I Solved A Homemade Functional Equation in Two Ways

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  • Опубліковано 20 лис 2023
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КОМЕНТАРІ • 41

  • @hmkl6813
    @hmkl6813 7 місяців тому +39

    Bruh just plug it in

    • @khengari77
      @khengari77 7 місяців тому +6

      That's what I was thinking of
      😂😂😂😂😂

    • @adityajayant8277
      @adityajayant8277 7 місяців тому

      Seriously man, bruh moment

  • @anoshgaming9924
    @anoshgaming9924 7 місяців тому +17

    3
    Seriously that was so easy

  • @adamrussell658
    @adamrussell658 7 місяців тому +7

    That 2nd method is so simple it feels like cheating!

  • @scottleung9587
    @scottleung9587 7 місяців тому +9

    The second method was pretty obvious.

  • @angelishify
    @angelishify 7 місяців тому +7

    Actually there is no need for f(x) to be linear; it could be anything. For instance, f(x) = √(x + 1) and then g(x) = x² − x − 1.

    • @pokemonjourneysfan5925
      @pokemonjourneysfan5925 7 місяців тому

      Q: Solve this functional equation. f(x)+f(1/x)=1, for x=/=0, x=/=1. The craziest part, is that I know of a solution, which is a rational function in x, but Idk how you would generally solve this problem. SyberMath has made videos on similar problems, but this one in particular I am stuck.

    • @MrGeorge1896
      @MrGeorge1896 Місяць тому

      @@pokemonjourneysfan5925 There are two restrictions: f(1) = 1/2 and some function of x which gives the negative value if applied to x ^ -1. This implies a logarithmic function so f(x) = 1/2 + C * ln(x).

    • @pokemonjourneysfan5925
      @pokemonjourneysfan5925 Місяць тому

      @@MrGeorge1896 Not exactly. And second, I assumed the function is a rational function of the form (ax+b)/(cx+d) for constants a, b, c and d.

    • @MrGeorge1896
      @MrGeorge1896 Місяць тому

      @@pokemonjourneysfan5925 Yeah, you are right, such solutions exist. I found two different solutions:
      f(x) = (ax + b) / [(a - b) (x - 1)] with a ≠ b and not both a = 0 and b = 0
      f(x) = (ax + b) / [(a + b) (x + 1)] with a ≠ -b and not both a = 0 and b = 0

  • @sunny_25
    @sunny_25 7 місяців тому +7

    Bruh...just put x=3 in the question..seriously that was easy...

    • @rickdesper
      @rickdesper 7 місяців тому +2

      Seriously. How is this a 9-minute video?

  • @LeDevBreton
    @LeDevBreton 7 місяців тому +4

    uh... 3 ? less than 5 seconds

  • @charlesgodswill6161
    @charlesgodswill6161 7 місяців тому

    Very well thought out 🥂🥂

  • @franciscolindume6361
    @franciscolindume6361 7 місяців тому

    Estupendo!

  • @furno_2761
    @furno_2761 7 місяців тому +1

    since nothing is known about g(x), its impossible to solve this problem. for all we know f(x) might not be have multiple values for some x, which im pretty sure makes it incorrectly defined.

    • @SyberMath
      @SyberMath  7 місяців тому

      we do have some info on g(x)

    • @furno_2761
      @furno_2761 7 місяців тому

      you're right, i didnt specify my complaint correctly. What i see as problematic, is that we dont specify anywhere that g(x) + x is injective, which would mean that defining the f function in this way might cause some problems. I do realise that this is somewhat petty for lack of a better word, but i believe that it's a important thing to state.

  • @tharunsankar4926
    @tharunsankar4926 7 місяців тому +1

    Sybermath: I have a challenge:
    Here is a system of equations:
    g(f(x+a) - f(x)) = x
    g(f(x)) -x/(x+1)g(f(x+1)) = 0
    Find what f(x) and g(x) is (class of solutions is okay as well) Bros please bump this comment!

  • @rakenzarnsworld2
    @rakenzarnsworld2 7 місяців тому

    Answer: 3

  • @mcwulf25
    @mcwulf25 7 місяців тому

    Surely just plug in x=3 as that's the value given.

  • @alextang4688
    @alextang4688 7 місяців тому

    The question should be revised to make that we can find the c. Problem will be more interesting. 😉😉😉😉😉😉

  • @yakupbuyankara5903
    @yakupbuyankara5903 7 місяців тому

    3

  • @monkeblazer3154
    @monkeblazer3154 7 місяців тому

    i have a doubt , can the second method be considered valid in math exams where problems are subjective and u needto give steps

    • @SyberMath
      @SyberMath  7 місяців тому

      In general, it should be ok but of course it depends on the professor (based on how picky and annoying they are)

    • @monkeblazer3154
      @monkeblazer3154 7 місяців тому

      i see tysm sir @@SyberMath

  • @user-fq4gu3gm9z
    @user-fq4gu3gm9z 7 місяців тому +3

    x=3 g(x)+x=5+3=8 so f(8)=3

  • @honestadministrator
    @honestadministrator 7 місяців тому +2

    f( 8 ) = f ( g(3) + 3) = 3

  • @szirsp
    @szirsp 7 місяців тому

    We know g(3)=5, so how does x=3 play out:
    f(g(3)+3)=3
    f(5+3)=3
    f(8)=3. We have the answer to the question.
    I don't know why this video is so long. No assumption of linearity required.

  • @easymoney1226
    @easymoney1226 7 місяців тому

    bro just x=3

  • @juicedatom
    @juicedatom 7 місяців тому

    So another interesting take could be to solve for f and g, but without the linear assumption.
    1] Use method 2 in the video to get f(8) = 3 like before.
    2] Given this, prove that g must be a constant function. This can be done by creating a very simple system of equations. Using the knowledge that f(8) = 3, we know that when g(x) + x = 8 then x = 3. Subtracting these gives us that g(x) = 5 for all x.
    3] By plugging in 5 for g(x) into the original equation we get another functional equation, f(5 + x) = x.
    4] Setting x = t - 5 the above equation becomes f(5 + t - 5) = t - 5, which simplifies to f(t) = t - 5.
    5] Finally, change of variables gives us f(x) = x - 5

    • @furno_2761
      @furno_2761 7 місяців тому

      thats incorrect.
      we dont get g(x) = 5 for all x. we only know that g(x) = 5 when x=3.
      What you did is exuivilant(pardon spelling) to saying that since x+2+x=8 when x=3, then x+2 =5 for all x.

  • @MrAmirabbass
    @MrAmirabbass 6 місяців тому

    Are you kidding me? Just put x=3 on both sides of the equation and obtain: f(8)=3

    • @SyberMath
      @SyberMath  6 місяців тому

      sure thing! no kidding 😁

  • @GreenMeansGOF
    @GreenMeansGOF 7 місяців тому +1

    Method 1: Long, complicated, uses an assumption that may or may not be true(f can be any invertible function)
    Method 2: Just plug in dummies😜