| Line 26: | Line 26: | ||
<math>=\sqrt{\frac{2}{\pi}}\left(\left[\frac{x}{w}\sin(wx)+\frac{1}{w^2}\cos(wx)\right]_0^k | <math>=\sqrt{\frac{2}{\pi}}\left(\left[\frac{x}{w}\sin(wx)+\frac{1}{w^2}\cos(wx)\right]_0^k | ||
| − | \right) | + | \right)= |
</math> | </math> | ||
| + | <math>\sqrt{\frac{2}{\pi}}\left(\frac{k}{w}\sin(kw)+\frac{1}{w^2}\cos(kw) | ||
| + | -\frac{1}{w^2}\right).</math> | ||
| + | 517: 5. See Bell's lecture near the top at | ||
| + | |||
| + | [http://www.math.purdue.edu/~bell/MA527/Lectures/lec11-10.pdf Lesson 33] | ||
[[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 09:06, 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 w\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 Bell's lecture near the top at
