Curiosities of the function $Q(x)=\sum_{n=1}^\infty \frac{P_n(x)}{n(2n+1)}$ where $P_n(x)$ is a sequence of all polynomials with unit coefficients

Question

Backround:

I have been studying the peculiar function $$Q(x)=\sum_{n=1}^\infty \frac{P_n(x)}{n(2n+1)}$$ where $P_n(x)$ is the set of all polynomials with unit coefficients, defined by the binary expansion of $n$. For example $$n=57=\color{red}{111001}_2\iff P_n(x)=\color{red}{1}x^5+\color{red}{1}x^4+\color{red}{1}x^3+\color{red}{0}x^2+\color{red}{0}x^1+\color{red}{1}x^0.$$

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Some properties:

This function seems intimately tied with the Euler-Mascheroni constant $\gamma$, $\pi$, and the natural logarithm. For instance, with some algebraic manipulation of some of the "easier" values of $n$, we can find that $$Q(0)=\sum_{n=0}^\infty \frac{1}{(2n+1)(2(2n+1)+1)}=\frac{1}{4}(\pi-\ln(4)),$$ as calculated by WolframAlpha, and $$Q(1)=\sum_{n=1}^\infty\frac{H(n)}{n(2n+1)}=\ln\left(\frac{4}{\pi}\right)+\gamma,$$ where $H(n)$ is the Hamming weight of the binary expansion of $n$, proven by combining some of the series expansions of $\gamma$.

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Questions:

Some questions arose while playing around with this function.

Can we find other "interesting values" of $Q$, for $Q(-1)$, $Q(2)$, $Q\left(\frac{1}{2}\right)$, for example?

Can we find a closed form of $Q$, in terms of other elementary / transcendental functions? Or at the very least,

the coefficients of its power series?

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My work on its power series:

When it comes to its power series

$$Q(x)=\sum_{n=0}^\infty c_n x^n,$$

using properties of binary, we can deduce that. $$c_n = \sum_{k=0}^\infty \sum_{m = 2^n} ^ {2^{n + 1} - 1} \frac{1}{(2^{n + 1} k + m)(2(2^{n + 1} k + m) + 1)}.$$ Plugging in values of $n=0,1,2$ into WolframAlpha, we find that $$c_0 = \frac{1}{4}(\pi-2\ln(2))\approx 0\approx 0.43883,$$ $$c_1 = \frac{1}{8}(\pi (2\sqrt{2}-1) - 6\ln 2)\approx 0.19816,$$ $$c_2 = \frac{1}{16}\left(\frac{\pi\left(-6-5\sqrt{2}+8\sqrt{2+\sqrt{2}}+4\sqrt{2(2+\sqrt{2})}\right)}{2+\sqrt{2}}-14\ln(2)\right)\approx 0.09301.$$ However I'm not sure what methods it used to calculate such, and if they can be used to generalize a closed form for any $c_n$. At the very least, my pattern recognition sees that $c_n$ is of the form $2^{-(n+2)}(A_n\pi - B_n \ln 2)$, with $A_n$ an algebraic number and $B_n$ a natural number.

Any and all insight would be greatly appreciated.

Answer 1

$\textbf{The coefficients:}$

It turns out that your conjecture about the closed form of $c_n$ is correct. Here is a derivation of an explicit closed form for $c_n$. As you have seen, the coefficients are given by

$$c_n=\sum_{k=1}^\infty\frac{a_{n,k}}{k(2k+1)}$$

where for each $n\geq 1$, $(a_{n,k})_k$ is a $2^{n+1}$-periodic sequence given by $a_{n,k}=\left\lfloor 2^{-n}k\right\rfloor\mod 2$. Letting, $\xi_n=e^{i\pi/2^n}$, we may use the discrete Fourier transform to write

$$a_{n,k}=\frac{1}{2^{n+1}}\sum_{m=0}^{2^{n+1}-1}B_{n,m}\xi_n^{mk}$$

where

\begin{equation} B_{n,m}:=\sum_{j=2^n}^{2^{n+1}-1}\xi_n^{-jm}= \begin{cases} 2^n&\text{if }m=0\\ -1+i\cot\left(\frac{m\pi}{2^{n+1}}\right)&\text{if $m$ is odd}\\ 0&\text{otherwise} \end{cases} \end{equation} We now make use of the following useful power series:

