   # Laguerre’s Equation Consider Laguerre’s Equation x y ' ' + ( 1 − x ) y ' + k y = 0 . Polynomial solutions of Laguerre’s equation are called Laguerre polynomials and are denoted by L k ( x ) . Use a power series of the form y = ∑ n = 0 ∞ a n x n to show that L k ( x ) = ∑ n = 0 k ( − 1 ) n k ! x n ( k − n ) ! ( n ! ) 2 Assume that a 0 = 1 . Determine the Laguerre polynomials L 0 ( x ) , L 1 ( x ) , L 2 ( x ) , L 3 ( x ) , and L 4 ( x ) . ### Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378

#### Solutions

Chapter
Section ### Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 16, Problem 20PS
Textbook Problem
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## Laguerre’s Equation Consider Laguerre’s Equation x y ' ' + ( 1 − x ) y ' + k y = 0 .Polynomial solutions of Laguerre’s equation are called Laguerre polynomials and are denoted by L k ( x ) . Use a power series of the form y = ∑ n = 0 ∞ a n x n to show that L k ( x ) = ∑ n = 0 k ( − 1 ) n k ! x n ( k − n ) ! ( n ! ) 2 Assume that a 0 = 1 .Determine the Laguerre polynomials L 0 ( x ) ,   L 1 ( x ) ,   L 2 ( x ) ,   L 3 ( x ) , and  L 4 ( x ) .

(a)

To determine

To prove: The Laguerre’spolynomial is given by Lk(x)=n=0k(1)nk!xn(kn)!(n!)2 using a power series of the form y=n=0anxn, when a Laguerre’s equation is given by xy''+(1x)y'+ky=0.

### Explanation of Solution

Given information:

The given Laguerre’s equation is xy''+(1x)y'+ky=0. And, it is to assume that a0=1.

Concept Used:

Power series is defined as an infinite series of the form,

n=0anxn=a0+a1x+a2x2+a3x3+.........+anxn+...., where x is a variable.

Calculation:

Assume y=n=0anxn is a solution of the given equation. Then, differentiating the assumed solution with respect to x,

y'=n=1nanxn1

Again differentiating above equation with respect to x,

y''=n=2n(n1)anxn2

Now, the given differential equation is,

xy''+(1x)y'+ky=0

Inserting the values of y and y'' in above equation,

xy''+(1x)y'+ky=0x(n=2n(n1)anxn2)+(1x)n=1nanxn1+kn=0anxn=0n=1n(n1)anxn1+n=1nanxn1xn=1nanxn1+kn=0anxn=0n=1n2anxn1n=1nanxn1+n=1nanxn1n=0nanxn+kn=0anxn=0n=1n2anxn1=(nk)n=0anxnn=0(n+1)2an+1xn=(nk)n=0anxn

Equating the coefficients of like terms, one can obtain the recursion formula

(b)

To determine

The values of Laguerre’s polynomials L0(x), L1(x), L2(x), L3(x), and L4(x) if Lk(x) is given by Lk(x)=n=0k(1)nk!xn(kn)!(n!)2.

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