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lalaly

  • 2 years ago

Fourier

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  1. lalaly
    • 2 years ago
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    @mr.math

  2. lalaly
    • 2 years ago
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    it s ok mr math, thanks anyways:)

  3. Mr.Math
    • 2 years ago
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    Oh I know the solution -.- My page went down twice!! :(

  4. Mr.Math
    • 2 years ago
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    I will write again using another browser. I love my solutions to be complete, so sorry for taking so long :)

  5. lalaly
    • 2 years ago
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    oh :-D take your time lol, i am sorry for bothering you xD

  6. Mr.Math
    • 2 years ago
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    I'm starting to hate Google Chrome. Lets see how Firefox works out for me :D

  7. Mr.Math
    • 2 years ago
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    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
    • 2 years ago
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    Now substitute \(u=x+\pi i \omega\) and use Gaussian integral to evaluate the integral above.

  9. lalaly
    • 2 years ago
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    :-D i will do that,, Thankyou soo much Mr math :D youre the best

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

  11. Mr.Math
    • 2 years ago
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    if you want! :)

  12. lalaly
    • 2 years ago
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    i know how to find it by normal distribution xD

  13. Mr.Math
    • 2 years ago
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    I have alwyas known that you're smart!

  14. lalaly
    • 2 years ago
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    haha not as smart as mr math xD

  15. Mr.Math
    • 2 years ago
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    If I remember well, you study Communication Engineering, right?

  16. lalaly
    • 2 years ago
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    yepp :-D lol you have a good memory

  17. Mr.Math
    • 2 years ago
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    Only the things about "important" people! ;)

  18. lalaly
    • 2 years ago
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    I feel special now :-$

  19. Mr.Math
    • 2 years ago
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    Fourier transform has many applications in your field, right?

  20. Mr.Math
    • 2 years ago
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    I have taken a course in Signals and Systems two semesters ago and I liked it.

  21. lalaly
    • 2 years ago
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    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
    • 2 years ago
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    Lol, Mathematicians are always the best. Engineers come second so you don't get upset :P

  23. lalaly
    • 2 years ago
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    lol no doubt:P I dont mind coming second after you xD

  24. Mr.Math
    • 2 years ago
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    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
    • 2 years ago
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    It is amazing,,, Thanks again :-D ... i wont drive u crazy after this dont worry hehe

  26. Mr.Math
    • 2 years ago
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    You look more "Arab" in this picture for some reason. I like all you pictures anyways :)

  27. Mr.Math
    • 2 years ago
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    You're welcome, and good luck!

  28. lalaly
    • 2 years ago
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    lol thats sweet:)

  29. Mr.Math
    • 2 years ago
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    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
    • 2 years ago
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    i appreciate ur help ^_^

  31. lalaly
    • 2 years ago
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    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
    • 2 years ago
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    @mr.math

  33. Mr.Math
    • 2 years ago
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    That's smarter but not easier :-)

  34. lalaly
    • 2 years ago
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    lol maybe :P ... i dont like having so many integrals ... so thats why i tried to find it in another way

  35. lalaly
    • 2 years ago
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    Mr math do u know bivariate random variables?

  36. Mr.Math
    • 2 years ago
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    I don't think the solution I gave has that many integrals. And I don't know bivariate random variables.

  37. lalaly
    • 2 years ago
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    lol its ok,,, i just wanted to ask a question xD

  38. Mr.Math
    • 2 years ago
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    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
    • 2 years ago
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    But I like the tricks you used.

  40. lalaly
    • 2 years ago
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    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
    • 2 years ago
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    Thanks! What you did is awesome too. Is this a homewrok asignment or what?

  42. Mr.Math
    • 2 years ago
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    assignment*

  43. lalaly
    • 2 years ago
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    lol yeah something like that :P

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