Explanation Formula used: δ f = f ( x + 1 ) − f ( x ) Calculation: We can use the definition of the first difference to rewrite: ⇒ δ ( x 3 ) = ( x + 1 ) 3 − x 3 ⇒ δ ( x 3 ) = x 3 + 1 3 + 3 x ( x + 1 ) − x 3 ⇒ δ ( x 3 ) = 1 + 3 x ( x + 1 ) ⇒ δ ( x 3 ) = 1 + 3 x 2 + 3 x ⇒ δ ( x 3 ) = 3 x 2 + 3 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 ⇒ δ ( x ) = ( x + 1 ) − x ⇒ δ ( x ) = x + 1 − x ⇒ δ ( x ) = 1 Now using these expressions, we need to find a third degree polynomial P such that δ P = ( x + 1 ) 2 . Basically, we need to use a combination of these three expressions to get such a P . This can be obtained as shown below: ⇒ ( x + 1 ) 2 = x 2 + 2 x + 1 ⇒ ( x + 1 ) 2 = 1 3 ( 3 x 2 + 3 x + 1 ) + 1 2 ( 2 x + 1 ) + 1 6 ( 1 ) ⇒ ( x + 1 ) 2 = 1 3 ( δ x 3 ) + 1 2 ( δ x 2

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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Question

Chapter 1.3, Problem 50E

To determine

Expressions for $δ (x^{3}), δ (x^{2}) and δ (x)$

A third degree polynomial P, such that $δ P = {(x + 1)}^{2}$

To prove:

$P (n) = 1^{2} + 2^{2} + 3^{2} + ..... + n^{2}$ .

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Chapter 1.3, Problem 49E

Chapter 1.3, Problem 51E

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Chapter 1 Solutions

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

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

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Knowledge Booster

Expressions for δ ( x 3 ) , δ ( x 2 ) and δ ( x ) A third degree polynomial P , such that δ P = ( x + 1 ) 2 To prove: P ( n ) = 1 2 + 2 2 + 3 2 + ..... + n 2 .

Expressions for δ(x3),δ(x2) and δ(x) A third degree polynomial P, such that δP=(x+1)2 To prove: P(n)=12+22+32+.....+n2.

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Chapter 1 Solutions

Expressions for $δ (x^{3}), δ (x^{2}) and δ (x)$

A third degree polynomial P, such that $δ P = {(x + 1)}^{2}$

To prove:

$P (n) = 1^{2} + 2^{2} + 3^{2} + ..... + n^{2}$ .