Complex Fourier Series

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  • Опубліковано 3 лют 2025

КОМЕНТАРІ • 61

  • @prettycillium
    @prettycillium 4 роки тому +23

    well, your impact on society after taking the decision of sharing videos about your knowledge is an important asset. Thank you very much for your brief and clear explanation of Fourier analysis. I have watched all videos from this playlist and founded decent as an electronics engineer from Turkey. Thank you again, Mr. Brunton

  • @zacharythatcher7328
    @zacharythatcher7328 4 роки тому +9

    This has been really helpful for my understanding of quantum mechanics. Professors always pretend to intuit that the easiest solutions to the hamiltonian are the complex periodic basis functions. Then when they say that they form a complete and orthogonal set of basis functions for all solutions I have a really hard time believing it. Breaking down that the Fourier series was invented as an eigenfunction/Hilbert space solution to PDEs on purpose really helps me digest that the basis is complete by grounding me in the fact that it is just a Fourier representation of any potential answer. Seeing you go through the proofs that quantum courses always go through, but in a more general sense for complex Fourier representations, really helps me understand that these aren’t magical properties of quantum mechanics that someone came up with by analyzing schrodinger’s equation, but rather intuitive properties of the mathematical tools that we are using to understand quantum phenomena.

  • @fnegnilr
    @fnegnilr 4 роки тому +26

    I like how you are using geometric figures to keep us grounded in the explanations.

  • @thomasdehaeze3422
    @thomasdehaeze3422 4 роки тому +118

    Minor comment about the integral : if j=k, then the exponential in the integral is equal to one, therefore the integral is equal to 2pi. Then we don't have to deal with limits.

    • @_chip
      @_chip 4 роки тому +4

      Isn't the issue the division by zero in the denominator of the normalization factor?

    • @KostasOreopoulos
      @KostasOreopoulos 4 роки тому +16

      @@_chip He said in the integral, in the previous step, before the integration.

    • @TheRandalf90
      @TheRandalf90 4 роки тому +1

      Where is the indetermination then? Why l'hospital rule?

    • @paulinfgo
      @paulinfgo 3 роки тому +6

      @@TheRandalf90 if you plug j=k prior to evaluating the integral, there's no indetermination

    • @ehabnasr6925
      @ehabnasr6925 2 роки тому +4

      Thank you! I was really gonna go crazy trying to reason why it's justified to use the limits or L'Hopital's rule and wasn't making any sense lol.

  • @christophs1801
    @christophs1801 4 роки тому +6

    I like this video series a lot. Very energetic and enthousiastic style of presenting! One thing i want to mention is that this only proves the orthogonality part, it does not show that the psi-functions form a basis.

  • @lioneloddo
    @lioneloddo 2 роки тому +1

    Thanks to this video, I realize that the concept of "orthogonality" is one of the most beautiful concept of physics !
    I see one corner in the room where I'm located and it's the same thing as the vibration of my guitar that I can see below the corner !
    It's incredible that, thanks to the maths, we can proove that there is a relationship, a link, an harmony between these 2 things : A room and a guitar.
    And this is the orthogonality !

  • @ElMalikHydaspes
    @ElMalikHydaspes Рік тому

    first time, really seeing a geometrical basis (no pun intended) explanation of Fourier series ... really well done and clear!

  • @ch3432
    @ch3432 3 роки тому +3

    Explanation and derivation of the concepts are so clear! Thanks for uploading the course :)

  • @erickappel4120
    @erickappel4120 11 місяців тому +1

    Steve Bruton explains so well. He deserves a statue!!!

  • @MathFinGuy
    @MathFinGuy Рік тому

    Very clear explanation ! that any function f(x) can be represented as projections on an infinite basis , given by e^(ikx)

  • @simong1666
    @simong1666 6 місяців тому

    g(x) are the basis functions that we are projecting onto. So g_k = e^ikx and the complex conjugate is the e^(-ikx). The integral is determining the coefficients (this is the same as doing a change of basis in linear algebra but with continuous functions)

  • @diveintoengineering6089
    @diveintoengineering6089 Рік тому

    Thank you for these lectures Prof. Brunton.

