4.5 KiB
Math416 Lecture 9
Review
Power Series
Let f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n be a power series.
Radius of Convergence
The radius of convergence of a power series is
R=\frac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}.
New Material on Power Series
Derivative of Power Series
Let f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n be a power series.
Let g(z)=\sum_{n=0}^{\infty}na_n(z-z_0)^{n-1} be another power series.
Then g is holomorphic on D(z_0,R) and g'(z)=f(z) for all z\in D(z_0,R). and f'(z)=g(z).
Proof:
Note radius of convergence of g is also R.
\limsup_{n\to\infty}|na_n|^{1/(n-1)}=\limsup_{n\to\infty}|a_n|^{1/n}.
Let z\in D(z_0,R).
let |z-z_0|<\rho<R.
Without loss of generality, assume z_0=0. Let |w|<\rho.
\begin{aligned}
\frac{f(z)-f(w)}{z-w}-g(z)&=\sum_{n=0}^{\infty}\left[\frac{1}{z-w}\left(a_n(z^n-w^n)\right)-na_nz^{n-1}\right] \\
&=\sum_{n=0}^{\infty}a_n\left[\frac{z^n-w^n}{z-w}-nz^{n-1}\right]
\end{aligned}
Notice that
\begin{aligned}
\frac{z^n-w^n}{z-w}&=\sum_{k=0}^{n-1}z^{n-1-k}w^k \\
&=z^{n-1}+z^{n-2}w+\cdots+w^{n-1}
\end{aligned}
Since
|w^k-z^k|=\left|(w-z)\left(\sum_{j=0}^{k-1}w^{k-1-j}z^j\right)\right|\leq|w-z|k\rho^{k-1}
\begin{aligned}
\frac{z^n-w^n}{z-w}-nz^{n-1}&=(z^{n-1}-z^{n-1})+(z^{n-2}w-z^{n-1})+\cdots+(z w^{n-1}-z^{n-1}) \\
&=z^{n-2}(w-z)+z^{n-3}(w^2-z^2)+\cdots+z^0(w^{n-1}-z^{n-1}) \\
&=\sum_{k=0}^{n-1}z^{n-1-k}(w^k-z^k)\\
&\leq\sum_{k=0}^{n-1}z^{n-1-k}|w-z|k\rho^{k-1} \\
&\leq|w-z|\sum_{k=0}^{n-1}k\rho^{k-1} \\
\end{aligned}
Apply absolute value,
\begin{aligned}
\left|\frac{f(z)-f(w)}{z-w}-g(z)\right|&\leq\sum_{n=0}^{\infty}|a_n||w-z|\left[\sum_{k=1}^{n-1}\rho^{n-1-k}k\rho^{k-1}\right] \\
&=|w-z|\sum_{n=0}^{\infty}|a_n|\left[\sum_{k=1}^{n-1}\rho^{n-2}k\right] \\
&=|w-z|\sum_{n=0}^{\infty}|a_n|\frac{n(n-1)}{2}\rho^{n-2} \\
\end{aligned}
Using Cauchy-Hadamard theorem, the radius of convergence of \sum_{n=0}^{\infty}\frac{ n(n-1)}{2}|a_n|z^{n-2} is at least
1/\limsup_{n\to\infty}\left[\frac{n(n-1)}{2}|a_n|\right]^{1/(n-1)}=R.
Therefore,
|w-z|\sum_{n=0}^{\infty}|a_n|\frac{n(n-1)}{2}\rho^{n-2} \leq C|w-z|
where C is dependent on \rho.
So \lim_{w\to z}\left|\frac{f(z)-f(w)}{z-w}-g(z)\right|=0. as desired.
QED
Corollary of power series
If f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n in D(z_0,R), then a_0=f(z_0), a_1=f'(z_0)/1!, a_2=f''(z_0)/2!, etc.
Definition (Analytic)
A function h on an open set U\subset\mathbb{C} is called analytic if for every z\in U, \exists \epsilon>0 such that on D(z,\epsilon)\subset U, h can be represented as a power series \sum_{n=0}^{\infty}a_n(z-z_0)^n.
Theorem (Analytic implies holomorphic)
If f is analytic on U, then f is holomorphic on U.
\sum_{n=0}^{\infty}\frac{1}{n!}f^{(n)}(z)^n
Radius of convergence is \infty.
So f(0)=1=ce^0=c
\sum_{n=0}^{\infty}\frac{1}{n}z^n
Radius of convergence is 1.
f'=\sum_{n=1}^{\infty}z^{n-1}=\frac{1}{1-z} (Geometric series)
So g(z)=c+\log(\frac{1}{1-z})=c+2\pi k i=\log(\frac{1}{1-z})+2\pi k i
Cauchy Product of power series
Let f(z)=\sum_{n=0}^{\infty}a_nz^n and g(z)=\sum_{n=0}^{\infty}b_nz^n be two power series.
Then f(z)g(z)=\sum_{n=0}^{\infty}=\sum_{n=0}^{\infty}c_nz^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}z^n
Theorem of radius of convergence of Cauchy product
Let f(z)=\sum_{n=0}^{\infty}a_nz^n and g(z)=\sum_{n=0}^{\infty}b_nz^n be two power series.
Then the radius of convergence of f(z)g(z) is at least \min(R_f,R_g).
Without loss of generality, assume z_0=0.
\begin{aligned}
\left(\sum_{j=0}^{N}a_jz^j\right)\left(\sum_{k=0}^{N}b_kz^k\right)-\sum_{l=0}^{N}c_lz^l&=\sum_{j=0}^{N}\sum_{k=N-j}^{N}a_jb_kz^{j+k}\\
&\leq\sum_{N/2\leq\max(j,k)\leq N}|a_j||b_k||z^{j+k}|\\
&\leq\left(\sum_{j=N/2}^{N}|a_j||z^j|\right)\left(\sum_{k=0}^{\infty}|b_k||z^k|\right)+\left(\sum_{j=0}^{\infty}|a_j||z^j|\right)\left(\sum_{k=N/2}^{\infty}|b_k||z^k|\right)\\
\end{aligned}
Since \sum_{j=0}^{\infty}|a_j||z^j| and \sum_{k=0}^{\infty}|b_k||z^k| are convergent, and \sum_{j=N/2}^{N}|a_j||z^j| and \sum_{k=N/2}^{\infty}|b_k||z^k| converges to zero.
So \left|\left(\sum_{j=0}^{N}a_jz^j\right)\left(\sum_{k=0}^{N}b_kz^k\right)-\sum_{l=0}^{N}c_lz^l\right|\leq\left(\sum_{j=N/2}^{N}|a_j||z^j|\right)\left(\sum_{k=0}^{\infty}|b_k||z^k|\right)+\left(\sum_{j=0}^{\infty}|a_j||z^j|\right)\left(\sum_{k=N/2}^{\infty}|b_k||z^k|\right)\to 0 as N\to\infty.
So \sum_{n=0}^{\infty}c_nz^n converges to f(z)g(z) on D(0,R_fR_g).