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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:
  • \( P_0(x) = 1 \)
  • \( P_1(x) = x \)
  • \( P_2(x) = \frac{1}{2} (3x^2 - 1) \)

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:
  • 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 \( \frac{2n+1}{2} \)
In practice, some coefficients may be zero due to the function's symmetry or properties, simplifying the series.
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.

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Most popular questions from this chapter

Solve the following differential equations by series and also by an elementary method and verify that your solutions agree. Note that the goal of these problems is not to get the answer (that's easy by computer or by hand) but to become familiar with the method of series solutions which we will be using later. Check your results by computer. $$y^{\prime \prime}=y$$

Solve the differential equations by the Frobenius method; observe that you get only one solution. (Note, also, that the two values of \(s\) are equal or differ by an integer, and in the latter case the larger \(s\) gives the one solution.) Show that the conditions of Fuchs's theorem are satisfied. Knowing that the second solution is \(\ln x\) times the solution you have, plus another Frobenius series, find the second solution. $$x^{2}(x-1) y^{\prime \prime}-x(5 x-4) y^{\prime}+(9 x-6) y=0$$

Solve the following differential equations by series and also by an elementary method and verify that your solutions agree. Note that the goal of these problems is not to get the answer (that's easy by computer or by hand) but to become familiar with the method of series solutions which we will be using later. Check your results by computer. $$\left(x^{2}+2 x\right) y^{\prime \prime}-2(x+1) y^{\prime}+2 y=0$$

Express each of the following polynomials as linear combinations of Legendre polynomials. Hint: Start with the highest power of \(x\) and work down in finding the correct combination. \(x-x^{3}\)

Expand the following functions in Legendre series. $$f(x)=\left\\{\begin{array}{lr} 0, & -1

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