# MATH3401

Lecture 29

### Poisson kernel

Taking the real part from 😀 defined in Lecture 28: Given "nice enough" $$\Phi(r_0, \phi)$$ defined on the boundary $$C_0$$ of $$B_{r_0}$$, a (infact, the) solution of the Dirichlet problem \text{(D) } \left\{ \begin{align} \Delta u = & \;0 \;\text{ in } \;B_{r_0}\\ u|_{\partial B_{r_0}} = & \;\Phi(r_0, \phi) \end{align} \right. is given by $$u(r, \theta) = \ds \frac{1}{2 \pi } \int_{0}^{2 \pi} \frac{r_0^2-r^2}{r_0^2-2r_0r \cos \left(\phi - \theta\right) +r^2} \Phi(r, \phi)d\Phi$$ $= \ds \frac{1}{2 \pi } \int_{0}^{2 \pi} P\left(r_0, r, \phi- \theta\right) \Phi(r, \phi)d\Phi$

where $$P(r_0, r, \phi- \theta)$$ is known as the Poisson kernel.

### Sequences

Compare the situation in $$\R$$.

Formally, a sequence is a function:

$$\N \ra \C$$ (or $$\N_0 \ra \C$$) $n \mapsto z_n$

Write: $$\{z_n\}$$.

### Sequences

Limit of a sequence:

$$\ds \lim_{n\ra \infty} z_n = z \iff \{z_n\}$$ converges to $$z$$.

$$\iff$$ Given $$\varepsilon >0$$ there exists $$N\in \N$$ such that $n\gt N \implies |z_n-z|\lt \varepsilon.$

### Series

Now for series: Formally we have $$\ds \sum_{n=0}^{\infty} z_n$$, $$\,z_n \in \C$$.

The series $$\sum z_n$$ converges as a series $$\iff$$ the associated sequence of partial sums $$\{s_n\}$$ converges as a sequence.

Here: $$s_n = \ds \sum_{k=0}^{n} z_k$$.

### Series

Typical question: Does $$\sum z_n$$ converge?

Test for NO: The $$n$$th term test $\sum z_n \text{ converges } \Rightarrow z_n \ra 0.$

$$\nLeftarrow$$ e. g. $$\sum 1/n$$ diverges.

### Series

Remark: $$\{z_n\}$$ is bounded says there exists $$M$$ such tat $$|z_n|\lt M$$ for all $$n$$.

$$*$$ convergent $$\Rightarrow$$ bounded, $$\nLeftarrow \{(-1)^n\}$$.

Definition: $$\sum z_n$$ converges absolutely $$\iff$$ $$\sum |z_n|$$ converges.

Absolute convergence $$\Rightarrow$$ convergence,

$$\nLeftarrow$$ $$\sum \frac{(-1)^n}{n}$$

### Series

Given $$\sum_{n=0}^{\infty} z_n$$, set $$S_n = \sum_{k=0}^{n} z_k$$ as above, and write $\rho_n = \sum_{k=n+1}^{\infty} z_k$

$$\rho$$ is the tail/remainder. Then $S_n \ra S \iff \sum z_n = S \;\text{ by definition.}$

Theorem: $$S_n \ra S \iff \rho_n =0$$.

### Series

Application

Claim $$\ds\sum_{n=0}^{\infty} z^n = \frac{1}{1-z}$$ for $$|z|\lt 1$$. $$\quad(*)$$

Proof: $$\qquad S_n = 1+ z+ \cdots + z^n$$

$$\qquad \qquad z S_n = z+ \cdots + z^n + z^{n+1}$$.

$$\Rightarrow (1-z) S_n = 1-z^{n+1}$$ $$\Rightarrow S_n = \dfrac{1-z^{n+1}}{1-z}\quad\qquad$$

$$\Rightarrow \rho_n = S - S_n$$ $$=\dfrac{z^{n+1}}{1-z}$$. Since $$|z|\lt 1$$, $$|\rho_n|\ra 0$$ as $$n\ra \infty$$,

$$\Rightarrow \rho_n\ra 0$$ as $$n\ra \infty$$, $$\Rightarrow S_n \ra S$$, which shows $$(*)$$.

### Series

Remark: As in $$\R$$, if $$\sum a_n$$ and $$\sum b_n$$ converge, then so do $\sum \left(a_n\pm b_n\right) \text{ to } \left(\sum a_n\right) \pm \left(\sum b_n\right).$

For $$\lambda \in \C$$ fixed

$$\ds\sum \left( \lambda a_n\right)$$ converges to $$\lambda \ds\sum a_n$$.

### Power Series

Centred at $$z_0$$, the series of a function $$f(z)$$ is $$\ds\sum_{n=0}^{\infty} a_n (z-z_0)^n.$$

Series will

• converge absolutely for $$|z-z_0|\lt R$$, where $$R$$ is the radius of convergence;
• diverge for $$|z-z_0|\gt R$$;
• check by hand for $$|z-z_0|= R$$;
• $$R=0$$, converge only at $$z_0$$;
• $$R=\infty$$, converge on $$\C$$.