Taking a relatively arbitrary combination of exponential and polynomial terms, for instance $$\sum_{n=0}^\infty \left(n^{2}\sin\left(n\right)+n\cos\left(3n-2\right)\right)\cos\left(5n+1\right)x^{n}$$ one will find that this series is 'miraculously' analytically continued by the same series except with $n$ replaced by $-n$ (additionally, with an extra negative sign and the index starting at $n=1$). In particular, the first series has an analytical continuation for $|x|>1$ given by $$-\sum_{n=1}^{\infty}\left(\left(-n\right)^{2}\sin\left(-n\right)-n\cos\left(3\left(-n\right)-2\right)\right)\cos\left(5(-n)+1\right)x^{-n}$$

In general, when $P(n) = \sum_{k=0}^N a_k n^k$ is a polynomial, then $\sum_{n=0}^\infty P(n)x^n$ is analytically continued by $- \sum_{n=1}^\infty P(-n)x^{-n}$. Likewise, when $F(n) = \sum_{k=0}^N a_k e^{i \theta_k n}$ is a Fourier-like series, then $\sum_{n=0}^\infty F(n) x^n$ is continued to $-\sum_{n=1}^\infty F(-n)x^{-n}$. The second result can be generalized to when $F(n)$ is almost periodic, and I imagine a similar result can be found when $P(n)$ is 'almost-polynomial' under a suitable definition.

One potential heuristic for why this relationship holds can be found by considering the residue theorem. If $(-1)^n f(n)$ is an entire function under sufficient growth conditions, then $\frac{1}{2 i} \int_{-1/2 - i\infty}^{-1/2 + i \infty} f(n) \csc(\pi n)$ is equal to both $\sum_{n=0}^\infty (-1)^n f(n)$ and $-\sum_{n=1}^\infty (-1)^n f(-n)$ by considering either a large rectangle covering everything to the right of the contour, or one covering everything to the left (the negative sign comes from switching the orientation of the contour).

This heuristic doesn't tell the whole story however. Consider the function $\frac{d}{dx}\frac{\Gamma'(x+1)}{\Gamma(x+1)}$ and its corresponding Taylor series $\sum_{k=0}^\infty (k+1) \zeta(k+2)(-x)^k$. $\zeta$ grows too large in the negative direction, so the contour does not converge and we can't support changing the direction of the contour. Nonetheless, it turns out that the series $$-\sum_{k=1}^\infty (-k+1) \zeta(-k+2)(-x)^{-k}$$ (where we understand $0\cdot \zeta(1)=1$) asymptotically extends the original series. In particular, if we fix $N$ then $\lim_{x \to +\infty} \sum_{k=1}^N (-k+1)\zeta(-k+2)(-x)^{-k} = \frac{d}{dx} \frac{\Gamma'(x+1)}{\Gamma(x+1)}$. In fact, taking only the first 10 terms of the sum gives an approximation where the difference between the series and the function at $x=3$ is less than $10^{-6}$ (and decreases to zero as $x$ becomes large).

Another case occurs with functions like $\sum_{n=0}^\infty \sin(\sin(n))x^n$, which is continued by $-\sum_{n=1}^\infty \sin(\sin(-n))x^{-n}$. In this case, analyzing the relationship between the two functions using the residue theorem seems even less clear, since $\sin(\sin(z))$ only remains small when $z$ has a small imaginary part.

Finally, I should note that its clear that there are cases where $\sum_{n=0}^\infty (-1)^nf(n)x^n \neq -\sum_{n=1}^\infty (-1)^nf(-n)x^{-n}$. For instance, if we choose $f(n) = \frac{1}{n^2+1}$, then $f(n) =f(-n)$. However, in simple cases like these we can salvage the relationship using the residue theorem to pick up the extra poles off the real line, so that, for example $$\sum_{n=0}^\infty \frac{(-1)^n x^n}{n^2+1} = -\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{-n}}{n^{2}+1}+\pi\cos\left(\ln\left(x\right)\right)\operatorname{csch}\left(\pi\right)$$


In light of these examples, I am curious about general characterizations on the relationship between $F(x)=\sum_{n=0}^\infty f(n)x^n$ and $F_2(x)=-\sum_{n=1}^\infty f(-n) x^{-n}$. Some questions that come to mind are

