Complex Fourier Series

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  • Опубликовано: 22 окт 2024

Комментарии • 61

  • @prettycillium
    @prettycillium 4 года назад +21

    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

  • @thomasdehaeze3422
    @thomasdehaeze3422 4 года назад +113

    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 года назад +4

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

    • @KostasOreopoulos
      @KostasOreopoulos 4 года назад +15

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

    • @TheRandalf90
      @TheRandalf90 3 года назад +1

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

    • @paulinfgo
      @paulinfgo 3 года назад +5

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

    • @ehabnasr6925
      @ehabnasr6925 2 года назад +3

      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.

  • @zacharythatcher7328
    @zacharythatcher7328 4 года назад +8

    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 года назад +25

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

  • @christophs1801
    @christophs1801 4 года назад +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 года назад +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 !

  • @ch3432
    @ch3432 3 года назад +3

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

  • @simong1666
    @simong1666 3 месяца назад

    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)

  • @erickappel4120
    @erickappel4120 8 месяцев назад +1

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

  • @Kasun_Ish
    @Kasun_Ish 3 месяца назад +1

    Such a cool explanation.

  • @Sau_UNCCMathFin
    @Sau_UNCCMathFin 11 месяцев назад

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

  • @ElMalikHydaspes
    @ElMalikHydaspes 8 месяцев назад

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

  • @diveintoengineering6089
    @diveintoengineering6089 Год назад

    Thank you for these lectures Prof. Brunton.

  • @Lumiat-b5o
    @Lumiat-b5o 20 дней назад +1

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

  • @RenormalizedAdvait
    @RenormalizedAdvait 3 года назад +1

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

  • @nishapawar7913
    @nishapawar7913 2 года назад

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

  • @zachzach8171
    @zachzach8171 3 года назад +1

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

    • @leopardus4712
      @leopardus4712 3 года назад

      why 1/2pi in the first place

    • @zachzach8171
      @zachzach8171 3 года назад

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

  • @SmithnWesson
    @SmithnWesson 3 месяца назад

    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

  • @thaigo972
    @thaigo972 4 года назад +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.

  • @individuoenigmatico1990
    @individuoenigmatico1990 9 месяцев назад

    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.

  • @jbangz2023
    @jbangz2023 2 года назад +1

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

    • @arisioz
      @arisioz Год назад

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

  • @arisioz
    @arisioz Год назад

    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'

  • @garekbushnell3454
    @garekbushnell3454 2 года назад

    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?

  • @matthewjames7513
    @matthewjames7513 2 года назад

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

  • @ISKportal
    @ISKportal 2 года назад

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

  • @guest_of_randomness
    @guest_of_randomness 4 года назад

    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 ∞?

  • @leopardus4712
    @leopardus4712 3 года назад +1

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

  • @MiguelGarcia-zx1qj
    @MiguelGarcia-zx1qj 4 года назад +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 ...

  • @xAlien95
    @xAlien95 4 года назад

    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 года назад

      The latter.

    • @thaigo972
      @thaigo972 4 года назад +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.

  • @lesnikow
    @lesnikow 4 года назад

    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 года назад

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

    • @lesnikow
      @lesnikow 4 года назад

      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.

  • @xiaoweidu4667
    @xiaoweidu4667 4 года назад +1

    powerful !

  • @kin_1997
    @kin_1997 8 месяцев назад

    ,,,, 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

  • @altbeb
    @altbeb Год назад

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

  • @fanwu8611
    @fanwu8611 4 года назад

    These videos are so wonderful and useful!

  • @sehailfillali615
    @sehailfillali615 4 года назад +2

    cool

  • @kaiseryet
    @kaiseryet 11 месяцев назад

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

  • @r1a933
    @r1a933 4 года назад

    made my day ❤️

  • @goodlack9093
    @goodlack9093 11 месяцев назад

    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.

  • @nishapawar7913
    @nishapawar7913 2 года назад

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

  • @drscott1
    @drscott1 Год назад

    👍🏼

  • @sansha2687
    @sansha2687 4 года назад

    9:19

  • @yeorinim2sida
    @yeorinim2sida 7 месяцев назад

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

  • @videofountain
    @videofountain 4 года назад

    ruclips.net/video/4cfctnaHyFM/видео.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 2 года назад

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

    • @putin_navsegda6487
      @putin_navsegda6487 Год назад

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

  • @mehershrishtinigam5449
    @mehershrishtinigam5449 2 года назад

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

  • @luismeron9815
    @luismeron9815 Год назад

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