29
$\begingroup$

The question is #$14$ from Chapter $2$ in Stein and Shakarchi's text Complex Analysis:

Suppose that $f$ is holomorphic in an open set containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $$\sum_{n=0}^\infty a_nz^n$$ denotes the power series expansion $f$ in the open unit disc, then $$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0.$$

I've shown that we can take $z_0=1$ without a loss of generality, but I'm having trouble showing the proof otherwise. One of the problems I'm having is because we aren't told the definition of a pole except that it is a place where the function isn't holomorphic. Disregarding this fact, the other problem I'm running into is that I don't know the order of the pole.

Making some additional assumptions, including that the pole is simple so we can write $F(z)=(z-1)f(z)$ as a holomorphic function, we see that $$F(z)=-a_0+z(a_0-a_1)+z^2(a_1-a_2)+\cdots$$ This almost gets me to the end with these added assumptions, but I don't think it's quite enough (why do we know some of the $a_i$'s aren't $0$, for example).

On another note, if we know $\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ exists, then it is easy to see that $\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}=1=|z_0|$; I, however, do not see why the limit must exist.

Are there any hints that someone can provide? Even a solution would be nice, especially if one can avoid making any assumptions about what a pole is or is not.

EDIT: So there isn't any confusion, I know the definition of a pole and I'm inclined to believe that the problem, as stated, necessarily has a pole at $z_0$. The problem is that the exercise is in Chapter $2$, and poles are introduced in Chapter $3$.

$\endgroup$
10
  • 2
    $\begingroup$ I don't have Stein and Shakarchi on me, but it's certainly not correct that a pole is the same thing as a place where the function is not holomorphic. Being a pole is a much more restrictive condition: it means that $(z-1)^m f$ is holomorphic at $z = 1$ for some positive integer $m$. $\endgroup$ Jun 14, 2013 at 23:32
  • $\begingroup$ I didn't actually sit down and work the whole thing out, but I think the approach you sketched above is at least approximately the right way to attack the problem. $\endgroup$ Jun 14, 2013 at 23:33
  • $\begingroup$ @Clayton There's more involved with being a pole than just failing to be holomorphic. For example $e^{1/z}$ fails to be holomorphic at $0$ but does not have a pole there. You will necessarily need this information to solve the problem. $\endgroup$
    – Potato
    Jun 14, 2013 at 23:52
  • $\begingroup$ @Clayton I'm fairly sure this is false if 'pole' is replaced by 'essential singularity,' so as I said, you will require the characterization of a pole Daniel mentioned to solve this. This means it cannot be solved without referring to material in chapter 3. (These books are good, but not the most well-edited things in the world. It does not surprise me there's a minor glitch like this.) $\endgroup$
    – Potato
    Jun 15, 2013 at 0:00
  • $\begingroup$ @Clayton So if the problem is really finding a counterexample in the case of an essential singularity, you should edit the question to say that. As it's written right not, it just looks like you're confused about what a pole is. $\endgroup$
    – Potato
    Jun 15, 2013 at 0:11

4 Answers 4

15
$\begingroup$

I will assume we can write

$$f(z) = \frac{c}{z_0-z} + \sum_{n=0}^{\infty} b_n z^n$$

for some value of $c \ne 0$, and $\lim_{n \to \infty} b_n = 0$. Then

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

where

$$a_n = b_n + \frac{c}{z_0^{n+1}}$$

Then

$$\begin{align}\lim_{n \to \infty} \frac{a_n}{a_{n+1}} &= \lim_{n \to \infty} \frac{\displaystyle b_n + \frac{c}{z_0^{n+1}}}{\displaystyle b_{n+1}+ \frac{c}{z_0^{n+2}}}\\ &= \lim_{n \to \infty} \frac{\displaystyle \frac{c}{z_0^{n+1}}}{\displaystyle \frac{c}{z_0^{n+2}}}\\ &= z_0\end{align}$$

as was to be shown. Note that the second step above is valid because $z_0$ is on the unit circle.

For a nonsimple pole, we may write

$$f(z) = \frac{c}{(z_0-z)^m} + \sum_{n=0}^{\infty} b_n z^n$$

for $m \in \mathbb{N}$. It might be known that

$$(1-w)^{-m} = \sum_{n=0}^{\infty} \binom{m-1+n}{m-1} w^n$$

Then

$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{n+1}{n+m} z_0$$

EDIT

@TCL observed that we can simply require that $b_n z_0^n$ goes to zero as $n \to \infty$. Then for a simple pole

$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{\displaystyle b_n z_0^n + \frac{c}{z_0}}{\displaystyle b_{n+1} z_0^n + \frac{c}{z_0^2}}$$

which you can see goes to $z_0$.

