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Chapter 12: Problem 5

Show that \(\int_{-1}^{1} P_{l}(x) d x=0, l>0 .\) Hint: Consider \(\int_{-1}^{1} P_{l}(x) P_{0}(x) d x.\)

Short Answer

Expert verified
The integral \( \int_{-1}^{1} P_{l}(x) \ dx = 0 \) for \( l > 0 \) due to the orthogonality property of Legendre polynomials.

Step by step solution

01

- Understand the Problem

We need to show that the integral of the Legendre polynomial \( P_{l}(x) \) from -1 to 1 is zero for \( l > 0 \). We will use the given hint involving the integral of \( P_{l}(x) P_{0}(x) \).
02

- Recall Legendre Polynomial Properties

Legendre polynomials \( P_{l}(x) \) are orthogonal over the interval \( [-1, 1] \) with respect to the weight function 1. Specifically, \( \int_{-1}^{1} P_{l}(x) P_{m}(x) \ dx = 0 \) if \( l \eq m \). This implies that for \( l > 0 \, \int_{-1}^{1} P_{l}(x) P_{0}(x) \ dx = 0 \).
03

- Identify \( P_{0}(x)\)

The zeroth Legendre polynomial \( P_{0}(x) \) is simply 1. Therefore, we can rewrite our integral of interest as \( \int_{-1}^{1} P_{l}(x) \ dx \).
04

- Use Orthogonality Property

Using the orthogonality property of Legendre polynomials with \( m = 0 \, \int_{-1}^{1} P_{l}(x)P_{0}(x) \ dx = 0 \), which simplifies to \( \int_{-1}^{1} P_{l}(x) \ dx = 0 \) for \( l > 0 \).
05

- Conclude the Proof

Since we have shown that the orthogonality condition holds, it follows directly that \( \int_{-1}^{1} P_{l}(x) \ dx = 0 \) for any \( l > 0 \).

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Key Concepts

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

Orthogonality of functions
Orthogonality is a concept that helps simplify complex problems by reducing interference between functions. Think of it like vectors: just as perpendicular vectors don’t affect each other’s directions, orthogonal functions don’t interfere with each other’s integrals.

For Legendre polynomials, the orthogonality property means that the integral of the product of different polynomials (over the interval \([-1, 1]\)) is zero. Mathematically, this is written as:
\[ \int_{-1}^{1} P_{l}(x) P_{m}(x) \ dx = 0 \ if \ l \eq \ m \]
Orthogonality simplifies many calculations, particularly in physics and engineering, where Legendre polynomials often show up in solutions to differential equations.
Integral properties
Integrals help in finding the area under a curve, but they also hold significance in understanding properties like orthogonality. Here, the key property we use is the orthogonality of Legendre polynomials over the interval \([-1, 1]\).

Due to this property:
\[ \int_{-1}^{1} P_{l}(x) P_{0}(x) \ dx = 0 \ for \ l > 0 \]
Knowing \[ P_{0}(x) = 1 \] simplifies our problem to: \[ \int_{-1}^{1} P_{l}(x) \ dx = 0 \ for \ l > 0 \]
This shows that the integral of any Legendre polynomial of order greater than zero is zero over the interval \([-1, 1]\), leveraging the orthogonality property.
Zeroth Legendre polynomial
The zeroth Legendre polynomial, \ P_{0}(x)\, is the simplest of the Legendre polynomials. It is defined as: \[ P_{0}(x) = 1 \]
This is a constant function. The simplicity of \[P_{0}(x)\] helps in utilizing many integral properties.

When dealing with the orthogonality of Legendre polynomials, \[P_{0}(x)\] plays a pivotal role because:
\[\int_{-1}^{1} P_{l}(x) P_{0}(x) \ dx = 0 \ for \ l > 0 \]
Simplifies to: \[\int_{-1}^{1} P_{l}(x) \ dx = 0 \ for \ l > 0 \]
Understanding \ P_{0}(x)\ makes it easier to grasp the behavior and properties of higher-order Legendre polynomials.

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Most popular questions from this chapter

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. $$36 x^{2} y^{\prime \prime}+\left(5-9 x^{2}\right) y=0$$

Consider each of the following problems as illustrations showing that, in a power series solution, we must be cautious about using the general recursion relation between the coefficients for the first few terms of the series. Solve \(y^{\prime \prime}+y^{\prime} / x^{2}=0\) by power series to find the relation $$a_{n+1}=-\frac{n(n-1)}{n+1} a_{n}.$$ If, without thinking carefully, we test the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) for convergence by the ratio test, we find $$\lim _{n \rightarrow \infty} \frac{\left|a_{n+1} x^{n+1}\right|}{\left|a_{n} x^{n}\right|}=\infty\quad (Show this.)$$ Thus we might conclude that the series diverges and that there is no power series solution of this equation. Show why this is wrong, and that the power series solution is \(y=\) const.

Expand the following functions in Legendre series. $$f(x)=\arcsin x$$

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. $$x y^{\prime \prime}-y^{\prime}+9 x^{5} y=0$$

(a) The generating function for Bessel functions of integral order \(p=n\) is $$ \Phi(x, h)=e^{(1 / 2) x\left(h-h^{-1}\right)}=\sum_{n=-\infty}^{\infty} h^{n} J_{n}(x) $$ By expanding the exponential in powers of \(x\left(h-h^{-1}\right)\) show that the \(n=0\) term is \(J_{0}(x)\) as claimed. (b) Show that $$ x^{2} \frac{\partial^{2} \Phi}{\partial x^{2}}+x \frac{\partial \Phi}{\partial x}+x^{2} \Phi-\left(h \frac{\partial}{\partial h}\right)^{2} \Phi=0 $$ Use this result and \(\Phi(x, h)=\sum_{n=-\infty}^{\infty} h^{n} J_{n}(x)\) to show that the functions \(J_{n}(x)\) satisfy Bessel's equation. By considering the terms in \(h^{n}\) in the expansion of \(e^{(1 / 2) x\left(h-h^{-1}\right)}\) in part (a), show that the coefficient of \(h^{n}\) is a series starting with the term \((1 / n !)(x / 2)^{n}\). (You have then proved that the functions called \(J_{n}(x)\) in the expansion of \(\Phi(x, h)\) are indeed the Bessel functions of integral order previously defined by (12.9) and (13.1) with \(p=n\).)
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