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

Short Answer

Expert verified
The Legendre series expansion of the function is \( f(x) ≈ -\frac{3}{32} + \frac{1}{2} x + \frac{15}{32}x^2 \).

Step by step solution

01

Understand the Legendre Series Expansion

Any function can be expanded in terms of Legendre polynomials. The Legendre series for a function can be written as: \[ f(x) = \frac{a_0}{2} + \boldsymbol{\text{sum}}_{n=1}^{\boldsymbol{\text{infinity}}} a_n P_n(x) \]Where \(a_n\) are the coefficients given by: \[ a_n = \frac{2n + 1}{2} \boldsymbol{\text{integral}}_{-1}^{1} f(x) P_n(x) dx \]
02

Identify the Function and its Domain

The given function is \[ f(x)=\begin{cases} 0, & -1 < x < 0 \ x, & 0 < x < 1 \ \text{undefined}, & \text{otherwise} \ \text{cancel undefined as this is not a part} \ \text{of the original function} \ \text{domain is limited from -1 to 1 only} \ \text{so we ignore outside part} \ \text{_the '_' represents a line_spaces shall be here}\end\{array\} \]The function is piece-wise and defined on the interval \( -1 \text{to} 1 \).
03

Calculate Coefficients \(a_n\)

The coefficients are calculated using the formula: \[ a_n = \frac{2n + 1}{2} \boldsymbol{\text{integral}}_{-1}^{1} f(x) P_n(x) dx \].Since the function is piecewise, split the integral into two parts: \[ a_n = \frac{2n + 1}{2} \boldsymbol{\text{integral}}_{-1}^{0} 0 \times P_n(x) dx + \frac{2n + 1}{2} \boldsymbol{\text{integral}}_{0}^{1} x \times P_n(x) dx \]The first integral is zero: \[ \boldsymbol{\text{integral}}_{-1}^{0} 0 \times P_n(x) dx = 0 \]Thus, \[ a_n = \frac{2n + 1}{2} \boldsymbol{\text{integral}}_{0}^{1} x P_n(x) dx \]
04

Evaluate the Integral

Evaluate the integral for each \(n\): \[ \boldsymbol{\text{integral}}_{0}^{1} x P_n(x) dx \].For example, \( P_0(x) = 1, P_1(x) = x, P_2(x) = \frac{1}{2}(3x^2 - 1), \text{and} so on \)
05

Calculate Example Coefficients

Calculate a few example coefficients. For \( a_0 \): \[ a_0 = \frac{1}{2} \boldsymbol{\text{integral}}_{0}^{1} x dx = \frac{1}{2} \times \frac{x^2}{2} \bigg|_0^1 = \frac{1}{4} \]For \( a_1 \): \[ a_1 = \frac{3}{2} \boldsymbol{\text{integral}}_{0}^{1} x^2 dx = \frac{3}{2} \times \frac{x^3}{3} \bigg|_0^1 = \frac{1}{2} \]For \( a_2 \): \[ a_2 = \frac{5}{2} \boldsymbol{\text{integral}}_{0}^{1} x \frac{1}{2}(3x^2 - 1) dx = \frac{5}{4} \times \boldsymbol{\text{integral}}_{0}^{1} (3x^3 - x) dx = \frac{5}{4} \times \bigg( \frac{3}{4} - \frac{1}{2} \bigg) = \frac{5}{16} \]
06

Formulate the Series

Using the coefficients calculated, the series becomes: \[ f(x) = \frac{a_0}{2} + a_1 P_1(x) + a_2 P_2(x) + \text{higher order terms} \].Substitute \(a_0, a_1, \text{and} a_2\): \[ f(x) ≈ \frac{1}{8} + \frac{1}{2} P_1(x) + \frac{5}{16} P_2(x) + \text{higher order terms} \]Recall \( P_1(x) = x \text{and} P_2(x) = \frac{1}{2}(3x^2 - 1) \)
07

Simplify the Final Expression

The simplified Legendre series using the first few terms: \[ f(x) ≈ \frac{1}{8} + \frac{1}{2} x + \frac{5}{16}\bigg( \frac{3}{2}x^2 - \frac{1}{2} \bigg) \]Final simplified form: \[ f(x) ≈ \frac{1}{8} + \frac{1}{2} x + \frac{15}{32}x^2 - \frac{5}{32} \]Grouping terms further: \[ f(x) ≈ -\frac{3}{32} + \frac{1}{2} x + \frac{15}{32}x^2 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Legendre Polynomials
Legendre polynomials are a type of orthogonal polynomials that are solutions to the Legendre differential equation. They are denoted by \( P_n(x) \), where \( n \) indicates the polynomial's degree. These polynomials are defined over the interval \([-1, 1]\) and have unique properties that make them very useful in approximating functions.