  1. How are these sums connected when $f(n)$ is an entire function, but diverges fast enough at $\pm i \infty$ so that the residue theorem can't be applied?
  2. Are there other broad classes of functions where $F(x)$ is analytically continued to $F_2(x)$? Are there broad classes of functions where the relationship fails?
  3. How are the sums related when $F(x)$ is a divergent series, for instance $f(n)=(\alpha n)!$ or $f(n) = (n^2)^n$.
  4. Are there cases where $F(x)$ and $F_2(x)$ are related in some way, but aren't analytical continuations of each other?
  5. Is there uniqueness, in the sense that if $\sum_{n=0}^\infty f(n) x^n = \sum_{n=0}^\infty g(n) x^n$, must we also have $-\sum_{n=1}^\infty f(-n)x^{-n} = -\sum_{n=1}^\infty g(-n)x^{-n}$

Edit: The Residue Approach

I've added this section to give a little bit more detail about what I am thinking about with the residue approach. In connection with Alexandre's answer, it is interesting to me that this approach seems to stop working at precisely the same point where $f(n)$ is no longer uniquely determined by by its values on the integers (by an application of Carlson's theorem).

We can apply the residue theorem to analyze sums of the form $\sum_{n=0}^\infty (-1)^n f(n)x^n$ where $x>0$ and and $f$ is an entire function of exponential type less than $\pi$ (i.e. $|f(r e^{i \theta})| < Me^{\pi r}$). With the latter constraint on the growth rate of $f$, through some algebriac manipulation we can obtain for the right half plane $\lim_{|z| \to \infty} f(z)x^z csc(\pi z) = 0$ when $x \leq \frac{1}{e^{\pi}}$. When $f(z)$ is an entire function, then, we have that $$\sum_{n=0}^\infty (-1)^n f(n)x^n = \frac{1}{2i}\int_C \csc(\pi z) f(z) x^z dz$$ Where $C$ is a rectangular contour whose height and width go to infinity. enter image description here However, all but the contour at $Re(z) = -\frac{1}{2}$ go to zero by the growth conditions, and so $\sum_{n=0}^\infty (-1)^n f(n) x^n = \frac{1}{2i}\int_{-\frac{1}{2}- i \infty}^{-\frac{1}{2}+ i \infty} f(z)x^z \csc(\pi z)dz$ for $x< \frac{1}{e^{\pi}}$. However, as long as the integral is analytic, we now have two analytic fucntions whose agree on a set with a limit point, so they represent the same analytic function. Moreover, the contour continues to converge even when the original sum doesn't, so the contour integral analytically continues the original function. Moreover, when $x> e^{\pi}$ then the $f(z) \csc(\pi z)x^z$ is small for values on the left-half plane. So, we obtain $$\frac{1}{2i} \int_{-\frac{1}{2}- i \infty}^{-\frac{1}{2}+ i \infty} f(z)x^z \csc(\pi z)dz = \frac{1}{2i} \int_{C_2} f(z)x^z \csc(\pi z)dz = -\sum_{n=1}^\infty (-1)^nf(-n)x^{-n} $$ Since the other pieces of the contour $C_2$ are zero (i.e. all the edges except the one where $Re(z)=\frac{1}{2}$). $C_2$ is a rectangle of infinite width and height facing in the opposite direction to $C$ given by this picture enter image description here

In the case where $f(n)$ is not entire, then we can usually shift the original contour to the line $Re(z) = c$ so that all the poles of $f$ are to the left of $c$. Then we obtain that, for $x< \frac{1}{e^{\pi}}$
$$\sum_{0 \leq n < c} (-1)^n f(n) x^n + \frac{1}{2i}\int_{c+C_{\rightarrow}} f(z) x^z \csc(\pi z) dz = \\ \sum_{0 \leq n < c} (-1)^n f(n) x^n + \frac{1}{2i}\int_{c-i \infty}^{c + i \infty} f(z) x^z \csc(\pi z) dz=\sum_{n=0}^\infty (-1)^n f(n) x^n$$ where $c+C_{\rightarrow}$ is the rectangular contour that covers everything to the right of $Re(z) = c$. And again for $x>e^{\pi}$ we can switch the direction of the contour, and this gives us that $$\sum_{0 \leq n < c} (-1)^n f(n) x^n + \frac{1}{2i}\int_{c+C_{\leftarrow}} f(z) x^z \csc(\pi z) dz = \\ \sum_{0 \leq n < c} (-1)^n f(n) x^n + \frac{1}{2i}\int_{c-i \infty}^{c + i \infty} f(z) x^z \csc(\pi z) dz=\\-\sum_{n=1}^\infty (-1)^n f(-n) x^{-n} + \frac{1}{2i} \sum_{} \text{Res}(f(z)\csc(\pi z)x^z, a_k)$$ Conviently, the extra terms $\sum_{0 \leq n < c} (-1)^n f(n) x^n$ that were added in initially cancel out since the $C_{\leftarrow}$ contour traverses those same points in the opposite direction.

