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lalaly Group TitleBest ResponseYou've already chosen the best response.1
it s ok mr math, thanks anyways:)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
Oh I know the solution . My page went down twice!! :(
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
I will write again using another browser. I love my solutions to be complete, so sorry for taking so long :)
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
oh :D take your time lol, i am sorry for bothering you xD
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
I'm starting to hate Google Chrome. Lets see how Firefox works out for me :D
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
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. \[F(\omega)=\int_{\infty}^{\infty}e^{x^2}e^{2\pi i x \omega}dx=\int_{\infty}^{\infty}e^{x^22\pi i x \omega}dx=\int_{\infty}^{\infty}e^{(x^2+2\pi i x \omega)}dx\] \[=\int_{\infty}^{\infty}e^{(x+\pi i \omega)^2\pi^2\omega^2}dx=e^{\pi^2\omega^2}\int_{\infty}^{\infty}e^{(x+\pi i w)^2}dx.\]
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
Now substitute \(u=x+\pi i \omega\) and use Gaussian integral to evaluate the integral above.
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
:D i will do that,, Thankyou soo much Mr math :D youre the best
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
You can see this for the integration of exp(x^2) https://www.youtube.com/watch?v=fWOGfzC3IeY
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
if you want! :)
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
i know how to find it by normal distribution xD
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
I have alwyas known that you're smart!
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
haha not as smart as mr math xD
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
If I remember well, you study Communication Engineering, right?
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
yepp :D lol you have a good memory
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
Only the things about "important" people! ;)
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
I feel special now :$
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
Fourier transform has many applications in your field, right?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
I have taken a course in Signals and Systems two semesters ago and I liked it.
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
Lol, Mathematicians are always the best. Engineers come second so you don't get upset :P
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
lol no doubt:P I dont mind coming second after you xD
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
It is amazing,,, Thanks again :D ... i wont drive u crazy after this dont worry hehe
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
You look more "Arab" in this picture for some reason. I like all you pictures anyways :)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
You're welcome, and good luck!
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
lol thats sweet:)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
i appreciate ur help ^_^
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
found the easy way :P thought id show it to u let f(x)=e^(x^2) and let f^ (fhat) = 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
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
That's smarter but not easier :)
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
lol maybe :P ... i dont like having so many integrals ... so thats why i tried to find it in another way
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
Mr math do u know bivariate random variables?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
I don't think the solution I gave has that many integrals. And I don't know bivariate random variables.
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
lol its ok,,, i just wanted to ask a question xD
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
If I was doing it with myself, I would just do it like this: \[F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}e^{x^2}e^{i\omega x}=\frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}e^{(x+\frac{i\omega}{2})^2{\omega^2\over 4}}dx\] \[=\frac{e^\frac{\omega^2}{4}}{\sqrt{2\pi}}\sqrt{\pi}=\large \frac{1}{\sqrt{2}} e^{\frac{\omega^2}{4}}.\]
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
But I like the tricks you used.
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
yeah thats shorter hehe,,, ill write down both,, what u did was awesome, i just wanted to share with u what i thought about
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
Thanks! What you did is awesome too. Is this a homewrok asignment or what?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.2
assignment*
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.1
lol yeah something like that :P
 2 years ago
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