Author Topic: Quiz5 T0101  (Read 2332 times)

Jilong Bi

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Quiz5 T0101
« on: March 08, 2018, 01:23:53 PM »
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter5/S5.2.P.html problem4 1.
given $$f(x) =e^{\frac{-ax^2}{2}}$$
$$\implies\widehat {f}(k) = (\frac{1}{\sqrt{2{\pi}a}})e^{\frac{-k^2}{2a}}$$
By theorem $$g(x) = f(x)e^{i{\beta}x} \implies \widehat {g}(k) = \widehat {f}(k-{\beta})$$
$$cos{\beta}x =\frac{ e^{i{\beta}x}+ e^{-i{\beta}x}}{2}$$
$$\implies \widehat {g}(k) = \frac{1}{2}\widehat {f}(k-{\beta})+\frac{1}{2}\widehat {f}(k+{\beta})$$
$$\implies \widehat {g}(k) = \frac{1}{2\sqrt{2{\pi}a}}[e^{\frac{-(k-{\beta})^2}{2a}}+e^{\frac{-(k+{\beta})^2}{2a}}]$$
same reson for $$sin{\beta}x =\frac{ e^{i{\beta}x}- e^{-i{\beta}x}}{2i}$$
$$\implies \widehat {g}(k) = \frac{1}{2i\sqrt{2{\pi}a}}[e^{\frac{-(k-{\beta})^2}{2a}}-e^{\frac{-(k+{\beta})^2}{2a}}]$$

« Last Edit: March 09, 2018, 06:49:13 AM by Victor Ivrii »

Yuxin Zhang

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Re: Quiz5 tut 0201
« Reply #1 on: March 08, 2018, 02:10:46 PM »
I will learn how to code as soon as possible.

Jingxuan Zhang

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Re: Quiz5 tut 0201
« Reply #2 on: March 08, 2018, 04:40:53 PM »
I just want to remind everyone who sees this so that he won't make the same mistake as I did:
$$F(\Re f)\neq\Re F(f)\text{ and } F(\Im f )\neq\Im F(f)$$

Victor Ivrii

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Re: Quiz5 T0101
« Reply #3 on: March 09, 2018, 06:49:00 AM »
Good job (Yuxin, Jilong). Jilong: please note that it is T0101, not T0201 (messing up tutorial sections not a big deal for us, but was a real monkey wrench thrown into  in TT1 for MAT244)


Please escape sin, cos by \ : \sin, \cos (and so on) it will make them upright

Jingxuan is correct. Also if $f$ is real, then $\hat{f}$ is real iff $f$ is even, $\hat{f}$ is purely imaginary iff $f$ is odd.

« Last Edit: March 09, 2018, 06:54:40 AM by Victor Ivrii »

Andrew Hardy

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Re: Quiz5 T0101
« Reply #4 on: March 11, 2018, 04:53:24 PM »
Considering the property that if $h(x)$ is of the form $f(x) \cdot g(x)$ that the Fourier transformation is the convolution, $\hat h(x) = \hat f(x) \ast \hat g(x)$ that gives us the same answer? So these properties of paired functions that we're memorizing are all special cases of this general fact about convolutions?

Victor Ivrii

Andrew, indeed it would give us the same answer, if we were able to calculate F.T. of $\cos(\beta x)$. The trouble is that  at this moment we cannot do it since integrals do not converge.
Later we will be able to do this in the sense of distributions $1\mapsto \delta(k)$, $e^{i\beta x}\mapsto \delta(k-\beta)$, $\cos(\beta x)\mapsto \frac{1}{2}(\delta(k-\beta)+\delta(k-\beta))$ etc and then indeed it would work.