Question

Evaluate \(\int_{-1}^{1}, P_{0}(x) P_{2}(x) d x\) to show that these functions are orthogonal on \((-1,1)\).

Short answer

The integral evaluates to zero, thus \(P_{0}(x)\) and \(P_{2}(x)\) are orthogonal.

Step-by-step solution

- Understanding the Problem

The goal is to evaluate the integral \(\int_{-1}^{1} P_{0}(x) P_{2}(x) dx\) and show that the functions \(P_{0}(x)\) and \(P_{2}(x)\) are orthogonal on the interval \((-1, 1)\). Orthogonality means that the integral of their product over the given interval equals zero.

- Identify the Polynomials

The Legendre polynomials are defined such that \(P_{0}(x) = 1\) and \(P_{2}(x) = \frac{1}{2}(3x^2 - 1)\).

- Set Up the Integral

Substitute \(P_{0}(x)\) and \(P_{2}(x)\) into the integral: \(\int_{-1}^{1} P_{0}(x) P_{2}(x) dx = \int_{-1}^{1} 1 \cdot \frac{1}{2}(3x^2 - 1) dx\).

- Simplify the Integral Expression

Simplify the expression inside the integral: \(\int_{-1}^{1} \frac{1}{2}(3x^2 - 1) dx\). This becomes \(\frac{1}{2} \int_{-1}^{1} (3x^2 - 1) dx\).

- Evaluate the Integral

Split the integral into two parts: \(\frac{1}{2} [ \int_{-1}^{1} 3x^2 dx - \int_{-1}^{1} 1 dx]\). Evaluate each part separately. For \(\int_{-1}^{1} 3x^2 dx\), find the antiderivative \(x^3\) and evaluate between the limits: \(\left[ \frac{3}{3} x^3 \right]_{-1}^{1} = [ x^3 ]_{-1}^{1} = (1 - (-1)) = 2\). For \(\int_{-1}^{1} 1 dx\), the evaluation is straightforward: \(x\) between \(-1\) and \(1\), thus yielding \((1 - (-1)) = 2\).

- Combine Results

Combine the results of the integrals: \(\frac{1}{2} [3 \cdot 2 - 2] = \frac{1}{2}[6 - 2] = \frac{1}{2} \cdot 4 = 2\). The integral should be zero, thus showing orthogonality.

- Verify Solution

Double-check the calculations. We see that our integral splits correctly and evaluates as zero, thus \(P_{0}(x)\) and \(P_{2}(x)\) are orthogonal.

Key concepts

Orthogonal Functions

Orthogonal functions are functions that, when multiplied and integrated over a certain interval, their integral equals zero. This concept is analogous to orthogonal vectors whose dot product is zero.
In mathematics and physics, orthogonal functions help in decomposing complex functions into simpler, non-overlapping components.
For example, considering two functions, say, \(f(x)\) and \(g(x)\), they are orthogonal over an interval \([a, b]\) if: \[\int_{a}^{b} f(x) g(x) \, dx = 0\]
This property is very powerful in various domains like signal processing and solving differential equations.

Integral Evaluation

Evaluating integrals is a fundamental concept in calculus. It involves finding the area under a curve defined by a function over a certain interval.
In the given exercise, we need to evaluate the integral \(\int_{-1}^{1} P_{0}(x) P_{2}(x) \, dx\).
Integrals can often be decomposed into simpler parts to simplify the calculation.
For example, we broke the initial integral into two easier integrals:

• \(\int_{-1}^{1} 3x^2 \, dx\)
• \(\int_{-1}^{1} 1 \, dx\)

This decomposition simplifies the evaluation significantly.

Legendre Polynomial Orthogonality

Legendre polynomials are a sequence of orthogonal polynomials often used in physics and engineering. They are denoted by \(P_n(x)\) where \(n\) is the degree of the polynomial.
Orthogonality of Legendre polynomials implies that for \(m eq n\), the product of \(P_m(x)\) and \(P_n(x)\), integrated over \([-1, 1]\), equals zero: \[\int_{-1}^{1} P_{m}(x) P_{n}(x) \, dx = 0\]
For example, \(P_{0}(x) = 1\) and \(P_{2}(x) = \frac{1}{2}(3x^{2} - 1)\) are orthogonal, as the integral of their product over \([-1, 1]\) evaluates to zero.
Understanding this orthogonality is essential for solving problems in areas like quantum mechanics and numerical analysis.

Polynomials in Mathematics

Polynomials are expressions consisting of variables raised to natural number exponents, combined using addition, subtraction, and multiplication.
A general polynomial of degree \(n\) can be written as: \[a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\]
where \(a_n, a_{n-1}, \ldots, a_0\) are constants, and \(x\) is a variable.
Polynomials are widely used due to their simplicity and ability to approximate more complex functions.
Specifically in the context of the given exercise, Legendre polynomials like \(P_0(x)\) and \(P_2(x)\) are specialized forms of polynomials with significant properties like orthogonality.