根函數討論的總結， 請不吝指正

（為方便起見，用英文敘述）

下麵是對 g(x)=f(f(x)) = x^2-x+1 求 f(1), f(0) 一題以及由其引發之討論，本著從特殊到一般的思路之總結，請KDE235大師以及各位大師過目指正，總結得是否到位和正確, 尤其是結論6。

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For x belong to domain X and for a given single valued function g(x) on G (X to G is an injective mapping), define f(x) so that f(f(x))=g(x); 

Name such defined f(x) as the square root function of g(x), then the following statements are true:

1. f(x) exists 

2. f(x) is a single valued function of x

3. f(g(x))=g(f(x)) 

4. If the set for all possible values of f(x) is F,  then F <=G

5. Both g(x) and f(x) have their own inverse functions.

6. For a given pair of x_p and x_q (>x_p) in X,

then the following sequence forms an equivalent class of [x_p, x_q] on domain X:

                          a(1)=g(x_p); a(2)=g(x_q)

                          a(i+2)=g(a(i)) for i =1 to infinity      (1)

Then the below defined function f(x) is the square root function of g(x), designate as g^1/2 (x)

                            f(x) = g^(1/2) (x) such that f(aj)=a(j+1)   (a(i) defined in (1) for j=1 to infinity)