Toronto Math Forum
APM346-2018S => APM346--Tests => Quiz-5 => Topic started by: Jilong Bi on March 08, 2018, 01:23:53 PM
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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 $$sin{\beta}x =\frac{ e^{i{\beta}x}- e^{-i{\beta}x}}{2i}$$
$$\implies \widehat {g}(k) = \frac{1}{2i\sqrt{2{\pi}a}}[e^{\frac{-(k-{\beta})^2}{2a}}-e^{\frac{-(k+{\beta})^2}{2a}}]$$
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I will learn how to code as soon as possible. :)
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I just want to remind everyone who sees this so that he won't make the same mistake as I did:
$$F(\Re f)\neq\Re F(f)\text{ and } F(\Im f )\neq\Im F(f)$$
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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)
$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
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.
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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?
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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.
PS
I noticed that Yuxin goes this way, but I cannot accept his arguments until he explains what does it mean. Probably he learned it in the Physics class, but these guys are sometimes reckless