### Author Topic: TT2 problem 5  (Read 2564 times)

#### Victor Ivrii ##### TT2 problem 5
« on: March 12, 2015, 07:22:40 PM »
Find Fourier transforms of the  function
\begin{equation*}
f(x)= \left\{\begin{aligned}
&1-x^2 &&|x|<1\\
&0 &&|x|>1
\end{aligned}\right.
\end{equation*}
and write this function as a Fourier integral.

#### Mark Nunez

• Full Member
•   • Posts: 16
• Karma: 0 ##### Re: TT2 problem 5
« Reply #1 on: March 12, 2015, 11:59:32 PM » Edited.
« Last Edit: March 13, 2015, 12:12:18 AM by Mark Nunez »

#### Yiyun Liu

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•   • Posts: 15
• Karma: 0 ##### Re: TT2 problem 5
« Reply #2 on: March 13, 2015, 12:08:13 AM »
$\begin{array}{l} part(1):\\ \\ \hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - 1}^1 {(1 - {x^2})} {e^{ - i\omega x}}dx\\ = \frac{1}{{2\pi }}[\int\limits_{ - 1}^1 {{e^{ - i\omega x}}dx - } \int\limits_{ - 1}^1 {{x^2}} {e^{ - i\omega x}}dx]\\ = \frac{1}{{2\pi }}(\frac{{ - 1}}{{i\omega }}{e^{ - i\omega x}}\mathop |\nolimits_{ - 1}^1 ) - ( - \frac{1}{{i\omega }}{x^2}{e^{ - i\omega x}}\mathop |\nolimits_{ - 1}^1 + \int\limits_{ - 1}^1 {\frac{{2x}}{{i\omega }}} {e^{ - i\omega x}}dx)\\ = - \frac{1}{{2\pi }}\int\limits_{ - 1}^1 {\frac{{2x}}{{i\omega }}} {e^{ - i\omega x}}dx\\ = - (\frac{2}{{\pi {\omega ^2}}}\cos \omega - \frac{2}{{\pi {\omega ^3}}}\sin (\omega ))\\ = \frac{2}{{\pi {\omega ^3}}}\sin (\omega ) - \frac{2}{{\pi {\omega ^2}}}\cos (\omega )\\ \\ part(2):\\ \\ f(x) = \int\limits_{ - 1}^1 {[\frac{2}{{\pi {\omega ^3}}}\sin (\omega ) - \frac{2}{{\pi {\omega ^2}}}\cos (\omega )]} {e^{i\omega x}}d\omega \end{array}$
« Last Edit: March 13, 2015, 12:49:48 AM by Yiyun Liu »

#### Victor Ivrii ##### Re: TT2 problem 5
« Reply #3 on: March 16, 2015, 06:08:41 AM »
Yiyun is right. The simplest way:
$$\hat{f}(k)=\frac{1}{2\pi}\int _{-1}^1 (1-x^2)e^{-ikx}\,dx = \frac{1}{2\pi} \int _0^1 (1-x^2)\bigl(e^{-ikx}+ e^{ikx}\bigr)\,dx= \frac{1}{\pi} \int _0^1 (1-x^2)\cos (kx)\,dx$$
where we used that the function is even. Integrating by parts
$$\hat{f}(k)= \frac{1}{\pi k} (1-x^2)\sin (kx)\bigr|_0^{1} + \frac{2}{\pi k} \int _0^1 x\sin (kx)\,dx$$
and the first term is $0$; integrating by parts again
$$\hat{f}(k)= \bigl[-\frac{2}{\pi k^2} \cos (kx) + \frac{2}{\pi k^3} \sin (kx)\bigr]\bigr|_0^{1}= -\frac{2}{\pi k^2} x\cos (k) + \frac{2}{\pi k^3} \sin (k)$$
« Last Edit: March 19, 2015, 07:18:56 PM by Victor Ivrii »