5.0 KiB
Math416 Lecture 9
Review
Power Series
Let f(\zeta)=\sum_{n=0}^{\infty}a_n(\zeta-\zeta_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(\zeta)=\sum_{n=0}^{\infty}a_n(\zeta-\zeta_0)^n be a power series.
Let g(\zeta)=\sum_{n=0}^{\infty}na_n(\zeta-\zeta_0)^{n-1} be another power series.
Then g is holomorphic on D(\zeta_0,R) and g'(\zeta)=f(\zeta) for all \zeta\in D(\zeta_0,R). and f'(\zeta)=g(\zeta).
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 \zeta\in D(\zeta_0,R).
let |\zeta-\zeta_0|<\rho<R.
Without loss of generality, assume \zeta_0=0. Let |w|<\rho.
\begin{aligned}
\frac{f(\zeta)-f(w)}{\zeta-w}-g(\zeta)&=\sum_{n=0}^{\infty}\left[\frac{1}{\zeta-w}\left(a_n(\zeta^n-w^n)\right)-na_n\zeta^{n-1}\right] \\
&=\sum_{n=0}^{\infty}a_n\left[\frac{\zeta^n-w^n}{\zeta-w}-n\zeta^{n-1}\right]
\end{aligned}
Notice that
\begin{aligned}
\frac{\zeta^n-w^n}{\zeta-w}&=\sum_{k=0}^{n-1}\zeta^{n-1-k}w^k \\
&=\zeta^{n-1}+\zeta^{n-2}w+\cdots+w^{n-1}
\end{aligned}
Since
|w^k-\zeta^k|=\left|(w-\zeta)\left(\sum_{j=0}^{k-1}w^{k-1-j}\zeta^j\right)\right|\leq|w-\zeta|k\rho^{k-1}
\begin{aligned}
\frac{\zeta^n-w^n}{\zeta-w}-n\zeta^{n-1}&=(\zeta^{n-1}-\zeta^{n-1})+(\zeta^{n-2}w-\zeta^{n-1})+\cdots+(\zeta w^{n-1}-\zeta^{n-1}) \\
&=\zeta^{n-2}(w-\zeta)+\zeta^{n-3}(w^2-\zeta^2)+\cdots+\zeta^0(w^{n-1}-\zeta^{n-1}) \\
&=\sum_{k=0}^{n-1}\zeta^{n-1-k}(w^k-\zeta^k)\\
&\leq\sum_{k=0}^{n-1}\zeta^{n-1-k}|w-\zeta|k\rho^{k-1} \\
&\leq|w-\zeta|\sum_{k=0}^{n-1}k\rho^{k-1} \\
\end{aligned}
Apply absolute value,
\begin{aligned}
\left|\frac{f(\zeta)-f(w)}{\zeta-w}-g(\zeta)\right|&\leq\sum_{n=0}^{\infty}|a_n||w-\zeta|\left[\sum_{k=1}^{n-1}\rho^{n-1-k}k\rho^{k-1}\right] \\
&=|w-\zeta|\sum_{n=0}^{\infty}|a_n|\left[\sum_{k=1}^{n-1}\rho^{n-2}k\right] \\
&=|w-\zeta|\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|\zeta^{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-\zeta|\sum_{n=0}^{\infty}|a_n|\frac{n(n-1)}{2}\rho^{n-2} \leq C|w-\zeta|
where C is dependent on \rho.
So lim_{w\to\zeta}\left|\frac{f(\zeta)-f(w)}{\zeta-w}-g(\zeta)\right|=0. as desired.
Corollary of power series
If f(\zeta)=\sum_{n=0}^{\infty}a_n(\zeta-\zeta_0)^n in D(\zeta_0,R), then a_0=f(\zeta_0), a_1=f'(\zeta_0)/1!, a_2=f''(\zeta_0)/2!, etc.
Definition (Analytic)
A function h on an open set U\subset\mathbb{C} is called analytic if for every \zeta\in U, \exists \epsilon>0 such that on D(\zeta,\epsilon)\subset U, h can be represented as a power series \sum_{n=0}^{\infty}a_n(\zeta-\zeta_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)}(\zeta)^n
Radius of convergence is \infty.
So f(0)=1=ce^0=c
\sum_{n=0}^{\infty}\frac{1}{n}\zeta^n
Radius of convergence is 1.
f'=\sum_{n=1}^{\infty}\zeta^{n-1}=\frac{1}{1-\zeta} (Geometric series)
So g(\zeta)=c+\log(\frac{1}{1-\zeta})=c+2\pi k i=\log(\frac{1}{1-\zeta})+2\pi k i
Cauchy Product of power series
Let f(\zeta)=\sum_{n=0}^{\infty}a_n\zeta^n and g(\zeta)=\sum_{n=0}^{\infty}b_n\zeta^n be two power series.
Then f(\zeta)g(\zeta)=\sum_{n=0}^{\infty}=\sum_{n=0}^{\infty}c_n\zeta^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}\zeta^n
Theorem of radius of convergence of Cauchy product
Let f(\zeta)=\sum_{n=0}^{\infty}a_n\zeta^n and g(\zeta)=\sum_{n=0}^{\infty}b_n\zeta^n be two power series.
Then the radius of convergence of f(\zeta)g(\zeta) is at least \min(R_f,R_g).
Without loss of generality, assume \zeta_0=0.
\begin{aligned}
\left(\sum_{j=0}^{N}a_j\zeta^j\right)\left(\sum_{k=0}^{N}b_k\zeta^k\right)-\sum_{l=0}^{N}c_l\zeta^l&=\sum_{j=0}^{N}\sum_{k=N-j}^{N}a_jb_k\zeta^{j+k}\\
&\leq\sum_{N/2\leq\max(j,k)\leq N}|a_j||b_k||\zeta^{j+k}|\\
&\leq\left(\sum_{j=N/2}^{N}|a_j||\zeta^j|\right)\left(\sum_{k=0}^{\infty}|b_k||\zeta^k|\right)+\left(\sum_{j=0}^{\infty}|a_j||\zeta^j|\right)\left(\sum_{k=N/2}^{\infty}|b_k||\zeta^k|\right)\\
\end{aligned}
Since \sum_{j=0}^{\infty}|a_j||\zeta^j| and \sum_{k=0}^{\infty}|b_k||\zeta^k| are convergent, and \sum_{j=N/2}^{N}|a_j||\zeta^j| and \sum_{k=N/2}^{\infty}|b_k||\zeta^k| converges to zero.
So \left|\left(\sum_{j=0}^{N}a_j\zeta^j\right)\left(\sum_{k=0}^{N}b_k\zeta^k\right)-\sum_{l=0}^{N}c_l\zeta^l\right|\leq\left(\sum_{j=N/2}^{N}|a_j||\zeta^j|\right)\left(\sum_{k=0}^{\infty}|b_k||\zeta^k|\right)+\left(\sum_{j=0}^{\infty}|a_j||\zeta^j|\right)\left(\sum_{k=N/2}^{\infty}|b_k||\zeta^k|\right)\to 0 as N\to\infty.
So \sum_{n=0}^{\infty}c_n\zeta^n converges to f(\zeta)g(\zeta) on D(0,R_fR_g).