f-R-R-f-f-x-x-2-x-1-f-x-Altered-Question-217541-

Question Number 217579 by Rasheed.Sindhi last updated on 16/Mar/25

f : R \to R

f (f (x)) = x^{2} - x + 1

f (x) = ?

You can't use 'macro parameter character #' in math mode

Commented by Ghisom last updated on 16/Mar/25

I thought of this, there might be no closed

term for f (x)

because :

g (g (x)) = x^{2} \Rightarrow g (x) = x^{\sqrt{2}}

[not possible for x < 0]

but if we try to add terms :

g (x) = x^{\sqrt{2}} + c \Rightarrow g (g (x)) = {(x^{\sqrt{2}} + c)}^{\sqrt{2}} + c

g (x) = x^{\sqrt{2}} + h (x) \Rightarrow weirdness

Commented by mr W last updated on 16/Mar/25

w i t h g i v e n c o n d i t i o n w e c a n f i n d

f (0), f (1), b u t w e c a n e v e n n o t f i n d

f (2) e t c ., n o t t o m e n t i o n f (x)

g e n e r a l l y . i t h i n k .

Commented by Rasheed.Sindhi last updated on 16/Mar/25

T han k s sirs!

Commented by Ghisom last updated on 17/Mar/25

just to explore some problems :

g (x) = {a x}^{2} + b x + c

G (x) = {A x}^{4} + {B x}^{3} + {C x}^{2} + D x + E = g (g (x)) =

A = a^{3}

B = 2 a^{2} b

C = a (2 a c + b (b + 1))

D = (2 a c + b) b

E = (a c + b + 1) c

this cannot be solved for every given quintuplet

A, B, C, D, E

even if we want G (x) = x^{4} + {C x}^{2} + E

\Rightarrow a = 1 \land b = 0

g (x) = x^{2} + c

G (x) = x^{4} + 2 {c x}^{2} + c (c + 1)

only very few are possible \dots

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