Author Topic: Problem3  (Read 4229 times)

Aida Razi

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Problem3
« on: November 07, 2012, 09:31:58 PM »
Solution is attached!

Calvin Arnott

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Re: Problem3
« Reply #1 on: November 07, 2012, 09:34:57 PM »
Problem 3

Let: $ \alpha > 0, \beta > 0 $. Using the Fourier transform of $ e^{-\frac{\alpha x^2}{2}} $ compute the Fourier transform for:

a. i) $ e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right)$

Answer:

We have that the Fourier transform for $f\left(x\right) = e^{-\frac{x^2}{2}} $ is given by $ F\left(k\right) = \sqrt{2 \pi} e^{-\frac{k^2}{2}} $, and that a property of the Fourier transform is $f\left(a x\right) \rightarrow \frac{1}{|a|}F\left(\frac{k}{a}\right) $. So for $ f\left(x\right) = e^{-\frac{x^2}{2}}, f\left(\sqrt{\alpha} x\right) = e^{-\frac{\alpha x^2}{2}} = g\left(x\right) \implies G\left(k\right) = \frac{1}{\sqrt{\alpha}}F\left(\frac{k}{\sqrt{\alpha}}\right) = \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha k^2}{2}} $ is the transform for $ e^{-\frac{\alpha x^2}{2}} $, using that $ \alpha > 0 \implies | \sqrt{\alpha} | = \sqrt{\alpha} $.

$$ \text{ Again; } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:} $$

$$ e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} e^{-\frac{\alpha x^2}{2}} + e^{- i \beta x} e^{-\frac{\alpha x^2}{2}}\right) $$

Using that for: $ g\left(x\right) = e^{i a x} f\left(x\right) $ the Fourier transform of $ g\left(x\right) $ is given by $ G\left(k\right) = F\left(k-a\right) $, and that the transform is linear: $ h\left(x\right) = a f\left(x\right) + b g\left(x\right) \implies H\left(k\right) = a F\left(k\right) + b G\left(k\right) $ our Fourier transform is then given by $ \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) $ where $ F\left(k\right) = \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha k^2}{2}} $

$$ \implies \frac{1}{2}\left(\sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} + \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right) $$

$$ = \sqrt{\frac{\pi}{2\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} +  e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right) \text{ is our transform } \blacksquare $$


ii) $e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right) $

Answer:

As in part i) $ \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) $ so we write:

$$ e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} e^{-\frac{\alpha x^2}{2}} - e^{- i \beta x} e^{-\frac{\alpha x^2}{2}}\right) $$

$$ \text{and our transform is given by: } \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: } F\left(k\right) = \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha k^2}{2}} $$

$$ \implies \frac{1}{2}\left(\sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} - \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right) $$

$$ = \sqrt{\frac{\pi}{2 i\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} -  e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right) \text{ is our transform } \blacksquare $$

b. i) $ x e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right)$

Answer:

We have for for any function $ f\left(x\right) $ with Fourier transform $ F\left(k\right), $ the transform of $ g\left(x\right) = x f\left(x\right) $  is given by: $ G\left(k\right) = i\frac{dF}{dk} $. Then for $ f\left(x\right) = e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right) $, $F\left(k\right) = \sqrt{\frac{\pi}{2\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} +  e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right) $, $ g\left(x\right) = x e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right) = x f\left(x\right) $, $ G\left(k\right) = i\frac{dF}{dk} = i \partial_k \left(\sqrt{\frac{\pi}{2\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} +  e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)\right)$

$$ \implies G\left(k\right) = i  \left(\left(k-\beta\right)\sqrt{\frac{\pi}{2\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} +  \left(k+\beta\right)e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)\right) \text{ is our transform} \blacksquare$$ 


ii) $ x e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right) $

Answer:

Again, for for any function $ f\left(x\right) $ with Fourier transform $ F\left(k\right), $ the transform of $ g\left(x\right) = x f\left(x\right) $  is given by: $ G\left(k\right) = i\frac{dF}{dk} $. Then for $ f\left(x\right) = e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right) $, $F\left(k\right) = \sqrt{\frac{\pi}{2 i \alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} -  e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right) $, $ g\left(x\right) = x e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right) = x f\left(x\right) $, $ G\left(k\right) = i\frac{dF}{dk} = i \partial_k \left(\sqrt{\frac{\pi}{2 i \alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} -  e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)\right)$

$$ \implies G\left(k\right) = i  \left(\left(k-\beta\right)\sqrt{\frac{\pi}{2 i \alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} -  \left(k+\beta\right)e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)\right) \text{ is our transform} \blacksquare$$ 
« Last Edit: November 18, 2012, 04:58:31 PM by Calvin Arnott »

Victor Ivrii

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Re: Problem3
« Reply #2 on: November 07, 2012, 09:59:48 PM »
Again, simpler to use (as Calvin did) F.T. properties rather than calculate integral (especially because one really needs to prove that one can replace $x$ by $x+i \omega$ (thus going into complex plane)

Kanita Khaled

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Re: Problem3
« Reply #3 on: November 15, 2012, 09:46:43 AM »
Aida, your solutions are always awesome and neat. Thanks :)