Explanation Formula used: δ f = f ( x + 1 ) − f ( x ) Calculation: We will use proof by induction to prove the given statement. We can use the definition of the first difference to write first few terms as: ⇒ δ ( x ) = ( x + 1 ) − x ⇒ δ ( x ) = x + 1 − x ⇒ δ ( x ) = 1 ⇒ δ ( x 2 ) = ( x + 1 ) 2 − x 2 ⇒ δ ( x 2 ) = x 2 + 1 2 + 2 x − x 2 ⇒ δ ( x 2 ) = 1 + 2 x ⇒ δ ( x 2 ) = 2 x + 1 Now, let us assume that the given statement is true for k To determine (b) To prove There exists a polynomial with degree ( k + 1 ) and leading coefficient 1 ( k + 1 ) : P ( x ) = 1 ( k + 1 ) x k + 1 + 1 k x k + ... Such that δ P = ( x + 1 ) k and P ( 0 ) = 0

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Try substituting 1. You will get 6 and not 0. Try other real values also. What About -2? Let us check.

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Question

Chapter 1.3, Problem 47E

To determine

(a)

To prove:

δ(xk+1)=(k+1)xk+kxk−1+(k−1)xk−2+.....

To determine

(b)

To prove

There exists a polynomial with degree (k+1) and leading coefficient 1(k+1) :

P(x)=1(k+1)xk+1+1kxk+...

Such that δP=(x+1)k and P(0)=0

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Chapter 1.3, Problem 46E

Chapter 1.4, Problem 1PQ

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Chapter 1 Solutions

Calculus - Standalone book

Ch. 1.1 - Prob. 1PQ Ch. 1.1 - Prob. 2PQ Ch. 1.1 - Prob. 3PQ Ch. 1.1 - Prob. 4PQ Ch. 1.1 - Prob. 5PQ Ch. 1.1 - Prob. 6PQ Ch. 1.1 - Prob. 7PQ Ch. 1.1 - Prob. 8PQ Ch. 1.1 - Prob. 9PQ Ch. 1.1 - Prob. 10PQ

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - 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(a) To prove: δ ( x k + 1 ) = ( k + 1 ) x k + k x k − 1 + ( k − 1 ) x k − 2 + .... .

Solution Summary: The author explains that the given statement has been proved using mathematical induction.

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(a)

To prove:

δ(xk+1)=(k+1)xk+kxk−1+(k−1)xk−2+.....

(b)

To prove

There exists a polynomial with degree (k+1) and leading coefficient 1(k+1) :

P(x)=1(k+1)xk+1+1kxk+...

Such that δP=(x+1)k and P(0)=0

Want to see the full answer?

Chapter 1 Solutions

(a) To prove: δ ( x k + 1 ) = ( k + 1 ) x k + k x k − 1 + ( k − 1 ) x k − 2 + .... .

Solution Summary: The author explains that the given statement has been proved using mathematical induction.

Concept explainers

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(a) To prove: δ(xk+1)=(k+1)xk+kxk−1+(k−1)xk−2+.....

(b) To prove There exists a polynomial with degree (k+1) and leading coefficient 1(k+1) : P(x)=1(k+1)xk+1+1kxk+... Such that δP=(x+1)k and P(0)=0

Want to see the full answer?

Chapter 1 Solutions

(a)

To prove:

δ(xk+1)=(k+1)xk+kxk−1+(k−1)xk−2+.....

(b)

To prove

There exists a polynomial with degree (k+1) and leading coefficient 1(k+1) :

P(x)=1(k+1)xk+1+1kxk+...

Such that δP=(x+1)k and P(0)=0