Solving for f in f(f(x)) = 2x^2 - 1

In summary, the conversation is about finding the function f(x) when f(f(x)) = 2x^2 - 1. The conversation involves multiple people suggesting different approaches and solutions, including using derivatives, inverse functions, and trigonometric substitutions. The final solution is given as f(x) = cos(sqrt(2)arccos(x)), with some restrictions on the domain. The conversation also includes a humorous suggestion to use imaginary numbers as the solution.
  • #1
Xamfy19
60
0
What is the f, when f(f(x)) = 2x^2 - 1,
Thanks alot!

:confused:
 
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  • #2
looks like homework. hint: take a guess.
 
  • #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
 
  • #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
 
  • #5
Just plug in f[x] for x. There's probably more to it than that, I'm sure.
 
  • #6
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
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
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
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
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
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
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
"cos(2t)=2x^2-1, if sin(t)=x."

try putting t=0 in there: 1=-1 is what you are claiming.
 
  • #14
you are right. Thanks
 
  • #15
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
only if the coefficient of x^2 were zero.
 
  • #17
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
matt grime said:
only if the coefficient of x^2 were zero.
It was definitely not my day! :cry:

-- AI
 
  • #19
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
 

1. What is the purpose of solving for f in f(f(x)) = 2x^2 - 1?

The purpose of solving for f in this equation is to find the function f that would result in the given output of 2x^2 - 1 when applied to an input of x. This is known as a functional equation, where the output of the function is dependent on the input and the function itself.

2. Is there a specific method for solving this type of equation?

Yes, there are several methods for solving functional equations like this one. One common approach is to use substitution, where you replace the f(x) in the equation with another variable, say y, and solve for y. Then, you can substitute y back in for f(x) and solve for f.

3. Can this equation have multiple solutions for f?

Yes, depending on the complexity of the equation, there can be multiple solutions for f. This means that there could be more than one function that satisfies the given equation. It is important to check your solutions to ensure that they satisfy the original equation.

4. What if I can't find a solution for f?

If you are unable to find a solution for f, it could mean that the equation is not solvable or that you may need to use more advanced mathematical techniques to solve it. It is also possible that the equation does not have a unique solution and there are infinitely many possible functions that could satisfy it.

5. How is solving this equation useful in science?

Solving functional equations like this one is useful in many areas of science, such as physics, engineering, and economics. It allows scientists to model and analyze complex systems by representing them as equations and solving for the unknown functions. This can help in understanding and predicting the behavior of these systems.

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