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Math416Complex Variables (Lecture 8)

Math416 Lecture 8

Review

Sequences of Functions

Let fn:G→Cf_n: G \to \mathbb{C} be a sequence of functions.

Convergence Pointwise

Definition:

Let z∈Gz\in G, ∀ϵ>0\forall \epsilon > 0, ∃N∈N\exists N \in \mathbb{N} such that ∀n≥N\forall n \geq N, ∣fn(z)−f(z)∣<ϵ|f_n(z) - f(z)| < \epsilon.

Convergence Uniformly

Definition:

∀ϵ>0\forall \epsilon > 0, ∀z∈G\forall z\in G, ∃N∈N\exists N \in \mathbb{N} such that ∀n≥N\forall n \geq N, ∣fn(z)−f(z)∣<ϵ|f_n(z) - f(z)| < \epsilon.

Convergence Locally Uniformly

Definition:

∀ϵ>0\forall \epsilon > 0, ∀z∈G\forall z\in G, ∃N∈N\exists N \in \mathbb{N} such that ∀n≥N\forall n \geq N, ∣fn(z)−f(z)∣<ϵ|f_n(z) - f(z)| < \epsilon.

Convergence Uniformly on Compact Sets

Definition: ∀C⊂G\forall C\subset G that is compact, ∀ϵ>0,∃N∈N s.t. ∀n≥N,∀z∈C,∣fn(z)−f(z)∣<ϵ\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, \forall z\in C, |f_n(z) - f(z)| < \epsilon

Power Series

Definition:

∑n=0∞cn(z−z0)n\sum_{n=0}^{\infty} c_n (z - z_0)^n

z0z_0 is the center of the power series.

Theorem of Power Series

If a power series converges at z0z_0, then it converges absolutely at every point of Br(z0)‾\overline{B_r(z_0)} that is strictly inside the disk of convergence.

Continue on Power Series

Review on lim sup⁡\limsup

The lim sup⁡(an)\limsup(a_n) an∈Ra_n\in\mathbb{R} is defined as the sup of subsequence of (an)(a_n) as nn approaches infinity.

It has the following properties that is useful for proving the remaining parts for this course.

Suppose (an)1∞(a_n)_1^\infty is a sequence of real numbers

  1. If ρ∈R\rho\in \mathbb{R} satisfies that ρ<lim sup⁡n→∞an\rho<\limsup_{n\to\infty}a_n, then {an:an>ρ}\{a_n : a_n > \rho\} is infinite.
  2. If ρ∈R\rho\in \mathbb{R} satisfies that ρ>lim sup⁡n→∞an\rho>\limsup_{n\to\infty}a_n, then {an:an>ρ}\{a_n : a_n > \rho\} is finite.

Limits of Power Series

Theorem 5.12

Cauchy-Hadamard Theorem:

The radius of convergence of the power series is given by ∑n=0∞an(z−z0)n\sum_{n=0}^{\infty} a_n (z - z_0)^n is given by

1R=lim sup⁡n→∞∣an∣1/n\frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}

Proof

Suppose (bn)n=0∞(b_n)^{\infty}_{n=0} is a sequence of real numbers such that lim⁡n→∞bn\lim_{n\to\infty} b_n may nor may not exists by (−1)n(1−1n)(-1)^n(1-\frac{1}{n}).

The limit superior of (bn)(b_n) is defined as

sn=sup⁡k≥nbks_n = \sup_{k\geq n} b_k

sns_n is a decreasing sequence, by completeness of R\mathbb{R}, every bounded sequence has a limit in R\mathbb{R}.

So sns_n converges to some limit s∈Rs\in\mathbb{R}.

Without loss of generality, this also holds for infininum of sns_n.

Forward direction:

We want to show that the radius of convergence of ∑n=0∞an(z−z0)n\sum_{n=0}^{\infty} a_n (z - z_0)^n is greater than or equal to 1lim sup⁡n→∞∣an∣1/n\frac{1}{\limsup_{n\to\infty} |a_n|^{1/n}}.

