lalaly
  • lalaly
Fourier
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lalaly
  • lalaly
@mr.math
lalaly
  • lalaly
it s ok mr math, thanks anyways:)
Mr.Math
  • Mr.Math
Oh I know the solution -.- My page went down twice!! :(

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Mr.Math
  • Mr.Math
I will write again using another browser. I love my solutions to be complete, so sorry for taking so long :)
lalaly
  • lalaly
oh :-D take your time lol, i am sorry for bothering you xD
Mr.Math
  • Mr.Math
I'm starting to hate Google Chrome. Lets see how Firefox works out for me :D
Mr.Math
  • Mr.Math
So you know the definition of Fourier transform: The Fourier Transform of an integrable function \(f(x)\) is define as \[\large F(\omega)=\int_{-\infty}^{\infty} f(x)e^{-2\pi i x\omega}.\] For our function \(f(x)=e^{-x^2}\), we have \[F(\omega)=\int_{-\infty}^{\infty}e^{-x^2}e^{-2\pi i x \omega}dx.\] The question becomes now how to evaluate this integral?! You probably know the famous Gaussian integral \(\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}\). We will manipulate our integrand to make it of a similar form. \[F(\omega)=\int_{-\infty}^{\infty}e^{-x^2}e^{-2\pi i x \omega}dx=\int_{-\infty}^{\infty}e^{-x^2-2\pi i x \omega}dx=\int_{-\infty}^{\infty}e^{-(x^2+2\pi i x \omega)}dx\] \[=\int_{-\infty}^{\infty}e^{-(x+\pi i \omega)^2-\pi^2\omega^2}dx=e^{-\pi^2\omega^2}\int_{-\infty}^{\infty}e^{-(x+\pi i w)^2}dx.\]
Mr.Math
  • Mr.Math
Now substitute \(u=x+\pi i \omega\) and use Gaussian integral to evaluate the integral above.
lalaly
  • lalaly
:-D i will do that,, Thankyou soo much Mr math :D youre the best
Mr.Math
  • Mr.Math
You can see this for the integration of exp(-x^2) https://www.youtube.com/watch?v=fWOGfzC3IeY
Mr.Math
  • Mr.Math
if you want! :)
lalaly
  • lalaly
i know how to find it by normal distribution xD
Mr.Math
  • Mr.Math
I have alwyas known that you're smart!
lalaly
  • lalaly
haha not as smart as mr math xD
Mr.Math
  • Mr.Math
If I remember well, you study Communication Engineering, right?
lalaly
  • lalaly
yepp :-D lol you have a good memory
Mr.Math
  • Mr.Math
Only the things about "important" people! ;)
lalaly
  • lalaly
I feel special now :-$
Mr.Math
  • Mr.Math
Fourier transform has many applications in your field, right?
Mr.Math
  • Mr.Math
I have taken a course in Signals and Systems two semesters ago and I liked it.
lalaly
  • lalaly
yeah but i am taking a course where fourier transform is solved in mathematics way... and its different from what ive taken in signals and systems and communication -.- i solved this question in a way ,, the professor said he wants it solved in mathematicians way not engineers :S
Mr.Math
  • Mr.Math
Lol, Mathematicians are always the best. Engineers come second so you don't get upset :P
lalaly
  • lalaly
lol no doubt:P I dont mind coming second after you xD
Mr.Math
  • Mr.Math
Well, I think as long as the solution provides all logical steps needed, it should be enough. I did take Fourier transform in two different courses one of which was this Engineering course I just told you about. I remember in that course we were allowed to use tables, but in a Math course we would have to derive them ourselves in one way or another.
lalaly
  • lalaly
It is amazing,,, Thanks again :-D ... i wont drive u crazy after this dont worry hehe
Mr.Math
  • Mr.Math
You look more "Arab" in this picture for some reason. I like all you pictures anyways :)
Mr.Math
  • Mr.Math
You're welcome, and good luck!
lalaly
  • lalaly
lol thats sweet:)
Mr.Math
  • Mr.Math
Important note: I should have used \(f\) instead of \(\omega\) there, because the transform I did was in terms of frequency not angular frequence. You can use \(\omega=2\pi f\) to write it in angular frequency, as you know.
lalaly
  • lalaly
i appreciate ur help ^_^
lalaly
  • lalaly
found the easy way :P thought id show it to u let f(x)=e^(-x^2) and let f^ (f-hat) = F \[f'(x)=-2xe^{-x^2}\] take fourier transform of both sides\[F[f'(x)]=-2F[xf(x)]\]we know that \[F[f']=iwF\]and \[\large{F[xf]=iF '}\]so now we have\[iwF=-2iF'\]the i cancels \[wF=-2F'\]now we seperate\[wdw=\frac{-2}{f}dF\]integrate both sides\[\frac{w^2}{2}=-2lnF+C\]take e of both sides\[\large{e^{\frac{w^2}{2}}=e^{lnF^{-2}+C}}\]so simplifying\[\huge{F=C_2e^{-\frac{w^2}{4}}}\]now we find the constant observe F(0)=C_2 so \[C_2=\frac{1}{\sqrt{2 pi}} \int\limits_{-\infty}^{\infty}e^{-x^2}dx\]we know that theat integral =sqrt(pi) so \[C_2=\frac{1}{\sqrt2}\] now \[\huge{F(w)=\frac{1}{\sqrt 2}e^{-\frac{w^2}{4}}}\]hoooooooooooooof lol
lalaly
  • lalaly
@mr.math
Mr.Math
  • Mr.Math
That's smarter but not easier :-)
lalaly
  • lalaly
lol maybe :P ... i dont like having so many integrals ... so thats why i tried to find it in another way
lalaly
  • lalaly
Mr math do u know bivariate random variables?
Mr.Math
  • Mr.Math
I don't think the solution I gave has that many integrals. And I don't know bivariate random variables.
lalaly
  • lalaly
lol its ok,,, i just wanted to ask a question xD
Mr.Math
  • Mr.Math
If I was doing it with myself, I would just do it like this: \[F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-x^2}e^{-i\omega x}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-(x+\frac{i\omega}{2})^2-{\omega^2\over 4}}dx\] \[=\frac{e^\frac{-\omega^2}{4}}{\sqrt{2\pi}}\sqrt{\pi}=\large \frac{1}{\sqrt{2}} e^{\frac{-\omega^2}{4}}.\]
Mr.Math
  • Mr.Math
But I like the tricks you used.
lalaly
  • lalaly
yeah thats shorter hehe,,, ill write down both,, what u did was awesome, i just wanted to share with u what i thought about
Mr.Math
  • Mr.Math
Thanks! What you did is awesome too. Is this a homewrok asignment or what?
Mr.Math
  • Mr.Math
assignment*
lalaly
  • lalaly
lol yeah something like that :P

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