1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    "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

    User Avatar
    Science Advisor
    Homework Helper

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

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Composite function
  1. Composite functions (Replies: 3)

Loading...