# Complex Variables - Legendre Polynomial

• MHB
• joypav
In summary, we have shown that $P_n(z)=\frac{1}{2\pi}\int_0^{2\pi}\left(z+\sqrt{z^2-1}\cos(\theta)\right)^{\!n}\,d\theta$, by using the previous problem and simplifying the integral.
joypav
We define the Legendre polynomial $P_n$ by
$$P_n (z)=\frac{1}{2^nn!}\frac{d^n}{dz^n}(z^2-1)^n$$
Let $\omega$ be a smooth simple closed curve around z. Show that
$$P_n (z)=\frac{1}{2i\pi}\frac{1}{2^n}\int_\omega\frac{(w^2-1)^n}{(w-z)^{n+1}}dw$$

What I have:

We know $(w^2-1)^n$ is analytic on and inside $\omega$ and z lies within $\omega$.
So, by Cauchy's Formula,
$$\frac{1}{2i\pi}\frac{1}{2^n}\int_\omega\frac{(w^2-1)^n}{(w-z)^{n+1}}dw$$
$$=\frac{1}{2i\pi}\frac{1}{2^n}\frac{2i\pi}{n!}\frac{d^n}{dw^n}(w^2-1)^n\rvert_{w=z}$$
$$=\frac{1}{2^n}\frac{1}{n!}\frac{d^n}{dw^n}(w^2-1)^n\rvert_{w=z}$$
$$=\frac{1}{2^nn!}\frac{d^n}{dz^n}(z^2-1)^n=P_n(z)$$

Now, take $\omega$ to be the circle of radius $\sqrt{|z^2-1|}$ centered at z. Show that
$$P_n(z)=\frac{1}{2\pi}\int_0^{2\pi}\left(z+\sqrt{z^2-1}\cos(\theta)\right)^{\!n}\,d\theta$$

What I have:

I am confused about how to apply the previous problem. If I could get a push in the right direction I would be very appreciative!

So, we have
\begin{align*}
\omega&= z+\sqrt{|z^2-1|} \, e^{i\theta} \\
\omega - z &=\sqrt{|z^2-1|} \, e^{i\theta} \\
d\omega &=\sqrt{|z^2-1|} \cdot i \cdot e^{i\theta} \, d\theta \\
\omega^2 &=z^2+2z\sqrt{|z^2-1|} \, e^{i\theta} + |z^2-1| \, e^{2i\theta}.
\end{align*}
Now we have the necessary ingredients to plug into what you showed before, namely, $\displaystyle P_n(z)=\frac{1}{2\pi i} \, \frac{1}{2^n} \, \oint_{\omega}\frac{(\omega^2 - 1)^n}{(\omega-z)^{n+1}} \, d\omega,$ to obtain
$$P_n(z)=\frac{1}{2\pi i} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, e^{i\theta}+|z^2-1| \, e^{2i\theta} - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, e^{i\theta}\right)^{n+1}} \cdot\sqrt{|z^2-1|}\cdot i \cdot e^{i\theta} \, d\theta.$$
There's definitely some simplification possible here. The hope is that, once you do simplify, you will get the desired result above. It does not seem impossible that you could get equality.

Thank you!

(for the sake of completion)
$$P_n(z)=\frac{1}{2\pi i} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, e^{i\theta}+|z^2-1| \, e^{2i\theta} - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, e^{i\theta}\right)^{n+1}} \cdot\sqrt{|z^2-1|}\cdot i \cdot e^{i\theta} \, d\theta$$
$$=\frac{1}{2\pi} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, e^{i\theta}+|z^2-1| \, e^{2i\theta} - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, e^{i\theta}\right)^{n}}d\theta$$
$$=\frac{1}{2\pi} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))+|z^2-1| \, (\cos(2\theta)+i\sin(2\theta)) - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))\right)^{n}}d\theta$$
$$=\frac{1}{2\pi} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))+|z^2-1| \, (2\cos^2(\theta)-1+2i\sin(\theta)\cos(\theta)) - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))\right)^{n}}d\theta$$
$$=\frac{1}{2\pi} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left((2)^n(\sqrt{|z^2-1|}) \, (\cos(\theta)+i\sin(\theta))(z+\sqrt{z^2-1}\cos(\theta) \, \right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))\right)^{n}}d\theta$$
$$=\frac{1}{2\pi} \,\int_{0}^{2\pi}{\displaystyle\left(z+\sqrt{z^2-1}\cos(\theta) \, \right)^{n}}d\theta.$$

## What are complex variables?

Complex variables are mathematical quantities that involve both real and imaginary components. They are typically represented as points on a complex plane, with the real component represented on the x-axis and the imaginary component represented on the y-axis.

## What are Legendre polynomials?

Legendre polynomials are a set of orthogonal polynomials that are used to solve differential equations involving complex variables. They are named after the French mathematician Adrien-Marie Legendre, who first studied them in the 18th century.

## What are the applications of complex variables and Legendre polynomials?

Complex variables and Legendre polynomials have many applications in mathematics, physics, engineering, and other fields. They are commonly used to solve problems involving electric fields, fluid dynamics, quantum mechanics, and signal processing, among others.

## How are Legendre polynomials calculated?

Legendre polynomials can be calculated using various methods, such as the Gram-Schmidt process, the Rodrigues formula, or the recurrence relation method. These methods involve manipulating the coefficients of the polynomials to generate higher-order terms.

## What is the significance of Legendre polynomials in complex analysis?

Legendre polynomials play a crucial role in complex analysis, as they are used to represent complex-valued functions as a series of polynomials. This allows for the evaluation of complex integrals and the solution of complex differential equations, which are essential tools in the study of complex variables.

Replies
2
Views
1K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
4
Views
1K
Replies
4
Views
955
Replies
2
Views
1K
Replies
2
Views
1K
Replies
9
Views
2K
Replies
3
Views
2K
Replies
7
Views
2K