I'm currently teaching myself power series and Taylor's theorem for complex analysis and I'm having trouble answering questions of the following form:

$1)$ Suppose the power series $\sum_{k=0}^{\infty}a_kz^k$has radius of convergence 2. Call the value of the sum $f(z)$ and let $g(z)=\frac{f(z)}{1-z}$.

$a)$ Explain how you know $g(z)$ has a representation of the form $g(z)=\sum_{n=0}^{\infty}b_nz^n$ valid at least for $|z|<1$.

$b)$ For each $n=0,1,2,3,$ find $b_n$ in terms of $a_0,a_1,a_2,a_3,...$

My approach for a:

$$g(z)=f(z)*\frac{1}{1-z}=f(z)*\sum_{n=0}^{\infty}z^n$$ But since $f(z)= \sum_{k=0}^{\infty}a_kz^k$

Then: $$g(z)=\sum_{k=0}^{\infty}a_kz^k *\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}b_nz^n $$

This is from the fact derived by Taylor's theorem that $$\sum_{n=0}^{\infty}c_nz^n=\sum_{k=0}^{\infty}a_nz^n *\sum_{n=0}^{\infty}b_nz^n$$

Not sure how to do part b

$2)$ Suppose $f:\mathbb{C} \to \mathbb{C}$ is analytic on all $\mathbb{C}$, and $$|f^{(3)}(z)|<M$$ for all $z$ in $\mathbb{C}$. Show that $f$ is a polynomial and what can you say about the degree of $f$.

Is this problem also a result of Taylor's theorem? How do I approach it? Thank you.

  • $\begingroup$ Have you met Liouville's theorem yet? It's a fairly straightforward corollary of Taylor's theorem. You can prove (2) by invoking Liouville's theorem, or directly from Taylor's theorem by reproducing some of the ideas in the proof of the former. $\endgroup$ – Joshua Pepper Mar 5 '14 at 1:21
  • $\begingroup$ so this says because $|f^{(3)}(z)|$ is bounded, then $f$ constant? $\endgroup$ – User69127 Mar 5 '14 at 1:34
  • $\begingroup$ No. You can conclude that $f^{(3)}$ is constant, however. $\endgroup$ – Joshua Pepper Mar 5 '14 at 1:34
  • $\begingroup$ Does #1 look right? $\endgroup$ – User69127 Mar 5 '14 at 1:55

Part 1

I don't think your answer to this is correct. You need to justify why we can be sure that $g$ has a power series that converges if $|z|<1$, and you haven't done that.

You have correctly shown that $g$ is a product of two power series: $$g(z)=\sum_{k=0}^{\infty}a_kz^k *\sum_{n=0}^{\infty}z^n$$ We also know that $f(z)=\sum_{k=0}^{\infty}a_kz^k$ has radius of convergence $R_1=2$, and $h(z) = \frac{1}{1-z}$ has radius of convergence $R_2=1$. Therefore, as a product of two power series, we know that the above expression for $g$ converges whenever $|z|< \min(R_1,R_2) = 1$. Note that we haven't yet shown that $g$ has a power series that converges when $|z|<1$ - that is something different.

However, power series converge uniformly within the radius of convergence, so, if $|z|<1$,we can multiply out the above expression to get: $$\begin{align} g(z) &= (a_0 + a_1z + a_2z^2 + \ldots)(1 + z + z^2 + z^3 + \ldots) \\ &= a_0 + (a_1 + a_0)z + (a_2 + a_1+a_0)z^2 + \ldots \end{align}$$ This shows that $g$ has a power series (that converges for $|z|<1$ by the above discussion), and also gives us an expression for the coefficients of the power series, i.e.: $$b_n=\sum^n_{k=0}a_k$$ This simultaneously answers part 1b of your problem. (Note that the problem statement only asked for the case $n=0,1,2,3$.)

Part 2

As hinted at in the comments, you can apply Liouville's theorem to $f^{(3)}$ and integrate three times to deduce that $f$ is a polynomial of degree at most $3$.

  • $\begingroup$ I should probably note that the statement "power series converge uniformly within the radius of convergence" glosses over some analytical details. Uniform convergence can fail on the full disc $|z|<R$, however we can use a sequence of nested, compact sets to recover the results about power series that we need for this kind of discussion. $\endgroup$ – Joshua Pepper Mar 5 '14 at 4:06
  • $\begingroup$ Great answer. Thanks for explaining part 1. Part 2 though how do you integrate $f^{(3)}$? Do i just write some general function? $\endgroup$ – User69127 Mar 5 '14 at 23:10
  • $\begingroup$ By Liouville's theorem, $f^{(3)}$ is constant, i.e. $f^{(3)}=a$ for some $a$. By integrating, $f^{(2)}=az+b$, $f^{(1)}=\frac{1}{2}z^2+bz+c$, $f=\frac{1}{6}az^3+\frac{1}{2}bz^2+cz+d$. However, some of $a,b,c,d$ could be equal to zero, and so we must be careful to note that the only thing we know about the degree of $f$ as a polynomial is that it is at most 3. $\endgroup$ – Joshua Pepper Mar 6 '14 at 7:16

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