# Composite function

What is the f, when f(f(x)) = 2x^2 - 1,
Thanks alot!!! matt grime
Homework Helper
looks like homework. hint: take a guess.

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

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

Just plug in f[x] for x. There's probably more to it than that, I'm sure.

Fredrik
Staff Emeritus
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

$$f(x)=2^\frac{1}{\sqrt 2 +1}x^\sqrt 2$$

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

$$f(f(x))=2x^2$$

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).

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?? -- AI

matt grime
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:
Fredrik
Staff Emeritus
Gold Member
TenaliRaman said:
Hmm why wouldn't it work?? -- AI
If you include an x^2 term in f(x) you get an x^4 term in f(f(x)).

Fredrik
Staff Emeritus
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.

matt grime
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.

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?

matt grime
Homework Helper
"cos(2t)=2x^2-1, if sin(t)=x."

try putting t=0 in there: 1=-1 is what you are claiming.

you are right. Thanks

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

matt grime
Homework Helper
only if the coefficient of x^2 were zero.

Fredrik
Staff Emeritus
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

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

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

Alkatran
Homework Helper
Xamfy19 said:
What is the f, when f(f(x)) = 2x^2 - 1,
Thanks alot!!! sqr(2)x+j

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

Yay for cheating! 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