(5 intermediate revisions by 2 users not shown)
Line 22: Line 22:
  
 
<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>
  
Line 32: Line 32:
 
-\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 12:06, 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


Back to the MA 527 start page

To Rhea Course List

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang