∑ n = 1 ∞ sin ( ( 2 n − 1 ) π x ) 2 n − 1 = π 2 ( f ( x ) − 1 2 ) for the function f ( x ) = { 0 , − L < x < 0 L , 0 < x < L which is the square wave with L = 1.

Question

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Answer 1

Question

Chapter 10.3, Problem 19P

a.

To determine

To show: ∑n=1∞sin((2n−1)πx)2n−1=π2(f(x)−12) for the function f(x)={0, −L<x<0L, 0<x<L which is the square wave with L = 1.

b.

To determine

To show: g(x)=π2(f(x)−12)−∑n=1∞sin((2n−1)πx)(2n−1)(1+(2n−1)2) converges much faster than the series simplify the function.

Expert Solution & Answer

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2. Suppose a LC circuit has the following differential equation: q'+4q=6etcos 4t, q(0) = 1 a. Find the function for q(t), use any method that we have studied in the course. b. What is the transient and the steady-state of the circuit?

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Answer 2

Question

Chapter 10.3, Problem 19P

a.

To determine

To show: ∑n=1∞sin((2n−1)πx)2n−1=π2(f(x)−12) for the function f(x)={0, −L<x<0L, 0<x<L which is the square wave with L = 1.

b.

To determine

To show: g(x)=π2(f(x)−12)−∑n=1∞sin((2n−1)πx)(2n−1)(1+(2n−1)2) converges much faster than the series simplify the function.

Expert Solution & Answer

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See solution

Answer 3

Question

Chapter 10.3, Problem 19P

a.

To determine

To show: ∑n=1∞sin((2n−1)πx)2n−1=π2(f(x)−12) for the function f(x)={0, −L<x<0L, 0<x<L which is the square wave with L = 1.

b.

To determine

To show: g(x)=π2(f(x)−12)−∑n=1∞sin((2n−1)πx)(2n−1)(1+(2n−1)2) converges much faster than the series simplify the function.

Expert Solution & Answer

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Answer 4

Question

Chapter 10.3, Problem 19P

a.

To determine

To show: ∑n=1∞sin((2n−1)πx)2n−1=π2(f(x)−12) for the function f(x)={0, −L<x<0L, 0<x<L which is the square wave with L = 1.

b.

To determine

To show: g(x)=π2(f(x)−12)−∑n=1∞sin((2n−1)πx)(2n−1)(1+(2n−1)2) converges much faster than the series simplify the function.

Expert Solution & Answer

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See solution

Answer 5

Chapter 10.3, Problem 19P

a.

To determine

To show: ∑n=1∞sin((2n−1)πx)2n−1=π2(f(x)−12) for the function f(x)={0, −L<x<0L, 0<x<L which is the square wave with L = 1.

b.

To determine

To show: g(x)=π2(f(x)−12)−∑n=1∞sin((2n−1)πx)(2n−1)(1+(2n−1)2) converges much faster than the series simplify the function.

Answer 6

Chapter 10.3, Problem 19P

a.

To determine

To show: ∑n=1∞sin((2n−1)πx)2n−1=π2(f(x)−12) for the function f(x)={0, −L<x<0L, 0<x<L which is the square wave with L = 1.

Answer 7

Chapter 10.3, Problem 19P

a.

To determine

To show: ∑n=1∞sin((2n−1)πx)2n−1=π2(f(x)−12) for the function f(x)={0, −L<x<0L, 0<x<L which is the square wave with L = 1.

Answer 8

b.

To determine

To show: g(x)=π2(f(x)−12)−∑n=1∞sin((2n−1)πx)(2n−1)(1+(2n−1)2) converges much faster than the series simplify the function.

Answer 9

b.

To determine

To show: g(x)=π2(f(x)−12)−∑n=1∞sin((2n−1)πx)(2n−1)(1+(2n−1)2) converges much faster than the series simplify the function.

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

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Answer 13

Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 2P Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 4P Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 8P Ch. 10.1 - Prob. 9P Ch. 10.1 - In each of Problems 1 through 13, either solve the...

∑ n = 1 ∞ sin ( ( 2 n − 1 ) π x ) 2 n − 1 = π 2 ( f ( x ) − 1 2 ) for the function f ( x ) = { 0 , − L &lt; x &lt; 0 L , 0 &lt; x &lt; L which is the square wave with L = 1.

To show: ∑n=1∞sin((2n−1)πx)2n−1=π2(f(x)−12) for the function f(x)={0, −L<x<0L, 0<x<L which is the square wave with L = 1.

To show: g(x)=π2(f(x)−12)−∑n=1∞sin((2n−1)πx)(2n−1)(1+(2n−1)2) converges much faster than the series simplify the function.

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Chapter 10 Solutions

∑ n = 1 ∞ sin ( ( 2 n − 1 ) π x ) 2 n − 1 = π 2 ( f ( x ) − 1 2 ) for the function f ( x ) = { 0 , − L < x < 0 L , 0 < x < L which is the square wave with L = 1.