Question

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) The Legendre polynomial \(P_{n}(x)\) is defined as the polynomial solution of the Legendre equation with \(\alpha=n\) that also satisfies the condition \(P_{n}(1)=1\). (a) Using the results of Problem 23 , find the Legendre polynomials \(P_{0}(x), \ldots . P_{5}(x) .\) (b) Plot the graphs of \(P_{0}(x), \ldots, P_{5}(x)\) for \(-1 \leq x \leq 1 .\) (c) Find the zeros of \(P_{0}(x), \ldots, P_{5}(x)\).

Short answer

Question: Find the first six Legendre polynomials P0(x), P1(x), ..., P5(x) using the given Legendre equation and plot their graphs within the range -1 ≤ x ≤ 1. Also find the zeros of these polynomials. Answer: Using the given recursion formula, we found the first six Legendre polynomials as follows: P0(x) = 1 P1(x) = x P2(x) = (3/2)x^2 - 1/2 P3(x) = (5/2)x^3 - (3/2)x P4(x) = (35/8)x^4 - (15/4)x^2 + 3/8 P5(x) = (63/8)x^5 - (35/4)x^3 + (15/8)x By solving for zeros, we found that: P0(x) has no zeros, P1(x) has a zero at x = 0, P2(x) has zeros at x = ±1/√3, P3(x) has zeros at x = 0, ±√(3/5), P4(x) has zeros at x = ±√[(3/7)±√(30/35)], P5(x) has zeros at x = 0, ±√[(5+√105)/14], ±√[(5-√105)/14]. To plot the graphs of these polynomials within the range -1 ≤ x ≤ 1, use a software application like Desmos, Geogebra, or a graphing calculator.

Step-by-step solution

(Step 1: Recall the recursion formula from Problem 23)

According to Problem 23, the recursion formula for Legendre polynomials is: $$ P_{n+1}(x)=\left(\frac{2 n+1}{n+1}\right) x P_{n}(x)-\frac{n}{n+1} P_{n-1}(x) $$ for \(n \geq 0\) and initial conditions: \(P_{0}(x)=1, P_{1}(x)=x\).

(Step 2: Use the recursion formula to find the Legendre polynomials)

Using the initial conditions and recursion formula, we can find the Legendre polynomials: $$ P_0(x)=1 \\ P_1(x)=x \\ P_2(x)=\frac{3}{2}x^2-\frac{1}{2} \\ P_3(x)=\frac{5}{2}x^3-\frac{3}{2}x \\ P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} \\ P_5(x)=\frac{63}{8}x^5-\frac{35}{4}x^3+\frac{15}{8}x $$ ##Part (b) - Plotting the Graphs## To plot the graphs of these polynomials for \(-1 \leq x \leq 1\), a software application like Desmos, Geogebra, or a graphing calculator can be used. This can help in visualizing the functions and their behavior within the specified range. ##Part (c) - Finding the Zeros## According to the definition and formulae presented above, the zeros of \(P_{0}(x), P_{1}(x), ..., P_{5}(x)\), can be found by solving the following equations: $$ P_0(x)=1: \text{No zeros} \\ P_1(x)=x: x=0 \\ P_2(x)=\frac{3}{2}x^2-\frac{1}{2}: x=\pm \frac{1}{\sqrt{3}} \\ P_3(x)=\frac{5}{2}x^3-\frac{3}{2}x: x=0, \pm \sqrt{\frac{3}{5}} \\ P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8}: x=\pm \sqrt{\frac{3}{7}\pm\sqrt{\frac{30}{35}}} \\ P_5(x)=\frac{63}{8}x^5-\frac{35}{4}x^3+\frac{15}{8}x: x=0, \pm \sqrt{\frac{5 \pm \sqrt{105}}{14}}, \pm \sqrt{\frac{5 - \sqrt{105}}{14}} $$

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are fundamental in understanding various natural phenomena and are used to model everything from simple springs to complex circuits. An ODE involves derivatives of a function with respect to one variable and doesn't rely on derivatives with respect to more than one variable. In the context of Legendre polynomials, we encounter the Legendre equation—a specific second-order differential equation.
This equation is pivotal in physics and engineering, especially in fields like electrostatics and quantum mechanics, where spherical symmetry is present. The standard Legendre differential equation is given by \[(1-x^2)y'' - 2xy' + \alpha(\alpha+1)y = 0.\]Solving this requires understanding initial conditions and guaranteeing that solutions exist within a specified radius.
For Legendre polynomials, solutions for specific values of \(\alpha\) form polynomial solutions that are orthogonal over the interval \([-1, 1]\). These functions play a critical role in expanding other functions through series in physics-related problems.

Series Solutions

Series solutions are a powerful tool for solving differential equations that might not be easily solvable by standard methods. These involve expressing the solution of a differential equation as an infinite sum of terms, specifically as a power series around a particular point. For Legendre polynomials, the series solution approach is applicable around the ordinary point, \(x=0\).
The solution will converge within a certain range, dictated by the equation itself. When dealing with a polynomial such as the Legendre equation, the power series allows us to find solutions that can then be expressed as polynomials of a specific degree—the Legendre polynomials themselves.
This technique extends to understanding how differential equations describe systems within defined regions, helping us compute solutions that meet specified boundary conditions. The fascinating aspect here is that when \(\alpha = n\), where \(n\) is an integer, the solution derives a polynomial of degree \(n\), essential for constructing spherical harmonics.

Recursion Formula

The recursion formula is a recurrent method used to simplify the calculation of sequences, particularly polynomials like the Legendre polynomials. It provides a systematic way to derive higher-order terms based on lower-order ones, making it an efficient method for computation.
The recursion formula for Legendre polynomials is expressed as:\[P_{n+1}(x) = \left(\frac{2n+1}{n+1}\right)xP_n(x) - \frac{n}{n+1}P_{n-1}(x),\]where we begin with the known initial conditions \(P_0(x) = 1\) and \(P_1(x) = x\).
Utilizing this formula, calculating polynomials \(P_n(x)\) becomes straightforward up the series, and helps in assembling a series solution that's potentially infinite. This is particularly helpful in fields requiring calculations of complex integrals, enabling us to derive orthogonal polynomials efficiently.

Radius of Convergence

The radius of convergence is a critical aspect of series solutions, defining the interval in which the series converges to a function. For differential equations like the Legendre equation, understanding the radius of convergence is essential to ensure that solutions are valid and representable by polynomials within that range.
For the Legendre equation, the radius of convergence is determined by examining the equation and identifying points where coefficients become singular or other significant changes occur. With Legendre polynomials, the distance to the nearest singular point is the radius of convergence.
Here, where the differential equation involves \[\left(1-x^{2}\right)y^{\prime\prime}-2xy^{\prime}+\alpha(\alpha+1)y=0,\]the nearest singularity occurs when \(1 - x^2 = 0\), or at \(x = \pm 1\). This implies a radius of convergence of 1 about the point \(x = 0\). Understanding this helps ensure the series solution accurately models phenomena in the desired interval without divergence.