### Author Topic: HA7-P3  (Read 5222 times)

#### Victor Ivrii

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##### HA7-P3
« on: November 01, 2015, 05:09:49 PM »

#### Xi Yue Wang

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##### Re: HA7-P3
« Reply #1 on: November 03, 2015, 11:14:57 PM »
For part (a), by inverse Fourier transform we get $$\int_{-\infty}^{\infty} \hat{f}(\omega)e^{i\omega x} d\omega = f(x)\\\int_{-\infty}^{\infty} e^{-\alpha |\omega|}e^{i\omega x } d\omega = \frac{2\alpha}{\alpha^2 + x^2}\\2\alpha[\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{\alpha^2 + x^2}e^{-i\omega x} dx] = e^{-\alpha |\omega|}\\\hat{f}( \omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{\alpha^2 + x^2}e^{-i\omega x} dx = \frac{1}{2\alpha}e^{-\alpha|\omega|}$$

For part (b), let $g(x) = xf(x) = \frac{x}{(x^2+\alpha^2)}$, where $f(x)$ is defined in part (a).
Then, $$\hat{g}(\omega) = i\hat{f'}(\omega) = \frac{-i\omega e^{-\alpha |\omega|}}{2|\omega|}\ ( |\omega|' = \frac{\omega}{|\omega|})$$

For part (c), is similar to problem 2, for $g(x) = f(x)\cos(\beta x)$, where $f(x)$ is defined in part (a). $\hat{g}(\omega) = \frac{1}{2}[\hat{f}(\omega - \beta) + \hat{f}(\omega + \beta)]$ $$=\frac{1}{4\alpha}[e^{-\alpha |\omega - \beta|}+e^{-\alpha |\omega + \beta|}]$$
for $g(x) = f(x)\sin(\beta x), \hat{g}(\omega) = \frac{1}{2i}[\hat{f}(\omega - \beta) - \hat{f}(\omega + \beta)]$
$$=\frac{1}{4i\alpha}[e^{-\alpha |\omega - \beta|}-e^{-\alpha |\omega + \beta|}]$$

For part (d), let $f(x) = \frac{\cos(\beta x)}{(\alpha^2 + x^2)}, g(x) = xf(x), \hat{g}(\omega) = i\hat{f'}(\omega)$ $$\hat{f}(\omega) = \frac{1}{4\alpha}[e^{-\alpha |\omega - \beta|}+e^{-\alpha |\omega + \beta|}]\\\hat{f'}(\omega) = -\frac{1}{4}[\frac{\omega - \beta}{|\omega-\beta|}e^{-\alpha|\omega-\beta|}+\frac{\omega + \beta}{|\omega+\beta|}e^{-\alpha|\omega+\beta|}]\\\hat{g}(\omega) = -\frac{i}{4}[\frac{\omega - \beta}{|\omega-\beta|}e^{-\alpha|\omega-\beta|}+\frac{\omega + \beta}{|\omega+\beta|}e^{-\alpha|\omega+\beta|}]$$
Similarly, let $f(x) = \frac{\sin(\beta x )}{(\alpha^2 + x^2)}, g(x) = xf(x), \hat{g}(\omega) = i\hat{f'}(\omega)$ $$\hat{f}(\omega) = \frac{1}{4i\alpha}[e^{-\alpha |\omega - \beta|}-e^{-\alpha |\omega + \beta|}]\\\hat{f'}(\omega) = -\frac{1}{4i}[\frac{\omega - \beta}{|\omega-\beta|}e^{-\alpha|\omega-\beta|}-\frac{\omega + \beta}{|\omega+\beta|}e^{-\alpha|\omega+\beta|}]\\\hat{g}(\omega) = -\frac{1}{4}[\frac{\omega - \beta}{|\omega-\beta|}e^{-\alpha|\omega-\beta|}-\frac{\omega + \beta}{|\omega+\beta|}e^{-\alpha|\omega+\beta|}]$$
« Last Edit: November 05, 2015, 12:58:52 PM by Xi Yue Wang »

#### Zaihao Zhou

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##### Re: HA7-P3
« Reply #2 on: November 12, 2015, 01:22:25 AM »
Professor, I was wondering if the coefficient before the integration of FT and IFT is important? Can we simply assume $1$? I have seen different forms as $\frac{1}{2\pi}$ and $\frac{1}{ \sqrt{2\pi}}$, when should we use which? It seems if we don't use the same coefficient we can't prove problem 1. Also for this question part (a), answer $\frac{\pi}{\alpha}e^{-\alpha|x|}$ makes sense as well.

#### Emily Deibert

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##### Re: HA7-P3
« Reply #3 on: November 12, 2015, 09:10:29 AM »
As far as I understand, there is technically a coefficient $\kappa$ in front of the FT and IFT---$\frac{\kappa}{2\pi}$ in front of the FT and $\frac{1}{\kappa}$ in front of the IFT. When I spoke to Professor Ivrii about this, he said that as long as we maintain the same value of $\kappa$ throughout our answers, we should get full marks. But perhaps Professor Ivrii can further clarify this.

In reference to problem one, we need the FT to be unitary, which would require us to choose a value of $\kappa=\sqrt{2\pi}$, I believe.

#### Zaihao Zhou

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##### Re: HA7-P3
« Reply #4 on: November 12, 2015, 11:03:39 AM »
Thanks Emily that helped a lot. Still hope prof can clarify this. I wish I can simply drop the coefficient.