  • $\begingroup$ I did not understand your statement about $\sum \sin(\sin n))z^n$. What are you saying about its analytic continuation? $\endgroup$ Nov 29, 2022 at 16:23
  • $\begingroup$ @AlexandreEremenko If I am not mistaken $\sin(\sin(n))$ is almost periodic, so $\sum \sin(\sin(n))z^n$ is analytically continued to $- \sum \sin(\sin(-n)) z^{-n}$. However, $\sin(\sin(n))$ grows very quickly off of the real line, so it doesn't seem possible to pick a contour that will still converge when $|z|>1$. It seems, at least with my naive approach, that the relationship between $\sum \sin(\sin(n)) z^n$ and $\sum \sin(\sin(-n)) z^{-n}$ cannot be explained using the residue theorem. $\endgroup$ Nov 29, 2022 at 19:15
  • $\begingroup$ Since $\sin$ is odd, $-\sin(\sin(-n))=\sin\sin(n)$ for all $n$ so what you say cannot be true. $\endgroup$ Nov 29, 2022 at 19:35
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    $\begingroup$ I like this question a lot, i always wondered if there is a more general rule that relates $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = \sum_{n=1}^{\infty} -x^{-n}$ and it seems like you've nailed it down. $\endgroup$ Nov 29, 2022 at 20:12
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    $\begingroup$ There might be some information at G. Pólya, Untersuchungen über Lücken and Singularitäten von Potenzreihen, Math. Zeit. 29 (1928–1929), 549–640. $\endgroup$ Jul 28, 2023 at 19:22

6 Answers 6


First of all, to make sense of $f(-n)$ we need some assumptions about $f$. For example, let $$\sum_{n=0}^\infty f(n)z^n\quad\quad\quad\quad (1)$$ be a series with positive radius of convergence. Then there is always an entire function of exponential type which interpolates $f(n)$. But such a function is usually not unique. One needs uniqueness to define $f(-n)$. Moreover, one needs the fact that (1) DOES HAVE an analytic continuation, to discuss a formula for it.

To make $f$ unique, and to assure an analytic continuation, one possible assumption is that the exponential type is less than $\pi$. More precisely, we have the following theorem of F. Carlson.

If $f$ is an entire function of exponential type, and $\overline{K}$ is its conjugate indicator diagram, then the power series (1) has an analytic continuation to the connected component of the complement of $e^{-K}$ which contains $0$.

So if $e^{-K}$ does not separate $0$ and $\infty$, our function has an analytic continuation to a neighborhood of $\infty$, with Laurent series $$-\sum_{1}^\infty f(-n)z^{-n}.$$ See, for example, L. Bieberbach, Analytische Fortsetzung, Springer 1955, section 1.3.

This theorem of Carlson does not cover all your examples, but this section of Bieberbach mentions a lot of versions and generalizations.

Remarks. Most likely the power series $$\sum_{n=1}^\infty z^n\sin\sin n$$ does not have any analytic continuation (the unit circle is the natural boundary). I write this by analogy with power series with random coefficients: the sequence $\sin\sin n$ behaves randomly. See https://arxiv.org/abs/1409.2736 for Taylor series with almost periodic coefficients. But notice: as I wrote above, there is ANOTHER entire function $f$ which is of exponential type and for which $f(n)=\sin\sin n$.

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    $\begingroup$ Interestingly, in the case where the sum takes the form $\sum_{n=0}^\infty (-1)^n f(n) x^n$, then the residue theorem tricks works precisely when $f(n)$ is exponential type less than $\pi$, since $ \csc(\pi z) = \frac{2 i}{e^{\pi z} - e^{-\pi z}}$ so the contour integral $\int_{c - i \infty}^{c + i \infty} f(z) x^z \csc(\pi z)$ does not diverge (when x>0). $\endgroup$ Nov 29, 2022 at 20:36

I'll convert my comment to an answer.

