# Discrete Algebraic Equations and Discrete Operator Equations(Presentations
for ICM 2010)

Wu Zi qian  Fangda group company,Shenzhen city,Guangdong province,China
$runton_{-}runton$$<EMAIL_ADDRESS>
###### Abstract.

We give constructive results of Hilbert’s 13th problem for discrete functions.
By them we give formula solution expressed by a superposition of functions of
one variable to equations constructed by discrete functions and equations with
parameterized discrete functions. Further more we give formula solution
expressed by a superposition of operators of one variable to equations
constructed by discrete operators and equations with parameterized discrete
operators. This is a Short communication, Section 9,Functional Analysis and
Application, Saturday, August 21, 2010,18:00-18:15, Room No. T3.

###### Key words and phrases:

Discrete function,commutation operator,tension-compression
operator,superposition operator,decomposition operator,discrete operator,high
operator

##  1\. Introduction

Problems about equations are very important and difficult. Solving quadratic
equation and cubic equation and quartic equation had cost the mathematicians
in history a great deal of time.

Babylonians solve quadratics in radicals in 2000 BC. Cubic equation and
quartic equation were solved by Italian mathematicians
Girolamo.Cardano(1501-1576) and Ludovico.Ferrari(1522-1565)in 16th
century,respectively.

But mathematicians met big troubles when they tried to solve quintic equation.
Leonhard.Euler(1707-783) believed quintic equation can be changed to a quartic
equation by transformation of variable. Niels.Henrik.Abel (1802-1829) got a
conclusion that there is no solution by radicals for a general polynomial
algebraic equation if n$\geq$5\. Evariste.Galois (1811-1832) built group
theory and got the same conclusion. His method come down to now and can be
found in any textbook about Galois group theory.

There is no solution by radicals. Are there any solutions of other forms such
as numerical solution and solution expressed in function of two variables or
of many variables or solution expressed in series or in integral?

We do not discuss numerical solutions because they belong to applied
mathematics. We prefer formula solution expressed in binary function to other
ones. What is a formula solution expressed in binary function? It contains
only function of two variables. We can give a expression of a alone binary
function at the beginning. We can replace any one of variables by a binary
function then we get a new expression. We can replace any one of variables of
this new expression by a binary function again and get a more complex
expression. We can repeat the procedure for any finite times. But it is not
easy to get solution expressed in binary function. It’s easier to get
solutions of other forms.History developed just like this.

Camille.Jordan (1838-1922) shows that algebraic equations of any degree can be
solved in terms of modular functions in 1870. Ferdinand.von.Lindemann (1852-
1939) expresses the roots of an arbitrary polynomial in terms of theta
functions in 1892. In 1895 Emory.McClintock (1840-1916) gives series solutions
for all the roots of a polynomial. Robert Hjalmal.Mellin (1854-1933) solves an
arbitrary polynomial equation with Mellin integrals in 1915. In 1925
R.Birkeland shows that the roots of an algebraic equation can be expressed
using hypergeometric functions in several variables. Hiroshi.Umemura expresses
the roots of an arbitrary polynomial through elliptic Siegel functions
in1984[1].

All of solutions mentioned above are not ones expressed in binary function.

By Tschirnhausen transformation a quintic equation or a sextic equation can be
changed to ones containing only two parameters so there are solutions
expressed in binary function for them. David.Hilbert presumed that there is no
solution expressed in binary function for polynomial equations of n when
n$\geq$7 and wrote his doubt into his famous 23 problems as the 13th one[2].

Hilbert published his last mathematical paper [3] in 1927 where he reported on
the status of his problems, he devoted 5 pages to the 13th problem and only 3
pages to the remaining 22 problems. We can see that so much attention Hilbert
paid to 13th problem. In 1957 V.I.Arnol’d proved that every continuing
function of many variables can be represented as a superposition of functions
of two variables and refuted Hilbert conjecture[4][5]. Furthermore,
A.N.Kolmogorov proved that every continuous function of several variables can
be represented as a superposition of continuous functions of one variable and
the operation of addition [6].

