Problem 19 - Complex Analysis for Mathematics and Engineering

Question

The Legendre polynomial \(P_{n}(z)\) is defined by \(P_{n}(z)=\frac{1}{2^{n} n !} \frac{d^{n}}{d z^{n}}\left[\left(z^{2}-1\right)^{n}\right]\) Use Cauchy's integral formula to show that \(P_{n}(z)=\frac{1}{2 \pi i} \int_{C} \frac{\left(\xi^{2}-1\right)^{n} d \xi}{2^{n}(\xi-z)^{n+1}}\) where \(C\) is a simple closed contour having positive orientation and \(z\) lies inside \(C\).

Step-by-Step Solution

Answer

Use Cauchy's formula to derive the \(n\)-th derivative, multiply by \(\frac{1}{2^n n!}\), and verify the integral form.

Step 1: Understand Cauchy's Integral Formula

Cauchy's Integral Formula states that if a function \(f\) is analytic inside and on some closed contour \(C\), then for any point \(z\) inside \(C\), \(f(z)\) is given by: \[f(z) = \frac{1}{2\pi i} \oint_{C} \frac{f(\xi)}{\xi-z} d\xi\] If taking derivatives, the \(n\)-th derivative \(f^{(n)}(z)\) is: \[f^{(n)}(z) = \frac{n!}{2\pi i} \oint_{C} \frac{f(\xi)}{(\xi-z)^{n+1}} d\xi\]

Step 2: Identify Function for Derivatives

Given the form of Legendre Polynomials, identify the function inside Cauchy's formula that needs differentiation: \(f(z) = (z^2 - 1)^n\). Our goal is to apply Cauchy's formula for \(f^{(n)}(z)\).

Step 3: Derive Using Cauchy's Integral Formula

Using Cauchy's formula for differentiation, the \(n\)-th derivative of \(f(z) = (z^2 - 1)^n\) is given by: \[f^{(n)}(z) = \frac{n!}{2\pi i} \oint_{C} \frac{(\xi^2 - 1)^n}{(\xi-z)^{n+1}} d\xi\] We're tasked with finding \( \frac{1}{2^n n!} f^{(n)}(z)\), where \((\xi^2 - 1)^n \) replaces \(f(\xi)\) in the integral expression.

Step 4: Simplify and Relate to Legendre Polynomial

Simplifying, we have: \[ \frac{1}{2^n n!} \cdot \frac{n!}{2 \pi i} \oint_{C} \frac{(\xi^2 - 1)^n}{(\xi-z)^{n+1}} d\xi = \frac{1}{2 \pi i} \oint_{C} \frac{(\xi^2 - 1)^n}{2^n(\xi-z)^{n+1}} d\xi\] which matches the given expression of \(P_n(z)\). This confirms the Legendre polynomial representation using Cauchy's formula.

Key Concepts

Legendre Polynomials

Legendre polynomials are a sequence of orthogonal polynomials that have applications across various fields of science and engineering, from physics to numerical analysis. They are particularly useful in solving problems involving spherical coordinates, because of their role in the expansion of functions. The formula for computing Legendre polynomials is given by:\[P_{n}(z)=\frac{1}{2^{n} n !} \frac{d^{n}}{dz^{n}}\left[\left(z^{2}-1\right)^{n}\right]\]This expression takes a function \((z^2 - 1)^n\), differentiates it \(n\) times, and scales it by the \( \left(\frac{1}{2^n n!}\right)\) constant. The differentiation gives us a polynomial of degree \(n\). These polynomials are orthogonal on the interval \([-1,1]\) with respect to the weight function \(1\).
Legendre polynomials are named after Adrien-Marie Legendre and are solutions to Legendre's differential equation, one of the most fundamental differential equations in mathematical physics.

Cauchy's Integral Formula

Cauchy's Integral Formula is a cornerstone of complex analysis. It provides a way to evaluate integrals of analytic functions over closed curves and highlights the beautiful properties of these functions. The formula states:\[f(z) = \frac{1}{2\pi i} \oint_{C} \frac{f(\xi)}{\xi-z} d\xi\]where \(f\) is analytic inside and on a positively oriented closed contour \(C\) and \(z\) is any point within \(C\). The key feature of this formula is its ability to relate the values of a function inside a contour to its values on the boundary, thus providing a way to recover the entire function from its behavior along the contour.
When extended to incorporate derivatives, it becomes:\[f^{(n)}(z) = \frac{n!}{2\pi i} \oint_{C} \frac{f(\xi)}{(\xi-z)^{n+1}} d\xi\]This version allows for the calculation of derivatives of all orders, connecting differentiation and integration in the context of analytic functions. This ability to calculate high order derivatives plays a crucial role in proving representations like that of the Legendre polynomials, where you differentiate and evaluate within the context of an integral.

Analytic Functions

Analytic functions are central to complex analysis, being functions that are locally represented by convergent power series. A function \(f\) is said to be analytic at a point \(z_0\) if it can be expressed as a power series:\[f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n\]for some neighborhood around \(z_0\). This means that analytic functions are infinitely differentiable, smooth, and possess derivatives of all orders. Key properties include:

• Their ability to conform to shapes, transmitting information from boundary values inward, as seen in Cauchy's Integral Formula.
• Being determined completely by their values in arbitrarily small regions.
• Following the identity theorem, where two analytic functions that agree on a set with an accumulation point agree everywhere their domains overlap.

These properties establish analytic functions as instrumental in both theoretical and applied mathematics, providing tools for evaluating complex integrals, conferring the power to expand functions into series, and ensuring the continuation of complex mathematical identities across their entire domains.