How to switch to a Laurent series' next convergence ring?

Solution 1:

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.

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.

Solution 2:

(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)

Solution 3:

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.