Result for Hilbert’s 13th problem is very important for us and it points us a
quite right direction to solve polynomial equations and general algebraic
equations. But method used in it is topological and the result is not a
constructive one. In this paper we will give a constructive result in discrete
situation. This result is very important. We can construct profuse discrete
algebraic equations and discrete operator equations and for this result we can
give any of them a formula solution. There is never such a mathematical
structure in the history of mathematics. This is the first time! A.G.Vitushkin
dissatisfies the current results about 13th problem and points out that the
algebraic core of the problem remains untouched[7]. We believe we have gotten
the algebraic core Vitushkin wanted.

##  2\. Constructive results for Hilbert’s 13th problem

A.N.Kolmogorov expresses function of several variables as a superposition of
functions of one variable like this:

(2.1) $\displaystyle
W(x_{1},x_{2}\cdots,x_{n})=\sum_{i=1}^{2n+1}f_{i}\Big{[}g_{i1}(x_{1})+g_{i2}(x_{2})+\cdots+g_{in}(x_{n})\Big{]}$

This is a existence result but it’s easy to give a constructive result for
discrete functions.

Definition 2.1 Let A={-1,0,1}, a three numbers function of M variables is
defined as:

g:$A^{M}$$\longrightarrow$A

There are 9 discrete points for a binary three numbers function. A binary
three numbers function can be indicated by a table with 4x4 elements. Its
first column indicates the first variable and the first row indicates the
second variable. To give a table with 4x4 elements is to define a binary three
numbers function and vice versa. For example:

| -1 | 0 | 1
---|---|---|---
-1 | 1 | -1 | 0
0 | -1 | 0 | 1
1 | 0 | 1 | -1
| -1 | 0 | 1
---|---|---|---
-1 | 0 | 0 | 0
0 | 0 | 0 | 0
1 | 0 | 0 | 0
| -1 | 0 | 1
---|---|---|---
-1 | 1 | 1 | 1
0 | 0 | 0 | 0
1 | -1 | -1 | -1

There are three functions in above tables. The first one is linear binary
three numbers function and the second one is identity function with o value
and value of the third one is not change with the second variable. There is
only one value for each discrete point in these three functions and they are
called single-valued binary three numbers function. It’s easy to know there
are $3^{9}$=19683 single-valued binary three numbers function. Can it be two-
valued or three-valued in a discrete point?certainly! There are three
combinations -1,0 and -1,1 and 0,1 for two-valued and only one
combination-1,0,1 for three-valued and numbers will be partitioned by symbol
’*’ if it’s a multi-valued. Can it be no-valued in a discrete point? Yes! We
will indicate it in ’N’. A binary three numbers function can be no-valued in
all 9 discrete points in uttermost like:

| -1 | 0 | 1
---|---|---|---
-1 | N | N | N
0 | N | N | N
1 | N | N | N

There is single-valued,two-valued,three-valued and no valued point in below
three numbers function.

| -1 | 0 | 1
---|---|---|---
-1 | -1 | 0 | 1
0 | -1*0 | -1*1 | 0*1
1 | N | -1*0*1 | N

It’s easy to know there are $8^{9}$ binary three numbers functions. A unary
three numbers function can be indicated by three value numbers partitioned by
symbol ’,’ in bracket and numbers will be partitioned by the symbol ’*’if it’s
many-valued, for example:(-1*0,N,-1*0*1).

The expression $H(x_{1},x_{2})=f[g_{11}(x_{1})+g_{12}(x_{2})]$ contains
$x_{1}$,$x_{2}$, but we intend to take H as a independent object not
containing $x_{1}$,$x_{2}$. We can’t express H in $f[g_{11}+g_{12}]$ because
we will get a unary function $[g_{11}+g_{12}]$ by adding $g_{11}$ and
$g_{12}$. $f[g_{11}+g_{12}]$ is also a unary function and is never equal to
binary the function H.

We express H by only f, $g_{11}$, $g_{12}$ without $x_{1},x_{2}$like this:

$H=f[g_{11}\widetilde{\alpha_{1}}+g_{12}\widetilde{\alpha_{2}}]$

To define a function is to give a rule to get it’s values. For such an
expression we are very clear the rule about getting values of the function if
we replace $\widetilde{\alpha_{1}}$ or $\widetilde{\alpha_{2}}$by
$x_{1}orx_{2}$ ,respectively. That is enough.

