(a) Explanation Given: P ( x ) = a n x n + a n − 1 x n − 1 + ... + a 1 x + a 0 Proof: Consider the following polynomial function P ( x ) = a n x n + a n − 1 x n − 1 + ... + a 1 x + a 0 First assume that if x = 0 , that is, if, p ( x ) = a 0 is a constant, then the following statement holds: If P ( 0 ) ( a ) = P ( a ) = 0 , then, a 0 = 0 Hence, P ( x ) = 0 Next, assume the statement is true for all polynomials of degree n − 1 or less. Let P ( x ) be a polynomial of degree at most n with P ( j ) ( a ) = 0 with j = 0 , 1 , 2 , ... n . By induction principle, if P ( x ) has degree less than n , then P ( x ) = 0 Assume the degree of P ( x ) is exactly n . P ( x ) = a n x n + a n − 1 x n − 1 + (b) To determine To prove that T n is the only polynomial of degree n or less that agrees with f at x = a to order n

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Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 9.4, Problem 71E

(a)

To determine

To show that if P(j)(a)=0,1,...,n, then P(x)=0 , that is, aj=0 for all j

(b)

To determine

To prove that Tn is the only polynomial of degree n or less that agrees with f at x=a to order n

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Chapter 9.4, Problem 70E

Chapter 9, Problem 1CRE

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Chapter 9 Solutions

Calculus - Standalone book

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To show that if P ( j ) ( a ) = 0 , 1 , ... , n , then P ( x ) = 0 , that is, a j = 0 for all j

Solution Summary: The author assumes that if P(0)(a) = P(x)=a_0 is a constant, then the following statement holds for all polynomials of degree

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To show that if P(j)(a)=0,1,...,n, then P(x)=0 , that is, aj=0 for all j

To prove that Tn is the only polynomial of degree n or less that agrees with f at x=a to order n

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Chapter 9 Solutions