  • @nishapawar7913
    @nishapawar7913 3 роки тому

    everything that u explaind just made me so clear...like some story :)...no questions left

  • @Kasun_Ish
    @Kasun_Ish 6 місяців тому +1

    Such a cool explanation.

  • @SmithnWesson
    @SmithnWesson 7 місяців тому

    8:10 You don't need L'Hopitals rule.
    When the exponent is zero, you are integrating a constant.
    Integrate[Exp[0], {x, -Pi, Pi}] = 2 Pi

  • @zachzach8171
    @zachzach8171 4 роки тому +1

    Great video! There is a minor issue @11:00: c_k should be 1/(2pi)* instead of

    • @leopardus4712
      @leopardus4712 4 роки тому

      why 1/2pi in the first place

    • @zachzach8171
      @zachzach8171 4 роки тому

      @@leopardus4712 It normalizes the projection by the square of the norm of the basis.

  • @jbangz2023
    @jbangz2023 2 роки тому +1

    Amazing, a mechanical engineer but with a deep knowledge of Advanced Mathematics

    • @arisioz
      @arisioz Рік тому

      He's got a math major and probably turned to mech for fluid dynamics in his PhD. Def not a mechanical engineer.

  • @RenormalizedAdvait
    @RenormalizedAdvait 4 роки тому +1

    Please upload a video on the convergence of Fourier series, especially piece-wise continuity and Dirichlet conditions.

  • @arisioz
    @arisioz Рік тому

    In Greek, ψ is pronounced as 'ps-ee' :)
    It's as if you would concatenate,
    - Only the consonant (ps) part of 'lapse'
    with the,
    - the first vowel of 'city'

  • @Lumiat-b5o
    @Lumiat-b5o 4 місяці тому +1

    Isn't that c_k should equals to $\frac{}{2\pi}$? Or the equation won't stand...

  • @individuoenigmatico1990
    @individuoenigmatico1990 Рік тому

    To be precise the proof that if f(x) is real then for all n, C_n=Č_-n depends on the fact that f(x) is real iff it is equal to its complex conjugate. So you just need to take the complex conjugate of the Fourier expansion of f(x) and equal it to the Fourier expansion itself. Since the functions e^inx are linearly independent one obtains C_n=Č_-n for all n.

  • @thaigo972
    @thaigo972 4 роки тому +1

    Instead of using limits or series expansions you could just say that the integral is given by that expression when j != k and by int[e^(i*0x)dx]=int(1dx) from -pi to pi when j=k.

  • @leopardus4712
    @leopardus4712 4 роки тому +1

    @10:35 why is there a 1/2pi outside ?

  • @ISKportal
    @ISKportal 2 роки тому

    Is it me or y'all find This playlist super exciting.

  • @fanwu8611
    @fanwu8611 4 роки тому

    These videos are so wonderful and useful!

  • @matthewjames7513
    @matthewjames7513 2 роки тому

    what is the relationship between C_k and the amplitude and phase of the k'th wave?

  • @garekbushnell3454
    @garekbushnell3454 3 роки тому

    Are these infinite orthogonal functions "pointing" in different directions in frequency space? And is there a comparable way of describing the directions in physical space in terms of a single function, the way e^ikx does?

  • @MiguelGarcia-zx1qj
    @MiguelGarcia-zx1qj 4 роки тому +1

    No, sir; by proving that a set of functions are mutually orthogonal in [-pi,pi], you do not prove that they form a basis of the square integrable functions on [-pi,pi]. And the proof is trivial: just drop some of the functions in a true basis (in fact the set of all psi[k] form one, but that's not the point), then you have a set of orthogonal functions unable to express the dropped ones.
    What I mean is that orthogonality alone is not enough to form a basis of a Hilbert space; you have to prove completitude (or density of the linear combinations, if you prefer).
    Ok, I see that the issue has been commented before. But still my counterexample could be of use ...

