Explanation Given: The series F ( x ) = 1 − x − x 2 + x 3 − x 4 − x 5 + x 6 − x 7 − x 8 + ... ... ( 1 ) and F ( x ) as a sum of three geometric series with common ratio x 3 . Formula Used: ∑ n = 0 ∞ x n = 1 + x + x 2 + x 3 + x 4 + ... Proof: The equation (1) can be rewritten as: F ( x ) = ( 1 + x 3 + x 6 + ... ) + ( x + x 4 + x 7 + ... ) − 2 ( x 2 + x 5 + x 8 + ... ) = ( 1 + x 3 + x 6 + ... ) + x ( 1 + x 3 + x 6 + ... ) − 2 x 2 ( 1 + x 3 + x 6 + ... ) These are the three geometric series with the common ratio x 3 . Now, F ( x ) = ( 1 + x 3 + x 6 + ... ) + x ( 1 + x 3 + x 6 + ... ) − 2 x 2 ( 1 + x 3 + x 6 + ... ) = [ 1 + x 3 + ( x 3 ) 2 + ... ] + x [ 1 + x 3 + ( x 3 ) 2 + ... ] − 2 x 2 [ 1 + x 3 + ( x 3 ) 2 + .....

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 11.6, Problem 58E

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To prove:

For $| x | < 1$ , $\frac{1 + 2 x}{1 + x + x^{2}} = 1 + x - 2 x^{2} + x^{3} + x^{4} - 2 x^{5} + x^{6} + x^{7} - 2 x^{8} + ...$

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Chapter 11.6, Problem 57E

Chapter 11.6, Problem 59E

Chapter 11 Solutions

CALCULUS (CLOTH)

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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To prove: For | x | < 1 , 1 + 2 x 1 + x + x 2 = 1 + x − 2 x 2 + x 3 + x 4 − 2 x 5 + x 6 + x 7 − 2 x 8 + ...

Solution Summary: The author explains how the equation can be rewritten as a sum of three geometric series with the common ratio x 3.

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To prove:

For $| x | < 1$ , $\frac{1 + 2 x}{1 + x + x^{2}} = 1 + x - 2 x^{2} + x^{3} + x^{4} - 2 x^{5} + x^{6} + x^{7} - 2 x^{8} + ...$

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Chapter 11 Solutions

To prove: For | x | &lt; 1 , 1 + 2 x 1 + x + x 2 = 1 + x − 2 x 2 + x 3 + x 4 − 2 x 5 + x 6 + x 7 − 2 x 8 + ...

Solution Summary: The author explains how the equation can be rewritten as a sum of three geometric series with the common ratio x 3.

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To prove: For |x|<1,1+2x1+x+x2=1+x−2x2+x3+x4−2x5+x6+x7−2x8+...

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Chapter 11 Solutions

To prove: For | x | < 1 , 1 + 2 x 1 + x + x 2 = 1 + x − 2 x 2 + x 3 + x 4 − 2 x 5 + x 6 + x 7 − 2 x 8 + ...

To prove:

For $| x | < 1$ , $\frac{1 + 2 x}{1 + x + x^{2}} = 1 + x - 2 x^{2} + x^{3} + x^{4} - 2 x^{5} + x^{6} + x^{7} - 2 x^{8} + ...$