Explanation Given: Let F ( x ) = ∑ n = 0 ∞ a n x n be a power series with radius of convergence R > 0. Proof: Here, F ( x ) = ∑ n = 0 ∞ a n x n is a power series with radius of convergence R > 0. As we know that x n − y n = ( x − y ) ( x n − 1 + x n − 2 y + x n − 3 y 2 + ... + x y n − 2 + y n − 1 ) As we know that if F ( x ) = ∑ n = 0 ∞ a n x n is a power series with radius of convergence R > 0, then F ( x ) = ∑ n = 0 ∞ a n x n is differentiable term by term F ′ ( x ) = ∑ n = 1 ∞ n a n x n − 1 Similarly, F ′ ( y ) = ∑ n = 1 ∞ n a n y n − 1 Now, as F ( x ) = ∑ n = 0 ∞ a n x n is differentiable, which infers that it is uniformly continuous also. | x n − y n | = | ( x − y ) ( x n − 1 + x n − 2 y + x n − 3 y 2 + ... + x y n − 2 + y n − 1 ) | = | ( x − y ) | | ( x n − 1 + x n − 2 y + x n − 3 y 2 + ... + x y n − 2 + y n − 1 ) | ≤ | ( x − y ) | | x n − 1 | + | x n − 2 y | + | x n − 3 y 2 | + ... To determine (b) To Show: The infinite series M = ∑ n = 0 ∞ 2 n | a n | R 1 n converges. To determine (c) To Show: That the inequality | F ( x ) − F ( x ) | ≤ M | x − y | To determine (d) To Prove: That the power series F ( x ) = ∑ n = 0 ∞ a n x n is continuous at x.

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Author: Rogawski

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Chapter 11.6, Problem 70E

To determine

(a)

To Prove:

The inequality $| x^{n} - y^{n} | \leq n | x - y | ({| x |}^{n - 1} + {| y |}^{n - 1})$

To determine

(b)

To Show:

The infinite series $M = \sum_{n = 0}^{\infty} 2 n | a_{n} | R_{1}^{n}$ converges.

To determine

(c)

To Show:

That the inequality $| F (x) - F (x) | \leq M | x - y |$

To determine

(d)

To Prove:

That the power series $F (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ is continuous at x.

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Chapter 11.6, Problem 69E

Chapter 11.7, Problem 1PQ

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Chapter 11 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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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(a) To Prove: The inequality | x n − y n | ≤ n | x − y | ( | x | n − 1 + | y | n − 1 )

Solution Summary: The author explains that if F ( x ) = n = 0, a power series with radius of convergence R > 0, is differentiable.

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(a)

To Prove:

The inequality $| x^{n} - y^{n} | \leq n | x - y | ({| x |}^{n - 1} + {| y |}^{n - 1})$

(b)

To Show:

The infinite series $M = \sum_{n = 0}^{\infty} 2 n | a_{n} | R_{1}^{n}$ converges.

(c)

To Show:

That the inequality $| F (x) - F (x) | \leq M | x - y |$

(d)

To Prove:

That the power series $F (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ is continuous at x.

Want to see the full answer?

Chapter 11 Solutions

(a) To Prove: The inequality | x n − y n | ≤ n | x − y | ( | x | n − 1 + | y | n − 1 )

Solution Summary: The author explains that if F ( x ) = n = 0, a power series with radius of convergence R > 0, is differentiable.

Videos

(a) To Prove: The inequality |xn−yn|≤n|x−y|(|x|n−1+|y|n−1)

(b) To Show: The infinite series M=∑n=0∞2n|an|R1n converges.

(c) To Show: That the inequality |F(x)−F(x)|≤M|x−y|

(d) To Prove: That the power series F(x)=∑n=0∞anxn is continuous at x.

Want to see the full answer?

Chapter 11 Solutions

(a)

To Prove:

The inequality $| x^{n} - y^{n} | \leq n | x - y | ({| x |}^{n - 1} + {| y |}^{n - 1})$

(b)

To Show:

The infinite series $M = \sum_{n = 0}^{\infty} 2 n | a_{n} | R_{1}^{n}$ converges.

(c)

To Show:

That the inequality $| F (x) - F (x) | \leq M | x - y |$

(d)

To Prove:

That the power series $F (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ is continuous at x.