Explanation Concept used: The n th Maclaurin polynomial can be expressed as, T n ( 0 ) = f ( 0 ) + f ′ ( 0 ) 1 ! x + f ″ ( 0 ) 2 ! x 2 + ⋯ + f n ( x ) n ! x n …… (1) Proof: The function is, f ( x ) = arcsin x The function f ( x ) = arcsin x can be written as, f ( x ) = sin − 1 x Substitute 0 for x in the function f ( x ) to obtain the value of f ( 0 ) as, f ( 0 ) = sin − 1 0 = 0 Differentiate the function f ( x ) with respect to x as, d d x f ( x ) = d d x sin − 1 x f ′ ( x ) = 1 1 − x 2 Substitute 0 for x in the above equation to obtain the value of f ′ ( 0 ) as, f ′ ( 0 ) = 1 1 − 0 2 = 1 Differentiate the function f ′ ( x ) with respect to x as, d d x f ′ ( x ) = d d x ( 1 1 − x 2 ) = 1 − x 2 ( 0 ) − 1 ⋅ 1 2 1 − x 2 ( − 2 x ) ( 1 − x 2 ) 2 f ″ ( x ) = x ( 1 − x 2 ) 3 2 Differentiate the function f ″ ( x ) with respect to x as, d d x f ″ ( x ) = d d x [ x ( 1 − x 2 ) 3 2 ] f ‴ ( x ) = ( 1 − x 2 ) 3 2 + 3 x 2 1 − x 2 ( 1 − x 2 ) 3 = ( 1 − x 2 ) [ 1 − x 2 + 3 x 2 ] ( 1 − x 2 ) 3 = 1 + 2 x 2 ( 1 − x 2 ) 5 2 Differentiate the function f ‴ ( x ) with respect to x as, d d x f ‴ ( x ) = d d x [ 1 + 2 x 2 ( 1 − x 2 ) 5 2 ] = ( 1 − x 2 ) 5 2 ( 4 x ) − ( 1 + 2 x 2 ) ( − 5 x ) ( 1 − x 2 ) 3 2 ( ( 1 − x 2 ) 5 2 ) 2 = ( 1 − x 2 ) 5 2 ( 4 x ) − ( 1 + 2 x 2 ) ( − 5 x ) ( 1 − x 2 ) 3 2 ( 1 − x 2 ) 5 = ( 1 − x 2 ) 3 2 [ ( 1 − x 2 ) 4 x + 5 x + 10 x 3 ] ( 1 − x 2 ) 5 Simplify the above equation as, f ( 4 ) ( x ) = d d x f ‴ ( x ) = ( 1 − x 2 ) 3 2 [ 4 x − 4 x 3 + 5 x + 10 x 3 ] ( 1 − x 2 ) 5 = [ 9 x + 6 x 3 ] ( 1 − x 2 ) 7 2 = 3 x ( 3 + 2 x 2 ) <

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Chapter 11.7, Problem 67E

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To show: That nth Maclaurin polynomial of f(x)=arcsinx is Tn(x)=x+12x33+1⋅32⋅4x55+⋯+1⋅3⋅5⋅⋯(n−2)2⋅4⋅6⋯(n−1)(xnn) .

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Chapter 11.7, Problem 66E

Chapter 11.7, Problem 68E

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Chapter 11 Solutions

CALCULUS (CLOTH)

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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That n th Maclaurin polynomial of f ( x ) = arcsin x is T n ( x ) = x + 1 2 x 3 3 + 1 ⋅ 3 2 ⋅ 4 x 5 5 + ⋯ + 1 ⋅ 3 ⋅ 5 ⋅ ⋯ ( n − 2 ) 2 ⋅ 4 ⋅ 6 ⋯ ( n − 1 ) ( x n n ) .

Solution Summary: The author explains that the nth Maclaurin polynomial is f(x)=mathrmarcsinx.

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To show: That nth Maclaurin polynomial of f(x)=arcsinx is Tn(x)=x+12x33+1⋅32⋅4x55+⋯+1⋅3⋅5⋅⋯(n−2)2⋅4⋅6⋯(n−1)(xnn) .

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Chapter 11 Solutions