Since ∑n=0∞1zn=11−z\sum_{n=0}^{\infty} 1z^n=\frac{1}{1-z} for ∣z∣<1|z|<1. Assume lim sup⁡n→∞∣an∣1/n\limsup_{n\to\infty} |a_n|^{1/n} is finite, then ∑n=0∞an(z−z0)n\sum_{n=0}^{\infty} a_n (z - z_0)^n converges absolutely at z0z_0.

Let ρ>lim sup⁡n→∞∣an∣1/n\rho>\limsup_{n\to\infty} |a_n|^{1/n}, then ∃N∈N\exists N \in \mathbb{N} such that ∀n≥N\forall n \geq N, ∣an∣1/n≤ρ|a_n|^{1/n}\leq \rho. (By property of lim sup⁡\limsup)

So 1R=lim sup⁡n→∞∣an∣1/n≤ρ\frac{1}{R}=\limsup_{n\to\infty} |a_n|^{1/n}\leq\rho

So R≥1ρR\geq\frac{1}{\rho}

Backward direction:

Suppose ∣z∣>R|z|>R, then ∃\exists number ∣z∣|z| such that ∣z∣>1ρ≥R|z|>\frac{1}{\rho}\geq R.

So ρ<lim sup⁡n→∞∣an∣1/n\rho<\limsup_{n\to\infty} |a_n|^{1/n}

This means that ∃\exists infinitely many njn_js such that ∣anj∣1/nj>ρ|a_{n_j}|^{1/n_j}>\rho

So ∣anjznj∣>ρnj∣z∣nj|a_{n_j}z^{n_j}|>\rho^{n_j}|z|^{n_j}

Series ∑n=1∞anzn\sum_{n=1}^{\infty} a_nz^n diverges, each individual term is not going to 00.

So ∑n=0∞an(z−z0)n\sum_{n=0}^{\infty} a_n (z - z_0)^n does not converge at zz if ∣z∣>1ρ≥R|z|> \frac{1}{\rho}\geq R

So R=1ρR=\frac{1}{\rho}.

What if ∣z−z0∣=R|z-z_0|=R?

For ∑n=0∞zn\sum_{n=0}^{\infty} z^n, the radius of convergence is 11.

It diverges eventually on the circle of convergence.

For ∑n=0∞1(n+1)2zn\sum_{n=0}^{\infty} \frac{1}{(n+1)^2}z^n, the radius of convergence is 11.

This converges everywhere on the circle of convergence.

For ∑n=0∞1n+1zn\sum_{n=0}^{\infty} \frac{1}{n+1}z^n, the radius of convergence is 11.

This diverges at z=1z=1 (harmonic series) and converges at z=−1z=-1 (alternating harmonic series).

Theorem 5.15

Differentiation of power series

Suppose ∑n=0∞an(z−z0)n\sum_{n=0}^{\infty} a_n (z - z_0)^n has a positive radius of convergence RR. Define f(z)=∑n=0∞an(z−z0)nf(z)=\sum_{n=0}^{\infty} a_n (z - z_0)^n, then ff is holomorphic on BR(0)B_R(0) and f′(z)=∑n=1∞nan(z−z0)n−1=∑k=0∞(k+1)ak+1(z−z0)kf'(z)=\sum_{n=1}^{\infty} n a_n (z - z_0)^{n-1}=\sum_{k=0}^{\infty} (k+1)a_{k+1} (z - z_0)^k.

Here below is the proof on book, which will be covered in next lecture.

Proof

Without loss of generality, assume z0=0z_0=0. Let RR be the radius of convergence for the two power series: ∑n=0∞anzn\sum_{n=0}^{\infty} a_n z^n and ∑n=1∞nanzn−1\sum_{n=1}^{\infty} n a_n z ^{n-1}. The two power series have the same radius of convergence ∣R∣|R|.

