Young's Inequality | A Geometric Proof of Young's Inequality

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  • Опубликовано: 9 ноя 2024

Комментарии • 33

  • @MyOneFiftiethOfADollar
    @MyOneFiftiethOfADollar 2 года назад +4

    I had seen the dry unmotivated algebraic proof of this inequality and it did not really stick for me.
    The inverse function integral interpretation way you proved is easier to grasp. Also connection to AM-GM a nice mnemonic. I liked your “hold your head this way” to be aware of inverse function without really graphing it 😀
    Do you know much about the mathematician Young?

  • @yashvardhan2093
    @yashvardhan2093 3 года назад +2

    Useful in deriving Hölder’s, nice video Prof!!

  • @shrivardhank.2418
    @shrivardhank.2418 2 года назад

    That graphical approach gives a good sense of what that inequality means, This can also be proved by weighted AM GM Inequality too, Thank you sir

  • @AxiomTutor
    @AxiomTutor 2 года назад +1

    Desus got fuck smart.

  • @yashvardhan6521
    @yashvardhan6521 3 года назад +1

    Well this can also be derived directly from weighted am gm with weights 1/p and 1/q which in turn comes from Jensen's.
    Nice alternate proof prof.

  • @annaluiza3592
    @annaluiza3592 3 года назад +1

    Thanks for sharing this! It helped a lot!

  • @zzz-hi3er
    @zzz-hi3er Месяц назад

    Thank you so much

  • @Szynkaa
    @Szynkaa 2 года назад

    very interesting and nice explained :)

  • @yeast4529
    @yeast4529 9 месяцев назад

    Really nice, this way is very easy

  • @yashvardhan6521
    @yashvardhan6521 3 года назад +1

    Well One question I wanted to ask
    Do we study these in higher mathematics as well or are these restricted to olympiads.
    Because whenever I searched for Holders or Minkowski or such inequalities for olympiad purposes, I used to see terms like Metric spaces and stuff which I don't know about.
    Could u illuminate plz.

    • @ProfOmarMath
      @ProfOmarMath  3 года назад +2

      Yes these inequalities are very much related to metric spaces, it’s cool stuff!

  • @ProfOmarMath
    @ProfOmarMath  4 года назад +4

    p and q should be > 1 rather than just positive

    • @charanthiru5761
      @charanthiru5761 4 года назад +2

      Yeah Sir it must be.
      And Sir your videos are too good! You make maths more attractive to me. And this Young's inequality proof was super cool!

    • @ProfOmarMath
      @ProfOmarMath  4 года назад +3

      @@charanthiru5761 Thanks! What are your favorite topics in mathematics?

    • @charanthiru5761
      @charanthiru5761 4 года назад +2

      @@ProfOmarMath Never thought that u would reply to me. Blessing.
      Calculus, Trig, Vector Algebra Sir.

    • @Oliver-cn5xx
      @Oliver-cn5xx 4 года назад

      Hej awesome, I knew something! Thanks for confirming in the comments. I love your stuff btw.

    • @ProfOmarMath
      @ProfOmarMath  4 года назад

      @@Oliver-cn5xx Thanks!

  • @pipilu3055
    @pipilu3055 Год назад

    cool! a picture proof is 1000 times better than a dry proof using AM GM

  • @ailemac9421
    @ailemac9421 3 года назад

    Thank you for the explanation 🤩🤩🤩🤩

  • @terryendicott2939
    @terryendicott2939 3 года назад

    Cool

  • @davie-uz1dl
    @davie-uz1dl 4 года назад

    hello there, thanks for ur video!
    why would you choose the function x^(p-1)? why not x^(p-2) or x^(p-3) or just a random function, containing the exponent p, somehow.
    Greetings!

    • @ProfOmarMath
      @ProfOmarMath  4 года назад +1

      Great question. I think the intuition is that when you integrate x^{p-1} you get a constant times x^p and the inequality in question is related to p-th powers of numbers.

    • @davie-uz1dl
      @davie-uz1dl 4 года назад

      @@ProfOmarMath Thanks a lot Sir, that helped!

    • @mathmaticalproblemandsolution
      @mathmaticalproblemandsolution 2 года назад

      i think if you choose p=1.00001 then x^p-2 or x^p-3 will not be increasing function

    • @violetasuklevska9074
      @violetasuklevska9074 Год назад

      You want the solution to the integral to be x^p/p, so you take its derivative.

  • @mayurjawane9463
    @mayurjawane9463 4 года назад

    Sir why we take here a^p = b^q,

    • @ProfOmarMath
      @ProfOmarMath  4 года назад +1

      If I'm understanding correctly, equality occurs when b=f(a) and they gives a^p=b^q

  • @imenlakrout6982
    @imenlakrout6982 3 года назад

    Why a , b 》0

    • @ProfOmarMath
      @ProfOmarMath  3 года назад

      Gets funny with negatives if p and q are odd