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

Thanks alot!!!

Thanks alot!!!

- Thread starter Xamfy19
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What is the f, when f(f(x)) = 2x^2 - 1,

Thanks alot!!!

Thanks alot!!!

- #2

matt grime

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looks like homework. hint: take a guess.

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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

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

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Fredrik

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[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).

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Hmm why wouldn't it work??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).

-- AI

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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.

I can solve it for a restricted domain.

Standard maths gripe:

A function needs its domain and range specifying.

Last edited:

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Fredrik

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If you include an x^2 term in f(x) you get an x^4 term in f(f(x)).TenaliRaman said:Hmm why wouldn't it work??

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Fredrik

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matt grime

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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.

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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?

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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.

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

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you are right. Thanks

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the coefficient of x^4 could be zero ....Fredrik said:If you include an x^2 term in f(x) you get an x^4 term in f(f(x)).

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matt grime

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only if the coefficient of x^2 were zero.

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Fredrik

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[tex]f(x)=\cos(\sqrt 2 \arccos x)[/tex]

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

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Alkatran

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sqr(2)x+jXamfy19 said:What is the f, when f(f(x)) = 2x^2 - 1,

Thanks alot!!!

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

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