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I concur with Shawn regarding problem 6. I have a question about the even extension in problem 29. I am getting that the fourier series is 2/pi-4/pi(1/3 cos(2x) + 1/(3*5) cos (4x) + 1/(5*7) *cos(6x)... The answer in the book has odd numbers instead. Their answer doesn't make sense to me. Any thoughts? | I concur with Shawn regarding problem 6. I have a question about the even extension in problem 29. I am getting that the fourier series is 2/pi-4/pi(1/3 cos(2x) + 1/(3*5) cos (4x) + 1/(5*7) *cos(6x)... The answer in the book has odd numbers instead. Their answer doesn't make sense to me. Any thoughts? | ||
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| + | From Shawn Whitman: My answer is also different than the one in the back of the book. an is 0 for odd n and -4/pi((n^2)-1) for even n. Note that the (n^2)-1 in the denominator is not a problem for n=1 if you start the summation from 2 instead of 1. This is okay since the odd n’s are zero. ((n^2)-1), for even n, yields 3, 15, 35 or 3*1, 3*5, 5*7 as the book suggests. | ||
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[[2013_Fall_MA_527_Bell|Back to MA527, Fall 2013]] | [[2013_Fall_MA_527_Bell|Back to MA527, Fall 2013]] | ||
Revision as of 13:23, 27 October 2013
Homework 9 collaboration area
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This is the place!
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From Jake Eppehimer:
I am not sure how to do number 6 on p. 494. I'm clueless, and there's no answer in the back to verify if I'm doing anything right. Any tips?
From Shawn Whitman: The method of undetermined coefficients for second order nonhomogeneous linear ODEs works well for this problem. See pages 81-84 and use the sum rule. Two of the constants will go to zero. Two others will result in 1/(omega^2-alpha^2) and 1/(omega^2-beta^2); thus the given constraints.
From Mnestero:
I concur with Shawn regarding problem 6. I have a question about the even extension in problem 29. I am getting that the fourier series is 2/pi-4/pi(1/3 cos(2x) + 1/(3*5) cos (4x) + 1/(5*7) *cos(6x)... The answer in the book has odd numbers instead. Their answer doesn't make sense to me. Any thoughts?
From Shawn Whitman: My answer is also different than the one in the back of the book. an is 0 for odd n and -4/pi((n^2)-1) for even n. Note that the (n^2)-1 in the denominator is not a problem for n=1 if you start the summation from 2 instead of 1. This is okay since the odd n’s are zero. ((n^2)-1), for even n, yields 3, 15, 35 or 3*1, 3*5, 5*7 as the book suggests. --- Back to MA527, Fall 2013
