A Homemade Functional Equation | Math Olympiads

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  • Опубликовано: 7 ноя 2024
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Комментарии • 22

  • @browhat6935
    @browhat6935 6 месяцев назад +2

    Absolute legend. Never stop making these ❤

  • @bugsfudd8295
    @bugsfudd8295 6 месяцев назад +2

    Man, this is like your third video today. Wish i had your energy

    • @yoav613
      @yoav613 6 месяцев назад

      And his passion too.

    • @SyberMath
      @SyberMath  6 месяцев назад

      Hehe thanks 😍

    • @SyberMath
      @SyberMath  6 месяцев назад

      😍

  • @MathsScienceandHinduism
    @MathsScienceandHinduism 6 месяцев назад +1

    I just wrote x/3+2/x as (x^2+6)/3x then expressed Right hand side as (x^4+36)/18x^2 and then used formula for (a+b)^2 to simplify and replaced (x^2+6)/3x by z then finally replaced z by x

  • @florianbuerzle2703
    @florianbuerzle2703 6 месяцев назад +3

    3rd 😇 See that x²/18 + 2/x² is "almost" (x/3 + 2/x)² so try to rewrite the rhs into a form that contains (x/3 + 2/x)².
    Like so: x²/18 + 2/x² = ( x²/9 + 4/x² ) / 2 = ( x²/9 + 4/x² + 2∙2/x∙x/3 - 2∙2/x∙x/3 ) / 2 = ( x/3 + 2/x )² / 2 - 2/3.
    So we have f( x/3 + 2/x ) = ( x/3 + 2/x )² / 2 - 2/3. Now we can finally re-label the argument and get f(x) = x²/2 - 2/3 😊

  • @scottleung9587
    @scottleung9587 6 месяцев назад

    Nice!

  • @ThAlEdison
    @ThAlEdison 6 месяцев назад

    The original expression has a reduced domain, |x/3+2/x|>=2sqrt(2/3)
    you can see that a few ways
    f(x/3+2/x)=(1/3)(x^2/6+6/x^2)=(2/3)cosh(ln(x^2/6))
    cosh(real)>=1
    so f(x/3+2/x)>=2/3

    • @tixanthrope
      @tixanthrope 6 месяцев назад

      Yup!
      This solution is valid only within (x/2 + 3/x)'s range:
      prnt.sc/65opaMmhKgJT
      Namely, any function f: R -> R that can be restrictable to the said solution within (x/2 + 3/x)'s range is a solution too.
      For example, f(0) may take any value as the original condition does not apply to zero.

    • @SirEdwardMadeupsurname
      @SirEdwardMadeupsurname 6 месяцев назад

      Exactly! There are actually an infinite number of functions satisfying the given property, unless f's domain is restricted to (x/3 + 2/x)'s range.

  • @weylguy
    @weylguy 6 месяцев назад

    Sybermath's second method is MUCH easier and quicker.

  • @ManjulaMathew-wb3zn
    @ManjulaMathew-wb3zn 6 месяцев назад

    Let u =x/3 and v=2/x then uv = 2/3
    Substituting f( u + v) = ( u^2 + v ^2)/2
    = ((u+v)^2)/2 -uv = ((u+v)^2)/2 - 2/3
    Replace u+v with x then f(x)=(x^2)/2 - 2/3

  • @Aditya_196
    @Aditya_196 6 месяцев назад

    Ok I guessed it will be x²/2 but don't know how tf constant came I will watch now
    Got it silly me forgot about the 2ab term after squaring

  • @Blabla0124
    @Blabla0124 6 месяцев назад

    Hold on: x/3 + 2/x is not 'onto'.If |t| < sqrt(6)/3 + 2/sqrt(6), f(t) is not defined. Your analysis holds only for |t| >= sqrt(6)/3 + 2/sqrt(6)
    F.i.; f(0) is not defined. Which x do I have to chose such that x/3 + 2/x = 0?

    • @ManjulaMathew-wb3zn
      @ManjulaMathew-wb3zn 6 месяцев назад

      I would love to see the responses to this.

    • @SirEdwardMadeupsurname
      @SirEdwardMadeupsurname 6 месяцев назад

      Exactly! There are actually an infinite number of functions satisfying the given property, unless f's domain is restricted to (x/3 + 2/x)'s range.

    • @YouTube_username_not_found
      @YouTube_username_not_found 6 месяцев назад

      >> "f(0) is not defined"
      Rather you should say _f(0) is unconstrained it could be anything_

    • @Blabla0124
      @Blabla0124 6 месяцев назад +1

      @@RUclips_username_not_found true, but im not writing a paper for the ams here ;-)

  • @Famoke
    @Famoke 6 месяцев назад

    I just solved this in like 10 seconds and you did a 10 minutes video on this??
    Let u = x/3 + 2/x then (u^2)/2 = (x^2)/18 + 2/(x^2) + 2/3
    So f(u) = (u^2)/2 - 2/3. Done

    • @tixanthrope
      @tixanthrope 6 месяцев назад

      congratulations! this channel is mainly for mortals :D

    • @SyberMath
      @SyberMath  6 месяцев назад

      😄