The true second derivative test

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  • Опубликовано: 9 фев 2025
  • In this video I present the second derivative test in multivariable calculus, which is used to find local maxima/minima/saddle points of a function. However, unlike the nonsense they usually teach in the books, here I'm presenting a more elegant way with some linear algebra blended in. I'm absolutely convinced that those two subjects go hand in hand! Enjoy!

Комментарии • 42

  • @omparghale
    @omparghale Год назад

    love the way you go over every step systematically, instead of rushing the tiny calculations.

  • @sergioh5515
    @sergioh5515 5 лет назад +1

    I'm over a year late but THIS IS AWESOME! THANKS FOR ENLIGHTENING ME W YOUR LINEAR ALGEBRA AS ALWAYS. Linear Algebra is my favorite and you make it even more fun and amazing for me!!!!!!!!!

  • @slavinojunepri7648
    @slavinojunepri7648 Год назад

    This video has taught me how to classify critical points of a multivariable function properly. I have not seen an intelligible proof for the classification based on eigen values, and I'd appreciate if you could provide one on a video. Thank you Dr. Peyam!

  • @911mire
    @911mire 7 лет назад +3

    Great bunny and great channel content, subbed long time ago, keep up with good work!

  • @davida2810
    @davida2810 4 года назад +1

    This is fantastic! I know a third method that I hope should be equivalent to this one. In this method, to classify the critical point of f at (x,y), you look at the second degree taylor polynomial of f at (x,y). Since the first derivatives are all zero, this polynomial is a quadratic form that approximates f around the point (x,y). Then we figure out whether the taylor polynomial is positive definite, negative definite, or neither. Each number in the signature of the quadratic form tells you which direction one variable "swings" in, I believe, and so its a min or max iff they all "swing" in the correct direction. Of course, this method also requires the secret ingredient that most Calc III textbooks don't have, Linear Algebra. If you're interested I could write it up and email you how it works!

  • @kwinvdv
    @kwinvdv 7 лет назад

    The nice thing about this definition is also that you can apply it to functions with even more variables.

  • @poetu2951
    @poetu2951 7 лет назад +11

    I would love an example of how to approach the problem if the test is inconclusive, e.g what to do next.

    • @drpeyam
      @drpeyam  7 лет назад +5

      I'll think about it, but unfortunately there's not much you can do! Because x^4 + y^4 (min), -x^4 - y^4 (max) and x^4 - y^4 (saddle) both have inconclusive tests at (0,0), but you have 3 very different behaviors! But I'll see what I can do about that.

    • @foreachepsilon
      @foreachepsilon 7 лет назад +1

      I imagine examining the function in the direction of the eigenvectors would show some behavior.

  • @japotillor
    @japotillor 7 лет назад +8

    If i get the opportunity to teach mvc, I'll be using this instead...beautiful

  • @adityadwivedi4412
    @adityadwivedi4412 4 года назад +1

    The first equation is similar to the coffin problem

  • @WoWSchockadin
    @WoWSchockadin 7 лет назад

    Just so much more beautiful than what I've learned at the University. Thanks for showing this. And btw

  • @aneeshsrinivas9088
    @aneeshsrinivas9088 3 года назад

    whats the 3d equivilent of a point of inflection?

    • @drpeyam
      @drpeyam  3 года назад

      det D^2 f = 0

    • @aneeshsrinivas9088
      @aneeshsrinivas9088 3 года назад

      The proof is Taylor series right?

    • @aneeshsrinivas9088
      @aneeshsrinivas9088 3 года назад

      At least In proving that having a positive or negative definite matrix results in convexity or concavity right?

  • @cedricp.4941
    @cedricp.4941 7 лет назад +1

    Very Nice !! What can we do when the test is inconclusive ? Also, how do we know that there will be only real eigenvalues ? :)

    • @gaukatore
      @gaukatore 7 лет назад +6

      The matrix will always be symmetric, as d^2f/dxdy=d^f/dydx. and symmetric (real) matrecies always have real eigenvalues.

    • @cedricp.4941
      @cedricp.4941 7 лет назад +1

      Yes, of course...Thank you !

  • @botondosvath2331
    @botondosvath2331 7 лет назад

    Would you mind making a video about the whole eigen-stuff, like calculating the eigenvalues, right and left eigenvectors, projection matrices, etc?

    • @drpeyam
      @drpeyam  7 лет назад +1

      There's one I made on blackpenredpen's channel, enjoy! ruclips.net/video/nQVh1FUrcTM/видео.html

    • @botondosvath2331
      @botondosvath2331 7 лет назад

      Dr. Peyam's Show Thanks! Haven't you done a video about the 3x3's nullspace too?

    • @drpeyam
      @drpeyam  7 лет назад +1

      Mmmmmh, I may have, but I don't quite remember! There's one on the Jordan Canonical Form on my channel, which may or may not help! :)

    • @botondosvath2331
      @botondosvath2331 7 лет назад

      Dr. Peyam's Show You were not showing the calculation there, because (I suppose) not that was the theme of the video, and the length of the calculation.

    • @drpeyam
      @drpeyam  7 лет назад +1

      Oh, my bad! I'll put it on my to-do list :)

  • @GonzieGr01
    @GonzieGr01 7 лет назад

    Could you do an example where you include the second derivative test into a constrained optimization problem using Lagrange? (Where you need to analyze a bordered Hessian)

    • @drpeyam
      @drpeyam  7 лет назад +1

      Do you mean like Lagrange multipliers?

    • @GonzieGr01
      @GonzieGr01 7 лет назад

      Dr. Peyam's Show Yes. Or a lagrangian

  • @MrRyanroberson1
    @MrRyanroberson1 7 лет назад

    From the very beginning I learned calculus for utilitarian reasons, what is the max or min of a graph? Just take the second derivative! The derivative is the slope, and when the slope is 0 that is a max or min. The slope, you know, is a minimum when the second derivative is negative at that point and a Max if it is positive.

  • @Zonnymaka
    @Zonnymaka 7 лет назад

    Very nice!

  • @Edddfm
    @Edddfm 4 года назад

    It actually reduces to what textbooks say, doesn't it?
    If a function, let's say f, has continous second partial derivatives, the hassian matrix (Hf) is a symmetric matrix (becuase D12f = D21f), so the fact that all of its eigenvalues are positive is equivalent to that fact that all of its principal minors are postive, i.e. D11f > 0 and detHf = (D11fD22f - (D12f)^2) > 0
    The other cases are similar.

    • @drpeyam
      @drpeyam  4 года назад

      It’s a sufficient condition and also it doesn’t give us any insight in the geometry of the hessian

    • @Edddfm
      @Edddfm 4 года назад

      @@drpeyam
      I just wondered why you called it garbage.
      Anyway, nice video.

  • @ArielSasson
    @ArielSasson 6 лет назад

    is it the same det(A-lambda*I) than det(lambda*I-A)?

    • @drpeyam
      @drpeyam  6 лет назад

      Technically not quite because the second one is +/- the first one, but it doesn’t matter because either both of them are zero or not zero.

  • @fracaralho
    @fracaralho 5 лет назад +1

    But... why is the other test garbage?

    • @drpeyam
      @drpeyam  5 лет назад +2

      It’s not a natural test, and a lot of the times it gives an inconclusive answer when there shouldn’t be one. Also it doesn’t generalize to higher dimensions

    • @fracaralho
      @fracaralho 5 лет назад +2

      @@drpeyam Thank you!