\left(s\right)\cos\left(\frac{\pi s}{2}\right)2^{-1-s}$$ we get $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\,s\right)=\Gamma\left(s\right)\cos\left(\frac{\pi s}{2}\right)\frac{2^{-1-s}}{3^{-s}-1},\,-1<\mathrm{Re}\left(s\right)<0.$$ So, inverting, we get $$\sum_{k\geq1}\sin^{2}\left(\frac{x}{3^{k}}\right)=\frac{1}{2\pi i}\int_{-1/2-i\infty}^{-1/2+i\infty}\Gamma\left(s\right)\cos\left(\frac{\pi s}{2}\right)\frac{2^{-1-s}}{3^{-s}-1}x^{-s}ds.$$ It is not difficult to prove that, if $\mathrm{Re}\left(s\right)\rightarrow-\infty$, then the integral goes to $0$ so, observing that the integrand has poles at $s=-2n,\,n\in\mathbb{N}^{*}$, by the residue theorem, we get $$\sum_{k\geq1}\sin^{2}\left(\frac{x}{3^{k}}\right)=\color{red}{\frac{1}{2}\sum_{n\geq1}\frac{\left(-1\right)^{n+1}\left(2x\right)^{2n}}{\left(3^{2n}-1\right)\left(2n\right)!}}$$ as wanted. This method can easily generalized to a generic $\left|a\right|>1.$ Another observation: taking $$x=\frac{a}{2}$$ the series in red taking a form of a Lambert series.