2
$\begingroup$

For a complex power series expansion $$ f(z)=\sum_{n=0}^{\infty} c_n(z-a)^n $$ with convergence radius $r$, we have that for $|z-a|<r$: $$ f'(z)=\sum_{n=1}^{\infty} nc_n (z-a)^{n-1}$$ (this is a theorem that I found in the book I use).

This means that the convergence radius of $\sum_{n=1}^{\infty} nc_n (z-a)^{n-1}$ is at least as big as that of $\sum_{n=0}^{\infty} c_n(z-a)^n $.

My question is: is the opposite also true? So, if $\sum_{n=1}^{\infty} nc_n (z-a)^{n-1}$ has convergence radius $\bar{r}$, can we state that $\sum_{n=0}^{\infty} c_n(z-a)^n $ has a convergence radius that is $\geq \bar{r}$. My guess is yes, because we can express a integration process as a differentiation process in the complex plane, but I'm not sure. And if yes, how would I go around proving this?

Thanks in advance.

$\endgroup$
1
$\begingroup$

Yes it is true. A way to prove it is using the root test to find the radius of convergence and to note that it gives the same result for $(c_n)_n$ and $((n+1)c_{n+1})_n$.

$\endgroup$
  • $\begingroup$ Thanks. So, this means that the expansions of $f$ and $f'$ always have the exact same convergence radius? $\endgroup$ – user161518 Mar 8 '15 at 15:53
  • $\begingroup$ Yes, if you have a function given by a powers series (with positive radius of convergence) than the function is differentiable its derivative is given by deriving the terms of the power series and this new power series has the same radius of convergence. $\endgroup$ – quid Mar 8 '15 at 15:57

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.