One issue that you're implicitly sweeping under the rug is that we need to be able to make sense of the evaluation of $f$ at negative integers. Given an arbitrary power series $\sum_{n=0}^{\infty} f(n) x^n$, a priori $f$ is just any function $f\colon \mathbb{N}\to\mathbb{C}$. Of course, if $f$ is given by some expression where it makes sense to plug in $-n$ like your $f(n) = (n^2\sin(n)+n\cos(3n-2))\cos(5n+1)$ then we can make sense of $f(-n)$.

A class of functions $f\colon \mathbb{N} \to \mathbb{C}$ which can sensibly be extended to the negative integers are those satisfying a (homogeneous) linear recurrence (with constant coefficients). This means that there is some $d \geq 0$ and complex numbers $\alpha_1,\ldots,\alpha_d \in \mathbb{C}$ with $\alpha_d \neq 0$ such that $$ f(n+d) + \alpha_1 f(n+d-1) + \cdots + \alpha_d f(n) = 0 $$ for all $n \geq 0$. Given such an $f$ we can define $f$ at negative values by "running the recurrence backwards," i.e., by setting $$ f(-n) = \frac{-1}{\alpha_d}( f(-n+d) + \alpha_1 f(-n+d-1) + \cdots + \alpha_{d-1} f(-n+1))$$ for all $n \geq 1$.

For such an $f$, it is known that formally (i.e., as formal power series, ignoring issues of convergence) we have $$ \sum_{n=0}^{\infty} f(n) x^n = - \sum_{n=1}^{\infty} f(-n) x^{-n}.$$ See for instance Proposition 5.2 in Stanley's Combinatorial Reciprocity Theorems.

Note that it is well-known that $f$ satisfying a linear recurrence is equivalent to the power series $F(x) = \sum_{n=0}^{\infty} f(n) x^n$ being a rational function $F(x)= P(x)/Q(x)$ where $\mathrm{deg}(P(x)) < \mathrm{deg}(Q(x))$. So that is the class of series $F(x)$ that this argument applies to.

Also note that if $f$ satisfies a linear recurrence then it is a "combination" of polynomials and exponential functions in the sense that there exist polynomials $P_1(x), \ldots, P_k(x) \in \mathbb{C}[x]$ and distinct nonzero complex numbers $\gamma_1,\ldots,\gamma_k \in \mathbb{C}^{*}$ such that $$ f(n) = P_1(n) \cdot \gamma_1^{n} + \cdots + P_k(n) \cdot \gamma_k^{n}$$ for all $n \geq 0$. So we could also define $f(-n)$ by naively plugging $-n$ into this formula.

EDIT: As Richard Stanley mentioned in the comments, continuing this idea of studying formal power series whose coefficients form a sequence satisfying some recurrence, this kind of reciprocity is known also for $D$-finite a.k.a. holonomic $F(x)$, i.e., for $f(n)$ that are $P$-recursive (under some mild conditions); see for instance Section 3 of his paper Differentially Finite Power Series.

  • $\begingroup$ But this suggests that maybe a place to look for more general $f(n)$ for which this reciprocity holds would be generalizations of homogeneous linear recurrences with constant coefficients: e.g. we could consider $f(n)$ that are $P$-recursive (meaning that they satisfy a recurrence with polynomial, instead of constant, coefficients). $\endgroup$ Nov 29, 2022 at 15:20
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    $\begingroup$ For more general situations than the rationality of $F(x)$, see Section 3 of math.mit.edu/~rstan/pubs/pubfiles/45.pdf. $\endgroup$ Nov 29, 2022 at 15:46
  • $\begingroup$ @RichardStanley: Thanks for that reference! So I see the extensions to the P-recursive/D-finite case is known. $\endgroup$ Nov 29, 2022 at 17:15

This is just a comment but I am not entitled and it will be too long anyway. If we start with a toy example where $f(n)=1$ for all $n$, then $F$ is the function $\frac 1{1-z}$ in the open disc, $F_2$ $\frac1{z-1}$ on its exterior. I want to suggest that the correct framework for this and for your question is that of distributions as boundary values of holomorphic functions. This was introduced by Gottfried Köthe in 1952. The most natural case is that of distributions on the unit circle as boundary values of functions which are analytic on its complement in the complex plane. (Another important case is that of distributions on the real line as boundary values of analytic functions on the upper and lower half planes).

