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<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)\, | + | \int_0^1(-1)\cos(wx)\,dx+ |
| − | \int_1^2(1)\cos(wx)\, | + | \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 08:51, 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}. $
