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<math>\hat{f}_c(w)=\sqrt{\frac{2}{\pi}}\left(
 
<math>\hat{f}_c(w)=\sqrt{\frac{2}{\pi}}\left(
\int_0^k w\cos(wx)\,dx\right)=
+
\int_0^k x\cos(wx)\,dx\right)=
 
</math>
 
</math>
  
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-\frac{1}{w^2}\right).</math>
 
-\frac{1}{w^2}\right).</math>
  
517: 5. See Bell's lecture near the top at
+
517: 5. See page 2 of Bell's 11/10/2010 lecture at
 
+
 
[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]
 +
 +
517: 7. See
 +
[http://www.math.purdue.edu/~bell/MA527/HWK/p517_7.pdf p. 517: 7 Solution]
 +
 +
And for solutions to the three problems on p. 528, go to
 +
[http://www.math.purdue.edu/~bell/MA527/jing Bell's Jing things]
 +
 +
[http://www.math.purdue.edu/~bell/MA527/prac2solns.pdf Exam 2 Practice Problem Solutions]
 +
  
 
[[2010 MA 527 Bell|Back to the MA 527 start page]]  
 
[[2010 MA 527 Bell|Back to the MA 527 start page]]  

Latest revision as of 11:06, 16 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

And for solutions to the three problems on p. 528, go to Bell's Jing things

Exam 2 Practice Problem Solutions


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