$$f(x)=\sum_{k=1}^\infty\frac{x^k}{k(2k+1)}=2-\log(1-x)-\frac{2}{\sqrt{x}}\tanh^{-1}(\sqrt{x})$$

for $|x|\leq 1$ and $x\ne 1$. We also have $f(1)=2-2\log(2)$. Through suffering and lots of miraculous cancellations, we may evaluate

\begin{equation} \mathfrak{R}[B_{n,m}f(\xi_n^m)]=-2+\log\left(2\sin\left(\frac{m\pi}{2^{n+1}}\right)\right)+\frac{m\pi}{2^{n+1}}\cot\left(\frac{m\pi}{2^{n+1}}\right)+\frac{\pi}{2}\tan\left(\frac{m\pi}{2^{n+2}}\right) \end{equation} for odd $m$. We therefore have that for $n\geq 1$,

\begin{equation} \begin{split} c_n&=\sum_{k=1}^\infty\frac{a_{n,k}}{k(2k+1)}\\ &=\frac{1}{2^{n+1}}\sum_{k=1}^\infty\frac{1}{k(2k+1)}\sum_{m=0}^{2^{n+1}-1}B_{n,m}\xi_n^{mk}\\ &=\frac{1}{2^{n+1}}\sum_{m=0}^{2^{n+1}-1}B_{n,m}\sum_{k=1}^{\infty}\frac{\xi_n^{mk}}{k(2k+1)}\\ &=\frac{1}{2^{n+1}}\sum_{m=0}^{2^{n+1}-1}B_{n,m}f(\xi_n^m)\\ &=\frac{B_{n,0}f(1)}{2^{n+1}}+\frac{1}{2^{n+1}}\sum_{m=1}^{2^{n+1}-1}\mathfrak{R}[B_{n,m}f(\xi_n^m)]\\ &=1-\log(2)+\frac{1}{2^n}\sum_{\substack{m=1\\\text{odd}}}^{2^n}\mathfrak{R}[B_{n,m}f(\xi_n^m)]\\ &=-\log(2)+\frac{1}{2^n}\sum_{\substack{m=1\\\text{odd}}}^{2^n}\left[\log\left(2\sin\left(\frac{m\pi}{2^{n+1}}\right)\right)+\frac{m\pi}{2^{n+1}}\cot\left(\frac{m\pi}{2^{n+1}}\right)+\frac{\pi}{2}\tan\left(\frac{m\pi}{2^{n+2}}\right)\right]\\ &=\left[\frac{1}{2^{n+1}}-1\right]\log(2)+\frac{\pi}{2^{n+1}}\sum_{\substack{m=1\\\text{odd}}}^{2^n}\left[\frac{m}{2^n}\cot\left(\frac{m\pi}{2^{n+1}}\right)+\tan\left(\frac{m\pi}{2^{n+2}}\right)\right]\\ \end{split} \end{equation} where the last equality is due to the finite product identity $$\sqrt{2}=\prod_{\substack{m=1\\\text{odd}}}^{2\ell}2\sin\left(\frac{m\pi}{4\ell}\right)$$

Finally, $c_0=\frac{\pi}{4}-\frac{1}{2}\log(2)$, can be calculated separately using the $5$-th line of the above chain of equalities (line $6$ uses the fact that $n\neq 0$, but line $5$ is still valid for $n=0$).

$\textbf{The analytic continuation:}$

I've been looking into producing a closed form for the analytic continuation of $Q(x)$. Though one can use the closed form for $c_n$ (along with the Taylor/Laurent expansions of tan/cot) to produce a closed form for $Q(x)$, this looks to be an unenlightening monster. However, due to a result of Gary, provided in his answer, we can cheat.

