Explanation Given: Let x be fixed and let f ( z ) = ∑ n = 0 ∞ P n ( x ) z n Approach: The Legendre polynomials satisfy the recurrence formula given by, ( n + 1 ) P n + 1 ( x ) = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ( x ) … ( 1 ) Calculation: Multiply each term of formula ( 1 ) by z n as, ( n + 1 ) P n + 1 ( x ) z n = ( 2 n + 1 ) x P n ( x ) z n − n P n − 1 ( x ) z n Take the summation from n = 1 to ∞ as, ∑ n = 1 ∞ ( n + 1 ) P n + 1 ( x ) z n = ∑ n = 1 ∞ ( 2 n + 1 ) x P n ( x ) z n − ∑ n = 1 ∞ n P n − 1 ( x ) z n Solve the above expression as, ∑ n = 1 ∞ ( n + 1 ) P n + 1 ( x ) z n = ∑ n = 1 ∞ ( 2 n + 1 ) x P n ( x ) z n − ∑ n = 1 ∞ n P n − 1 ( x ) z n = 2 ∑ n = 1 ∞ n x P n ( x ) z n + ∑ n = 1 ∞ x P n ( x ) z n − ∑ n = 1 ∞ ( n + 1 ) P n ( x ) z n + 1 = 2 x ∑ n = 1 ∞ n P n ( x ) z n + x ∑ n = 1 ∞ P n ( x ) z n − ∑ n = 1 ∞ n P n ( x ) z n + 1 − ∑ n = 1 ∞ P n ( x ) z n + 1 = 2 x ∑ n = 1 ∞ n P n ( x ) z n + x ∑ n = 1 ∞ P n ( x ) z n − z ∑ n = 1 ∞ n P n ( x ) z n − z ∑ n = 1 ∞ P n ( x To determine (b) To obtain: The solution to the differential equation in part (a) as, f ( z ) = ( 1 − 2 x z + z 2 ) − 1 / 2

Fundamentals of Differential Equations [With CDROM] - 7th Edition

7th Edition

ISBN: 9780321410481

Author: Saff, Edward B., Snider, Arthur David, Nagle, R. Kent

Publisher: Addison Wesley

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Chapter 8.8, Problem 35E

To determine

(a)

To show:

The recurrence formula (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x) when multiplied by zn gives the differential equation as,

dfdz=x−z1−2xz+z2f

To determine

(b)

To obtain:

The solution to the differential equation in part (a) as,

f(z)=(1−2xz+z2)−1/2

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Chapter 8.8, Problem 34E

Chapter 8.8, Problem 36E

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Chapter 8 Solutions

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(a) To show: The recurrence formula ( n + 1 ) P n + 1 ( x ) = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ( x ) when multiplied by z n gives the differential equation as, d f d z = x − z 1 − 2 x z + z 2 f

Solution Summary: The author explains that the recurrence formula (n+1)P_n-1 (x ) when multiplied by

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(a)

To show:

The recurrence formula (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x) when multiplied by zn gives the differential equation as,

dfdz=x−z1−2xz+z2f

(b)

To obtain:

The solution to the differential equation in part (a) as,

f(z)=(1−2xz+z2)−1/2

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Chapter 8 Solutions

(a) To show: The recurrence formula ( n + 1 ) P n + 1 ( x ) = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ( x ) when multiplied by z n gives the differential equation as, d f d z = x − z 1 − 2 x z + z 2 f

Solution Summary: The author explains that the recurrence formula (n+1)P_n-1 (x ) when multiplied by

Videos

(a) To show: The recurrence formula (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x) when multiplied by zn gives the differential equation as, dfdz=x−z1−2xz+z2f

(b) To obtain: The solution to the differential equation in part (a) as, f(z)=(1−2xz+z2)−1/2

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Chapter 8 Solutions

(a)

To show:

The recurrence formula (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x) when multiplied by zn gives the differential equation as,

dfdz=x−z1−2xz+z2f

(b)

To obtain:

The solution to the differential equation in part (a) as,

f(z)=(1−2xz+z2)−1/2