The Properties of Vector Differentiation

Sdílet
Vložit
  • čas přidán 24. 05. 2024
  • Complete playlist: • The MOST Fundamental C...
    grinfeld.org
    / pavelgrinfeld
    grinfeld.org/books/An-Introdu...
    grinfeld.org/books/A-Tensor-D...

Komentáře • 22

  • @APaleDot
    @APaleDot Před měsícem +4

    How can you say that you don't need functions with vector arguments? All of linear algebra relies on such functions.

    • @tdchayes
      @tdchayes Před měsícem +2

      I think he means vector arguments that don't depend further on a scalar value. Remember that we are doing differential geometry. I suspect he would be fine with a function like a reflection that has a vector argument that depends on time (or some other parameter). Those are the linear transformations from linear algebra, and you can certainly take a derivative with respect to whatever parameter you have used.

    • @samueldeandrade8535
      @samueldeandrade8535 Před měsícem

      Man, don't even try to understand. This is a weird place.

    • @MathTheBeautiful
      @MathTheBeautiful  Před měsícem +4

      Excellent point and I should make another video that clarifies this point. In this course, I use the term "vector" in the narrow sense of directed segment. It is explained in this video: czcams.com/video/N32KI6qoeRA/video.html.
      Even with that definition, you could conceive functions of vector argument - for example "length" is a function of a vector. So when I say that "there is no such thing as a function of a vector argument", I mean "for the purposes of this narrative, there is no such thing as a function of a vector argument". And that's good news - fewer types of objects makes for a simpler framework.

  • @OtterMorrisDance
    @OtterMorrisDance Před měsícem

    Amazing to see these new videos, thanks so much for sharing them, always a delight to watch you bring this wonderful topic to life.

  • @OtterMorrisDance
    @OtterMorrisDance Před měsícem

    Thanks!

  • @PedroDuqueBR
    @PedroDuqueBR Před měsícem

    Of course you need f:R^n->R^m, and of course you can take its directional derivative (supposing it’s differentiable). f’(x_1, …,f_n) in the y (vector in R^n) direction will be a vector in R^m whose i-th component is the dot product . The total derivative will be a linear operator. The subject is comprehensively treated in Apostol’s volume 2.

  • @Phi1618033
    @Phi1618033 Před měsícem

    5:32 I assume you're not talking about functions like, for example, the function that defines the length of a vector -- f( *x* ) = sqrt(x1^2+x2^2+x3^2) -- which does take in a vector value argument and puts out a scalar value number.

  • @FineFlu
    @FineFlu Před měsícem

    Wouldnt you be able to treat the division as an inversion, like a matrix inverse?

    • @MathTheBeautiful
      @MathTheBeautiful  Před měsícem

      I'm always open to ideas! Go ahead an propose an operation!

  • @paokaraforlife
    @paokaraforlife Před měsícem +1

    i mean, electric potential for example is a scalar function with a vector argument so i don't understand what you mean when you say it doesn't exist or there is no need for it

  • @pippotopazio2400
    @pippotopazio2400 Před měsícem

    👍

  • @evankalis
    @evankalis Před měsícem

    What you teach on this youtube channel is almost like a conspiracy in the sense that it is all cohesive but to convince other that im not a crazy person id have to teach them all you have taught me (not an easy feat)

  • @Pluralist
    @Pluralist Před měsícem

  • @davidhand9721
    @davidhand9721 Před měsícem

    Uh, are you not familiar with fields? We use a gradient instead of a derivative. Functions that don't have derivatives exist.

    • @MathTheBeautiful
      @MathTheBeautiful  Před měsícem

      See my reply to the pinned comment. A later video will deal with the gradient.