One of the most important properties of Legendre polynomials is that they are orthogonal with respect to the inner product:
\[ \int_{-1}^{1} P_m(x) P_n(x) \, dx = 0, \, \text{if} \, m \e n \]\
This orthogonality means we can decompose functions into sums of these polynomials, which we call a Legendre series expansion. Here are the first few Legendre polynomials:

  • \( P_0(x) = 1 \)
  • \( P_1(x) = x \)
  • \( P_2(x) = \frac{1}{2}(3x^2 - 1) \)
Piecewise Functions
Piecewise functions are functions defined by different expressions in different intervals of the domain. For example, in the given exercise:
\[ f(x)=\begin{cases} \, \ \0, & -1 < x < 0 \ \ x, & \, 0 < x < 1 \ \ \end{cases} \]

This function is defined as 0 for \(-1 < x < 0\) and as \( x \) for \( 0 < x < 1 \). This means the function takes different values depending on the interval of \( x \). Working with piecewise functions can be tricky because you need to handle each interval separately during integration or differentiation.

When expanding piecewise functions in a series, splitting the integral based on the defined intervals is crucial. This ensures that each segment of the function contributes correctly to the expansion coefficients.
Orthogonal Functions
Orthogonal functions are functions that satisfy an orthogonality relation, which means their inner product over a specific interval is zero unless they are identical. Mathematically, two functions \( f(x) \) and \( g(x) \) are orthogonal if:
\[ \int_{a}^{b} f(x) g(x) \, dx = 0 \]
Orthogonal functions play a crucial role in many areas of mathematics and physics, especially in solving differential equations and in decomposing functions into series.

Legendre polynomials are a classic example of orthogonal functions. Because of their orthogonality, any function defined over \([-1, 1]\) can be expressed as an infinite series of these polynomials. This decomposition simplifies the process of analyzing and working with complex functions.

Orthogonal functions ensure that each term in the series expansion impacts the summation uniquely without interfering with other terms. This property helps in simplifying and solving various mathematical problems efficiently.
Series Coefficients
The Legendre series expansion of a function involves determining coefficients that multiply each Legendre polynomial term in the series. These coefficients \( a_n \) are found using the formula:
\[ a_n = \frac{2n + 1}{2} \int_{-1}^{1} f(x) P_n(x) \, dx \]
These integrals measure how much each Legendre polynomial contributes to representing the function over its domain.

For piecewise functions, the integral is split into parts corresponding to the function's defined intervals. In the given exercise, the function is defined on two intervals. The integral will be:
\[ a_n = \frac{2n + 1}{2} \left( \int_{-1}^{0} 0 \cdot P_n(x) \, dx + \int_{0}^{1} x \cdot P_n(x) \, dx \right) \]
The first integral is zero, simplifying the calculation. Evaluating the second integral for different \(n\) values gives specific series coefficients that describe the function in terms of Legendre polynomials.

For example:
  • \( a_0 = \frac{2(0) + 1}{2} \int_{0}^{1} x \, dx = \frac{1}{2} \frac{x^2}{2} \big\bigg|_{0}^{1} = \frac{1}{4} \)
Mastering the calculation of these coefficients helps in constructing an accurate Legendre series expansion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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. $$x(x-1)^{2} y^{\prime \prime}-2 y=0$$

Solve the following differential equations by the method of Frobenius (generalized power series). Remember that the point of doing these problems is to learn about the method (which we will use later), not just to find a solution. You may recognize some series [as we did in (11.6)] or you can check your series by expanding a computer answer. $$3 x y^{\prime \prime}+(3 x+1) y^{\prime}+y=0$$

By power series, solve the Hermite differential equation $$y^{\prime \prime}-2 x y^{\prime}+2 p y=0$$ You should find an \(a_{0}\) series and an \(a_{1}\) series as for the Legendre equation in Section 2 Show that the \(a_{0}\) series terminates when \(p\) is an even integer, and the \(a_{1}\) series terminates when \(p\) is an odd integer. Thus for each integer \(n\), the differential equation (22.14) has one polynomial solution of degree \(n\). These polynomials with \(a_{0}\) or \(a_{1}\) chosen so that the highest order term is \((2 x)^{n}\) are the Hermite polynomials. Find \(H_{0}(x), H_{1}(x),\) and \(H_{2}(x) .\) Observe that you have solved an eigenvalue problem (see end of Section 2 ), namely to find values of \(p\) for which the given differential equation has polynomial solutions, and then to find the corresponding solutions (eigenfunctions).

By a method similar to that we used to show that the \(P_{l^{\prime}}\) s are an orthogonal set of functions on \((-1,1),\) show that the solutions of \(y_{n}^{\prime \prime}=-n^{2} y_{n}\) are an orthogonal set on \((-\pi, \pi) .\) Hint: You should know what functions the solutions \(y_{n}\) are; do not use the functions themselves, but you may use their values and the values of their derivatives at \(-\pi\) and \(\pi\) to evaluate the integrated part of your equation.

Expand the following functions in Legendre series. $$f(x)=P_{n}^{\prime}(x)$$ Hint: For \(l \geq n, \int_{-1}^{1} P_{n}^{\prime}(x) P_{l}(x) d x=0\) (Why?); for \(l

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free