Chapter 12: Problem 4
Expand the following functions in Legendre series. $$f(x)=\arcsin x$$
Short Answer
Expert verified
\( \arcsin(x) = \sum_{n=0}^{\infty/2} \frac{2^n (n!)^2}{(2n+1)(2n!)} P_{2n+1}(x) \)
Step by step solution
01
- Understanding Legendre Series
A Legendre series expansion expresses a function in terms of Legendre polynomials. The series for a function \( f(x) \) takes the form \( f(x) = \sum_{n=0}^{\infty} a_n P_n(x) \), where \( P_n(x) \) are Legendre polynomials and \( a_n \) are coefficients to be computed.
02
- Definition of Coefficients
The coefficients \( a_n \) in the Legendre series are given by \( a_n = \frac{2n+1}{2} \int_{-1}^{1} f(x) P_n(x) \mathrm{d}x \).
03
- Using Orthogonality Property
Use the orthogonality property of Legendre polynomials: \( \int_{-1}^{1} P_m(x) P_n(x) \mathrm{d}x = \frac{2}{2n+1} \delta_{mn} \), where \( \delta_{mn} \) is the Kronecker delta.
04
- Compute the Coefficients
Compute \( a_n \) using the definition: \( a_n = \frac{2n+1}{2} \int_{-1}^{1} \arcsin(x) P_n(x) \mathrm{d}x \). This integral can be computed using integration techniques or referring to tables of integrals.
05
- Specific Result for \(f(x) = \arcsin(x)\)
For \( f(x) = \arcsin(x) \), some coefficients vanish due to symmetry properties and some basic calculations. The specific result is: \( \arcsin(x) = \sum_{n=0}^{\frac{\infty}{2}} \frac{2^n (n!)^2}{(2n+1)(2n!)} P_{2n+1}(x) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Legendre polynomials
Legendre polynomials, often denoted as \( P_n(x) \), are a sequence of orthogonal polynomials that arise in solving certain types of differential equations. These polynomials are essential in physics and engineering, particularly in problems involving spherical coordinates. One of their key properties is orthogonality over the interval \( [-1, 1] \). This means that the integral of the product of two different Legendre polynomials over this interval is zero.
Legendre polynomials can be generated using Rodrigues' formula:
\[ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n \]
This formula allows us to derive any \( P_n(x) \) polynomial. For example:
The Legendre polynomials form a complete set of functions on the interval \( [-1, 1] \), enabling the expansion of any reasonable function \( f(x) \) into a Legendre series.
Legendre polynomials can be generated using Rodrigues' formula:
\[ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n \]
This formula allows us to derive any \( P_n(x) \) polynomial. For example:
- \( P_0(x) = 1 \)
- \( P_1(x) = x \)
- \( P_2(x) = \frac{1}{2} (3x^2 - 1) \)
The Legendre polynomials form a complete set of functions on the interval \( [-1, 1] \), enabling the expansion of any reasonable function \( f(x) \) into a Legendre series.
coefficients
The coefficients in a Legendre series expansion are critical for representing a function \( f(x) \) as a sum of Legendre polynomials. They are denoted as \( a_n \) and calculated using the integral formula:
\[ a_n = \frac{2n+1}{2} \int_{-1}^{1} f(x) P_n(x) \,\mathrm{d}x \]
This formula uses the function \( f(x) \), Legendre polynomial \( P_n(x) \), and the orthogonality property. The term \( \frac{2n+1}{2} \) ensures proper scaling of the coefficients.
For instance, to find the coefficient \( a_n \) for a given function \( f(x) \), follow these steps:
\[ a_n = \frac{2n+1}{2} \int_{-1}^{1} f(x) P_n(x) \,\mathrm{d}x \]
This formula uses the function \( f(x) \), Legendre polynomial \( P_n(x) \), and the orthogonality property. The term \( \frac{2n+1}{2} \) ensures proper scaling of the coefficients.
For instance, to find the coefficient \( a_n \) for a given function \( f(x) \), follow these steps:
- Identify the function \( f(x) \)
- Choose the Legendre polynomial \( P_n(x) \)
- Integrate \( f(x) P_n(x) \) over the interval \( [-1, 1] \)
- Multiply by \( \frac{2n+1}{2} \)
orthogonality property
The orthogonality property of Legendre polynomials is a foundational concept. Mathematically, it is expressed as:
\[ \int_{-1}^{1} P_m(x) P_n(x) \,\mathrm{d}x = \frac{2}{2n+1} \delta_{mn} \]
Here, \( \delta_{mn} \) is the Kronecker delta, which is 1 when \( m = n \) and 0 otherwise. This property ensures that the Legendre polynomials are orthogonal over \( [-1, 1] \), meaning the integral of the product of two different polynomials from the sequence is zero.
This orthogonality simplifies the computation of coefficients in a Legendre series. When calculating \( a_n \), the integrals involving different Legendre polynomials vanish, leaving only the integrals of \( P_n(x) \) squared. This dramatically reduces the complexity of finding each coefficient.
Orthogonality also guarantees the uniqueness of the Legendre series expansion. For any function \( f(x) \) that can be expressed as a series, the coefficients \( a_n \) are uniquely determined by the orthogonality property.
\[ \int_{-1}^{1} P_m(x) P_n(x) \,\mathrm{d}x = \frac{2}{2n+1} \delta_{mn} \]
Here, \( \delta_{mn} \) is the Kronecker delta, which is 1 when \( m = n \) and 0 otherwise. This property ensures that the Legendre polynomials are orthogonal over \( [-1, 1] \), meaning the integral of the product of two different polynomials from the sequence is zero.
This orthogonality simplifies the computation of coefficients in a Legendre series. When calculating \( a_n \), the integrals involving different Legendre polynomials vanish, leaving only the integrals of \( P_n(x) \) squared. This dramatically reduces the complexity of finding each coefficient.
Orthogonality also guarantees the uniqueness of the Legendre series expansion. For any function \( f(x) \) that can be expressed as a series, the coefficients \( a_n \) are uniquely determined by the orthogonality property.