I understand that one can in theory analytically continue a function by repeatedly computing new Taylor series. Suppose for example we have an analytic function $f$ defined on some open set $U$ and compute


for some $z_0\in U$ close to the boundary of $U$. If this converges on $V$ where $U\cap V$ is non-empty, we can then compute another Taylor series to extend further:


for some $z_1\in V\setminus U$ etc.

However, it is impossible to compute infinitely many terms and higher derivatives quickly become prone to large amounts of cancellation error.

Furthermore, one must repeated drop the degree of the next series expansion, as demonstrated here, in order for the result to be useful. Otherwise, with the same degree at the new point $z_1$, you will end up recovering the original $T_0$ and fail to approximate $f$ further away.

So how can one actually numerically compute the analytic continuation of a function?

In my specific case, I have a set of data points over a subinterval of $\mathbb R$ and I know some basic behavior about the function $f$'s derivatives (all derivatives are positive over the subinterval and to the right, which is the area I want to continue to) and that it has no singularities to the right of the given subinterval.

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    $\begingroup$ You might look at this recent paper of Trefethen: Quantifying the ill-conditioning of analytic continuation. $\endgroup$ Mar 20, 2020 at 15:32
  • $\begingroup$ It would seem to me this kind of problem is harder the more I look into it. x.x $\endgroup$ Mar 20, 2020 at 15:48
  • $\begingroup$ No, the analytic continuation is not continuous in the coefficients, looking at finitely coefficients doesn't work, you need explicit prior (such as "the function is analytic on those two disks and it is bounded by that function on each") to do numerical approximation. The key is that a bound for $f(z)$ on $|z-a|=r$ gives an upper bound for the accuracy of its Taylor polynomials at $a$ when evaluated for $|z-a| < r$. $\endgroup$
    – reuns
    Mar 20, 2020 at 16:27
  • $\begingroup$ It seems you have an Extrapolation problem which is very difficult in general. If the data is well behaved you are very lucky. $\endgroup$
    – Somos
    Mar 20, 2020 at 17:16
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    $\begingroup$ @reuns Yes, I already understand that, there is no need to yell. As I pointed out at the end, I know there are singularities in the direction I am continuing, so I know the radii of convergence. I didn't mention that I knew a bound for the area I am searching, though from the context of where the function comes from, I can expect bounds (sublinear as $\Re(z)\to\infty$). $\endgroup$ Mar 20, 2020 at 18:52

2 Answers 2


It's often possible to increase the radius of convergence by applying a suitable conformal mapping to the series. The transform then moves singularities away to a larger distance from the point you are expanding around. For example, the conformal mapping:

$$z = \frac{p w}{p+1-w}\tag{1}$$

Keeps the points $z=0$ and $z=1$ fixed while it moves the point $z = -p$ to infinity, while the point at infinity, which is often a singular point, is moved to $w = p+1$.

Even if only a few terms of the series expansion of a function are known, then we can apply such a conformal transform to evaluate the function outside the radius of convergence. Take e.g. the series for $\log(1+2 z)$:

$$\log(1+2 z) = 2 z - 2 z^2 +\frac{8 z^3}{3}-4 z^4+\frac{32 z^5}{5}-\frac{32 z^6}{3}+\frac{128 z^7}{7} -32 z^8 +\frac{512 z^9}{9} -\frac{512 z^{10}}{5}\cdots$$

This series has a radius of convergence of $\frac{1}{2}$, so it seems that we cannot use this series to evaluate $\log(1+2 z)$ at $z = 1$. However, we can use this series by applying the conformal mapping (1) for e.g. $p = 1$. This yields the series:

$$w + \frac{w^3}{12} + \frac{w^5}{80} + \frac{w^7}{448} + \frac{w^9}{2304}+\cdots$$

The point $z = 1$ corresponds to $w = 1$, and it's clear that the series converges very fast for $w = 1$, while the original series was divergent at $z = 1$. Note that the computation of this series involves just the substitution of (1) in the series and then a re-expansion in powers of $w$ to order 10.

  • $\begingroup$ That is actually really interesting and simple. Do you have anything I could read up on about it? $\endgroup$ Mar 21, 2020 at 4:07
  • $\begingroup$ @SimplyBeautifulArt I'll expand this answer and add some references to the literature. These sorts of methods were used in a rudimentary form already by Euler as a means to give meaning to divergent summations. There are many different approaches to this problem, Mathematicians have more focused on getting to a notion of summability that would then also apply to divergent series, while in physics people were more concerned with doing practical computations. Perturbation theory is a frequently used tool in physics, resummation methods are a popular tool within this field. $\endgroup$ Mar 22, 2020 at 12:43
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    $\begingroup$ @lcv You want to maximize the radius of convergence, so that means that the singularity closest to the origin after applying the transform should be at a maximum distance, But this is not so easy to implement, in most cases we don't know where the poles of the function are, Zinn-Justin has shown that if you use the transform $x = p z/(1-z)$ to find the expansion around infinity from the expansion around zero then equating the largest power to zero and choosing the largest value for p yields the fastest converging series, because larger p corresponds to a larger radius of convergence. $\endgroup$ Dec 9, 2022 at 6:57
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    $\begingroup$ @lcv Thanks for asking me. I have an idea that looks like working for that case. I'll write it up later today. $\endgroup$ Dec 20, 2022 at 23:31
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    $\begingroup$ @SaibalMitra thank you, that seems amazing. $\endgroup$
    – lcv
    Dec 21, 2022 at 8:01

Too longfor a comment.

Using @Saibal Mitra's example $$\log(1+2z)=2 z-2 z^2+\frac{8 z^3}{3}-4 z^4+\frac{32 z^5}{5}-\frac{32 z^6}{3}+\frac{128 z^7}{7}+O\left(z^8\right)$$ does not allow to do anything if $z > \frac 12$. For $z=1$, it would give the ridiculous $\frac{444}{35}$ which is $11.5$ times too large.

Using the same coefficients, consider the $[4,3]$ corresponding Padé approximant $$\log(1+2z)=\frac{2 z+\frac{34 z^2}{7}+\frac{8 z^3}{3}+\frac{4 z^4}{35} } { 1+\frac{24 z}{7}+\frac{24 z^2}{7}+\frac{32 z^3}{35}}+O\left(z^8\right)$$ For $z=1$, it gives $\frac{1012}{921}$ which corresponds to a relative error of $0.018$%. For $z=2$, it leads to a relative error of $0.22$%.


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