Composite function

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  • #1
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What is the f, when f(f(x)) = 2x^2 - 1,
Thanks alot!!!

:confused:
 

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  • #2
matt grime
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looks like homework. hint: take a guess.
 
  • #3
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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
 
  • #4
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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
 
  • #5
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Just plug in f[x] for x. There's probably more to it than that, I'm sure.
 
  • #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).
 
  • #7
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Fredrik said:
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).
Hmm why wouldn't it work?? :confused:

-- AI
 
  • #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:
  • #9
Fredrik
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TenaliRaman said:
Hmm why wouldn't it work?? :confused:

-- AI
If you include an x^2 term in f(x) you get an x^4 term in f(f(x)).
 
  • #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.
 
  • #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.
 
  • #12
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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?
 
  • #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.
 
  • #14
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you are right. Thanks
 
  • #15
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Fredrik said:
If you include an x^2 term in f(x) you get an x^4 term in f(f(x)).
the coefficient of x^4 could be zero ....

-- AI
 
  • #16
matt grime
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only if the coefficient of x^2 were zero.
 
  • #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]
 
  • #18
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matt grime said:
only if the coefficient of x^2 were zero.
It was definitely not my day! :cry:

-- AI
 
  • #19
Alkatran
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Xamfy19 said:
What is the f, when f(f(x)) = 2x^2 - 1,
Thanks alot!!!

:confused:
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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