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Anas.P

  • one year ago

fourier transform question first question please as soon as possible.

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  1. anas.p
    • one year ago
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    @dan815

  2. anas.p
    • one year ago
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    Find the fourier Transform of \[e^{-a^{2}x^{2}} ,a>0\]Given \[\int\limits_{-\infty}^{\infty} e^{-t} dt = \sqrt{\pi}\]

  3. anas.p
    • one year ago
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    @SolomonZelman

  4. anonymous
    • one year ago
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    I think you meant \(e^{-\color{red}{t^2}}\) in that last integral?

  5. anonymous
    • one year ago
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    \[\large\begin{align*}\mathcal{F}\{e^{-a^2x^2}\}&=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-a^2x^2}e^{-i\xi x}\,dx\\[2ex] &=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-a^2\left(x^2+\frac{i\xi}{a^2}x\right)}\,dx \end{align*}\] Complete the square: \[\begin{align*}-a^2\left(x^2+\frac{i\xi}{a^2}x\right)&=-a^2\left(x^2+\frac{i\xi}{a^2}x+\left(\frac{i\xi}{2a^2}\right)^2-\left(\frac{i\xi}{2a^2}\right)^2\right)\\[2ex] &=-a^2\left(\left(x+\frac{i\xi}{2a^2}\right)^2+\frac{\xi^2}{4a^4}\right)\\[2ex] &=-\left(ax+\frac{i\xi}{2a}\right)^2-\frac{\xi^2}{4a^2}\end{align*}\] A substitution will do the rest: \(y=ax+\dfrac{i\xi}{2a}\) with \(dy=a\,dx\).\[\large\begin{align*}\mathcal{F}\{e^{-a^2x^2}\}&=\frac{1}{2a\pi}\int_{-\infty}^\infty e^{-y^2-\frac{\xi^2}{4a^2}}\,dy \end{align*}\]

  6. anas.p
    • one year ago
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    yes it was \[e^{t^{-2}}\] isn't it \[\frac{ 1 }{ \sqrt{2\pi} }\] instead of \[\frac{ 1 }{ {2\pi} }\] otherwise thank you very much... it had been bugging me for days...

  7. anonymous
    • one year ago
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    I've seen a few variations of the transform's definition, things like \(\dfrac{1}{2\pi}\int e^{-i\xi x}\,dx\) and \(\int e^{-2\pi i\xi x}\,dx\) (which are identical) as well as \(\dfrac{1}{\sqrt{2\pi}}\) in place of \(\dfrac{1}{2\pi}\). (see here: http://mathworld.wolfram.com/FourierTransform.html ) I'm not sure myself about the details of the advantage to using one definition over another, but I think it has something to do with symmetry. As long as you're consistent, either way is fine.

  8. anas.p
    • one year ago
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    cool..... thanks again

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