$\endgroup$
14
  • $\begingroup$ @Potato: thanks! $\endgroup$
    – Ron Gordon
    Jun 14, 2013 at 23:52
  • $\begingroup$ I thought $a_n = b_n + \frac{c}{z_0^{n+1}}$? Why is there a $(-1)^n$? $\endgroup$
    – TCL
    Jun 15, 2013 at 0:04
  • $\begingroup$ @RonGordon: The procedure of taking the limit of $b_n$ and not of the rest is suspect and probably wrong. $\endgroup$ Jun 15, 2013 at 0:04
  • $\begingroup$ @Clayton: good point, I mistakenly used the wrong expansion and saw abs value signs. Fixed. $\endgroup$
    – Ron Gordon
    Jun 15, 2013 at 0:05
  • $\begingroup$ @TCL: a mistake. Fixed. $\endgroup$
    – Ron Gordon
    Jun 15, 2013 at 0:06
11
$\begingroup$

This is just an attempt to give a rigorous proof of Ron Gordon's answer.

Suppose the order of the pole at $z_0$ is $m$. Then for some constants $c_0,\ldots,c_m, c_m\neq 0,$ $$g(z)=f(z)-\left(\frac{c_m}{(z_0-z)^m}+\cdots +\frac{c_1}{(z_0-z)}\right)$$ is analytic at $z_0$. Since $f(z)$ is analytic on an open set containing the unit disk except at $z_0$, we see that $g(z)$ is analytic on an (possibly different) open set containing the unit disk. So $$g(z)=\sum_{n=0}^\infty b_nz^n$$ has radius of convergence greater than 1, in particular the series converges at $z_0$ and $\lim_n b_nz_0^n=0$. So for $|z|<1$, $$f(z)=\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty b_nz^n+\left(\frac{c_m}{(z_0-z)^m}+\cdots +\frac{c_1}{(z_0-z)}\right)$$ For $1\le k\le m$, we have $$\frac{c_k}{(z_0-z)^k}=\sum_{n=0}^\infty \frac{(k)_nc_k}{n!z_0^{n+k}} z^n$$ where $(k)_n=k(k+1)\cdots(k+n-1)$ and $(k)_0=1$. It follows that $$a_n=b_n+\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{n+k}}$$ and \begin{align*}\frac{a_n}{a_{n+1}}&=\frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{1+k}}+b_{n+1}z_0^n}\\ &= z_0 \frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{k}}+b_{n+1}z_0^{n+1}} \end{align*} Now divide the numerator and denominator of the last expression by $$\frac{(m)_{n+1}}{(n+1)!}$$, which is equal to 1 when $m=1$ and approaches infinity as $n\to\infty$ if $m>1$. Furthermore, for $1\le k\le m$, $$\frac{(k)_n}{n!}\cdot \frac{(n+1)!}{(m)_{n+1}}=\frac{(k)_n (n+1)}{(m)_n (n+m)}$$ approaches 0 as $n\to\infty$ if $k<m$ and approaches 1 if $k=m$. It follows that the fraction $$\frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{k}}+b_{n+1}z_0^{n+1}}$$ approaches 1 as $n\to\infty$, and the proof is complete.

EDIT. If $k<m$, $$\frac{(k)_n }{(m)_n }=\prod_{i=0}^{n-1} \frac{k+i}{m+i}=\prod_{i=0}^{n-1} \left(1-\frac{m-k}{m+i}\right)\to 0$$ because $\sum_{i=0}^\infty \frac{m-k}{m+i}=\infty$.

$\endgroup$
7
$\begingroup$

Suppose that $$ \begin{align} f(z)(z-z_0)^n &=(z-z_0)^n\sum_{k=0}^\infty a_kz^k\\ &=\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}z^jz_0^{n-j}\sum_{k=0}^\infty a_kz^k\\ &=\sum_{k=0}^\infty\left(\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right)z^k \end{align} $$ converges in a neighborhood of $z_0$. Thus, for some $r\lt1$, we have $$ \left|\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right|\le c\,r^k $$ That is, the the $n^{\text{th}}$ finite difference of the sequence $a_kz_0^k$ satisfies $$ \Delta^na_kz_0^k=O(r^k) $$ Inverting the finite difference operator yields that there is an $n-1$ degree polynomial $P$ so that $$ a_kz_0^k=P(k)+O(r^k) $$ Taking the limit of the ratio of terms yields $$ \lim_{k\to\infty}\frac{a_kz_0^k}{a_{k+1}z_0^{k+1}}=\lim_{k\to\infty}\frac{P(k)+O(r^k)}{P(k+1)+O(r^{k+1})}=1 $$ That is, $$ \lim_{k\to\infty}\frac{a_k}{a_{k+1}}=z_0 $$


Inverting Finite Difference Operators

If we define the finite difference operator as $$ \Delta a_k=a_k-a_{k-1} $$ then, as with indefinite integrals, when inverting $\Delta$, we need to include a constant: $$ \Delta^{-1}a_k=c+\sum_{j=1}^k a_j $$ Suppose that $\Delta a_k=P_n(k)+O(r^k)$, where $P_n$ is a degree $n$ polynomial and $0\le r\lt1$. Since the sum of a degree $n$ polynomial is a degree $n+1$ polynomial and the sum of $O(r^k)$ is $c+O(r^k)$, $a_k=P_{n+1}(k)+O(r^k)$.

