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Table 1 in 11.10 on Page 534 is a table of Fourier Cosine Transforms.
 
Table 1 in 11.10 on Page 534 is a table of Fourier Cosine Transforms.
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Form [[User:Park296|Eun Young]]:
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From the lesson 30 lecture note, we know that
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<math>\int_0^{\infty} \frac{2 \sin w }{\pi w} \cos wv  \ \ dw = 1 </math> if -1<v<1 and 0 otherwise.
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Consider only positive v. Then,  <math>\int_0^{\infty} \frac{2 \sin w }{\pi w} \cos wv  \ \ dw = 1 </math> if 0 <v<1 and 0 if v>1.
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From this, we can compute <math>\int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv  \ \ dw . </math>
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Let 2w = t. Then,
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<math>\int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv  \ \ dw = \int_0^{\infty} \frac{ 2 \sin t }{ \pi  \frac t 2}  cos (t \frac v 2 ) \frac{dt}{2}
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= \int_0^{\infty} \frac{2 \sin t}{ \pi t} cos(\frac v 2  t)  \ \ dt  = 1 \ \ \text{if} \ \ 0 < \frac v 2  < 1  \  \ \ \text{and }  0  \ \ \ \text{if} \ \  \frac v 2 >1 </math>.
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Thus,
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<math>\int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv  \ \ dw  =1 \ \ \text{if} \ \ 0 <v < 2  \  \ \ \text{and }  0  \ \ \ \text{if} \ v >2 </math>.
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Combine the second and the last equations. Then, you will get the answer.
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----
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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 11:21, 4 November 2013

Homework 10 collaboration area


From Jake Eppehimer:

I must be doing something wrong on problem 2 of Lesson 31. There doesn't seem to be a way to solve the integral if you use formula (1b) with the answer from problem 1. Any tips?

Also, I don't understand how to do number 5 on Lesson 29. Any help would be appreciated.

Response from Mickey Rhoades Mrhoade

I get 2/pi times a bunch of sinc function integrals which have to be evaluated with the Dirichlet Integral. I can use Table 11.10 relationship #10 and get the answer without chugging out all the integrals. ---

From Jake:

What page is this table on? I'm not seeing anything. Thanks.

From Andrew:

Table 1 in 11.10 on Page 534 is a table of Fourier Cosine Transforms.


Form Eun Young:

From the lesson 30 lecture note, we know that $ \int_0^{\infty} \frac{2 \sin w }{\pi w} \cos wv \ \ dw = 1 $ if -1<v<1 and 0 otherwise. Consider only positive v. Then, $ \int_0^{\infty} \frac{2 \sin w }{\pi w} \cos wv \ \ dw = 1 $ if 0 <v<1 and 0 if v>1.

From this, we can compute $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw . $

Let 2w = t. Then, $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw = \int_0^{\infty} \frac{ 2 \sin t }{ \pi \frac t 2} cos (t \frac v 2 ) \frac{dt}{2} = \int_0^{\infty} \frac{2 \sin t}{ \pi t} cos(\frac v 2 t) \ \ dt = 1 \ \ \text{if} \ \ 0 < \frac v 2 < 1 \ \ \ \text{and } 0 \ \ \ \text{if} \ \ \frac v 2 >1 $.

Thus, $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw =1 \ \ \text{if} \ \ 0 <v < 2 \ \ \ \text{and } 0 \ \ \ \text{if} \ v >2 $.

Combine the second and the last equations. Then, you will get the answer.



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