I was looking for ways to continue gap series, and it seemed to be that they could be continued outside of the boundary by simply turning $$f(x)= \sum_{n=0}^\infty x^{n^k}$$ into $$g(x) =- \sum_{n=1}^\infty x^{-n^k}$$ for odd values of k.

These two functions seem to agree at any angle that is a rational multiple of $\pi$. Plugging in $x = \cos\left(\frac{p}{q} \pi\right) + i\sin\left(\frac{p}{q} \pi\right)$ gives: $$f(x) =\sum_{n=0}^\infty x^{n^k} = \sum_{n=0}^\infty \cos\left(\frac{p}{q} \pi n^k\right) + i\sin\left(\frac{p}{q} \pi n^k\right)$$ $$g(x) =-\sum_{n=1}^\infty x^{-n^k} = \sum_{n=1}^\infty -\cos\left(\frac{p}{q} \pi n^k\right) + i\sin\left(\frac{p}{q} \pi n^k\right)$$ The imaginary parts are equal since $\sin(0)=0$, so both series are exactly the same. The imaginary part also appears to converges when we are at a rational multiple of $\pi$, and seems to agree with the the method of using Ceasero summation. For instance, at the angle $\frac{2}{7}\pi$, the function and its continuation looks like this: enter image description here Zooming in on the point $x = 1$: enter image description here

The red line is the value assigned by Cesàro summation for $x=1$. (Here is the link to the desmos graph if you would like to test out different angles: https://www.desmos.com/calculator/fkjjctmuqf )

Similar arguments give that the real part is equal to $\frac{1}{2}$ when it converges. Numerical testing also seems to suggest that all orders of the derivatives are also equal for $f(x)$ and $g(x)$ at rational multiples of $\pi$.

In general, its seems to be true that the analytical continuation of $\sum_{n=0}^\infty f(n)x^{g(n)}$ is $-\sum_{n=1}^\infty f(-n)x^{g(-n)}$, where I think $f(n)$ must be analytic for all n, and for $g(n)$, the highest power of $n^k$ must be odd (I'm less sure if this restriction is right).

Is the formula valid, for instance, for $\sum_{n=0}^\infty x^{n^3}$, and in general? I'm unsure if I can apply the Identity theorem here since the two functions aren't defined on the boundary unless I regularize the sums. Any help or insight on this problem would be appreciated!

  • $\begingroup$ You misspell Cesaro's name. $\endgroup$ Jun 16, 2021 at 16:05
  • $\begingroup$ You truncate your series at 305, for larger truncation values I get entirely different graphs. $\endgroup$ Jun 16, 2021 at 16:57
  • $\begingroup$ @მამუკაჯიბლაძე What values did you input (for angle, and the number of terms)? I haven't found any cases where the function changes by increasing the number of terms. $\endgroup$ Jun 16, 2021 at 17:32
  • $\begingroup$ @AlexandreEremenko I've changed his name to the correct spelling, thanks for pointing that out. $\endgroup$ Jun 16, 2021 at 17:34
  • $\begingroup$ I am using Mathematica. Issuing Table[Sum[Cos[.99999 2/7 \[Pi] n^3]+I Sin[.99999 2/7 \[Pi] n^3],{n,0,M}],{M,50,500,50}]//TableForm gives$$\begin{array}{r}31.6954\, -8.4058 i\\26.7249\, -16.9325 i\\24.3491\, -12.6764 i\\22.9538\, -5.2478 i\\26.5947\, -6.4563 i\\30.9138\, -1.14736 i\\30.2552\, -7.71583 i\\35.886\, -8.65421 i\\40.7469\, -12.9488 i\\49.5485\, -20.9481 i\end{array}$$ $\endgroup$ Jun 16, 2021 at 17:58

2 Answers 2


The Fabry gap theorem demonstrates that $f$ does not have an analytic continuation beyond the unit disk (as was observed by Alexandre Eremenko), but one can generalize the notion of analytic continuation. I however claim that $g$ cannot be the generalized analytic continuation of the function $f$ for any reasonable notion of generalized analytic continuation.

Define functions $f,g$ by letting $$f(z)=\sum_{n=0}^{\infty}z^{n^{3}},g(z)=-\sum_{n=1}^{\infty}z^{-n^{3}}.$$

Conflicting behavior on the boundary

Observe that $f(w)=-g(w^{-1})+1$ and $g(z)=-f(z^{-1})-1$ for appropriate $w,z$. Therefore, if we set $h=f\cup g$, then $h:\mathbb{C}\cup\{\infty\}\setminus S^{1}\rightarrow\mathbb{C}\cup\{\infty\}$ and $h$ satisfies the functional equation $(h(z)+h(z^{-1}))^{2}=1$ where $h(z)+h(z^{-1})=1$ whenever $|z|<1$ and $h(z)+h(z^{-1})=-1$ whenever $|z|>1$. This is a problem since if $g$ were truly a good generalized analytic continuation of the function $f$, then the generalized analytic continuation of $h=1$ inside the circle $S^{1}$ cannot be $h=-1$.

