Does this method analytically continue gap series series?

Question

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: Zooming in on the point $x = 1$:

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!

Answer 1

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.

$$\sum_{k=1}^{\infty}\mu(k)(f(z^{k^{3}})-1)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\mu(k)z^{(kn)^{3}}=z.$$

$$\sum_{k=1}^{\infty}\mu(k)g(z^{k^{3}})=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}-\mu(k)z^{-(nk)^{3}}=-z^{-1}.$$

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

while

$$\sum_{k=1}^{\infty}\mu(k)k^{3}z^{k^{3}-1}g'(z^{k^{3}})=\frac{-1}{z^{2}}.$$

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 $\mathbb{C}\cup\{\infty\}\setminus S^{1}$).

Answer 2

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.