Binary three numbers function is called single term binary three numbers
function if it can be represented as
$H=f[g_{11}\widetilde{\alpha_{1}}+g_{12}\widetilde{\alpha_{2}}]$ in which f
,$g_{ij}$ is unary three numbers function and it will be called L term binary
three numbers function if it can be expressed as $\sum
f_{i}[g_{i1}\widetilde{\alpha_{1}}+g_{i2}\widetilde{\alpha_{2}}](i=1,L)$.
Expressing a function of many variables as this form is also called
representing it as a superposition of functions of one variable or decomposing
it to functions of one variable.

For example F is a single term binary three numbers function:

$F=(0,0,1)\Big{[}(0,0,1)\widetilde{\alpha_{1}}+(-1,-1,0)\widetilde{\alpha_{2}}\Big{]}$

Theorem 2.1 Every binary three numbers function can be represented as a
superposition of three numbers functions of one variable.

A binary three numbers function is called singular binary three numbers
function if it’s zero in all discrete points but except one. It’s called
standard singular three numbers binary function if non-zero point is in lower-
right location. Definitions for singular three numbers function of three
variables and for standard singular three numbers function of three variables
are similar.

First we prove that the standard singular three numbers binary function is a
single term one. It’s clear the standard singular binary three numbers
function is F above. (0,0,1) in(0,0,1)$\widetilde{\alpha_{1}}$ in the
expression of F is called raw function. Raw of none-zero point will change if
we adjust the location of ‘1’ in (0,0,1). (-1,-1,0) of(-1,-1,0)
$\widetilde{\alpha_{2}}$ in it is called column fuction.Column of none-zero
point will change if we adjust the location of ‘0’in (-1,-1,0). The first
(0,0,1) in it is called value function. Value which may be single-valued or
multi-valued or no-valued of non-zero point will change if we modify ‘1’ in
(0,0,1). Thus we know that every singular binary three numbers function can be
represented as a superposition of three numbers functions of one variable.

Because every binary three numbers function can be transformed to sum of 9
singular binary three numbers functions then we get our theorem.

So every binary three numbers function can be represented as:

(2.2)
$\displaystyle\Psi_{2}=\sum_{i=1}^{L}f_{i}[g_{i1}\widetilde{\alpha_{1}}+g_{i2}\widetilde{\alpha_{2}}]$

Here L is not greater than 9. Thus we can express and can construct a binary
three numbers function by unary three numbers functions.

We can extend all these result to N numbers function of several variables.In
the decomposition of standard singular binary three numbers function if we
replace raw function (0,0,1) by (0,0,$\cdots$0,1),column function (-1,-1,0) by
(-1,-1,$\cdots$-1,0) and value function (0,0,1)by(0,0,$\cdots$0,1)
,respectively.Then we can extend this expression to N numbers functions of two
variables. Situation for N numbers functions of M variables is similar. So we
get conclusion below.

If N$\geq$M+1 a general N numbers function of M variables can be decomposed
as:

(2.3)
$\displaystyle\psi=\sum_{i=1}^{L}fi\sum_{j=1}^{M}g_{ij}\widetilde{\alpha_{i}}$

If N$<$M+1, the number of unary function in expression of singular discrete
function will be bigger than M+1. For example a standard singular three
numbers function of three variables 3 can be represented as:

(2.4)
$\displaystyle\Psi_{3}=(0,0,1)\\{(0,0,1)[(0,0,1)\widetilde{\alpha_{1}}+(-1,-1,0)\widetilde{\alpha_{2}}]+(-1,-1,0)\widetilde{\alpha_{3}}\\}$

Here are more location functions (-1,-1,0) and (0,0,1) than one of the
standard singular binary three numbers function. Expressions for singular
three numbers function of three variables and for general three numbers
function of three variables are similar to ones of binary three numbers
functions.

All conclusions here are not suit to two numbers function. So we have:

Theorem 2.2 Every N numbers (N$\geq$3) function of M variables can be
represented as a superposition of N numbers functions of one variable.

