### Author Topic: Problem3  (Read 5807 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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• OK
##### 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) $e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right)$

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

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

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

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