Letting $(a;q)_n$ be the $q$-Pochhammer symbol, we see that

$$(x/2;1/2)_\infty=\prod_{k=1}^\infty\left(1-\frac{x}{2^k}\right)=\sum_{n=0}^\infty\frac{(-1)^nx^n}{2^{\frac{n(n+1)}{2}}(1/2;1/2)_n}$$

has roots precisely at $2^n$ for $n\geq 1$. Gary's result tells us that $Q(x)$ admits a meromorphic continuation with simple poles at $2^n$ for $n\geq 1$, so

$$(x/2;1/2)_\infty Q(x)=\sum_{n=0}^\infty \left[\sum_{k=0}^n\frac{(-1)^kc_{n-k}}{2^{\frac{k(k+1)}{2}}(1/2;1/2)_k}\right]x^n$$

is an entire function, and the above Taylor series converges everywhere. We therefore have the closed form

$$Q(x)=\frac{1}{(x/2;1/2)_\infty}\sum_{n=0}^\infty \left[\sum_{k=0}^n\frac{(-1)^kc_{n-k}}{2^{\frac{k(k+1)}{2}}(1/2;1/2)_k}\right]x^n$$

which allows us to calculate $Q(x)$ everywhere it's defined.

$\textbf{Final thoughts:}$

Now that we have a way to evaluate the analytic continuation of $Q(x)$, I would be interested to know about the existence/structure of the roots of $Q(x)$, as well as the order of $(x/2;1/2)_\infty Q(x)$. Perhaps this is wishful thinking, but if we're very lucky, there might exist a clean Weierstrass factorization of $(x/2;1/2)_\infty Q(x)$, or a related function.

Answer 2

Elaboration on the formula of Claude Leibovici. If we indeed start the summation at $k=0$ in the formulae for $c_n$, we obtain $$ c_n = \sum\limits_{k = 1}^\infty {( - 1)^{k + 1} \left[ {\psi\! \left( {2^n k + \frac{1}{2}} \right) - \psi (2^n k)} \right]} $$ where $\psi$ is the digamma function. From the Laplace transform representation of $\psi(x+a)$ (cf. Lemma $2.2$ here), one can show that $$ \psi\! \left( {x + \frac{1}{2}} \right) - \psi (x) = \frac{1}{{2x}} + \int_0^{ + \infty } {\tanh \left( {\frac{t}{2}} \right){\rm e}^{ - 2xt} {\rm d}t} , $$ which leads to the rapidly converging integral formula $$ c_n = \frac{{\log 2}}{{2^{n + 1} }} + \int_0^{ + \infty } {\tanh \left( {\frac{t}{2}} \right)\frac{{\rm d}t}{{{\rm e}^{2^{n + 1} t} + 1}}} . $$ This should allow you to at least numerically compute the coefficients. We can also use this integral to obtain some asymptotic inequalities for large $n$. It is known (see, e.g., Theorem $2.2$ of this paper) that for any $N\geq 1$ and $t>0$, $$ \tanh \left( {\frac{t}{2}} \right) = \sum\limits_{k = 1}^{N - 1} {\frac{{2(2^{2k} - 1)B_{2k} }}{{(2k)!}}t^{2k - 1} } + \theta _N (t)\frac{{2(2^{2N} - 1)B_{2N} }}{{(2N)!}}t^{2N - 1} , $$ with an appropriate $0<\theta _N (t)<1$ ($B_k$ denotes the $k$th Bernoulli number). Hence, using $(24.7.1)$ and the mean value theorem for improper integrals, we find \begin{align*} c_n = \frac{{\log 2}}{{2^{n + 1} }} & + \sum\limits_{k = 1}^{N - 1} {( - 1)^{k + 1} \frac{{(2^{2k} - 1)(2^{2k} - 2 )\pi ^{2k} B_{2k}^2 }}{{2k \cdot (2k)!}}\frac{1}{{(2^{n + 1} )^{2k} }}} \\ & + \Theta _N (n)( - 1)^{N + 1} \frac{{(2^{2N} - 1)(2^{2N}-2 )\pi ^{2N} B_{2N}^2 }}{{2N \cdot (2N)!}}\frac{1}{{(2^{n + 1} )^{2N} }}. \end{align*} for any $n,\,N\geq 1$ and with an appropriate $0<\Theta _N (n)<1$. The least term occurs around $N \approx \pi 2^n$ providing an absolute error of size $$ \mathcal{O}\!\left( 2^{ - n/2} \exp ( - 2^{n + 1} \pi ) \right). $$ The procedure outlined in the comments suggests that $Q(x)$ extends meromorphically to the complex $x$-plane with simple poles at the positive even powers of $2$ and $$ \mathop {{\mathop{\rm Res}\nolimits} }\limits_{x = 2} Q(x) = -\log 2,\qquad \mathop {{\mathop{\rm Res}\nolimits} }\limits_{x = 2^{2k} } Q(x) = ( - 1)^{k} \frac{{(2^{2k} - 1)(2^{2k} - 2)\pi ^{2k} B_{2k}^2 }}{{2k \cdot (2k)!}} $$ for $k\ge 1$.

