How to expand an analytical function on a given curve?

Question

Since the Taylor expansion of a function that happens to have singularities for nearby values (including complex ones) does not globally converge, I am looking for a different way to expand a function that is analytical in a given interval $[a,b]$, or more generally, along a (finite) curve $\gamma:[0,1]\to\Gamma([0,1]) \subset \mathbb C$, that is valid for the whole interval/curve. Does such an expansion exist?

Answer 1

I'm not sure what kind of expansion you are looking for. But for example, if the curve is simple (so that the complement of the curve is connected), Runge's theorem shows that $f$ can be written as a uniform limit (on the curve) of polynomials. In general though, there is no explicit fornula for the approximating polynomials.

Answer 2

For a simple closed curve

$$\gamma:[a, b] \to \Gamma, t\mapsto \gamma(t), \\ \gamma(b)=\gamma(a), \\ \forall c\neq d\in(a,b): \gamma(c)\neq\gamma(d)$$

note that since there are no singularities on the curve, we can transform the curve to a circle along which a Laurent series can be obtained and converges in at least a small annulus around that circle. So let's consider

$$g(e^{i\phi}) := f(\gamma(a+\tfrac{b-a}{2\pi}\phi))$$

and calculate its Laurent series coefficients $g_k$:

$$g_k = \frac1{2\pi}\int\limits_0^{2\pi} g(e^{i\phi})e^{-ik\phi}d\phi$$

(that looks like a Fourier transform, doesn't it?)

Therefore we have

$$\begin{align*} f(\gamma(a+\tfrac{b-a}{2\pi}\phi)) &= \sum_{k=-\infty}^\infty g_k e^{ik\phi} \\\text{or}\quad f(\gamma(t)) &= \sum_{k=-\infty}^\infty g_k e^{2\pi ik\frac{t-a}{b-a}} \end{align*}.$$

So the representation of a function that is analytical along a closed simple curve is a Fourier series (along that curve). Of course convergence on the unit circle limits the coefficients stronger than all valid Fourier series:

$$r<1<R\quad\text{where}\quad r = \limsup_{k\to\infty}|g_{-k}|^{\frac1k}, \quad \frac1R = \limsup_{k\to\infty}|g_k|^{\frac1k}$$

It is also interesting to note that the convergence radii of $g$'s Laurent series determine how far away from $\gamma$ you can go (by analytical continuation of $\gamma$ to complex values via $\phi\to \phi+i\ln\rho$ where $\rho$ is withing the convergence annulus radii) while this series for $f(\gamma(t))$ still converges.

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Concerning open curves, I have some ideas that I have yet to investigate:

• There exists a final curve which completes the open curve to one that fulfills the conditions above (otherwise there would be infinite rays of singularities) so just guess one completion until things work
• Generalize the above method for self-intersecting curves, extend to a curve that walks back on itself
• Transform the curve such that it is covered by some annulus within which a Laurent series converges

To be continued...