NoteNextra-origin/content/Math416/Math416_L7.md

Math416 Lecture 7

Review

Exponential function

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

Logarithm Reviews

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

Proof

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

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

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

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

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

Definition 4.16

I don't know if this material is covered or not, so I will add it here to prevent confusion for future readers

If a and c are complex numbers, with a\neq 0, then by the values of a^c one means the value of e^{c\log a}.

For example, 1^i=e^{i (2\pi n i)}

If you accidentally continue on this section and find it interesting, you will find Riemann zeta function

z(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}

And analytic continuation for such function for number less than or equal to 1.

And perhaps find trivial zeros for negative integers on real line. It is important to note that the Riemann zeta function has non-trivial zeros, which are located in the critical strip where the real part of s is between 0 and 1. The famous Riemann Hypothesis conjectures that all non-trivial zeros lie on the critical line where the real part of s is \frac{1}{2}.

Chapter 5. Power series

Convergence

Necessary Condition for Convergence

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

Geometric series

Let c be a complex number

\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.

Theorem 5.4 (Triangle Inequality for Series)

If the series \sum_{n=0}^{\infty}c_n converges, then \left|\sum_{n=0}^\infty c_n\right|\leq \sum_{n=0}^{\infty}|c_n|.

Definition 5.5

\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 5.7

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

The N-th partial sum of the series \sum_{n=0}^\infty f_n is \sum_{n=0}^{N}f_n

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

Corollary from definition of a^b in complex plane

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(z-z_0)^n.

Definition 5.9 Region of Convergence

For every power series, there exists a radius of convergence r such that the series converges absolutely and locally uniformly on B_r(z_0).

And it diverges pointwise outside B_r(z_0).

Proof

Without loss of generality, we can assume that z_0=0.

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

We want to show that the series converges absolutely and uniformly on \overline{B_r(0)} (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 z\in\overline{B_r(0)}, |c_nz^n|\leq |c_n| |z|^n \leq M \left(\frac{|z|}{r}\right)^n.

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

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

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

We denote this r captialized by te radius of convergence

Possible Cases for the Convergence of Power Series

1. Convergence Only at $z = 0$:

   • Proof: If the power series \sum_{n=0}^{\infty} c_n (z - z_0)^n converges only at z = 0, it means that the radius of convergence R = 0. This occurs when the terms c_n (z - z_0)^n do not tend to zero for any z \neq 0. The series diverges for all z \neq 0 because the terms grow without bound.

2. Convergence Everywhere:

   • Proof: If the power series converges for all z \in \mathbb{C}, the radius of convergence R = \infty. This implies that the terms c_n (z - z_0)^n tend to zero for all z. 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 |z - z_0| < R and diverges for |z - z_0| > R. On the boundary |z - z_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.