Explanation Calculation: Given that S is a base for P n − 1 ( ℝ ) . ( i . e . ) S = { 1 , x , x 2 , … , x n − 1 } is the basis for P n − 1 ( ℝ ) . So, all polynomials of degree m ≤ n − 1 is written as a linear combination of elements of S . ( i . e . ) p ( x ) = a 0 + a 1 x + ... + a n − 1 x n − 1 . Now consider an arbitrary polynomial of degree n such that p ( x ) = a 0 + a 1 x + ... + a n x n ∈ P n . Let S ' = S ∪ { x n }

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Question

Chapter 4.6, Problem 52P

To determine

To prove:

The S∪{xn} is a basis for Pn(ℝ).

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Chapter 4.6, Problem 51P

Chapter 4.6, Problem 53P

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Chapter 4 Solutions

Differential Equations and Linear Algebra (4th Edition)

Ch. 4.1 - True-False Review For items a-I, decide if the...Ch. 4.1 - Prob. 2TFR Ch. 4.1 - For items a-I, decide if the given statement is...Ch. 4.1 - For items a-I, decide if the given statement is...Ch. 4.1 - For items a-I, decide if the given statement is...Ch. 4.1 - For items a-I, decide if the given statement is...Ch. 4.1 - Prob. 7TFR Ch. 4.1 - Prob. 8TFR Ch. 4.1 - For items a-I, decide if the given statement is...Ch. 4.1 - Prob. 10TFR

Knowledge Booster

To prove: The S ∪ { x n } is a basis for P n ( ℝ ) .

To prove: The S∪{xn} is a basis for Pn(ℝ).

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Chapter 4 Solutions

To prove:

The S∪{xn} is a basis for Pn(ℝ).