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Verifying that a Series Converges In Exercises 21-24, use the power series solution of the differential equation to verify that the series converges to the given function on the indicated interval. ∑ n = 0 ∞ 2 n ! x 2 n + 1 ( 2 n n ! ) 2 ( 2 n + 1 ) = arcsin x , ( − 1 , 1 ) Differential equation: ( 1 − x 2 ) y ″ − x y ′ = 0

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Multivariable Calculus

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

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Section
BuyFindarrow_forward

Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 16.4, Problem 24E
Textbook Problem
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Verifying that a Series Converges In Exercises 21-24, use the power series solution of the differential equation to verify that the series converges to the given function on the indicated interval.

n = 0 2 n ! x 2 n + 1 ( 2 n n ! ) 2 ( 2 n + 1 ) = arcsin x , ( 1 , 1 )

Differential equation: ( 1 x 2 ) y x y = 0

To determine

To prove: That the series n=0(2n)!x2n+1(2nn!)2(2n+1) converges to arcsinx on the interval (1,1) by using the power series solution of the differential equation (1x2)yxy=0.

Explanation of Solution

Given information:

The differential equation (1x2)yxy=0.

n=02n!x2n+1(2nn!)2(2n+1)=arcsinx,(1,1)

Proof:

Let y=n=0anxn be the solution of the differential equation (1x2)yxy=0.

Then, y=n=1nanxn1 and y=n=2n(n1)anxn2.

Substituting the value of y and y in the differential equation (1x2)yxy=0.

(1x2)yxy=0(1x2)n=2n(n1)anxn2xn=1nanxn1=0n=2n(n1)anxn2n=2n(n1)anxnn=1nanxn=0n=2n(n1)anxn2=n=2n(n1)anxn+n=1

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Chapter 16 Solutions

Multivariable Calculus
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