Functional equations

The functions $f,g$ satisfy the following functional equations.



Now, to make these functional equations more comparable to each other, we shall take the derivative of both sides of both equations in order to get rid of constant $-1$.

We have $$\sum_{k=1}^{\infty}\mu(k)k^{3}z^{k^{3}-1}f'(z^{k^{3}})=1$$



If $f$ and $g$ are generalized analytic continuations of each other, then such a notion of generalized analytic continuation would have to break down so that one would have to obtain incompatible functional equations (where the sum converges uniformly on compact subsets of $\mathbb{C}\cup\{\infty\}\setminus S^{1}$).

  • $\begingroup$ No part of the exploration in “Conflicting Behavior on the Boundary” depended on $z^{n^3}$ and could’ve been for $z^n$ instead which would lead to the conclusion $-\sum 1/z^n$ isn’t a generalized analytic continuation of $\sum z^n $ $\endgroup$ Dec 2, 2022 at 14:10
  • $\begingroup$ Ah theres also a typo it should be $g(z) = -f(z^{-1})+1$ $\endgroup$ Dec 2, 2022 at 14:27
  • $\begingroup$ Yea I don't feel the second argument involving functional equations is valid either but not because of any typos. It's not clear how that same argument doesn't prove $\sum_{n=1}^{\infty} - \frac{1}{x^n}$ isn't the continuation. of $\sum_{n=0}^{\infty} x^n$. Since you have $$ \sum_{k=1, n=1}^{\infty, \infty} \mu(k)z^{kn} = \sum_{k=1}^{\infty} \frac{\mu(k)z^k}{1-z^k} = z$$ and similarly form $z^{-1}$ going the other way and differentiate the two. $\endgroup$ Dec 2, 2022 at 16:17
  • $\begingroup$ Additionally theres a typo in the derivative the $\frac{-1}{z^2}$ should be $\frac{1}{z^2}$ im still checking if this causes any problems. $\endgroup$ Dec 2, 2022 at 16:58

For $k=1$ your formula indeed gives an analytic continuation, but for $n\geq 3$, it is known that your function $f$ has no analytic continuation (the unit circle is the natural boundary of your function $f$). This follows from Fabry's gap theorem.

  • $\begingroup$ For what reason does a natural boundary stop an analytical continuation? Since the function and the proposed continuation agree at angles which are a rational multiples of $\pi$, then they agree in a dense set. Isn't this enough-- or do the functions have to agree on a larger set to be a continuation? $\endgroup$ Jun 16, 2021 at 17:27
  • 1
    $\begingroup$ @CalebBriggs The functions have to agree on some connected open set. Without this condition the idea of analytic continuation is useless, because pretty much any function can be analytically continued to any other $\endgroup$
    – Wojowu
    Jun 16, 2021 at 17:36
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    $\begingroup$ @Caleb Briggs: by definition of an analytic continuation. The functions must agree on an open set. $\endgroup$ Jun 17, 2021 at 13:29
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    $\begingroup$ There are notions of generalized analytic continuation where for example, if $A$ is a countable subset of $S^{1}$, $\sum_{a\in A}|c_{a}|<\infty$, and $f(z)=\sum_{a\in A}\frac{c_{a}}{x-a}$, and $g$ is an entire function, then one can coherent extend $f+g$ from inside $S^{1}$ to outside $S^{1}$ since one can recover $\{(a,c_{a})\mid a\in A\}$ from the restriction of $f+g$ to the inside of $S^{1}$ since $c_{a}=\lim_{r\rightarrow 1^{-}}(1-r)(f+g)(ra)$ for each $a\in A$. See this Master's thesis for more details. scholarspace.manoa.hawaii.edu/bitstream/10125/29513/1/… $\endgroup$ Jul 17, 2021 at 22:48
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    $\begingroup$ Then $F$ is a generalized analytic continuation of $f$ to a holomorphic function with domain $\mathbb{C}\setminus S^{1}$, but $F$ is the continuation of $f$ defined by the functional equation $F(z)=F(1/z)$. This means that without being careful, one can generalized analytically continue one function to any other function. $\endgroup$ Jul 18, 2021 at 0:34

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