Iterating, we get that if $\Delta^n a_k=O(r^k)$, then $a_k=P_{n-1}(k)+O(r^k)$.

$\endgroup$
7
  • $\begingroup$ I don't see why $ \left|\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right|\le c\,r^k $ $\endgroup$
    – TCL
    Jun 15, 2013 at 19:56
  • $\begingroup$ @TCL: The power series for an analytic function will converge out to the nearest pole. For example, the radius of convergence for the power series for $\frac1{1-z}$ is $1$ because of the pole at $z=1$. It is given that $(z-z_0)^nf(z)$ is holomorphic in an open set containing the closure of the unit disc for some $n$ ($z_0$ is not an essential singularity since it is a pole). That means that the radius of convergence of $$\sum_{k=0}^\infty\left(\sum_{j=0}^n(-1)^{n-j} \binom{n}{j}a_{k-j}z_0^{n-j}\right)z^k$$ is greater than $1$. $\endgroup$
    – robjohn
    Jun 15, 2013 at 22:12
  • $\begingroup$ Thanks.+1. Where can I find info about inverting finite difference operator? $\endgroup$
    – TCL
    Jun 16, 2013 at 1:57
  • 1
    $\begingroup$ @TCL: I have added a small section on inverting finite difference operators on the relevant functions. $\endgroup$
    – robjohn
    Jun 16, 2013 at 4:38
  • $\begingroup$ I think it should be $$f(z)(z-z_0)^n=\sum_{k=0}^\infty\left(\sum_{j=0}^k(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right)z^k,$$ after you commutate the double summation. $\endgroup$
    – Riemann
    May 20, 2021 at 8:00
1
$\begingroup$

We can do this by induction on $m$ where $m$ is the order of the pole.

For a simple pole assume we have proved the result.

Then suppose $m>1$, then $(z-z_0)^mf(z)=g(z)$ where $g(z)$ is holomorphic on a disk of radius bigger than $1$.

Consider $h_i(z):= (z-z_0)^{m-i}f(z)$ for $1\leq i \leq m$.
Each $h_i$ is holomorphic on $B(0,1)$ and has a pole of order $i$ at $1$.

$h_i(z)= \sum_{n=0}^{\infty} a_{in} z^n, |z|<1$

Since $h_m(z)=f(z)$ we would be done if we show $\lim_{n\to \infty} \frac{a_{in}}{a_{i(n+1)}}=z_0 \Rightarrow \lim_{n\to \infty} \frac{a_{(i+1)n}}{a_{(i+1)n}}=z_0$ Note: We use the induction hypothesis to get this for $h_1$.

To this end, we make the following observation.

$(z-z_0)h_{i+1}(z)=h_{i}(z)$

Comparing coefficients we deduce

$a_{in}=a_{(i+1)(n-1)}-z_0a_{(i+1)n}$

and hence the following

$a_{(i+1)n}z_0^{n+1}= -(a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n)$

Now, $\lim_{n\to \infty} \frac{a_{(i+1)n}}{a_{(i+1)n}} = z_0 \lim_{n\to \infty} \frac{a_{(i+1)n}z_0^n}{a_{(i+1)n}z_0^{n+1}}$.

Thus it is enough to show that

$\lim_{n\to \infty} \frac{a_{(i+1)n}z_0^n}{a_{(i+1)n}z_0^{n+1}}= 1$

$\lim_{n\to \infty} \frac{a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n}{a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n + a_{i(n+1)}z_0^{n+1}}=1$

$\lim_{n \to \infty} \frac{S_n(z_0)}{S_{n+1}(z_0)}=1$

where $S_n(z)$ is the $n^{th}$ partial sum at $z$ of $h_i$.

But the above limit is same as $\lim_{n \to \infty} \lim_{z\to z_0} \frac{(z-z_0)^iS_n(z)}{(z-z_0)^iS_{n+1}(z)} $

Then since the inner term converges uniformly in $z$ (independent of $n$) we can interchange the limits, which gives

$ \lim_{z\to z_0} \lim_{n \to \infty} \frac{(z-z_0)^iS_n(z)}{(z-z_0)^iS_{n+1}(z)}=\lim_{z \to z_0} \frac{g(z)}{g(z)}=1$

Please let me know if anything is unclear.

$\endgroup$

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.