Line 4: Line 4:
  
 
<math>\hat{f}_c(w)=\sqrt{\frac{2}{\pi}}\left(
 
<math>\hat{f}_c(w)=\sqrt{\frac{2}{\pi}}\left(
\int_0^1(-1)\cos(wx)\,dw+
+
\int_0^1(-1)\cos(wx)\,dx+
\int_1^2(1)\cos(wx)\,dw
+
\int_1^2(1)\cos(wx)\,dx
\right)
+
\right)=
 
</math>
 
</math>
  
 +
<math>=\sqrt{\frac{2}{\pi}}\left([-\frac{1}{w}\sin(wx)]_0^1
 +
+[\frac{1}{w}\sin(wx)]_1^2\right)=
 +
</math>
 +
 +
<math>=\sqrt{\frac{2}{\pi}}\frac{1}{w}\left(
 +
-(\sin(w)-0)+(\sin(2w)-\sin(w))
 +
\right)=</math>
 +
 +
<math>=\sqrt{\frac{2}{\pi}}\frac{\sin(2w)-2\sin(w)}{w}.</math>
  
 
[[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 07:51, 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}. $

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