Author Topic: TT2-P5  (Read 443 times)

Victor Ivrii

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TT2-P5
« on: November 17, 2016, 03:28:26 AM »
Find Fourier transforms of the  functions
\begin{equation}
f_\pm (x)= e^{-\varepsilon |x|}\theta(\pm x)
\end{equation}
and write these function as a Fourier integrals, where $\theta$ is a Heaviside function: $\theta(t)=1$ for $t>0$ and $\theta(t)=0$ for $t<0$.

Bonus (1pt).
Write Fourier transforms of the  functions $g(x)=f_+(x)+ f_-(x)$ and $h(x)= f_+(x)- f_-(x)$.

Shentao YANG

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Re: TT2-P5
« Reply #1 on: November 17, 2016, 09:39:15 AM »
First of all, I think the description of the problem is a little bit problematic since it does not define $\theta(0)$. I would use the convention that $\theta(0)=0$

Here is a reference: https://en.wikipedia.org/wiki/Heaviside_step_function

For
$${f_ + }(x) = {e^{ - \varepsilon |x|}}\theta ( + x) = \left\{ {\matrix{
   {{e^{ - \varepsilon x}}} & {x \ge 0}  \cr
   0 & {x < 0}  \cr

 } } \right.$$

$${{\hat f}_ + }(k) = {1 \over {2\pi }}\int_0^\infty  {{e^{ - (\varepsilon  + ik)x}}dx = } \left. { - {{{e^{ - (\varepsilon  + ik)x}}} \over {2\pi (\varepsilon  + ik)}}} \right|_0^\infty  = {1 \over {2\pi (\varepsilon  + ik)}}$$

Similarly,
$${{\hat f}_ - }(k) = {1 \over {2\pi }}\int_{ - \infty }^0 {{e^{(\varepsilon  - ik)x}}dx = } \left. {{{{e^{(\varepsilon  - ik)x}}} \over {2\pi (\varepsilon  - ik)}}} \right|_{ - \infty }^0 = {1 \over {2\pi (\varepsilon  - ik)}}$$
Therefore,
$${f_ + }(x) = \int_{ - \infty }^\infty  {{1 \over {2\pi (\varepsilon  + ik)}}} {e^{ikx}}dk$$
$${f_ - }(x) = \int_{ - \infty }^\infty  {{1 \over {2\pi (\varepsilon  - ik)}}} {e^{ikx}}dk$$
And
$$\hat g(x) = {{\hat f}_ + }(k) + {{\hat f}_ - }(k) = {1 \over {2\pi (\varepsilon  + ik)}} + {1 \over {2\pi (\varepsilon  - ik)}} = {{2\varepsilon } \over {2\pi ({\varepsilon ^2} + {k^2})}}$$
$$\hat h(x) = {{\hat f}_ + }(k) - {{\hat f}_ - }(k) = {1 \over {2\pi (\varepsilon  + ik)}} - {1 \over {2\pi (\varepsilon  - ik)}} = {{ - 2ik} \over {2\pi ({\varepsilon ^2} + {k^2})}}$$

XinYu Zheng

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Re: TT2-P5
« Reply #2 on: November 17, 2016, 11:13:51 AM »
Just a remark: if a function is different or undefined at a single point (or any finite number of points) it does not change the value of the integral. This is because a finite number of points has Jordan measure zero (https://en.wikipedia.org/wiki/Jordan_measure). So there is no problem in not defining $\theta(0)$.

Shentao YANG

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Re: TT2-P5
« Reply #3 on: November 17, 2016, 08:25:50 PM »
So there is no problem in not defining $\theta(0)$.
Well...I think the main problem is that Fourier transformation is defined on the whole real line. So, if $\theta$ is not defined at $0$, then it may be questionable to do all the later calculation.
Also...the domain of Heaviside function does contain $0$...

Victor Ivrii

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Re: TT2-P5
« Reply #4 on: November 18, 2016, 07:31:06 AM »
If we are talking about functions we do not care about their values in a few particular points (distribution will be another matter but for them "value at some particular point" is not defined). Heaviside function could be defined at $0$ as 0 (to make it semi-continuous from the left), $1$ (to make it semi-continuous from the right), or $\frac{1}{2}$ (as a half-sum of the limits).