Given the Laurent series $\sum\limits_{k=-\infty}^\infty a_k^{(l)} z^k = f(z)|_{r_l<|z|<R_l}$ of a meromorphic function $f$ on $\mathbb C$ with convergence region $r_l< |z|< R_l$, one can use analytical continuation to obtain the values in the regions $r_m<|z|<R_m$ where either $r_m=R_l$ or $R_m=r_l$, i.e. switch between series converging in neighbouring rings the boundaries of which are touching isolated singularities. But how can the new coefficients $a_k^{(m)}$ be obtained from the old ones?

As an example to what I mean, take the geometric series $$1+q+q^2+... = \sum_{k=0}^\infty q^k = \frac1{1-q}\Bigg|_{|q|<1}$$ which conveges for $|q|<1$, and its "counterpart" for $|q|>1$, $$\frac1{1-q}\Big|_{|q|>1} = \frac1q\frac{1}{\tfrac1q-1}\Bigg|_{\big|\tfrac1q\big|<1} = -\frac1q\sum_{k=0}^\infty \left(\frac1q\right)^k = -\sum_{k=-\infty}^1 q^k$$

So the question for that example would be, how to get from the $r_1=0\le|q|<1=R_1$ coefficients $$a_k^{(1)} = \begin{cases}1 & k\ge0 \\ 0 & k<0\end{cases}$$ to the $r_2=R_1=1<|q|<\infty=R_2$ ones $$a_k^{(2)} = \begin{cases}0 & k > 1 \\ -1 & k\le 1\end{cases}$$ without the trick I used, i.e. only from the coefficients (and convergence radii)?

  • $\begingroup$ related: How to properly translate the coefficients of a Taylor series? $\endgroup$ Commented Sep 24, 2013 at 13:13
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    $\begingroup$ this is actually an excellent question. I'm teaching complex analysis at the moment, I'd love to be able to offer even a partial answer to my class. My general idea is rather low-tech. I just replace $z-z_o$ with $(z-z_1)+z_1-z_o$ and work out the arithmetic. If $z_1$ is within the domain of the original expansion then I think this will provide a continuation. But, I'm not so sure this approach is ammenable to your problem. $\endgroup$ Commented Oct 11, 2014 at 11:37
  • $\begingroup$ @JamesS.Cook Thanks :) Yes, that should work. My other linked question might basically does that $\endgroup$ Commented Oct 11, 2014 at 11:42
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    $\begingroup$ I think it would be good to work an example where there is a singularity on the edge. Perhaps working on something specific would be enlightening. That said, textbooks are suspiciously void of examples with details and the topic of Monodromy occupied the efforts of mathematicians far my superior the better part of the nineteenth century. I must think on your question. $\endgroup$ Commented Oct 11, 2014 at 11:51
  • $\begingroup$ Maybe the representation $c\prod_k(z-z_k)^{n_k}$ of a meromorphic function is "better" for global validity... $\endgroup$ Commented Oct 27, 2014 at 19:14

3 Answers 3


To begin, the question is what is known. If we are given $f(z)$ on $\mathbb{C} \cup \{\infty \}$, except at finitely many isolated singular points, then Laurent's Theorem provides formulas for the coefficients in terms of the integral of $f(z)$ around a circle in each annulus centered about the point in question. So, the method to find the Laurent series in the next ring has nothing to do with the coefficients in the given ring. So, certainly the global formula for $f(z)$ is not given as that is not the spirit of the question.

Suppose we study Laurent series about $z_o$ then if $f(z)$ is holomorphic (analytic) for $r < |z-z_o| < t$ then there exists a Laurent series decomposition of $f(z)$ for each $z$ in the annulus: $$ f(z) = \sum_{n=-\infty}^{\infty}a_n (z-z_o)^n$$ where for any $s \in (r,t)$ $$ a_n = \frac{1}{2\pi i} \int_{|z-z_o|=r} \frac{f(z)}{ (z-z_o)^{n+1}}\, dz $$ Notice, these coefficients are unique for a given center and a given annulus. The choice of $s$ does not alter the integral above because the integrand is holomorphic within the annulus so the deformation from one radius to another does not alter the value.

So, what I take from this, whatever method you use to discover the coefficients of the Laurent series in the new annulus, once you find them at any point, you've found them everywhere. They are unique to each ring.

It follows that we can use the recentering trick somewhat indirectly. This picture gives the idea: the red dots are the isolated singular points. We begin with the Laurent series on the yellow ring and we want to discover the coefficients on the blue annulus.

Laurent Recentering

If we can obtain values for $f(z)$ on a contour in the blue ring which is smoothly deformable to a circle in the blue ring then that means by Laurent's Theorem we can derive the coefficient which uniquely prescribe the Laurent decomposition with center $z_o$ in the blue annulus. So, Weierstrauss' technique of recentering would seem to get us the values for a continuation to the brown annulus which gives us some of the values we need. Now, what I am not sure of at the moment, is, if the singularities block us from obtaining the closed contour of values which we need. In any event, actually carrying this out for a specific example seems rather involved... unless... we obtain the global formula for $f(z)$ as in your geometric series example. Then the expansion naturally flows from our knowledge of the global formula which is not really what you're after.

