0
$\begingroup$

I'm having trouble with the concept of analytic continuation of power series beyond the radius of convergence. For example for:

$$f(z)=z-z^2+z^3-z^4+\cdots=\sum_{n=0}^\infty(-1)^nz^{n+1}$$

I get the convergence radius

$$R=\frac{1}{\lim\sup\sqrt[n]{|(-1)^n|}}=1$$

I've seen the proof that there's at least a singular point on the frontier, but I'm not sure how to show to what extent $f$ can be analytically continued.

Any explanation or hint would be appreciated!

$\endgroup$
  • 4
    $\begingroup$ This is a geometric series, so you should be able to compute its sum explicitly. (Then you can see how to do the analytic continuation.) $\endgroup$ – mrf Jan 4 '16 at 14:52
  • 3
    $\begingroup$ In this case, you can write down the closed form of $f$, and that immediately gives you an analytic continuation to almost the whole plane. $\endgroup$ – Daniel Fischer Jan 4 '16 at 14:53
4
$\begingroup$

Using the formula for the Geometric series you get $$f(z)=z-z^2+z^3-z^4+...=z(1-z+z^2-z^3+....)=z \cdot \frac{1}{1-(-z)}=\frac{z}{1+z}$$ for $|z| <1$.

Now, $\frac{z}{1+z}$ is analytic on $\mathbb C \backslash \{-1 \}$ and agrees with your power series in your disk of convergence. This is what we mean by analytic continuation.

$\endgroup$
  • $\begingroup$ Thanks! But what happens if $|z|>1$? $\endgroup$ – G. Schiele Jan 5 '16 at 14:07
  • $\begingroup$ @G.Schiele The RHS is a function which is still Analytic. But it is not equal with your given series, as the series is divergent there. This is why it is called a continuation ;) $\endgroup$ – N. S. Jan 5 '16 at 15:35
  • $\begingroup$ Yes, I mean what happens to the original function. If $|z|<1$ we can continue it through the closed form, but we haven't said anything about the $|z|>1$ case. $\endgroup$ – G. Schiele Jan 5 '16 at 15:51
  • 1
    $\begingroup$ @G.Schiele The original function is only defined for $|z| <1$... To keep things simple, let us use $f$ for the original function and $g$ for the closed form. Then $f(z)$ is only defined for $|z| <1$, and nowhere else (as the series is divergent for all $|z| \geq 1$). Meanwhile $g(z)$ is defined on $\mathbb C \backslash \{ -1 \}$ and, on the domain of $f$ we have $$f=g $$ $\endgroup$ – N. S. Jan 5 '16 at 16:16
  • $\begingroup$ There is nothing else you can say about $f$ on $|z| \geq 1$ other than $f$ is NOT DEFINED. $\endgroup$ – N. S. Jan 5 '16 at 16:17
2
$\begingroup$

Multiplying $$f(z)=z-z^2+z^3-z^4+\cdots$$ by $z$ gives $$zf(z)=z^2-z^3+z^4-z^5+\cdots.$$ Adding the two gives $$zf(z)+f(z)=z,$$ so that $$f(z)(z+1)=z,$$ from which we obtain $$f(z)=\frac{z}{z+1}\tag{$\mid z\!\!\mid<1$}.$$ But this expression is analytic on all $\mathbb{C}\setminus\{-1\}$ and agrees with $f(z)$ for $|z|<1$ so the right hand side is the analytic continuation of $f(z)$ to the punctured complex plane.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.