# Composite function

1. Oct 13, 2004

### Xamfy19

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

2. Oct 13, 2004

### matt grime

looks like homework. hint: take a guess.

3. Oct 13, 2004

### Xamfy19

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. Oct 13, 2004

### TenaliRaman

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. Oct 13, 2004

### Chrono

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

6. Oct 13, 2004

### Fredrik

Staff Emeritus
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).

7. Oct 14, 2004

### TenaliRaman

Hmm why wouldn't it work??

-- AI

8. Oct 14, 2004

### matt grime

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
9. Oct 14, 2004

### Fredrik

Staff Emeritus
If you include an x^2 term in f(x) you get an x^4 term in f(f(x)).

10. Oct 14, 2004

### Fredrik

Staff Emeritus
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. Oct 14, 2004

### matt grime

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. Oct 14, 2004

### Xamfy19

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. Oct 14, 2004

### matt grime

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

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

14. Oct 14, 2004

### Xamfy19

you are right. Thanks

15. Oct 14, 2004

### TenaliRaman

the coefficient of x^4 could be zero ....

-- AI

16. Oct 14, 2004

### matt grime

only if the coefficient of x^2 were zero.

17. Oct 14, 2004

### Fredrik

Staff Emeritus
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

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

18. Oct 14, 2004

### TenaliRaman

It was definitely not my day!

-- AI

19. Oct 14, 2004

### Alkatran

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