Parseval's Identity, Fourier Series, and nice applications.

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Опубликовано:

30 сен 2024

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Комментарии : 47

@MichaelPennMath 11 месяцев назад

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@davidcroft95 11 месяцев назад

"a real basis requires finite sums" Hilbert spaces: *cries in infinite dimensions*

@ingobojak5666 11 месяцев назад

17:00 That's not a Dirac delta, it's a Kronecker delta.

@FadiAkil 11 месяцев назад

17:00 Kronecker, not Dirac.

@zh84 11 месяцев назад

Yes, I was going to say that. Dirac's delta is a function of one variable.

@Arty_x_g 8 месяцев назад

Uh, i May be wrong, but that Is a generalised Fourier transform of 1, so it's correct to Say It is indeed the Dirac Delta.

@Arty_x_g 8 месяцев назад

Nevermind, i see what you are trying to say. Yes it's over n not over the frequency f. I was recalling Plancherel's identity, which Is a generalization of the same theorem

@TomFarrell-p9z 11 месяцев назад

This is a foundation of signal processing in undergraduate electrical engineering. (In graduate EE, we build on it by considering the Fourier analysis of probabilistic functions). Great presentation!

@joelklein3501 11 месяцев назад

7:00 This is the most outrageous column vector I've ever seen. How dare you write the indices in a decreasing order??

@emanuellandeholm5657 11 месяцев назад

The complex exps are eigenfunctions and the weights are eigenvalues. It's analogous to ie. "basis", "span" etc, but this is an infinite dimensional space. A vector space. The most general form of a vector space is uncountably infinite (aleph 0). It's quite interesting that we can construct a countably infinite set of eigenfunctions that "span" this space (in an L2 sense). There's nothing magical about the Fourier basis, we can also use polynomials like Legendre, Chebychev, Bessel etc.

@emanuellandeholm5657 11 месяцев назад

The eigenfunctions used to solve the Schrödinger wave eq in spherical coords are known as the spherical harnonics Y(m, n). They have properties similar to the complex exps and the orthogonal polys in L2. These eigenfunctions are all solutions to second order linear diff eqs.

@orenfivel6247 11 месяцев назад

I think for evaluation of a_n and b_n via the integrals you need a minus sign in the exponent in the integrand i.e. e^{-i \pi nx/L}

@michaelguenther7105 11 месяцев назад

For f(x) = sin(x) = (e^ix - e^-ix)/(2i) we can determine a_n by inspection. |a_-1| = |a_1| = 1/2, and all other a_n = 0, so Parseval's identity becomes 1/4 + 1/4 = 1/2 = 1/(2 pi) integral of sin^2(x) over -pi to pi. Although pretty cool, it's a bit like using a sledgehammer to crack a nut.

@idjles 11 месяцев назад

Nice to see Basel hidden in Fourier!!

@Mystery_Biscuits 11 месяцев назад

I thought when doing Fourier series with complex exponentials, you need the conjugate in the formula for the coefficients, so in this case it would be a_n = 1/2L \inf_{-L}^L f(x) e^{-i n \pi x/L} dx ?

@andrewkarsten5268 11 месяцев назад

No, you can go either way, but whichever way you do you need to be consistent with. It’s a preference thing, but if you keep it consistent you’ll get the same thing

@r2k314 11 месяцев назад

@@andrewkarsten5268 please provide a reference for your claim.

@andrewkarsten5268 11 месяцев назад

@@r2k314 it’s well known among mathematicians, and I’ve had professors that have done it both ways. I don’t have a link that states explicitly you can do it either way, but I’ve seen it in practice done both ways and it works fine. The only difference you get is the sign of the imaginary component of the output function after doing the transform, but that doesn’t really matter because when pulling the coefficients for the Fourier series, but that sign flip is canceled out in the infinite sum if you also keep the complex exponential consistent as I said.

@mrAngelo212play 11 месяцев назад

@@r2k314 It's due to the symmetry of the imaginary number, as it's most consistent definition is that it is one of the solutions of x²+1=0, so you can usually make the substitution of i>-i & -i->i, as -i is the other solution of the equation. Another way to understand it is the fact that if a polynomial equation has only real coefficients, any complex solutions are accompanied with it's complex conjugate, which means that in the real polynomial space R[x], it's associated Null space is independent of the sign associated with the imaginary constant, with the important fact being that they're conjugates.

@r2k314 11 месяцев назад

@@mrAngelo212play I'm looking at it from this perspective: By using the conjugate you show that Fourier series "Work" because of orthogonality. If you don't, I don't get how you extract the coefficients. But I don't have a deep understanding, because I don't understand the relevance of your response.

@eiseks3410 3 месяца назад

Really nice!

@depiction3435 5 месяцев назад

That was really intuitive

@jesusalej1 11 месяцев назад

I was never told about four year' series.

@thomasjefferson6225 8 месяцев назад

im not your biggest fan, i dont say that to be rude but to emphasise this is a good way to deliver this material.

@davidgillies620 11 месяцев назад

I first encountered Parseval's theorem in physics when we used it to derive the Shannon-Nyquist sampling theorem.

@mekbebtamrat817 9 месяцев назад

Very cool. Any way you could share a link or an article with a similar derivation? And happy new year!

@davidgillies620 9 месяцев назад

@@mekbebtamrat817 I don't have the exact proof to hand but if I recall correctly it involves showing that the exponential terms in the Fourier expansion of a square-integrable function span (i.e. form a complete basis set over) a vector space. This forms a series of sinc functions and by swapping the order of integration (Fubini) we can equate the energy of the representation in the time domain and frequency domain.

@joetursi9573 3 месяца назад

Great, Thanks.

@maxxis4035 11 месяцев назад

I have a question for anyone really. So I have to determine if the following integral converge or diverge: from 0 to (pi)^2 of (1/[1- cos(sqrt(x)]), any help is appriciated.

@MartinPerez-oz1nk 11 месяцев назад

THANKS PROFESOR !!!!!, VERY, VERY INTERESTING !!!!!, PERFECT !!!!

@daskhard 9 месяцев назад

Are there similar channels with tasks on differential geometry and topology?

@SUMONPAL-re2pv 6 месяцев назад

Class very nice

@nitroh7745 Месяц назад

🔥🔥

@Happy_Abe 11 месяцев назад

Where were we able to take the sums outside of the integral here?

@MichaelRothwell1 8 месяцев назад

Yeah! I really enjoyed this video and the applications, but was disappointed that the integral-infinite sum swap was glossed over.

@Happy_Abe 8 месяцев назад

@@MichaelRothwell1likewise!

@goodplacetostop2973 11 месяцев назад

30:48

@aidarosullivan5269 11 месяцев назад

What is the difference between Fourier series and and Fourier transform?

@andrewkarsten5268 11 месяцев назад

The fourier series returns the original function or coefficients you started with, while the fourier transform takes the function as an input and outputs a different function that evaluates to the coefficients of the fourier series at given inputs. Hope I was clear enough with that explaination.

@bjornfeuerbacher5514 11 месяцев назад

The Fourier series is used for periodic functions, the Fourier transform for non-periodic ones.

@tomholroyd7519 11 месяцев назад

but why is = sum v w CONJUGATE where did the conjugate come from

@coc235 11 месяцев назад

From the definition

@courbelm5111 11 месяцев назад

Because you want to be positive

@StephenHamer-w5b 11 месяцев назад

The conjugate is needed so that will be pure real when w = v - one of the rules for an "inner product" on a real or complex vector space. (See the Wikipedia article on "inner product spaces")