Note we not only prove the existence of representation by superposition of
functions of one variable and give a constructive procedure. We just only gave
the method to decomposing a function but expression is not the shortest one.
Decomposition with terms being equal to its discrete points is called a
trivial decomposition. Actual terms are more less. Decomposition with less
terms than trivial decomposition is called non-trivial decomposition. It’s an
important topic to study non-trivial decompositions and will not be stated
here.

##  3\. Equations constructed by three numbers functions

There are $3^{9}$ single-valued binary three numbers function and $8^{9}$ ones
if they contain many-valued or no-valued ones. How many equations can we
construct with these functions? So many! How many things need to study about
group of the order 3 ? Too poor! So we know there are ample mineral resources
in this task.

Theorem 3.1 Every algebraic equation constructed by three numbers functions of
two variables can be represented as a superposition of three numbers functions
of one variable.

It’s simple to improve it. Solution of any equation is always function of
several variables. By substituting -1,0,1 to the equation respectively we can
get this function easily because field of definition of it is only three
numbers -1,0,1. We can get the solution expressed by function of one variable
by decomposing this function. That is wonderful that we can construct
equations and solve them freely in a mathematics system! In this paper we just
only solve the equation though there are multitudinous equations:

$(x\psi_{1}a)\psi_{3}(x\psi_{2}b)=c$

Here and in this paper we do’nt write functions of two variables in prefix
form like $\psi_{3}[\psi_{1}(x,a),\psi_{2}(x,b)]]=c$ for clearness.This
equation is called two branches equation. $\psi_{i}$ is parameterized function
and can be any one of $8^{9}$ three numbers functions of two variables. So
actually we solve not one equation but a kind of equation and the method
possesses universality.

Assume function $\psi_{1}$,$\psi_{2}$ and $\psi_{3}$ in the two branches
equation is $\Omega_{1}$,$\Omega_{2}$ and $\Omega_{3}$ ,respectively:

| -1 | 0 | 1
---|---|---|---
-1 | -1 | 1 | 0
0 | 0 | -1 | 1
1 | 1 | 0 | -1
| -1 | 0 | 1
---|---|---|---
-1 | 0 | -1 | 1
0 | -1 | 0 | -1
1 | 1 | 1 | 0
| -1 | 0 | 1
---|---|---|---
-1 | 1 | -1 | 0
0 | 0 | 1 | -1
1 | -1 | 0 | 1

When a=b=c=-1 we get numerical equation
$[x\Omega_{1}(-1)]\Omega_{3}[x\Omega_{2}(-1)]=-1$. We know only -1 is the
solution of this equation by substituting -1,0,1 to it. So we can know that
W(-1,-1,-1)=-1. By the same way we can get other values of W(a,b,c). W(a,b,c)
can be expressed by table below.

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | -1 | N | N |  | 0 | N | -1*0*1 |  | 1 | -1*0*1 | N
0 | 1 | -1*0*1 | N |  | -1 | N | N |  | 0 | N | -1*0*1
1 | 0 | N | -1*0*1 |  | 1 | -1*0*1 | N |  | -1 | N | N

In this table the first column indicates the first function number a and the
first row indicates the second function number b and c indicates the third
function number. Decomposing this function of three variables we get the
solution expressed by a superposition of functions of one variable.

$x=(0,0,-1)\Bigg{\\{}(0,0,1)\Big{[}(1,0,0)a+(0,-1,-1)b\Big{]}+(0,-1,-1)c\Bigg{\\}}$

+$(0,0,1)\Bigg{\\{}(0,0,1)\Big{[}(1,0,0)a+(0,-1,-1)b\Big{]}+(-1,-1,0)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(1,0,0)a+(-1,0,-1)b\Big{]}+(0,-1,-1)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(1,0,0)a+(-1,0,-1)b\Big{]}+(-1,0,-1)c\Bigg{\\}}$

+$(0,0,$-1*0*1$)\Bigg{\\{}(0,0,1)\Big{[}(1,0,0)a+(-1,0,-1)b\Big{]}+(-1,-1,0)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(1,0,0)a+(-1,-1,0)b\Big{]}+(0,-1,-1)c\Bigg{\\}}$

