# [SOLVED]Limit of Poisson kernel

#### dwsmith

##### Well-known member
Prove:
$$\lim_{r\to 1}P(r,\theta) = \begin{cases} \infty, & \theta = 0\\ 0, & \text{otherwise} \end{cases}$$
For the first piece, take the summation
$$P(1,0) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} 1^n\right).$$
Then $\sum\limits_{n = 1}^{\infty} 1^n = \infty$.
Therefore, we have a positive number plus infinity which is infinity when $r\to 1$ and $\theta = 0$.
For the second piece, take the fractional representation of the Poisson kernel,
$$P(1,\theta) = \frac{1}{2\pi}\frac{0}{2 - 2\cos\theta} = 0.$$
Therefore, $P(r,\theta) = 0$ for all $\theta\neq 0$.
That is,
$$\lim_{r\to 1}P(r,\theta) = \begin{cases} \infty, & \theta = 0\\ 0, & \text{otherwise} \end{cases}$$

#### girdav

##### Member
Could you write the definition of Poisson kernel?

#### dwsmith

##### Well-known member
Could you write the definition of Poisson kernel?
$$P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}$$

#### Sudharaka

##### Well-known member
MHB Math Helper
$$P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}$$
Can you please tell me where you found this definition?

#### dwsmith

##### Well-known member
Can you please tell me where you found this definition?
A class on Fourier series

A class on Engineering Analysis

The book Elementary Partial Differential Equations by Berg and McGregor

My Engineering Analysis book that I can't remember the name.

Separate handout notes by my Fourier Analysis professor.

#### Sudharaka

##### Well-known member
MHB Math Helper
A class on Fourier series

A class on Engineering Analysis

The book Elementary Partial Differential Equations by Berg and McGregor

My Engineering Analysis book that I can't remember the name.

Separate handout notes by my Fourier Analysis professor.
I suggest you to check the definition again. The correct one is given >>here<<.

\begin{eqnarray}

P_r(\theta)&=&\sum_{n=-\infty}^\infty r^{|n|}e^{in\theta}=\frac{1-r^2}{1-2r\cos\theta +r^2}\mbox{ where }0 \le r < 1.\\

&=&\sum_{n=-\infty}^\infty r^{|n|}\cos(n\theta)+i\sum_{n=-\infty}^\infty r^{|n|}\sin(n\theta)\\

\end{eqnarray}

Since, $$r^{|n|}\sin(n\theta)$$ is an odd function it is clear that the second sum is equal to zero.

\begin{eqnarray}

\therefore P_r(\theta)&=&\sum_{n=-\infty}^\infty r^{|n|}\cos(n\theta)\\

&=&1+2\sum_{n=1}^\infty r^{n}\cos(n\theta)\\

\end{eqnarray}

Hence we finally get,

$P_r(\theta)=1+2\sum_{n=1}^\infty r^{n}\cos(n\theta)= \frac{1-r^2}{1-2r\cos\theta +r^2}\mbox{ where }0 \le r < 1.$

#### dwsmith

##### Well-known member
I suggest you to check the definition again. The correct one is given >>here<<.

\begin{eqnarray}

P_r(\theta)&=&\sum_{n=-\infty}^\infty r^{|n|}e^{in\theta}=\frac{1-r^2}{1-2r\cos\theta +r^2}\mbox{ where }0 \le r < 1.\\

&=&\sum_{n=-\infty}^\infty r^{|n|}\cos(n\theta)+i\sum_{n=-\infty}^\infty r^{|n|}\sin(n\theta)\\

\end{eqnarray}

Since, $$r^{|n|}\sin(n\theta)$$ is an odd function it is clear that the second sum is equal to zero.

\begin{eqnarray}

\therefore P_r(\theta)&=&\sum_{n=-\infty}^\infty r^{|n|}\cos(n\theta)\\

&=&1+2\sum_{n=1}^\infty r^{n}\cos(n\theta)\\

\end{eqnarray}

Hence we finally get,

$P_r(\theta)=1+2\sum_{n=1}^\infty r^{n}\cos(n\theta)= \frac{1-r^2}{1-2r\cos\theta +r^2}\mbox{ where }0 \le r < 1.$
The books are already opened to the pages. I see it clearly.

Last edited:

#### Sudharaka

##### Well-known member
MHB Math Helper
Yes it seems that there is a slight difference in the definition of the Poisson Kernel. In some books it's defined as,

$P(r,\theta)=\frac{1-r^2}{1-2r\cos\theta +r^2}$

whereas in others,

$P(r,\theta)=\frac{1}{2\pi}\frac{1-r^2}{1-2r\cos\theta +r^2}$

You seem to be using this second definition. However notice that you are missing a $$n$$ in the summation of post #3.

$P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos({\color{red}n}\theta)\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}$

#### dwsmith

##### Well-known member
$P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos({\color{red}n}\theta)\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}$

Typo.
Is my soln correct?

#### Sudharaka

##### Well-known member
MHB Math Helper
Prove:
$$\lim_{r\to 1}P(r,\theta) = \begin{cases} \infty, & \theta = 0\\ 0, & \text{otherwise} \end{cases}$$
For the first piece, take the summation
$$P(1,0) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} 1^n\right).$$
Then $\sum\limits_{n = 1}^{\infty} 1^n = \infty$.
Therefore, we have a positive number plus infinity which is infinity when $r\to 1$ and $\theta = 0$.
For the second piece, take the fractional representation of the Poisson kernel,
$$P(1,\theta) = \frac{1}{2\pi}\frac{0}{2 - 2\cos\theta} = 0.$$
Therefore, $P({\color{red}1},\theta) = 0$ for all $\theta\neq 0$.
That is,
$$\lim_{r\to 1}P(r,\theta) = \begin{cases} \infty, & \theta = 0\\ 0, & \text{otherwise} \end{cases}$$
Yeah it's correct.