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Consider $z$ is complex and

$$f(z) = \sum_{n=1}^{\infty} \frac{z^{n^2}}{n^n}$$

This function has a natural boundary at $|z| = 1$.

Now I was thinking about summability methods or continuations beyond the natural boundary.

Define $$f(z,x) = \sum_{n=1}^{\infty} \frac{z^{n^2}}{n^n} n^{-n^2 (1-x)}x^n$$

Notice $f(z,1) = f(z)$.

Now let $f_+(z)$ be the proposed continuation of $f(z)$ that converges within some annulus $1 < |z| < r$.

Can we say

$$f_+(z) = \lim_{x \to 1-} f(z,x) = \lim_{x \to 1-} \sum_{n=1}^{\infty} \frac{z^{n^2}}{n^n}n^{-n^2 (1-x)} x^n $$

If so, what is the value of $r$ ?

And if not, why not ?

How do we even do such limits ?

edit

Some are confused and think this is about analytic continuation, but that is not the case !

Analytic continuation through a natural boundary does not exist.

But it might be definable on the other side of the boundary.

Such as for instance

$$t(z) = \sum_{n>0} \frac{z^n}{(z^n - 1) n^3}$$

Is defined within and outside of the unit circle and it has a taylor series that only works within the unit circle.

See also : generalized analytic continuation and Continuation of functions beyond natural boundaries.

And this paper :

https://arxiv.org/pdf/1301.1175.pdf

mick
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  • It's a natural boundary. The singularities on the boundary are too dense to try to analytically extend through it. – Brevan Ellefsen Dec 15 '23 at 03:10
  • @BrevanEllefsen the boundary is a natural boundary but it is bounded and $C^{\infty}$. The continuation is not analytic continuation. That would ofcourse be inpossible. Not every continuation is analytic continuation. We might have an analytic function beyond the boundary. Such functions exist. – mick Dec 15 '23 at 03:33
  • @BrevanEllefsen quote : The theory of generalized analytic continuation studies continuations of meromorphic functions in situations where traditional theory says there is a natural boundary. – mick Dec 15 '23 at 03:36
  • See my edit too – mick Dec 15 '23 at 11:53
  • I changed the question. – mick Jan 12 '24 at 23:55
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    I will come back to this ... – mick Jun 09 '25 at 00:13
  • Even functions in exponents are often quite tricky to deal with, the related $ \sum_{n=1}^{\infty} \frac{z^{n^{3}}}{n^n}$ on the other hand is naturally continued as $ -\sum_{n=0}^{\infty} (-1)^n n^n z^{-n^{3}}$ which you might find interesting. – Sidharth Ghoshal Jun 09 '25 at 00:30
  • @SidharthGhoshal plz explain that one ! – mick Jun 09 '25 at 11:27
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    Here’s a good place to start – Sidharth Ghoshal Jun 10 '25 at 02:36
  • @SidharthGhoshal you made a typo then what turned me in the wrong assumption. its $n^{-n}$ what you meant probably ? Yes making the relations $f(-1/x) = f(x)$ or similar are nice. But I feel maybe a bit to general for all lacunary functions. I would be convincing if both truncated series agree somewhat on zero's and such. I need to think ... Then again maybe you did actually meant what you said considering your link ... it does converge for z outside the radius 1 yes ... hmm – mick Jun 11 '25 at 23:44
  • oh no that wasn't a typo the connection is a bit weird. i will definitely come back and explain whats going on there but ill cut to the chase and say "even these advanced techniques do not help your situation because your series has an $n^2$ in the exponent of $z$" – Sidharth Ghoshal Jun 12 '25 at 03:19
  • @SidharthGhoshal Yeah I figured that by your link. Well the idea that $\sum f(n) x^n $ has "continuation" $\sum ( f(n) x^n )^{-t}$ for some $t$ usually is a powerful one. Usually $t=1$. Fractional derivatives are also a thing. I will come back to this. And I will post another question about continuations. – mick Jun 12 '25 at 10:53
  • Small typo you have: it’s $\sum_{n=0} f(n) = -\sum_{n=1} f(-n)$. Thus when the $n$ is in the exponent conveniently creates the reciprocal behavior when the exponent is odd. And for $n^n$ that also conveniently has this reciprocal flipping behavior. – Sidharth Ghoshal Jun 12 '25 at 14:35
  • @SidharthGhoshal perhaps of interest to you is tommy's idea of using polylog as continuations. So $f(z) = \sum_n a_n z^n = f(z) = \sum_n \sum_m (b_m n^m) x^n = \sum b_m polylog(... $ valid if the order of summation do not disagree on finite values. – mick Jun 14 '25 at 11:08
  • do u have a link to it? I do know that the permutation $\sum_n \sum_m = \sum_m \sum_n $ often necessitates the euler-maclaurin formula as a correction (when the sums are infinite and of the original form sum_n f(nx) and order of summation does matter). I haven't seen a more general theory before for that which it seems like your polylog example covers. – Sidharth Ghoshal Jun 14 '25 at 23:44
  • @SidharthGhoshal unfortunately the written stuff he wrote has been lost in a fire that some gangster made. – mick Jun 15 '25 at 19:22
  • If there's a metaphor there it has flown over my head, so i will interpret your sentence literally here. Damn that's sad to hear. How did u first encounter tommy's results? also who is tommy? are they a user here by any chance? – Sidharth Ghoshal Jun 15 '25 at 22:12
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    @SidharthGhoshal tommy1729, my friend and mentor. Its in my bio. We play chess. He posted on the tetration forum where I read. – mick Jun 15 '25 at 22:21

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