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 + ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

See similar textbooks

Concept explainers

Minimization

In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in ter…

Maxima and Minima

The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.

Derivatives

A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which …

Concavity

In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the …

Videos

Question

Chapter 11.6, Problem 69E

To determine

To prove:

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.6, Problem 68E

Chapter 11.6, Problem 70E

Students have asked these similar questions

Suppose fn → f and the functions fn all satisfy the Lipschitz condition |fn(x) – fn(y)| < M|x – y| for some constant M in- dependent of n. Prove that f also satisfies the same Lipschitz condition.

For any positive integer n show that 1 n + 1 holds for any f e C'[0, 1] satisfying f(0) = 0. Verify that the equality holds if and only if f(x) is a linear function.

Use extrapolation to derive an O(h) formula for f'(xo). O Suppose that N(h) is an approximation to M for every h > 0 and that M = N(h) + K,h + K2h + K,h+ , for some constants K1, K2, K3, Use the values N (h), N (), and N () to produce an O(h) .... approximation to M.

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

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

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 |

Concept explainers

Videos

To prove:

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

Want to see the full answer?

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 |

Concept explainers

Videos

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

Want to see the full answer?

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 $F (x) = \frac{a_{0} + a_{0} x + ... + a_{M - 1} x^{M - 1}}{1 - x^{M}}$ .