  • $\begingroup$ You understood my question perfectly - were the global formula known there'd be no need to attempt obtaining the blue coefficients from the yellow ones (nice illustration btw). So far I got the impression no analytical formula is possible, merely an algorithm: Slightly move the yellow annulus centre to obtain values in a small part $B_1$ of the blue annulus (if the radius shrunk instead of growing, you moved towards a singularity and need to choose a different direction). Now, choose a circle inside $B_1$ and Laurent-expand there to obtain values in a larger part $B_2$. Rinse and repeat... $\endgroup$ Commented Oct 27, 2014 at 19:11
  • $\begingroup$ ...until you got values along a closed curve in the blue annulus (the outer radius of which you actually don't know yet) and use that path to Laurent-expand around the original centre. However there doesn't seem to be a guarantee that one doesn't accidentally skip the blue annulus and obtain the coefficients from the second-to-next annulus or even further out... $\endgroup$ Commented Oct 27, 2014 at 19:13
  • $\begingroup$ I started an answer elaborating on my previous comments, though I'm not sure when/if I'll complete it $\endgroup$ Commented Oct 29, 2014 at 13:14

(Expanding my comment on James' answer)

James already provided a nice picture, which I'll build upon here:

At the moment, I don't know of an analytical formula, but the algorithm is this:

  • Given: the Laurent-series in the yellow zone, delimited by $r<|z-z_0|<R$
  • Sought: the one in the next-outer blue zone, delimited by $R<|z-z_0|<{\color{blue}{R_b}}$ (it's straightforward to modify all this for switching to a smaller annulus)
  • Shift the centre to a point $z_1$ which lies not in direction of a singularity that causes $R$ to be $R$ (which can be verified by checking the new $R_1$)
  • Use the $z_1$-series to obtain values in a small section of the blue annulus around $\color{lime}{z_3}$, put them into the formula for a Laurent-series expansion (actually, Taylor, since we avoided the singularity - to do expand around the singularity instead for fewer steps?) in that small section of blue, with convergence radius $\color{lime}{R_2}$
  • move along the yellow-blue interface towards the convergence boundary at $\color{orange}{z_3}$, Taylor-expand there
  • go on until a full circle is available (when approaching singularities on the interface, go round them via the blue zone)
  • use the Laurent-expansion with that bumped-circle around $z_0$

Ok, now let's try to get this more analytical:

The series representation around $z_l$ is

\begin{align} f_l(z) &= \sum_{k=-\infty}^\infty c_k^{(l)} (z-z_l)^k = f(z)\Big|_{r_l<|z-z_l|<R_l}, \\ &\quad\text{and in its convergence zone we have} \\ c_k^{(l)} &= \frac1{2\pi i}\oint\limits_{r_l<|z-z_l|<R_l}\frac{f(z)\,dz}{(z - z_l)^{k+1}} \quad \Big| \quad z = z_l + re^{i\phi},\ dz = ir e^{i\phi}\,d\phi,\ r\in(r_l,R_l) \\ &= \frac1{2\pi r^k}\int_0^{2\pi}f(z_l+re^{i\phi})\cdot e^{-ik\phi}\,d\phi. \tag{c}\label{c} \end{align}

(Note how $\eqref{c}$ is a Fourier transform along a circle, see this question for more on that)

Now, let's use a slightly different centre $z_m = z_l - d_m$ and choose $r\in(r_l+|d_m|,R_l-|d_m|)$ (which implicitly requires $|d_m|<\frac{R_l-r_l}2$ to make sense) so we remain in the yellow annulus: \begin{align} c_k^{(l)} &= \frac1{2\pi r^k}\int_0^{2\pi}\Big[\sum_p c_p^{(m)}\underbrace{(re^{i\phi} + d_m)^p}_{=\sum\limits_{n=0}^p \binom pn d_m^{p-n}r^ne^{in\phi}}\Big]\cdot e^{-ik\phi}\,d\phi \quad\Bigg|\quad \int_0^{2\pi}e^{i(n-k)\phi}\,d\phi = 2\pi\delta_{nk} \\ &= \sum_{p=k}^\infty \binom pk c_p^{(m)}d_m^{p-k}. \end{align}

Let's take $d_m$ to be infinitesimal, so we have

$$c_k^{(l)} \dot= c_k^{(m)} + (k+1)c_{k+1}^{(m)}d_m$$

where $\dot=$ denotes equal up to $\mathcal O(d_m^2)$.

