根函数讨论的总结

下面是对 g(x)=f(f(x)) = x^2-x+1 求 f(1), f(0) 一题以及由其引发之讨论，本着从特殊到一般的思路之总结。主要思路和方法源自KDE235大师.

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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)

 

以上方法是对kde235大师方法的推广。kde235大师原文在此：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.

 

特别指出，6，7中的方法实质是函数的迭代。迭代是解决所有非线性问题最基本的方法。迭代还是混沌现象的形成机理。

 

练习：

Find f(x) so that f(f(x))=g(x) :

1 g(x)=x  求 f(x)

2. g(x)=1/x (x不等于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)