+$(0,0,$-1*0*1$)\Bigg{\\{}(0,0,1)\Big{[}(1,0,0)a+(-1,-1,0)b\Big{]}+(-1,0,-1)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(1,0,0)a+(-1,-1,0)b\Big{]}+(-1,-1,0)c\Bigg{\\}}$

+$(0,0,1)\Bigg{\\{}(0,0,1)\Big{[}(0,1,0)a+(0,-1,-1)b\Big{]}+(0,-1,-1)c\Bigg{\\}}$

+$(0,0,-1)\Big{]}\Bigg{\\{}(0,0,1)\Big{[}(0,1,0)a+(0,-1,-1)b\Big{]}+(-1,0,-1)c\Bigg{\\}}$

+$(0,0,$-1*0*1$)\Bigg{\\{}(0,0,1)\Big{[}(0,1,0)a+(-1,0,-1)b\Big{]}+(0,-1,-1)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(0,1,0)a+(-1,0,-1)b\Big{]}+(-1,0,-1)c\Bigg{\\}}$

+$(0,0,N)\Big{]}\Bigg{\\{}(0,0,1)\Big{[}(0,1,0)a+(-1,0,-1)b\Big{]}+(-1,-1,0)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(0,1,0)a+(-1,-1,0)b\Big{]}+(0,-1,-1)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(0,1,0)a+(-1,-1,0)b\Big{]}+(-1,0,-1)c\Bigg{\\}}$

+$(0,0,$-1*0*1$)\Bigg{\\{}(0,0,1)\Big{[}(0,1,0)a+(-1,-1,0)b\Big{]}+(-1,-1,0)c\Bigg{\\}}$

+$(0,0,1)\Bigg{\\{}(0,0,1)\Big{[}(0,0,1)a+(0,-1,-1)b\Big{]}+(-1,0,-1)c\Bigg{\\}}$

+$(0,0,-1)\Bigg{\\{}(0,0,1)\Big{[}(0,0,1)a+(0,-1,-1)b\Big{]}+(-1,-1,0)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(0,0,1)a+(-1,0,-1)b\Big{]}+(0,-1,-1)c\Bigg{\\}}$

+$(0,0,$-1*0*1$)\Bigg{\\{}(0,0,1)\Big{[}(0,0,1)a+(-1,0,-1)b\Big{]}+(-1,0,-1)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(0,0,1)a+(-1,0,-1)b\Big{]}+(-1,-1,0)c\Bigg{\\}}$

+$(0,0,$-1*0*1$)\Bigg{\\{}(0,0,1)\Big{[}(0,0,1)a+(-1,-1,0)b\Big{]}+(0,-1,-1)c\Bigg{\\}}$

+$(0,0,N)\Bigg{\\{}(0,0,1)\Big{[}(0,0,1)a+(-1,-1,0)b\Big{]}+(-1,0,-1)c\Bigg{\\}}$

+$(0,0,N)\Big{]}\Bigg{\\{}(0,0,1)\Big{[}(0,0,1)a+(-1,-1,0)b\Big{]}+(-1,-1,0)c\Bigg{\\}}$

There are 24 but not 27 terms because there are three discrete point with 0
value.

##  4\. Four special operators

How to get a new function from a known one? Operator is correspondence between
functions. To give a correspondence between known functions and new functions
is to give an operator. Four special operators mentioned here are easy to be
understood intuitively and are important to solve equations with parameterized
functions however so we must pay attention to them.

Definition 4.1Commutation operators. Assume there is an function of two
variables $\psi$, $a_{1}\psi a_{2}=a_{0}$,its commutation functions
$\psi$(1,2,0),$\psi$(1,0,2),$\psi$(0,2,1),$\psi$(
2,1,0),$\psi$(2,0,1),$\psi$(0,1,2) will be defined by following formulas and
we introduce commutation operators of one variable C[1,2,0], C[1,0,2],
C[0,2,1], C[2,1,0], C[2,0,1], C[0,1,2] then new functions can be expressed by
$\psi$ and commutation operators.

