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[http://www.math.purdue.edu/~bell/MA527/Lectures/lec11-10.pdf Lesson 33]
 
[http://www.math.purdue.edu/~bell/MA527/Lectures/lec11-10.pdf Lesson 33]
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517: 7. See
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[http://www.math.purdue.edu/~bell/MA527/HWK/p517_7.pdf p. 517: 7 Solution]
  
 
[[2010 MA 527 Bell|Back to the MA 527 start page]]  
 
[[2010 MA 527 Bell|Back to the MA 527 start page]]  

Revision as of 10:36, 11 November 2010

Homework 12 Solutions

517: 1.

$ \hat{f}_c(w)=\sqrt{\frac{2}{\pi}}\left( \int_0^1(-1)\cos(wx)\,dx+ \int_1^2(1)\cos(wx)\,dx \right)= $

$ =\sqrt{\frac{2}{\pi}}\left([-\frac{1}{w}\sin(wx)]_0^1 +[\frac{1}{w}\sin(wx)]_1^2\right)= $

$ =\sqrt{\frac{2}{\pi}}\ \frac{1}{w}\left( -(\sin(w)-0)+(\sin(2w)-\sin(w)) \right)= $

$ =\sqrt{\frac{2}{\pi}}\ \frac{\sin(2w)-2\sin(w)}{w}. $

517: 2.

$ \hat{f}_c(w)=\sqrt{\frac{2}{\pi}}\left( \int_0^k x\cos(wx)\,dx\right)= $

$ =\sqrt{\frac{2}{\pi}}\left(\left[\frac{x}{w}\sin(wx)+\frac{1}{w^2}\cos(wx)\right]_0^k \right)= $

$ \sqrt{\frac{2}{\pi}}\left(\frac{k}{w}\sin(kw)+\frac{1}{w^2}\cos(kw) -\frac{1}{w^2}\right). $

517: 5. See page 2 of Bell's 11/10/2010 lecture at

Lesson 33

517: 7. See

p. 517: 7 Solution

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