In the above case we are using the representation of the delta distribution on the unit circle—the fact that the two functions are analytic continuations of each other across this boundary between the two regions of analyticity (with the exception of the singularity) corresponds to the vanishing of the distribtion on this open subset of the circle,i.e., the complement of $1$.

The case where $f$ is a power of $n$ can be dealt with similarly, using higher derivatives of $\delta$. Taking linear combinations then deals with the situation where it is a polynomial.

As I said, this isn‘t an answer but I hope that it might help to provide a suitable point of view for a systematic approach.

The initial article is „Die Randverteilung Analytischer Funktionen“, Math. Zeit. 57 (1952), 13-33, by G. Köthe.

  • $\begingroup$ I am interested in where I can read more about this $\endgroup$ Nov 30, 2022 at 2:55

This is also just a comment but large:

It might be worthwhile to try poking around with this formula on functions which cannot be analytically continued to see what arises. Trying to break the formula might reveal insights about how it works / where it matters.

An example that comes to mind is the classical lacunary function: $$ \sum_{n=0}^{\infty} x^{2^n} $$ which cannot be analytically continued outside the unit disk.

This function can be written as $$ \sum_{n=1}^{\infty} \left[ \frac{1}{n\ln(2)} \frac{\sin(2\pi n)}{\sin(2\pi \log_2(n))} x^n \right]$$

So it's natural then to consider the expression

$$ -\sum_{n=0}^{\infty} \left[ \frac{1}{-n \ln(2)} \frac{- \sin(2\pi n)}{\sin(2\pi \log_2(-n))} x^{-n} \right] = -\sum_{n=0}^{\infty} \left[ \frac{1}{n \ln(2)} \frac{\sin(2\pi n)}{\sin(2\pi \log_2(n)+\frac{2i\pi^2}{\ln(2)})} x^{-n}\right] $$

The reformulation of this series is actually 0 for all coefficients with $n < 0$ suggesting no analytic continuation is possible using the simple reformulation this way so thats kind of cool (it at least confirms what we already know) [perhaps by applying your residue tricks we might still be able to extract some more information out of here]

We can try another Lacunary Function:

Lets consider the function $ \sum_{n=0}^{\infty} x^{n^3} $. This series should be generated by

$$ M(x) \int_C e^{i \pi z} csc(\pi \sqrt[3]{z} ) x^z dz = \sum_{n=0}^{\infty} x^{n^3}$$

Where I believe $M(x) = \frac{1}{2ix^{\frac{2}{3}}}$ (although I still need to understand the residue trick better).

Anyways ignoring the multiplier and just summing the residues of the poles on the LHS and RHS suggests then that

$$ - \sum_{n=1}^{\infty} x^{-n^3} $$

Is its "continuation" which we graph below

The graphs do seem to line up kinda nicely.

graph of functions

We can differentiate this thing and there really seems to be something nice cooking up here even at the level of the derivatives:

enter image description here

We can even take a 2nd Derivative... and at about 80 terms it starts to look like convergence:

