(a) Explanation Result used: The Binomial series If k is any number and | x | < 1 then, ( 1 + x ) k can be expressed as ( 1 + x ) k = ∑ n = 0 ∞ ( k n ) x n = 1 + k x + k ( k − 1 ) 2 ! x 2 + k ( k − 1 ) ( k − 2 ) 3 ! + ⋯ Given: The series is g ( x ) = ∑ n = 0 ∞ ( k n ) x n . Proof: The derivative of the function g ( x ) computed as follows, g ′ ( x ) = d d x ( ∑ n = 0 ∞ ( k n ) x n ) = ∑ n = 0 ∞ ( k n ) n x n − 1 Multiply by ( 1 + x ) on both sides, ( 1 + x ) g ′ ( x ) = ( 1 + x ) ∑ n = 0 ∞ ( k n ) n x n − 1 = ∑ n = 0 ∞ ( k n ) n x n − 1 + ∑ n = 0 ∞ ( k n ) n x n Replace n with n + 1 in the first series, ( 1 + x ) g ′ ( x ) = ∑ n = 0 ∞ ( k n + 1 ) ( n + 1 ) x n + ∑ n = 0 ∞ ( k n ) n x n = ∑ n = 0 ∞ k ! ( n + 1 ) ! ( k − ( n + 1 ) ) ! ( n + 1 ) x n + ∑ n = 0 ∞ k ! n ! ( k − n ) ! n x n = [ ∑ n = 0 ∞ ( n + 1 ) k ( k − 1 ) ( k − 2 ) ⋯ ( k (b) To determine To show: The derivative of the function h ( x ) is 0. (c) To determine To deduce: The series g ( x ) = ( 1 + x ) k .

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Chapter 11.10, Problem 85E

(a)

To determine

To show: The derivative of the series g(x) is g′(x)=kg(x)1+x.

(b)

To determine

To show: The derivative of the function h(x) is 0.

(c)

To determine

To deduce: The series g(x)=(1+x)k.

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Chapter 11.10, Problem 84E

Chapter 11.10, Problem 86E

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SINGLE VAR.CALC. TEXT W/WEBASSIGN >C<

Ch. 11.1 - (a) What is a sequence? (b) What does it mean to...Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

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The derivative of the series g ( x ) is g ′ ( x ) = k g ( x ) 1 + x .

Solution Summary: The author explains the derivative of the function g(x).

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To show: The derivative of the series g(x) is g′(x)=kg(x)1+x.

To show: The derivative of the function h(x) is 0.

To deduce: The series g(x)=(1+x)k.

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Chapter 11 Solutions