Explanation Given: Suppose that the coefficient of F ( x ) = ∑ n = 0 ∞ a n x n are periodic; that is, for some whole number M > 0 , we have a M + n = a n . Proof: Suppose that F ( x ) is periodic with a M + n = a n for some whole number M and all n . The F ( x ) can be written as the sum of M geometric series: F ( x ) = a 0 ( 1 + x M + x 2 M + ... ) + a 1 ( x + x M + 1 + x 2 M + 1 + ... ) + a 2 ( x 2 + x M + 2 + x 2 M + 2 + ... ) + ... + a M − 1 ( x M − 1 + x 2 M − 1 + x 3 M − 1 + ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 11.6, Problem 69E

To determine

To prove:

$F (x)$ converges absolutely for $| x | < 1$ and that $F (x) = \frac{a_{0} + a_{0} x + ... + a_{M - 1} x^{M - 1}}{1 - x^{M}}$ .

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Chapter 11.6, Problem 68E

Chapter 11.6, Problem 70E

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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: F ( x ) converges absolutely for | x | < 1 and that F ( x ) = a 0 + a 0 x + ... + a M − 1 x M − 1 1 − x M .

Solution Summary: The author proves that F (x) converges absolutely for | x |

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To prove:

$F (x)$ converges absolutely for $| x | < 1$ and that $F (x) = \frac{a_{0} + a_{0} x + ... + a_{M - 1} x^{M - 1}}{1 - x^{M}}$ .

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Chapter 11 Solutions

To prove: F ( x ) converges absolutely for | x | &lt; 1 and that F ( x ) = a 0 + a 0 x + ... + a M − 1 x M − 1 1 − x M .

Solution Summary: The author proves that F (x) converges absolutely for | x |

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To prove: F(x) converges absolutely for |x|<1 and that F(x)=a0+a0x+...+aM−1xM−11−xM.

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Chapter 11 Solutions

To prove: F ( x ) converges absolutely for | x | < 1 and that F ( x ) = a 0 + a 0 x + ... + a M − 1 x M − 1 1 − x M .

To prove:

$F (x)$ converges absolutely for $| x | < 1$ and that $F (x) = \frac{a_{0} + a_{0} x + ... + a_{M - 1} x^{M - 1}}{1 - x^{M}}$ .