Answer 3

This probably won't help much, (didn't want to make it an "answer", yet it wouldn't fit in a comment) but by noting that: $$\begin{split}\frac{1}{(2^{n + 1} k + m)(2(2^{n + 1} k + m) + 1)} &= 2\cdot\left(\frac{1}{2(2^{n + 1} k + m)}-\frac{1}{2(2^{n + 1} k + m) + 1}\right)\\ &= 2\cdot\left(\frac{(-1)^{2m}}{2^{n + 2} k + 2m}+\frac{(-1)^{2m+1}}{2^{n + 2} k + 2m + 1}\right)\end{split}$$ you can first rewrite each $c_n$ as: $$c_n = 2\cdot\sum_{k=1}^{\infty} \sum_{m'=2^{n+1}}^{2^{n+2}-1} \frac{(-1)^{m'}}{2^{n + 2} k + m'}$$ and then by observing that $(-1)^{m'} = (-1)^{2^{n+2}k+m'}$, you can re-rewrite $c_n$ as: $$c_n = 2 \cdot \sum_{s=1}^{\infty} \frac{f_n(s)}{s}$$ where: $f_n(s) := \begin{cases} (-1)^s \,\,&\text{if }\, s = 2^{n+2}k + m',\,\, k \in \mathbb{N}^*,\, m' \in \{2^{n+1},2^{n+2}-1\}\\ 0 \,\, &\text{otherwise}\end{cases}$, thus making each $\frac{1}{2}c_n$ a subseries of the alternating harmonic series $\displaystyle S := \sum_{s \geq 1} \frac{(-1)^s}{s}$.
This should explain at least intuitively the correlation with $\pi$ and $\ln(2)$ since $S$ converges to $-\ln(2)$ and the series $\displaystyle\sum_{s \geq 0} \frac{(-1)^{s}}{2s+1}$, which is kinda "close" to $S$, converges to $\displaystyle\frac{\pi}{4}$, but past that I'm afraid I can't really help.

There is however one other thing that I couldn't help but notice while writing this: is it normal that your sums begin with $k = 1$ and not $k = 0$, and $n = 1$ instead of $n = 0$ in $Q(0)$? If it is normal, sorry for bringing that up, otherwise you'd also want to change things accordingly in my post.

Answer 4

Not very helpful.

Using the same approach as @Bruno B $$c_n= \sum_{k=1}^\infty \sum_{m = 2^n} ^ {2^{n + 1} - 1}\Bigg[\frac{1}{2^{n+1}\,k+m} -\frac{2}{ 2^{n+2}\,k+2 m+1}\Bigg]$$ leads to

$$c_n=\sum_{k=1}^\infty\Bigg[\psi \left(2^n (2 k+1)+\frac{1}{2}\right)-\psi \left(2^n (2 k+1)\right) \Bigg]$$ $$-\sum_{k=1}^\infty\Bigg[\psi \left(2^{n+1} (k+1)+\frac{1}{2}\right)-\psi \left(2^{n+1} (k+1)\right)\Bigg]$$

They all write $$c_n=\alpha_n+\beta_n\,\log(2)+\gamma_n\, \pi$$ where $(\alpha_n,\beta_n)$ are rational and $\gamma_n$ irrational.

Computing $c_3$ seems to require an incredibly long time

To obtain the numbers you posted, I suppose that the summation starts at $k=0$ and not $k=1$.