Explanation Given: The recurrence relation is, ( n + 1 ) P n + 1 + n P n − 1 = ( 2 n + 1 ) x P n , n = 1 , 2 , 3 , ... Calculation: Consider the recurrence relation, ( n + 1 ) P n + 1 + n P n − 1 = ( 2 n + 1 ) x P n … ( 1 ) Substitute 1 for n in the above relation and simplify. ( 1 + 1 ) P 1 + 1 + ( 1 ) P 1 − 1 = ( 2 ( 1 ) + 1 ) x P 1 2 P 2 + P 0 = 3 x P 1 Substitute x for P 1 and 1 for P 0 in above equation to find P 2 . 2 P 2 + ( 1 ) = 3 x ( x ) 2 P 2 = 3 x 2 − 1 P 2 = 3 x 2 − 1 2 P 2 = 1 2 ( 3 x 2 − 1 ) Substitute 2 for n in the equation ( 1 ) relation and simplify. ( ( 2 ) + 1 ) P 2 + 1 + ( 2 ) P 2 − 1 = ( 2 ( 2 ) + 1 ) x P 2 3 P 3 + 2 P 1 = 5 x P 2 Substitute x for P 1 and 1 2 ( 3 x 2 − 1 ) for P 2 in above equation to find P 3 .

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 11.3, Problem 2P

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The values of P2,P3 and P4.

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Chapter 11.2, Problem 21P

Chapter 11.3, Problem 3P

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Starting with P0(x) = 1 and P1(x) = x, use the recurrence relation (n+1)Pn+1+nPn−1 =(2n+1)xPn, n =1,2,3,... to determine P2, P3, and P4.

Chapter 11 Solutions

Differential Equations and Linear Algebra (4th Edition)

Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 2TFR Ch. 11.1 - Prob. 3TFR Ch. 11.1 - Prob. 4TFR Ch. 11.1 - Prob. 5TFR Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 7TFR Ch. 11.1 - Prob. 8TFR Ch. 11.1 - Prob. 9TFR Ch. 11.1 - Prob. 10TFR

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The values of P 2 , P 3 and P 4 .

Solution Summary: The author explains the recurrence relation, which is (n+1)P_n+n

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The values of P2,P3 and P4.

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Chapter 11 Solutions