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。主要思路和方法源自KDE235大師.
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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. Please note here the square root is regarding the functional relationship such that g=f^2 or f =g^(1/2), not regarding the square root of the numerical value of g(x).
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. For a given single valued g(x), all functions satisfying f(f(x))=g(x) form an equivalent class {f(x)}
i.e if both p(p(x))=g(x) and q(q(x))=g(x), the p~q
This implies two corollaries:
4.1 A given g(x) might have multiple square root functions
4.2 If e(x) not equal to g(x), then their square root functions will be different as well.
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)
以上方法是對kde235大師方法的推廣。kde235大師原文在此：https://bbs.wenxuecity.com/netiq/139381.html
7. Along the same vein, for h(x)=f(f(f(x))=f^3(x), if h(x) is single valued on domain X, then the
following sequence forms an equivalent class of [x_p, x_q, x_r] on domain X:
b(1)=h(x_p); b(2)=h(x_q); b(3)=h(x_r)
b(i+3)=h(b(i)) for i =1 to infinity (2)
Then the below defined function f(x) is the cubic root function of h(x), designate as h^1/3 (x)
f(x) = h^(1/3) (x) such that f(bj)=b(j+1) (b(i) defined in (2) for j=1 to infinity)
8. The higher order of root functions can be defined in a similar manner.
特別指出，6，7中的方法實質是函數的迭代。迭代是解決所有非線性問題最基本的方法。迭代還是混沌現象的形成機理。
練習：
Find f(x) so that f(f(x))=g(x) :
1 g(x)=x 求 f(x)
2. g(x)=1/x (x不等於0）， 求f(x)
3. g(x)= x^p 求 f(x）
4. Prove that x^-p and x^p belong to the same equivalent class for a given g(x), and find such g(x)
5。 若X is between 0 and pi，g(x)=f(f(x))=cos(x); 求 f(0), f(1), f(pi)
6. Let g(x) = x^2+x-1, 求 f(0）and f(1)
