The Properties of Vector Differentiation

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  • Опубликовано: 16 ноя 2024

Комментарии • 24

  • @APaleDot
    @APaleDot 5 месяцев назад +5

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

    • @tdchayes
      @tdchayes 5 месяцев назад +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 5 месяцев назад

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

    • @MathTheBeautiful
      @MathTheBeautiful  5 месяцев назад +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: ruclips.net/video/N32KI6qoeRA/видео.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 5 месяцев назад

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

  • @PedroDuqueBR
    @PedroDuqueBR 5 месяцев назад

    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.

    • @MathTheBeautiful
      @MathTheBeautiful  5 месяцев назад

      You're correct. Please see the pinned comment.

  • @FineFlu
    @FineFlu 5 месяцев назад

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

    • @MathTheBeautiful
      @MathTheBeautiful  5 месяцев назад

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

  • @Phi1618033
    @Phi1618033 5 месяцев назад

    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.

    • @MathTheBeautiful
      @MathTheBeautiful  5 месяцев назад

      Good point! See my reply to the pinned comment.

  • @TeslaTesla-c6i
    @TeslaTesla-c6i 2 месяца назад

    In support of @APaleDot’s comment: the dot product is a scalar function of vector arguments. And we can differentiate dot products.

  • @paokaraforlife
    @paokaraforlife 5 месяцев назад +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

    • @MathTheBeautiful
      @MathTheBeautiful  5 месяцев назад

      You're correct. Please see the pinned comment.

  • @evankalis
    @evankalis 5 месяцев назад

    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)

  • @OtterMorrisDance
    @OtterMorrisDance 5 месяцев назад

    Thanks!

  • @pippotopazio2400
    @pippotopazio2400 5 месяцев назад

    👍

  • @Pluralist
    @Pluralist 5 месяцев назад

  • @davidhand9721
    @davidhand9721 5 месяцев назад

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

    • @MathTheBeautiful
      @MathTheBeautiful  5 месяцев назад

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