Ok, so the next step is obtaining the new convergence radii to make sure we didn't accidentally move towards a singularity:

\begin{align} \frac1{R^{(l)}} &= \limsup_{k\to\infty}|c_k^{(l)}|^{\frac1k} \\ &= \limsup_{k\to\infty}\underbrace{\Big|c_k^{(m)} + (k+1)c_{k+1}^{(m)}d_m\Big|^{\frac1k}}_{\dot=\Big(|c_k^{(m)}|^2 + 2(k+1)\Re\big[(c_k^{(m)})^*c_{k+1}^{(m)}d_m\big] \Big)^{\frac1{2k}}} \\ &\dot= \limsup_{k\to\infty} |c_k^{(m)}|^{\frac1k}\Bigg(1+\tfrac{k+1}{k}\underbrace{\frac{\Re\big[(c_k^{(m)})^*c_{k+1}^{(m)}d_m\big]}{|c_k^{(m)}|^2}}_{=\Re\frac{c_{k+1}^{(m)}d_m}{c_k^{(m)}}}\Bigg) \\ &\le \frac1{R^{(m)}}\Big(1+\Re\Big[d_m \limsup_{k\to\infty}\tfrac{k+1}k\tfrac{c_{k+1}^{(m)}}{c_k^{(m)}}\Big]\Big) \\\Rightarrow R^{(l)} & \ge R^{(m)}\underbrace{\Big(1+\Re\Big[d_m \limsup_{k\to\infty}\tfrac{k+1}k\tfrac{c_{k+1}^{(m)}}{c_k^{(m)}}\Big]\Big)^{-1}}_{\dot=1-\Re\Big[d_m \limsup_{k\to\infty}\tfrac{k+1}k\tfrac{c_{k+1}^{(m)}}{c_k^{(m)}}\Big]} \end{align}

I'm sure that can be improved somehow...

To be continued (next step: calculate new convergence radius to make sure one doesn't approach the interface-singularities)

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    $\begingroup$ curses. I had hoped the bounty might precipitate some additional comments from some of the experts lurking about... $\endgroup$ Commented Nov 4, 2014 at 4:05
  • $\begingroup$ @JamesS.Cook thanks anyway :) it got some attention, but apparently the problem is not that straightforward. which makes it all the more interesting. maybe it'll just take some more time, I'll definitely improve my answer at some point $\endgroup$ Commented Nov 4, 2014 at 9:11

Here is another approach.

Let $A_1 = \{ r_1<|z|<r_2 \}$, $A_2 = \{ r_2<|z|<r_3 \}$, $A_3 = \{ r_1<|z|<r_3 \}$. Let $f(z)$ be a meromorphic function on $A_3$ which is holomorphic on $A_1$ and $A_2$. Write the Laurent expansion of $f$ on $A_1$: $$f(z) = \sum_{n \in \mathbb{Z}} a_n z^n.$$ Let $z_1,\dots,z_r \in \{ |z| =r_2 \}$ be the poles of $f$ and $n_1,\dots,n_r$ their order. Let $$g(z)=\prod_{i=1}^r (z-z_1)^{-n_i}.$$ Calculate the Laurent expansion of $g$ on $A_1$: $$g(z) = \sum_{n \in \mathbb{Z}} \alpha_n z^n.$$ Then $f-g$ is holomorphic on $A_3$ and has Laurent expansion $$f(z)-g(z) = \sum_{n \in \mathbb{Z}} (a_n-\alpha_n) z^n.$$ Calculate the Laurent expansion of $g$ on $A_2$: $$g(z) = \sum_{n \in \mathbb{Z}} \beta_n z^n.$$ We conclude that the Laurent expension of $f(z)$ on $A_2$ is $$f(z) = \sum_{n \in \mathbb{Z}} (a_n-\alpha_n) z^n + \sum_{n \in \mathbb{Z}} \beta_n z^n = \sum_{n \in \mathbb{Z}} (a_n-\alpha_n+\beta_n) z^n.$$

This seems quite trivial, but the trick is I don't know if there is a formula for $(z_i)_i$ and $(n_i)_i$ in function of the $(a_n)_n$. Of course if you can compute $f$ on $A_2$, then you can easily get the $(z_i)_i$ and $(n_i)_i$ with an integral.

  • $\begingroup$ Unless I'm missing something, I think you need to do a little more work in defining $g$: it'll have the same poles as $f$ to the same order, but it need not have the same negative Laurent coefficients at those poles so $f-g$ might not be holomorphic. Alternately, you could use your $g$ and look at $f(z)/g(z)$ instead of $f(z)-g(z)$, but then the formal computation gets a lot messier... $\endgroup$
    – Micah
    Commented Oct 31, 2014 at 20:37
  • $\begingroup$ Interesting Ansatz, thanks :) But I agree with @Micah, your definition of $g$ won't compensate the singularities entirely. Either you know all singularities (instead of just the ones at the interface) and invoke Mittag-Leffler, or you directly craft $g$'s Laurent-series (actually, in $A_1$, Taylor-series) fittingly, though I don't yet know how one could achieve that. I'll try and give this some more thought $\endgroup$ Commented Oct 31, 2014 at 21:01
  • $\begingroup$ @Micah you are right, I have messed up here. If you write $f_i(z)=\sum_{n=-n_i}^{-1} u^i_n (z-z_i)^n$ the singular part of $f$ at $z_i$ then take $g(z):= \sum_{i} f_i = \sum_i \sum_{n=-n_i}^{-1} u^i_n (z-z_i)^n$. $\endgroup$
    – user10676
    Commented Oct 31, 2014 at 21:11

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