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(For those whove thought about convergence issues) Check that the power series expansion

Linear Algebra: A Geometric Approach | 2nd Edition | ISBN: 9781429215213 | Authors: Ted Shifrin, Malcolm Adams ISBN: 9781429215213 438

Solution for problem 15 Chapter 7.3

Linear Algebra: A Geometric Approach | 2nd Edition

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Linear Algebra: A Geometric Approach | 2nd Edition | ISBN: 9781429215213 | Authors: Ted Shifrin, Malcolm Adams

Linear Algebra: A Geometric Approach | 2nd Edition

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Problem 15

(For those whove thought about convergence issues) Check that the power series expansion f (x) = _ k=0 xk k! converges for any real number x and that f (x) = ex , as follows. a. Fix x _= 0 and choose an integer K so that K 2|x|. Then show that for k > K, we have |x|k k! C _ 12 _ kK, where C = |x| K . . . |x| 2 |x| 1 is a fixed constant. b. Conclude that the series k=K+1 |x|k k! is bounded by the convergent geometric series C j=1 1 2j and therefore converges and, thus, that the entire original series converges absolutely. c. It is a fact that every convergent power series may be differentiated (on its interval of convergence) term by term to obtain the power series of the derivative (see Spivak, Calculus, Chapter 24). Check that f _ (x) = f (x) and deduce that f (x) = ex .

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Week 7 Ch. 5 Tying in data with parameters Simple random sample (SRS): a sample of n elements obtained by a method so that any two collections of n elements are equally likely of being selected as the sample - n denotes sample size - you get in big trouble in statistics if you use biased samples - Two types: 1. Sampling with replacement a. Means that...

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Chapter 7.3, Problem 15 is Solved
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Textbook: Linear Algebra: A Geometric Approach
Edition: 2
Author: Ted Shifrin, Malcolm Adams
ISBN: 9781429215213

Linear Algebra: A Geometric Approach was written by and is associated to the ISBN: 9781429215213. The answer to “(For those whove thought about convergence issues) Check that the power series expansion f (x) = _ k=0 xk k! converges for any real number x and that f (x) = ex , as follows. a. Fix x _= 0 and choose an integer K so that K 2|x|. Then show that for k > K, we have |x|k k! C _ 12 _ kK, where C = |x| K . . . |x| 2 |x| 1 is a fixed constant. b. Conclude that the series k=K+1 |x|k k! is bounded by the convergent geometric series C j=1 1 2j and therefore converges and, thus, that the entire original series converges absolutely. c. It is a fact that every convergent power series may be differentiated (on its interval of convergence) term by term to obtain the power series of the derivative (see Spivak, Calculus, Chapter 24). Check that f _ (x) = f (x) and deduce that f (x) = ex .” is broken down into a number of easy to follow steps, and 162 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 31 chapters, and 547 solutions. The full step-by-step solution to problem: 15 from chapter: 7.3 was answered by , our top Math solution expert on 03/15/18, 05:30PM. This textbook survival guide was created for the textbook: Linear Algebra: A Geometric Approach, edition: 2. Since the solution to 15 from 7.3 chapter was answered, more than 209 students have viewed the full step-by-step answer.

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