  • @guest_of_randomness
    @guest_of_randomness 4 роки тому

    it's a good view... thx for it. but how can i proof that there isn't any direction got missed out to form any function?. how can i know that all the direction i need to make a function is inside ψk of -∞ to ∞?

  • @xAlien95
    @xAlien95 4 роки тому

    The only thing I still need to figure out is: is the Professor actually writing in reverse behind a glass blackboard or is he just being video-captured normally and then flipped horizontally?

    • @AntiProtonBoy
      @AntiProtonBoy 4 роки тому

      The latter.

    • @thaigo972
      @thaigo972 4 роки тому +1

      Looking at old videos ou can see he writes with his left hand and has right-parted hair. Here he is writing with his right hand and has left-parted hair, so it's pretty safe to say the video has been flipped. You would only have to write backwards on one of these if you were doing it live, but that'd be a pain lmao.

  • @altbeb
    @altbeb Рік тому

    so.. why the complex conjugate of the phi j vector..

  • @r1a933
    @r1a933 4 роки тому

    made my day ❤️

  • @lesnikow
    @lesnikow 4 роки тому

    What do you use to justify that a countably infinite collection of mutually orthogonal functions on the right kind of function space (which you show) actually spans and is hence a basis of that function space (which you claim)? Thanks.

    • @lesnikow
      @lesnikow 4 роки тому

      @Mohamed Abdulla Saalim No, see for instance Example 2 here: www.math.lsa.umich.edu/~kesmith/infinite.pdf

    • @lesnikow
      @lesnikow 4 роки тому

      That example is the space of infinite sequences of real numbers, where you can construct an infinite set of linearly independent orthonormal elements which don't span.

  • @kin_1997
    @kin_1997 Рік тому

    ,,,, so the f is the function we are trying to approximate, what the heck is the g(X) ? it doesnt seem to be defined anywhere ( other than graph few vidoes back maybe ) still not clear of its purpose

  • @xiaoweidu4667
    @xiaoweidu4667 4 роки тому +1

    powerful !

  • @kaiseryet
    @kaiseryet Рік тому

    Should have proved it’s complete to show that Fourier basis is indeed a basis?

  • @drscott1
    @drscott1 Рік тому

    👍🏼

  • @sehailfillali615
    @sehailfillali615 4 роки тому +2

    cool

  • @nishapawar7913
    @nishapawar7913 3 роки тому

    i feeingl like i wann learn everything just from you,....lol

  • @goodlack9093
    @goodlack9093 Рік тому

    I am sorry, can anyone point me to the explanation of why we are summing from k=-inf? It seems like a bottleneck for me now, the rest and everything that follows I understand.

  • @yeorinim2sida
    @yeorinim2sida 11 місяців тому

    I've lost my way from 09:56 ...😭😭😭😭😭

  • @sansha2687
    @sansha2687 4 роки тому

    9:19

  • @videofountain
    @videofountain 4 роки тому

    ua-cam.com/video/4cfctnaHyFM/v-deo.html in the video. When j == k ...Would it not be valid and simple to reason the following? The value (j-k) is 0. Value e to the power (i * 0 * x) is ... e to the power (0) is ... 1. The integral is just integrating the value (1). This is an alternative to L'Hôpital's rule and Taylor series.

    • @Pyroguy92
      @Pyroguy92 3 роки тому

      yo that seems way faster, dont see why it wouldn't yield the same piecewise

    • @putin_navsegda6487
      @putin_navsegda6487 Рік тому

      But on this step we already integrated, so if there is no x we just have 0.

  • @luismeron9815
    @luismeron9815 Рік тому

    If someone ever tells you , you are a bad UA-camr plz tell me. Im willing to got to war to defend you

  • @mehershrishtinigam5449
    @mehershrishtinigam5449 3 роки тому

    so that's how you're supposed to say L'Hopital ...