Explanation Given information: c k = f ( k ) ( a ) k ! Proof: We are given a polynomial T n which has same values as the function f at x = a and the derivatives of the polynomial equals the derivatives of the function f at x = a . T n = c 0 + c 1 ( x − a ) + c 2 ( x − a ) 2 + c 3 ( x − a ) 3 + c 4 ( x − a ) 4 + ... + c n ( x − a ) n T n ( 1 ) = c 1 + 2 c 2 ( x − a ) + 3 c 3 ( x − a ) 2 + 4 c 4 ( x − a ) 3 + ... + n c n ( x − a ) n − 1 T n ( 2 ) = 2 c 2 + 2 × 3 c 3 ( x − a ) + 4 c 4 ( x − a ) 2 + ... + n ( n − 1 ) c n ( x − a ) n − 2 . . . T n ( n ) = n ( n − 1 ) ( n − 2 ) ( n − 3 )

15th Edition

ISBN: 9781133109631

Author: Stewart, JAMES, Day, Troy

Publisher: Cengage Learning,

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Chapter 3.8, Problem 44E

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To show: if a polynomial Tn has the same value at a and the same derivatives at x = a as a function f , then its coefficients are given by the formula ck=f(k)(a)k! .

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Chapter 3.8, Problem 43E

Chapter 3.8, Problem 1P

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Chapter 3 Solutions

Biocalculus

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if a polynomial T n has the same value at a and the same derivatives at x = a as a function f , then its coefficients are given by the formula c k = f ( k ) ( a ) k ! .

Solution Summary: The author explains that if a polynomial T_n has the same values and derivatives as the function f, then its coefficients are given by the formula.

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To show: if a polynomial Tn has the same value at a and the same derivatives at x = a as a function f , then its coefficients are given by the formula ck=f(k)(a)k! .

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