| Line 13: | Line 13: | ||
</math> | </math> | ||
| − | <math>=\sqrt{\frac{2}{\pi}}\frac{1}{w}\left( | + | <math>=\sqrt{\frac{2}{\pi}}\ \frac{1}{w}\left( |
-(\sin(w)-0)+(\sin(2w)-\sin(w)) | -(\sin(w)-0)+(\sin(2w)-\sin(w)) | ||
\right)=</math> | \right)=</math> | ||
| − | <math>=\sqrt{\frac{2}{\pi}}\frac{\sin(2w)-2\sin(w)}{w}.</math> | + | <math>=\sqrt{\frac{2}{\pi}}\ \frac{\sin(2w)-2\sin(w)}{w}.</math> |
| + | |||
| + | 517: 2. | ||
| + | |||
| + | <math>\hat{f}_c(w)=\sqrt{\frac{2}{\pi}}\left( | ||
| + | \int_0^k w\cos(wx)\,dx\right)= | ||
| + | </math> | ||
| + | |||
| + | <math>=\sqrt{\frac{2}{\pi}}\left(\left[\frac{x}{w}\sin(wx)+\frac{1}{w^2}\cos(wx)\right]_0^k | ||
| + | \right) | ||
| + | </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 09:01, 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) $
