Lecture 5 | The Fourier Transforms and its Applications

[grunder20]

Spectacular lecture.

[zephyxj]

The way he ends his lectures halfway through explaining an important point reminds me of the Arabian Nights

[MrGreySuit]

Amazing videos

ha ha, I wonder how many people have written "Not Great" on their notes

[Reicouross]

Why does he integrate by dy when calculating the coefficients at 12:52? he is still doing an integral from 0 to 1 (1 period of the disk radial) so should he still be in x? what does the y variable represent?

[Reicouross]

Why does he integrate by dy when calculating the coefficients? he is still doing an integral from 0 to 1 (1 period of the disk radial) so should he still be in x? what does the y variable represent?

[ibreakkidslegs]

"the rigor police are off-duty" haha

[ralucada123]

excellent!!



[hai2410]

can anyone explain to me the part at 49:20 why the absolute value of the full integral is only less than or equal to the integral of the two separate absolute products- why isn't it precisely equal? Thanks

[FrankClautier]

@hai2410 It's because the absolute value of a sum is always less than or equal to the sum of the absolute values of the individual terms, e.g. abs(a+b)<=abs(a)+abs(b). Since a definite integral is a sum, the same principle applies. Also observe that the absolute value (the modulus) of the complex exponential equals 1 since exp(a*t*i) (with a=-2*pi*k*t/T) can be rewritten as cos(a*t)+i*sin(a*t), resulting in an absolute value of 1. Cheers!

[pimpbenben]

OMG SHOW ME THE MAGICAL THINGGGG!!!!

[zjamuing]

the only non-nerd-type professor out there; ...i.e. the coolest one!!

[knighttango]

ha ha...i did not have to wait until wednesday...

[kevinatucla]

Finally we get some application. I wished he told us at very beginning about application. I am 39 and i can't watch something unless I know it has some application. Anyways Thank you for the lectures. I already owe 25,000 to UCLA, actually to Sallie mae for a worthless degree in International Developmnet Studies. Without these lectures I could have never satisfied my thurst for enginerring. Thank you.

[afederdk]

@kevinatucla How is it YouTube's problem that you're some old fck with learning disabilities?

[iBradleyAllen]

@kevinatucla for me its the opposite, i have lots of application of the Fourier transform, years but need to see the underlying theory. Im SO grateful he started talking about the series first and starts to switch over in #5

u can imagine a series of analog samples in some memory array; now tell me the frequency components that series of numbers represents? impossible! well the Ft does it. how? you send thru different frequencies and see what resonates. show the math? I cant. yet.

[maurostsouk]

He is one of the best teachers I've ever seen! I only have a minor comment. At 9:50 he claims that as t->infty temperature->0. I think that he does not take into account the term for k=0 which is non-zero and therefore the temperature does not (necessarily) go to 0 but rather becomes independent of x and goes to the average initial temperature. If for example the initial temperature is homogenous u(x,0)=a then u(x,t)=a sattisfies the diff. eq. since both u_t=0 and u_xx=0.

[tHe0nLyNeXuS]

The way I see it he did not specify the initial temperature so it is assumed to be 0.

That is, it the the initial average temp (name it t_0) were non zero we would have (pardon the LaTeX notation):

u(x,t) = \sum_{k}{ t_0 + \hat{f(x)} e^{-2 k^2 \pi^2 t} e^{2 \pi k x} ) } = t_0 + \sum_{k}{ \hat{f(x)} e^{-2 k^2 \pi^2 t} e^{2 \pi k x} ) }

As you said, over time the temperature becomes independent of x averaging everywhere to t_0.

[ArtinEmil]

You are absolutely correct here! I thought the same thing. He forgot what happens at k=0.

What really happens is, u(x,+oo) = f^ (0). But f^ (0) is the 0-th Fourier coefficient which calculates to integral from 0 to 1 of f(x) dx. But this integral is the AVERAGE uniform temperature.

So at t goes to infitity we have shown that the temperature becomes uniformly the same everywhere. This is true because as you heat something it gets the same temperature eveywhere eventually.

[leocmen]

Very good, very good...

Great support for whoever is investigating this subject...

[akiforu]

excellent

[noobmartin]

This is a great complement to my regular studies in series and transforms.