For z,w∈C,n∈Nz,w\in \mathbb{C}, n\in \N, zn−wn=(z−w)∑k=0n−1zkwn−k−1z^n-w^n=(z-w)\sum_{k=0}^{n-1} z^k w^{n-k-1}

Let z1∈BR(0)z_1\in B_R(0), ∣z1∣<ρ<R|z_1|<\rho<R for some ρ∈R\rho\in\mathbb{R}.

f(z)−f(z1)z−z1−g(z1)=1z−z1[∑n=0∞anzn−∑n=0∞anz1n]−∑n=1∞nanz1n−1=∑n=1∞an[zn−z1nz−z1−nz1n−1]=∑n=1∞an[(∑k=0n−1zkz1n−k−1)−nz1n−1]=∑n=2∞an[∑k=1n−1z1n−k−1(zk−z1k)]\begin{aligned} \frac{f(z)-f(z_1)}{z-z_1}-g(z_1)&=\frac{1}{z-z_1}\left[\sum_{n=0}^\infty a_n z^n -\sum_{n=0}^\infty a_n z_1^n\right]-\sum_{n=1}^{\infty} n a_n z_1 ^{n-1}\\ &=\sum_{n=1}^{\infty} a_n \left[\frac{z^n-z_1^n}{z-z_1}-nz_1^{n-1}\right]\\ &=\sum_{n=1}^{\infty} a_n \left[\left(\sum_{k=0}^{n-1}z^kz_1^{n-k-1}\right)-nz_1^{n-1}\right]\\ &=\sum_{n=2}^{\infty} a_n \left[\sum_{k=1}^{n-1}z_1^{n-k-1}(z^k-z^k_1)\right] \end{aligned}

Using the lemma again we get

∣zk−z1k∣=∣z−z1∣∣∑j=0k−1zjz1k−j−1∣≤∣z−z1∣∑j=0k−1∣zj∣∣z1k−j−1∣≤kρk−1∣z−z1∣\begin{aligned} |z^k-z_1^k|&=|z-z_1|\left|\sum_{j=0}^{k-1}z_jz_1^{k-j-1}\right|\\ &\leq |z-z_1| \sum_{j=0}^{k-1}|z_j||z_1^{k-j-1}|\\ &\leq k\rho^{k-1}|z-z_1| \end{aligned}

Then,

∣f(z)−f(z1)z−z1−g(z1)∣=∣∑n=2∞an[∑k=1n−1z1n−k−1(zk−z1k)]∣≤∑n=2∞∣an∣[∑k=1n−1∣z1∣n−k−1∣zk−z1k∣]≤∑n=2∞∣an∣[∑k=1n−1ρn−k−1(kρk−1∣z−z1∣)]=∣z−z1∣∑n=2∞∣an∣[n(n−1)2ρn−2]\begin{aligned} \left|\frac{f(z)-f(z_1)}{z-z_1}-g(z_1)\right|&=\left|\sum_{n=2}^{\infty} a_n \left[\sum_{k=1}^{n-1}z_1^{n-k-1}(z^k-z^k_1)\right]\right|\\ &\leq \sum_{n=2}^{\infty} |a_n| \left[\sum_{k=1}^{n-1}|z_1|^{n-k-1}|z^k-z_1^k|\right]\\ &\leq \sum_{n=2}^{\infty} |a_n| \left[ \sum_{k=1}^{n-1} \rho^{n-k-1} (k\rho^{k-1}|z-z_1|) \right]\\ &=|z-z_1|\sum_{n=2}^\infty|a_n|\left[\frac{n(n-1)}{2}\rho^{n-2}\right] \end{aligned}

One can use ratio test to find that ∑n=2∞∣an∣[n(n−1)2ρn−2]\sum_{n=2}^\infty|a_n|\left[\frac{n(n-1)}{2}\rho^{n-2}\right] converges, we denote the sum using MM

So ∣f(z)−f(z1)z−z1−g(z1)∣≤M∣z−z1∣\left|\frac{f(z)-f(z_1)}{z-z_1}-g(z_1)\right|\leq M|z-z_1| for ∣z∣<ρ|z|<\rho.

So lim⁡z→z1f(z)−f(z1)z−z1=g(z1)\lim_{z\to z_1}\frac{f(z)-f(z_1)}{z-z_1}=g(z_1).

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