5
$\begingroup$

Consider the branch of log defined on $\mathbb{C}$ with the negative real axis and origin removed. I was told that its Taylor series about the point $z_0 = -2 + i$ converges in a radius $\sqrt5$, which means the Taylor series actually converges for points on the negative real axis (not in its domain).

I can understand why this could be possible: the real function $f(x) = x^2$ defined on the real interval $|x| < 1$ has a Taylor series that converges everywhere to $g(x) = x^2$, where $g$ is defined on the entire real line.

But how do I actually prove that there is a branch of log that the Taylor series about $z_0 = -2 + i$ converges to? Can the Taylor series converge to a function that is not a branch of log? Why can't the radius of convergence $\sqrt5$ be increased to include the origin?

$\endgroup$
2
  • 3
    $\begingroup$ It might be clearer to think about the branches of $\log$ as sections of its Riemann surface. It doesn't know where you've chosen to make the branch cut; it just knows that the nearest singularity it sees is the origin. If you expand it in power series then the radius of convergence will be as large as possible on the Riemann surface, though this may not agree with where you've chosen the branch cut to lie. To ensure that the power series at $-2+i$ agrees with your branch of $\log$ (and your branch cut), you only consider its values on the relevant side of the branch cut. $\endgroup$ Nov 20, 2013 at 16:41
  • $\begingroup$ (In your case, you would only consider the values of the power series where $\operatorname{Re}(z) > 0$). $\endgroup$ Nov 20, 2013 at 16:44

1 Answer 1

4
$\begingroup$

Why can't the radius of convergence $\sqrt{5}$ be increased to include the origin?

The real part of logarithm (no matter what branch you take) is $\log |z|$. This is unbounded as $z\to 0$. But a power series is bounded on compact subsets of its disk of convergence. Therefore, no power series representing the logarithm (whatever branch) can have $0$ inside of its disk of convergence.

Can the Taylor series converge to a function that is not a branch of $\log$?

By definition, anything you get from Taylor series of $f$ is some branch of $f$, because it's related to $f$ by analytic continuation.

how do I actually prove that there is a branch of log that the Taylor series about $z_0=−2+i$ converges to?

There is nothing magical about negative real axis: you can cut the plane along any other half-line from $0$ to $\infty$ and define a branch of $\log$ in the slit plane (by interpreting $\arg z$ there). For example, the half-line $\{(2-i)t:t\ge 0\}$ would do. Or simply define $$\operatorname{Log}z =\log( z/(-2+i))+\log (-2+i)$$ where $\log$ is the principal branch.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .