- lalaly

Fourier

- schrodinger

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- lalaly

@mr.math

- lalaly

it s ok mr math, thanks anyways:)

- Mr.Math

Oh I know the solution -.- My page went down twice!! :(

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## More answers

- Mr.Math

I will write again using another browser. I love my solutions to be complete, so sorry for taking so long :)

- lalaly

oh :-D take your time lol, i am sorry for bothering you xD

- Mr.Math

I'm starting to hate Google Chrome. Lets see how Firefox works out for me :D

- 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

Now substitute \(u=x+\pi i \omega\) and use Gaussian integral to evaluate the integral above.

- lalaly

:-D i will do that,, Thankyou soo much Mr math :D youre the best

- Mr.Math

You can see this for the integration of exp(-x^2) https://www.youtube.com/watch?v=fWOGfzC3IeY

- Mr.Math

if you want! :)

- lalaly

i know how to find it by normal distribution xD

- Mr.Math

I have alwyas known that you're smart!

- lalaly

haha not as smart as mr math xD

- Mr.Math

If I remember well, you study Communication Engineering, right?

- lalaly

yepp :-D lol you have a good memory

- Mr.Math

Only the things about "important" people! ;)

- lalaly

I feel special now :-$

- Mr.Math

Fourier transform has many applications in your field, right?

- Mr.Math

I have taken a course in Signals and Systems two semesters ago and I liked it.

- 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

Lol, Mathematicians are always the best. Engineers come second so you don't get upset :P

- lalaly

lol no doubt:P
I dont mind coming second after you xD

- 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

It is amazing,,, Thanks again :-D ... i wont drive u crazy after this dont worry hehe

- Mr.Math

You look more "Arab" in this picture for some reason. I like all you pictures anyways :)

- Mr.Math

You're welcome, and good luck!

- lalaly

lol thats sweet:)

- 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

i appreciate ur help ^_^

- 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

@mr.math

- Mr.Math

That's smarter but not easier :-)

- lalaly

lol maybe :P ... i dont like having so many integrals ... so thats why i tried to find it in another way

- lalaly

Mr math do u know bivariate random variables?

- Mr.Math

I don't think the solution I gave has that many integrals. And I don't know bivariate random variables.

- lalaly

lol its ok,,, i just wanted to ask a question xD

- 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

But I like the tricks you used.

- 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

Thanks! What you did is awesome too. Is this a homewrok asignment or what?

- Mr.Math

assignment*

- lalaly

lol yeah something like that :P

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