Problem 52 The Legendre polynomials are als... [FREE SOLUTION]

Chapter 6: Problem 52

The Legendre polynomials are also generated by Rodrigues' formula $$ P_{n}(x)=\frac{1}{2^{2} n !} \frac{d^{n}}{d x^{n}}\left(x^{2}-1\right)^{n} . $$ Verify the results for $n=0,1,2,3$.

Short Answer

Expert verified

Legendre polynomials for n=0, 1, 2, 3 are: 1, x, $\frac{3}{2}x^2 - \frac{1}{2}$, and $\frac{5}{2}x^3 - \frac{3}{2}x$.

Step by step solution

Understanding Rodrigues' formula

Rodrigues' formula generates Legendre polynomials: $P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n$. We'll compute this for $n=0, 1, 2, 3$. This involves differentiating the expanded form of $(x^2 - 1)^n$, where $n$ is the order of the polynomial, and simplifying.

Compute for n=0

Substituting $n=0$ into Rodrigues' formula, we get $P_0(x) = \frac{1}{1} \frac{d^0}{dx^0} (x^2 - 1)^0 = \frac{d^0}{dx^0} 1 = 1$. Hence, $P_0(x) = 1$.

Compute for n=1

For $n=1$, the formula becomes $P_1(x) = \frac{1}{2^1 1!} \frac{d^1}{dx^1} (x^2 - 1)^1$. Expand $(x^2 - 1)^1 = x^2 - 1$. Now differentiate once to get $2x$. Therefore, $P_1(x) = \frac{1}{2} \cdot 2x = x$.

Compute for n=2

For $n=2$, we use $P_2(x) = \frac{1}{2^2 \cdot 2!} \frac{d^2}{dx^2} (x^2 - 1)^2$. First, expand $(x^2 - 1)^2 = x^4 - 2x^2 + 1$. Take the second derivative: $12x^2 - 4$. Thus, $P_2(x) = \frac{1}{8} \cdot (12x^2 - 4) = \frac{3}{2}x^2 - \frac{1}{2}$. Correct it with normalization factor: $P_2(x)= \frac{3}{2}x^2 - \frac{1}{2}$.

Compute for n=3

For $n=3$, we evaluate $P_3(x) = \frac{1}{2^3 \cdot 3!} \frac{d^3}{dx^3} (x^2 - 1)^3$. First, expand $(x^2 - 1)^3 = x^6 - 3x^4 + 3x^2 - 1$. Differentiate three times to get $120x^3 - 24x$. Thus, $P_3(x) = \frac{1}{48} (120x^3 - 24x) = \frac{1}{2}(5x^3 - 3x) = \frac{5}{2}x^3 - \frac{3}{2}x$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rodrigues' formula

Rodrigues' formula is a powerful tool used to generate Legendre polynomials. These are a sequence of polynomials that are solutions to certain types of differential equations. The formula is written as:\[P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n\]Here's what it means:

$P_n(x)$ is the nth Legendre polynomial.
The term $(x^2 - 1)^n$ is the base expression whose nth derivative is taken.
$\frac{d^n}{dx^n}$ involves differentiating $(x^2 - 1)^n$ n times.
$n!$ is the factorial of $n$, used for normalization.

This formula highlights a methodical approach to constructing each polynomial by applying calculus principles.

Differential equations

Legendre polynomials often appear as solutions to differential equations, more specifically, the Legendre differential equation:\[(1 - x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + n(n+1)y = 0\]In this equation:

You'll find it involves second-order differentiation, as indicated by $\frac{d^2y}{dx^2}$.
The expression $n(n+1)$ is crucial, connecting the polynomial order to the differential equation.
This equation is commonly encountered in physics, particularly in problems involving spherical coordinates due to its symmetry properties.

Solving these equations often leads to discovering orthogonal polynomials like Legendre polynomials.

Polynomial order

The "order" of a polynomial is a term used to describe the highest degree of its terms. In Legendre polynomials, the order $n$ determines both the degree of the polynomial and its complexity:

Lower-order polynomials are simpler with fewer terms. For instance, $P_0(x) = 1$ is just a constant.
As the order increases, polynomials become more complex. For example, $P_3(x) = \frac{5}{2}x^3 - \frac{3}{2}x$ includes cubic and linear terms.
The order also dictates how many times $(x^2 - 1)^n$ is differentiated, making higher-order calculations more involved.

Understanding the polynomial order helps in predicting the structure and behavior of Legendre polynomials in applications.

Orthogonal polynomials

Legendre polynomials are a classic example of orthogonal polynomials. Orthogonality here implies a special kind of independence. Two functions (or polynomials) are orthogonal if their inner product over a specific interval is zero:\[\int_{-1}^{1} P_m(x) P_n(x) \, dx = 0, \quad \text{for} \quad m eq n \]Key properties:

This orthogonality simplifies computations in areas such as numerical analysis and approximation theory.
It allows for decompositions of more complex functions into simple polynomial components over the interval $[-1, 1]$.
Legendre polynomials often appear in expansions of functions in terms of a series, much like Fourier series.

Understanding this orthogonality helps leverage the mathematical properties of Legendre polynomials in practical and theoretical scenarios.

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91影视

The Legendre polynomials are also generated by Rodrigues' formula $$ P_{n}(x)=\frac{1}{2^{2} n !} \frac{d^{n}}{d x^{n}}\left(x^{2}-1\right)^{n} . $$ Verify the results for \(n=0,1,2,3\).

Short Answer

Step by step solution

Understanding Rodrigues' formula

Compute for n=0

Compute for n=1

Compute for n=2

Compute for n=3

Key Concepts

Rodrigues' formula

Differential equations

Polynomial order

Orthogonal polynomials

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