Line 33: Line 33:
  
 
517: 5. See page 2 of Bell's 11/10/2010 lecture 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
 
517: 7. See
 +
[http://www.math.purdue.edu/~bell/MA527/HWK/p517_7.pdf p. 517: 7 Solution]
  
[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]
  
 
[[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 11:28, 12 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

Back to the MA 527 start page

To Rhea Course List

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva