Is there a closed form expression for the following formal power series $$\large\sum_{k\ge 0}\dfrac{z^{2^k}}{1+z^{2^k}}$$ Till now I have tried in vain finding any progress to simplify this expression. Please help.

  • $\begingroup$ what is the limit of $k$??? infinity or not??? $\endgroup$ – E.H.E Sep 16 '15 at 20:39
  • 2
    $\begingroup$ I thought $k\ge 0$ implies $0\le k<\infty$. Is that not standard? $\endgroup$ – Samrat Mukhopadhyay Sep 16 '15 at 20:41
  • $\begingroup$ @SamratMukhopadhyay It does. Your notation is fine. $\endgroup$ – Mark Viola Sep 16 '15 at 20:43
  • $\begingroup$ Hmm. Thanks @Dr.MV. $\endgroup$ – Samrat Mukhopadhyay Sep 16 '15 at 20:43
  • $\begingroup$ The equivalent problem is to evaluate the series $$\sum_{k=0}^\infty \frac{1}{1+x^k}$$for $|x|>1$. WA gives a result that depends on the $x$-digamma function $\psi_x$. So, if you consider that closed-form enough, then you have a way forward. $\endgroup$ – Mark Viola Sep 16 '15 at 20:52

If we set: $$f(z)=\sum_{k\geq 0}\frac{z^{2^k}}{1+z^{2^k}}\tag{1}$$ we have: $$ f(z^2)=\sum_{k\geq 0}\frac{z^{2\cdot 2^k}}{1+z^{2\cdot 2^k}}=\sum_{k\geq 0}\frac{z^{2^{k+1}}}{1+z^{2^{k+1}}}=f(z)-\frac{z}{1+z}.\tag{2}$$ If we further assume: $$ f(z)=\sum_{m\geq 1} a_m\,z^{m}\tag{3} $$ $(2)$ gives: $$ a_k = a_{2k}+1,\qquad 0 = a_{2k+1}-1\tag{4} $$ hence:

$$ f(z) = \sum_{m\geq 1}\left(1-\nu_2(m)\right) z^m\tag{5} $$

where $$\nu_2(m)=\max\{r\in\mathbb{N}:2^r\mid m\}.$$

  • $\begingroup$ What function is $v_2$ representing here? $\endgroup$ – Brevan Ellefsen Sep 16 '15 at 20:56
  • $\begingroup$ @BrevanEllefsen: the $2$-adic height, now written. $\endgroup$ – Jack D'Aurizio Sep 16 '15 at 20:57
  • $\begingroup$ For short, the coefficient of $z^m$ in the Taylor expansion of $f(z)$ just depends on the number of trailing zeroes in the binary representation of $m$. $\endgroup$ – Jack D'Aurizio Sep 16 '15 at 20:59
  • $\begingroup$ Nice simplification! $\endgroup$ – Samrat Mukhopadhyay Sep 16 '15 at 21:17

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