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Composite function

  1. Oct 13, 2004 #1
    What is the f, when f(f(x)) = 2x^2 - 1,
    Thanks alot!!!

    :confused:
     
  2. jcsd
  3. Oct 13, 2004 #2

    matt grime

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    looks like homework. hint: take a guess.
     
  4. Oct 13, 2004 #3
    I have tried different answers, none of them working. I even tried to use derivatives or inverse functions, still no correct answer came out from my guessing
     
  5. Oct 13, 2004 #4
    matt prolly has some really ingenious trick up his sleeve!

    what i have is pretty ugly!!

    assume f(x) = ax^2+bx+c
    f(f(x)) = a(ax^2+bx+c)^2 + b(ax^2+bx+c) + c
    simplify this...
    calculate f(f(x)) for any three values of x say -1,0,1
    and find corresponding values of a,b,c

    -- AI
     
  6. Oct 13, 2004 #5
    Just plug in f[x] for x. There's probably more to it than that, I'm sure.
     
  7. Oct 13, 2004 #6

    Fredrik

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    Wow, this problem is much more difficult than it looks. I've been trying for almost an hour and I haven't figured it out yet. The closest I got was

    [tex]f(x)=2^\frac{1}{\sqrt 2 +1}x^\sqrt 2[/tex]

    but that's not the correct answer, since this f would satisfy

    [tex]f(f(x))=2x^2[/tex]

    What TenaliRaman suggested doesn't work. f is definitely not a polynomial. (Try f(x)=x and f(x)=x² and you'll see the problems immediately).
     
  8. Oct 14, 2004 #7
    Hmm why wouldn't it work?? :confused:

    -- AI
     
  9. Oct 14, 2004 #8

    matt grime

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    Nope, no ingenious trick springs to my mind. I thought I saw a non-ingenious one but it wasn't right.

    I can solve it for a restricted domain.

    Standard maths gripe:
    A function needs its domain and range specifying.
     
    Last edited: Oct 14, 2004
  10. Oct 14, 2004 #9

    Fredrik

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    If you include an x^2 term in f(x) you get an x^4 term in f(f(x)).
     
  11. Oct 14, 2004 #10

    Fredrik

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    Does this problem really have a solution? I don't see how it could. If anyone figures this out (either finds the correct f, or proves that the problem has no solution), I would appreciate if they could post the solution here.
     
  12. Oct 14, 2004 #11

    matt grime

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    Here's a 'solution', haven't thought about it too closely to be honest:

    let cosy = x

    then f^2(cosy) = cos2y

    so the function f(cosy) = cos(sqrt(2)y) will do, since cos(y)=x we have

    f(x) = cos(sqrt(2)arccos(x))

    Note that unless one wants to pass to complex arguments then one needs to restrict the domain a lot.
     
  13. Oct 14, 2004 #12
    I just tried using trig to substitue into the composite function. However, I need more help. The following was my thought:

    Since f(f(x))=2x^2-1, cos(2t)=2x^2-1, if sin(t)=x.

    If I can find how to derive to f(f(x))=cos(2t) from f(x)=?, then it may be solved. Any suggestion?
     
  14. Oct 14, 2004 #13

    matt grime

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    "cos(2t)=2x^2-1, if sin(t)=x."

    try putting t=0 in there: 1=-1 is what you are claiming.
     
  15. Oct 14, 2004 #14
    you are right. Thanks
     
  16. Oct 14, 2004 #15
    the coefficient of x^4 could be zero ....

    -- AI
     
  17. Oct 14, 2004 #16

    matt grime

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    only if the coefficient of x^2 were zero.
     
  18. Oct 14, 2004 #17

    Fredrik

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    Funny...I also tried the substitution x=cos t yesterday, but I got stuck at f(f(cos t))=cos 2t. I didn't see that it was easy to continue from there. I agree that the correct answer is

    [tex]f(x)=\cos(\sqrt 2 \arccos x)[/tex]
     
  19. Oct 14, 2004 #18
    It was definitely not my day! :cry:

    -- AI
     
  20. Oct 14, 2004 #19

    Alkatran

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    sqr(2)x+j

    j is special:
    j*j = -1, j*x = 0, j <> 1*j

    Yay for cheating! :rolleyes:

    Or....
    f(x):

    f(x) = r(x+i)
    I returns the imaginary part of a number, R returns the real part
    if I(f(x)) > 1 then f(x) = R(f(x))

    Function not defined when x = .5
     
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