# 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

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 $$n=57=\color{red}{111001}_2\iff P_n(x)=\color{red}{1}x^5+\color{red}{1}x^4+\color{red}{1}x^3+\color{red}{0}x^2+\color{red}{0}x^1+\color{red}{1}x^0.$$

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 $$\gamma$$.

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?

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.

• Doesn't $Q(2)$ trivially diverge since $P_n(2)=n$ by definition, so $Q(2)=\sum_{n\geq 1}\frac{1}{2n+1}$? Oct 27, 2022 at 19:54
• @C-RAM I realized this too after posting, although $x>2$ might have some analytic continuation related to the 2-adic numbers, given its heavy association with binary. Oct 27, 2022 at 22:23
• @C-RAM That's a good question. For a while I've been interested in the connection between binary expansions and unit-coefficient polynomials, so that lead me to creating some series of the form $\sum r_n \cdot P_n(x)$, so it was a matter of picking an appropriate $r_n$ which behaved "nicely". I ended up choosing $r_n = \frac{1}{n(2n+1)}$ for the reason you suggested however, indeed. Oct 29, 2022 at 20:31
• Also, (and this is just an opinion) you might wnat to consider being more explicit about the fact that $P_n(x)$ are related to the binary expansion of $n$ in your title. Something like "Curiosities of the function $Q(x)=\sum_{n=1}^\infty \frac{P_n(x)}{n(2n+1)}$ where $P_n(x)$ are polynomials related to the binary expansion of $n$." If you disagree, you should at least consider changing "the set of all polynomials" to "a sequence of all polynomials", which is a bit more correct in this context. Oct 30, 2022 at 21:01
• @C-RAM Thanks for reminding me! I've hopefully fixed the indexing. As for the title, although I would prefer to have changed it to what you suggested about binary, It's already right around the character limit. I did have room to change it to "a sequence of all polynomials", however. Nov 1, 2022 at 1:09

$$\textbf{The coefficients:}$$

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

$$$$B_{n,m}:=\sum_{j=2^n}^{2^{n+1}-1}\xi_n^{-jm}= \begin{cases} 2^n&\text{if }m=0\\ -1+i\cot\left(\frac{m\pi}{2^{n+1}}\right)&\text{if m is odd}\\ 0&\text{otherwise} \end{cases}$$$$ 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

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

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

• I wish I could +2 or +3 haha. Impressing that a closed form could be found ""easily""! Oct 29, 2022 at 13:46
• Ah this is a fantastic deduction. I too am impressed with the result (and the methods / identities used here). When the activity on this question has died down I'll more-than-likely accept this as answer, but as each of these answers so far have been of great insight, I'll also delegate some bounties for the sake of fairness and my appreciation. Oct 29, 2022 at 20:25
• Wow! I really like how yours and Gary's answers came together so elegantly. The analytic continuation is quite the achievement. I must say that I am happy how there has been more interest in this function $Q(x)$ than I could have expected. This pretty much answers all my big questions, but be my guest if you want to continue exploring it, as I know I will certainly be playing around with your solution! Nov 7, 2022 at 20:21
• @Graviton Many thanks for the bounty! I am a big fan of weird functions, series, and integrals, and I particularly enjoyed working on this problem. Even though I authored a large part of the proofs for the results here, I'm just as surprised as you are that any such results were possible. I doubt I'll have any thing else to add, but if I do come up with a anything, I'll be sure to post an update. (PS: I greatly enjoyed your Leech/E8 Lattice collab.) Nov 9, 2022 at 6:09
• Nice to see that my formula could be used.
– Gary
Nov 12, 2022 at 4:24

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 $$k\ge 1$$.

• Very interesting analysis! So the convergence radius of $Q(x)$ is indeed precisely $2$... Oct 31, 2022 at 3:12
• @C-RAM Yes and we can subtract functions with singularities at $x=2$, $x=4$, etc. to continue analytically $Q(x)$ to larger and larger circles. For instance since $$c_n = \frac{{\log 2}}{2}\frac{1}{{2^n }} + \frac{{\pi ^2 }}{{96}}\frac{1}{{4^k }} + \mathcal{O}\!\left( {\frac{1}{{16^n }}} \right),$$ the function $$Q(x) - \frac{{\log 2}}{{2 - x}} - \frac{{\pi ^2 }}{{24}}\frac{1}{{4 - x}}$$ will be analytic in $|x|<16$.
– Gary
Oct 31, 2022 at 3:19
• I was just about to ask about that. Unfortunately, it seems that the asymptotics of $B_{2k}^2$ prevents us from taking the limit of this process to directly obtain a pole expansion of $Q(x)$. I wonder if something else could be done to give a closed form for the analytic continuation of $Q(x)$? Oct 31, 2022 at 3:48
• @C-RAM Good observation.
– Gary
Oct 31, 2022 at 3:52

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.

• Thanks for adding some insight! I'll look over this more carefully in a moment, but to briefly answer your question at the end - Edit: actually now I'm doubting myself so I'll have to double-check the indexing in $Q(0)$, but I can guarantee the value and general equation for $Q(x)$ is correct. Oct 27, 2022 at 22:26

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 $$k=1$$.