## lalaly Group Title Fourier 2 years ago 2 years ago

1. lalaly

@mr.math

2. lalaly

it s ok mr math, thanks anyways:)

3. Mr.Math

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

4. Mr.Math

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

5. lalaly

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

6. Mr.Math

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

7. 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. $F(\omega)=\int_{-\infty}^{\infty}e^{-x^2}e^{-2\pi i x \omega}dx=\int_{-\infty}^{\infty}e^{-x^2-2\pi i x \omega}dx=\int_{-\infty}^{\infty}e^{-(x^2+2\pi i x \omega)}dx$ $=\int_{-\infty}^{\infty}e^{-(x+\pi i \omega)^2-\pi^2\omega^2}dx=e^{-\pi^2\omega^2}\int_{-\infty}^{\infty}e^{-(x+\pi i w)^2}dx.$

8. Mr.Math

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

9. lalaly

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

10. Mr.Math

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

11. Mr.Math

if you want! :)

12. lalaly

i know how to find it by normal distribution xD

13. Mr.Math

I have alwyas known that you're smart!

14. lalaly

haha not as smart as mr math xD

15. Mr.Math

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

16. lalaly

yepp :-D lol you have a good memory

17. Mr.Math

Only the things about "important" people! ;)

18. lalaly

I feel special now :-\$

19. Mr.Math

Fourier transform has many applications in your field, right?

20. Mr.Math

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

21. 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

22. Mr.Math

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

23. lalaly

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

24. 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.

25. lalaly

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

26. Mr.Math

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

27. Mr.Math

You're welcome, and good luck!

28. lalaly

lol thats sweet:)

29. 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.

30. lalaly

i appreciate ur help ^_^

31. 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

32. lalaly

@mr.math

33. Mr.Math

That's smarter but not easier :-)

34. lalaly

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

35. lalaly

Mr math do u know bivariate random variables?

36. Mr.Math

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

37. lalaly

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

38. Mr.Math

If I was doing it with myself, I would just do it like this: $F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-x^2}e^{-i\omega x}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-(x+\frac{i\omega}{2})^2-{\omega^2\over 4}}dx$ $=\frac{e^\frac{-\omega^2}{4}}{\sqrt{2\pi}}\sqrt{\pi}=\large \frac{1}{\sqrt{2}} e^{\frac{-\omega^2}{4}}.$

39. Mr.Math

But I like the tricks you used.

40. 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

41. Mr.Math

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

42. Mr.Math

assignment*

43. lalaly

lol yeah something like that :P