2
$\begingroup$

My professor gave this question to the class for practice, to not be turned in for grading. I tried doing what he said in the hint, but am just getting nowhere. Could you please give any pointers, tips or suggestions? I want to do what the tip says, to use the power series expansion and then apply Liouville's Theorem. Thank you.
enter image description here

$\endgroup$
  • $\begingroup$ Or just write down Cauchy's integral formula for the kth derivative of f(z) at z = 0 as a contour integral over a circle with radius R. $\endgroup$ – Count Iblis Apr 4 '17 at 23:24
  • $\begingroup$ Could you explain in more detail please? $\endgroup$ – kemb Apr 4 '17 at 23:27
  • $\begingroup$ I want to use the power series expansion. $\endgroup$ – kemb Apr 4 '17 at 23:27
4
$\begingroup$

Let us write the Taylor series of $f$ as

$$ f(z) = \sum_{n=0}^{\infty} a_n z^n. $$

Define $p(z) = \sum_{k=0}^{n-1} a_k z^k$ to be the polynomial which is the sum of the first $n$ terms in the Taylor series of $f$. Then we have

$$ f(z) = p(z) + a_n z^n + \sum_{k=1}^{\infty} a_{n+k} z^{n+k}. $$

Consider the function $h(z) := \frac{f(z) - p(z)}{z^n}$. This function is well-defined on $\mathbb{C} \setminus \{ 0 \}$ and in fact extends to an entire function on $\mathbb{C}$ using the power series expansion

$$ h(z) = a_n + \sum_{k=1}^{\infty} a_{n+k} z^{k}. $$

In addition,

$$ |h(z)| = \frac{|f(z) - p(z)|}{|z|^n} \leq M + \frac{|p(z)|}{|z|^n} \leq M + 1 $$

for $z$ large enough. Hence, by Liouville's theorem we have $h(z) \equiv h(0) = a_n$ and so $f(z) = p(z) + a_n z^n$ for all $z \in \mathbb{C}$.

$\endgroup$
  • $\begingroup$ Hi, but could you use the constants M, r, and n in the question and use the format of the question. $\endgroup$ – kemb Apr 4 '17 at 23:39
  • $\begingroup$ This explanation is a bit advanced for me $\endgroup$ – kemb Apr 4 '17 at 23:39
  • $\begingroup$ Could you please simply it, going along with the hint given in the question and using the exact format, variables given in question, just trying to understand. $\endgroup$ – kemb Apr 4 '17 at 23:40
  • $\begingroup$ @BOB: I've followed the hint. The first part consists of manipulating $f(z)$ using the Taylor series. We subtract from $f$ the first $n$ terms of the series (this is $p$) and divide by $z^n$. Then we show that the resulting function is in fact entire and so is constant and then deduce that $f$ is equal to the first $n + 1$ terms of the power series - i.e it is a polynomial of degree $\leq n$. The $n$ and the $M$ are the same as in the question, and "for $z$ large enough" means $|z| > \max(R, C)$ where $C$ is a constant for which $\frac{|p(z)}{|z|^n} \leq 1$ for all $|z| > C$. $\endgroup$ – levap Apr 4 '17 at 23:45
  • $\begingroup$ @BOB: In any case, I've edited my answer and I hope the edited version will be easier for you to follow. $\endgroup$ – levap Apr 4 '17 at 23:50

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.