NoteNextra-origin/pages/Math416/Math416_L7.md

Lecture 7

Review

Exponential function

e^z=e^{x+iy}=e^x(\cos y+i\sin y)

Logarithm

Definition 4.9 Logarithm

A logarithm of a is any b such that e^b=a.

Branch of Logarithm

A branch of logarithm is a continuous function f on a domain D such that e^{f(z)}=\exp(f(z))=z for all z\in D.

Continue on Chapter 4 Elementary functions

Logarithm

Theorem 4.11

\log(\zeta) is holomorphic on \mathbb{C}\setminus\{0\}.

Proof:

We proved that \frac{\partial}{\partial\overline{z}}e^{\zeta}=0 on \mathbb{C}\setminus\{0\}.

Then \frac{d}{dz}e^{\zeta}=\frac{\partial}{\partial x}e^{\zeta}=0 if we know that e^{\zeta} is holomorphic.

Since \frac{d}{dz}e^{\zeta}=e^{\zeta}, we know that e^{\zeta} is conformal, so any branch of logarithm is also conformal.

Since \exp(\log(\zeta))=\zeta, we know that \log(\zeta) is the inverse of \exp(\zeta), so \frac{d}{dz}\log(\zeta)=\frac{1}{e^{\log(\zeta)}}=\frac{1}{\zeta}.

EOP

We call \frac{f'}{f} the logarithmic derivative of f.

Chapter 5. Power series

If \sum_{n=0}^{\infty}c_n converges, then \lim_{n\to\infty}c_n=0 exists.

Geometric series

\sum_{n=0}^{N}c^n=\frac{1-c^{N+1}}{1-c}

If |c|<1, then \lim_{N\to\infty}\sum_{n=0}^{N}c^n=\frac{1}{1-c}.

otherwise, the series diverges.

Proof:

The geometric series converges if \frac{c^{N+1}}{1-c} converges.

(1-c)(1+c+c^2+\cdots+c^N)=1-c^{N+1}

If |c|<1, then \lim_{N\to\infty}c^{N+1}=0, so \lim_{N\to\infty}(1-c)(1+c+c^2+\cdots+c^N)=1.

If |c|\geq 1, then c^{N+1} does not converge to 0, so the series diverges.

EOP

Convergence

Definition 5.4

\sum_{n=0}^{\infty}c_n

converges absolutely if \sum_{n=0}^{\infty}|c_n| converges.

Note: Some other properties of converging series covered in Math4111, bad, very bad.

Definition 5.6 Convergence of sequence of functions

A sequence of functions f_n converges pointwise to f on a set G if for every z\in G, \forall\epsilon>0, \exists N such that for all n\geq N, |f_n(z)-f(z)|<\epsilon.

(choose N based on z)

A sequence of functions f_n converges uniformly to f on a set G if for every \epsilon>0, there exists a positive integer N such that for all n\geq N and all z\in G, |f_n(z)-f(z)|<\epsilon.

(choose N based on \epsilon)

A sequence of functions f_n converges locally uniformly to f on a set G if for every z\in G, \forall\epsilon>0, \exists r>0 such that for all z\in B(z,r), \forall n\geq N, |f_n(z)-f(z)|<\epsilon.

(choose N based on z and \epsilon)

A sequence of functions f_n converges uniformly on compacta to f on a set G if it converges uniformly on every compact subset of G.

Theorem 5.?

If the subsequence of a converging sequence of functions converges (a), then the original sequence converges (a).

You can replace (a) with locally uniform convergence, uniform convergence, pointwise convergence, etc.

UNKNOWN

We defined a^b=\{e^{b\log a}\} if b is real, then a^b is unique, if b is complex, then a^b=e^{b\log a}\{e^{2k\pi ik b}\},k\in\mathbb{Z}.

Power series

Definition 5.8

A power series is a series of the form \sum_{n=0}^{\infty}c_n(\zeta-\zeta_0)^n.

Theorem 5.10

For every power series, there exists a radius of convergence R such that the series converges absolutely and locally uniformly on B(\zeta_0,R).

And it diverges pointwise outside B(\zeta_0,R).

Proof:

Without loss of generality, we can assume that \zeta_0=0.

Suppose that the power series is \sum_{n=0}^{\infty}c_n (\zeta)^n converges at \zeta=re^{i\theta}.

We want to show that the series converges absolutely and uniformly on \overline{B(0,r)} (closed disk, I prefer to use this notation, although they use \mathbb{D} for the disk (open disk)).

We know c_n r^ne^{in\theta}\to 0 as n\to\infty.

So there exists M\geq|c_n r^ne^{in\theta}| for all n\in\mathbb{N}.

So \forall \zeta\in\overline{B(0,r)}, |c_n\zeta^n|\leq |c_n| |\zeta|^n \leq M \left(\frac{|\zeta|}{r}\right)^n.

So \sum_{n=0}^{\infty}|c_n\zeta^n| converges absolutely.

So the series converges absolutely and uniformly on \overline{B(0,r)}.

If |\zeta| > r, then |c_n \zeta^n| does not tend to zero, and the series diverges.

EOP

Possible Cases for the Convergence of Power Series

1. Convergence Only at $\zeta = 0$:

   • Proof: If the power series \sum_{n=0}^{\infty} c_n (\zeta - \zeta_0)^n converges only at \zeta = 0, it means that the radius of convergence R = 0. This occurs when the terms c_n (\zeta - \zeta_0)^n do not tend to zero for any \zeta \neq 0. The series diverges for all \zeta \neq 0 because the terms grow without bound.

2. Convergence Everywhere:

   • Proof: If the power series converges for all \zeta \in \mathbb{C}, the radius of convergence R = \infty. This implies that the terms c_n (\zeta - \zeta_0)^n tend to zero for all \zeta. This can happen if the coefficients c_n decrease rapidly enough, such as in the exponential series.

3. Convergence Within a Finite Radius:

   • Proof: For a power series with a finite radius of convergence R, the series converges absolutely and uniformly for |\zeta - \zeta_0| < R and diverges for |\zeta - \zeta_0| > R. On the boundary |\zeta - \zeta_0| = R, the series may converge or diverge depending on the specific series. This is determined by the behavior of the terms on the boundary.