enter image description here

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    $\begingroup$ Actually, part of my initial interest in this question comes from trying to analyze lacunary functions. In particular, series like $\sum_{n=0}^\infty \frac{(-1)^n x^n}{n!} w^{n^k}$ can be continued (not analytically) along the real line by taking a contour starting from near the origin with rays in the directions $\sqrt[k]{i}$ and $\sqrt[k]{-i}$. The contour here continues to converge when $w$ is real and $|w|>1$, and this extension continuously extends the derivatives beyond the boundary. Thus, some of my motivation is understanding ill-conditioned applications of the residue theorem. $\endgroup$ Nov 30, 2022 at 22:14
  • $\begingroup$ Interesting, these are theta function like objects. What happens if we drop the taylor term and just look at $\sum_{n=0}^{\infty} w^{n^2}$ does the residue trick yield anything well behaved function "reciprocal"-series? $\endgroup$ Dec 1, 2022 at 18:45
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    $\begingroup$ Actually we can get even smarter than that... we can use $\csc(\pi \sqrt{x})$ to filter terms and then your residue theorem trick. So I imagine something like $$ \frac{1}{2i} \int_{C} \csc(\pi \sqrt{z}) w^z dx = \sum_{n=0}^{\infty} (-1)^n w^{n^2}$$ (note we have a branch cut here so that makes things a little complex) $\endgroup$ Dec 1, 2022 at 19:53
  • $\begingroup$ Hmm that is a very interesting idea, I will surely think about that (note that one has to multiply by an extra $\frac{1}{2 \sqrt{z}}$ to cancel out the residues). Interestingly, this gives the same (unappealing) value as my method along the real line $w>1$, which is a constant $\frac{1}{2}$, and I plan to look in detail later about the relationship when $w$ is a complex number $\endgroup$ Dec 1, 2022 at 20:27
  • $\begingroup$ See my addition to the answer^ $\endgroup$ Dec 1, 2022 at 22:20

Some extended comments:

Investigations of this type appear in the theory of the Mellin transform and, in particular, the use of Ramanujan's master heuristic (theorem, formula) as illustrated in the MO-Q "Ramanujan's Master Formula: A proof and relation to umbral calculus" and its Related Links.

The Mellin transform is also integral in exploring the properties of functions satisfying $f(x) = \frac{1}{x} f(\frac{1}{x})$ as illustrated in the MSE-Q "Does the functional equation f(1/r)=rf(r) have any nontrivial solutions besides f(r)=1/√r?".

There's a book written by some subset of the authors Hardy, Littlewood, and Titchmarsh that addresses these relations, I believe. (I've meant to find it again lately after coming across it years ago. If my laptop remains stable today, I'll try to find it again.)


Ok so I just realized there is a very simple operator calculus reason for this (I apologize if my answers are annoying since they are very closely related but I am using this to document my ideas and also learning for myself how this works).

Consider the operator $H: f(s) \rightarrow xf(s+1)$. It then follows quite naturally that

$$ f(s) + xf(s+1) + x^2 f(s+2) + ... = \sum_{n=0}^{\infty} H^n[f(s)] = \frac{1}{1-H}$$

But recall the geometric series $\frac{1}{1-x}$ has two series of convergence $1+x+x^2 + x^3 ... $ and the laurent series $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... $


$$ \frac{1}{1-H} = -H^{-1} - H^{-2} - H^{-3} .... = -\frac{f(s-1)}{x}-\frac{f(s-2)}{x^2} - \frac{f(s-3)}{x^3} ... $$

Evaluating both series at $s=0$ then gives us that

$$ \sum_{n=0}^{\infty} f(n) = \left[\frac{1}{1-H}\right]_{@s=0} = -\sum_{n=1}^{\infty} f(-n)x^{-n} $$

And that is the connection that you found. I'm not sure what the ramifications of this formula entail but there is a (non optimistic) chance it could be deep.

For example the Jacobi theta function $\sum_{n=0}^{\infty} x^{n^2} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \frac{e^{2i\pi n} - 1}{e^{2i\pi \sqrt{n}} - 1}x^{-n} $ so we could try to bash it with this thing. What's curious is that once $n$ goes negative then $\sqrt{n}$ goes imaginary and $e^{2i\pi \sqrt{n}} $ starts to grow exponentially fast so this series somewhat shockingly actually will converge and produce something (up to a choice of branch cut). Is that the "natural" continuation of the Jacobi theta function? Edit: i spoke too soon. I have the resultant series is just 0 everywhere and so isn’t a very satisfying continuation at all.

  • $\begingroup$ This is an interesting answer, but I’m worried that it doesn’t illuminate what’s actually happening. In particular, the rule that $\sum f(n)=-\sum f(-n)$ doesn’t always hold, but I don’t see how this approach can recover the error terms between the two sums. Do you have some notion on how to apply this in cases like $\sum \frac{1}{n-1/2} x^n$, or other cases where $f(n)$ has a pole? $\endgroup$ Jul 28, 2023 at 3:14
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    $\begingroup$ And, as for the Jacobi functions, I’ve noticed that extensions tend to obtain a constant when they neglect to consider the pole at infinity that comes from $e^{n^2}$ $\endgroup$ Jul 28, 2023 at 3:16

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