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Given $f : \mathbb{R} \to \mathbb{R}$ defined as $$f(f(x))=x^2-x+1$$ Find value of $f(0)$

I assumed $g(x)=f(f(x))$

we gave

$$g(x)=(x-1)^2+(x-1)+1$$

Also $$g(x-1)=(x-1)^2-(x-1)+1$$

subtracting both we get

$$g(x)-g(x-1)=2x-2$$

i have no clue from here any hint will suffice

daw
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Ekaveera Gouribhatla
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2 Answers2

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We have that $$f(f(0))=1$$ Then $$f(1)=f(f(f(0)))=f(0)^2-f(0)+1$$ On the other hand, $$f(f(1))=1$$ and hence, $$f(1)=f(f(f(1)))=f(1)^2-f(1)+1$$ And so we get that $f(1)=1$. Then $$f(0)^2-f(0)=0$$ So, $f(0)$ is $0$ or $1$.

Following a comment by @Calvin, $f(0)=0$ must be rejected.

ajotatxe
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    Note that $f(0)$ cannot be 0, so your solution is incomplete. To complete it, observe that if $ f(0) = 0$, then we get $ 0 = f(0) = f(f(0)) = f(0)^2 - f(0) + 1 = 1 $. Hence, $f(0) = 1 $. – Calvin Lin Aug 27 '19 at 02:22
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Notice that $0^2-0+1=1^2-1+1=1$. Then if $f(0)=f(1)=1$, we indeed have $f(f(0))=1=f(f(1))$.

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    in case of interest: owing to the fixpoint having slope $1,$ there is a method, due to Ecalle, for constructing $C^\infty$ $f(x)$ for $x > 1/2,$ where $f$ is real analytic except for $x=1$ itself. The critical point at $x=1/2$ becomes a boundary because we need to use the inverse function of $x^2 - x + 1$ to get an attracting point. – Will Jagy Dec 16 '17 at 19:13
  • For completeness, you need to show that $f(0)$ can take no other value. – Calvin Lin Aug 27 '19 at 02:27