f-f-x-x-2-x-1-f-x-Modification-of-Q-1147-

Question Number 1229 by Rasheed Soomro last updated on 16/Jul/15

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

f (x) = ?

(M o d i f i c a t i o n o f Q 1147)

Commented by 123456 last updated on 17/Jul/15

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

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

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

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

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

f (x) (f (x) - 1) = f (1 - x) (f (1 - x) - 1)

Commented by Rasheed Ahmad last updated on 18/Jul/15

D e a r 123456

I u n d e r s t o o d y o u r l o g i c w i t h

s o m e d i f f i c u l t y . A c t u a l l y y o u r

e x p l a n a t i o n l a c k s i n - b e t w e e n i n s t r u c t i o n s

b a d l y . P l p u t s o m e i n s t r u c t i o n s

f o r c l a r i f i c a t i o n a n d a l s o i n s e r t

s o m e m o r e s t e p s i n o r d e r t o

t o b e u n d e r s t o o d e a s i l y .

A n y w a y n o w a f t e r u n d e r s t a n d i n g

I a p p r e c i a t e y o u r l o g i c .

Commented by Rasheed Ahmad last updated on 17/Jul/15

V e r y g o o d r u l e a n d v e r y n i c e

f e a t u r e y o u h a v e d i s c o v e r e d :

R u l e : d (f ∙ f) = {[d (f)]}^{2}

f e a t u r e : f (f (x)) = f (f (1 - x))

A p p r e c i a t i o n f o r y o u!

Commented by Rasheed Ahmad last updated on 17/Jul/15

I t c a n b e p r o v e d t h a t f (x) i s n o t a

p o l y n o m i a l .

Commented by 123456 last updated on 17/Jul/15

yes, because if d (f) is the degree of f

then the degree of f ∙ f is d (f ∙ f) = {[d (f)]}^{2}

then

{[d (f)]}^{2} = 2 \Rightarrow d (f) = \sqrt{2}

wich is impossible

however its have some symetry, you easy

can see that

f (f (x)) = f (f (1 - x))

Commented by 123456 last updated on 19/Jul/15

f (0) = 1 (from Q 1147)

f (1) = 1 (from Q 1147)

claim 1 : f (x) = 1 only if x = 0 or x = 1

proof :

suppose there are some x \notin {0, 1} such that f (x) = 1, then

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

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

1 = x^{2} - x + 1

x^{2} - x = 0

x (x - 1) = 0

x = 0 \lor x = 1

contradiction with the fact that x \notin {0, 1}

claim 2 : f (0) = 1

suppose that f (0) = a with a \notin {0, 1}

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

f (f (0)) = 1 (x = 0)

f (a) = 1

contradction with claim 1 .

but a \neq 0, because if a = 0, then

f (f (0)) = 1

f (0) = 1

contradiction with f (0) = a = 0

claim 3 : ∄ x, f (x) = 0

suppose that there is some x that f (x) = 0, then

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

f (0) = x^{2} - x + 1

by claim 2

1 = x^{2} - x + 1

x = 0 \lor x = 1

contradction with f (0) = f (1) = 1

claim 4 : f (x) = x only for x = 1

suppose that f (x) = x for some x \neq 1, then

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

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

x = x^{2} - x + 1

x^{2} - 2 x + 1 = 0

{(x - 1)}^{2} = 0

x = 1

wich contradict the fact that x \neq 1

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x^{2} - x + 1 = 0

Δ = {(- 1)}^{2} - 4 (1) (1) = 1 - 4 = - 3

x = \frac{- (- 1) \pm \sqrt{- 3}}{2 (1)} = \frac{1 \pm i \sqrt{3}}{2} = e^{\pm i π / 3}

- - - - - - - - - - - - - - - -

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

f (x) = y

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

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

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

f (y) = z

f (z) = y^{2} - y + 1

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f (f (1 / 2)) = 3 / 4

f (3 / 4) = {(f (1 / 2))}^{2} - f (1 / 2) + 1

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