# Toronto Math Forum

## APM346-2015S => APM346--Home Assignments => HA6 => Topic started by: Victor Ivrii on March 05, 2015, 08:36:46 AM

Title: HA6 problem 3
Post by: Victor Ivrii on March 05, 2015, 08:36:46 AM
Let $\alpha>0$. Based on Fourier transform of $e^{-\alpha x^2/2}$ find Fourier transforms of

a.  $e^{-\alpha x^2/2}\cos (\beta x)$, $e^{-\alpha x^2/2}\sin (\beta x)$;

b.  $x e^{-\alpha x^2/2}\cos (\beta x)$, $x e^{-\alpha x^2/2}\sin (\beta x)$.
Title: Re: HA6 problem 3
Post by: Yiyun Liu on March 05, 2015, 09:07:14 PM
$\begin{gathered} part(a): \hfill \\ \hfill \\ f(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}} \hfill \\ \hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - \infty }^\infty {{e^{\frac{{ - \alpha {x^2}}}{2}}}} {e^{ - i\omega x}}dx = \frac{1}{{\sqrt {2\pi \alpha } }}{e^{\frac{{ - {\omega ^2}}}{{2\alpha }}}} \hfill \\ g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x) = \frac{1}{{2i}}f(x)({e^{i\beta x}} - {e^{ - i\beta x}}) \hfill \\ thus,\hat g(\omega ) = \frac{1}{{2i}}\left[ {(\hat f(\omega - \beta ) - (\hat f(\omega + \beta )} \right] \hfill \\ = \frac{1}{{2i\sqrt {2\pi \alpha } }}\left( {{e^{\frac{{ - {{(\omega - \beta )}^2}}}{{2\alpha }}}} - {e^{\frac{{ - {{(\omega + \beta )}^2}}}{{2\alpha }}}}} \right) \hfill \\ similar, \hfill \\ g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x) = \frac{1}{2}f(x)({e^{i\beta x}} + {e^{ - i\beta x}}) \hfill \\ \hat g(x) = \frac{1}{2}\left[ {(\hat f(\omega - \beta ) + (\hat f(\omega + \beta )} \right] = \frac{1}{{2\sqrt {2\pi \alpha } }}\left( {{e^{\frac{{ - {{(\omega - \beta )}^2}}}{{2\alpha }}}} + {e^{\frac{{ - {{(\omega + \beta )}^2}}}{{2\alpha }}}}} \right) \hfill \\ \hfill \\ part(b); \hfill \\ \hfill \\ f(x) = x{e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x) \hfill \\ let,f(x) = xg(x),g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x),hence, \hfill \\ \hat f(\omega ) = i\frac{{d\hat g(x)}}{{d\omega }} = \frac{i}{{2\sqrt {2\pi \alpha } }}(\frac{{ - (\omega - \beta )}}{\alpha }{e^{^{\frac{{ - {{(\omega - \beta )}^2}}}{{2\alpha }}}}} - \frac{{(\omega + \beta )}}{\alpha }{e^{\frac{{ - {{(\omega + \beta )}^2}}}{{2\alpha }}}}) \hfill \\ likewise, \hfill \\ f(x) = x{e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x) \hfill \\ f(x) = xg(x),where,g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x),hence, \hfill \\ \hat f(\omega ) = i\frac{{d\hat g(x)}}{{d\omega }} = \frac{1}{{2\sqrt {2\pi \alpha } }}(\frac{{ - (\omega - \beta )}}{\alpha }{e^{^{\frac{{ - {{(\omega - \beta )}^2}}}{{2\alpha }}}}} + \frac{{(\omega + \beta )}}{\alpha }{e^{\frac{{ - {{(\omega + \beta )}^2}}}{{2\alpha }}}}) \hfill \\ \hfill \\ \end{gathered}$
Title: Re: HA6 problem 3
Post by: Victor Ivrii on March 07, 2015, 05:13:12 AM
Yiyun, you should use MathJax only in math mode (not for a text).