(4.1a) $\displaystyle
a_{1}\psi[1,2,0]a_{2}=a_{0}\qquad\qquad\psi(1,2,0)=C(1,2,0)(\psi)$ (4.1b)
$\displaystyle
a_{1}\psi[1,0,2]a_{0}=a_{2}\qquad\qquad\psi(1,0,2)=C(1,0,2)(\psi)$ (4.1c)
$\displaystyle
a_{0}\psi[0,2,1]a_{2}=a_{1}\qquad\qquad\psi(0,2,1)=C(0,2,1)(\psi)$ (4.1d)
$\displaystyle
a_{2}\psi[2,1,0]a_{1}=a_{0}\qquad\qquad\psi(2,1,0)=C(2,1,0)(\psi)$ (4.1e)
$\displaystyle
a_{2}\psi[2,0,1]a_{0}=a_{1}\qquad\qquad\psi(2,0,1)=C(2,0,1)(\psi)$ (4.1f)
$\displaystyle
a_{0}\psi[0,1,2]a_{1}=a_{2}\qquad\qquad\psi(0,1,2)=C(0,1,2)(\psi)$

Note $\psi$(1,2,0)is $\psi$ itself.

Numbers in brackets indicates new locations of function numbers and of
function result after commutating. That is say original function doesn’t
satisfy the new relation gotten by commuting location of function numbers and
of function result but new one satisfies it. New relation with new function
and new location is equivalent to original one in despite of their forms are
different. For example: if $\Omega$ is the first table below then
$C(1,0,2)(\Omega)$, $C(0,2,1)(\Omega)$ , $C(2,1,0)(\Omega)$,
$C(2,0,1)(\Omega)$, $C(0,1,2)(\Omega)$will be other tables ,respectively.

| -1 | 0 | 1
---|---|---|---
-1 | -1 | 1 | 0
0 | 0 | -1 | 1
1 | 1 | 0 | -1
| -1 | 0 | 1
---|---|---|---
-1 | -1 | 1 | 0
0 | 0 | -1 | 1
1 | 1 | 0 | -1
| -1 | 0 | 1
---|---|---|---
-1 | -1 | 0 | 1
0 | 0 | 1 | -1
1 | 1 | -1 | 0

| -1 | 0 | 1
---|---|---|---
-1 | -1 | 0 | 1
0 | 1 | -1 | 0
1 | 0 | 1 | -1
| -1 | 0 | 1
---|---|---|---
-1 | -1 | 0 | 1
0 | 0 | 1 | -1
1 | 1 | -1 | 0
| -1 | 0 | 1
---|---|---|---
-1 | -1 | 0 | 1
0 | 1 | -1 | 0
1 | 0 | 1 | -1

We can get any combination of function numbers and of function result for
$C(1,0,2)(\Omega)$ by commuting the second function number and function result
for $C(1,2,0)(\Omega)$. Situations for other commutation functions are similar
to it. We don’t limit function at all when we do commutation operator. May be
we get a many-valued function by a not monotonic function or get an function
with no values in some discrete points by a not surjective function. The same
situation may be exists in other three special operators. We have shown our
opinion above. An mathematics system is extensive and open if it involves
solving equation so it’s impossible to limit functions in it. I have ever
tried to limit function in ones of single-valued but failed because function
of many-valued or of no-valued can be introduced from function of single-
valued by special operators. This problem had troubled me for a long time
until I read materials about extension of group. I known functions of many-
valued or of no-valued are not difficult to be accepted by mathematicians.
Commutation operator for binary functions can be extended to function of many
variables. Showing all commutation functions is integrity in logical and not
all of them will be used in solving equations. There are only two commutation
functions for a unary function:

(4.2a) $\beta_{e}(a)=a_{0}\qquad\qquad\qquad\beta_{e}=\beta$ (4.2b)
$\beta_{t}(a_{0})=a\qquad\qquad\qquad\beta_{t}=C(\beta)$

Definition 4.2Tension-compression operator.Assume there is a binary function
$\psi$ and an unary function $\beta$, $\beta(a_{1})\psi a_{2}=a_{0}$, we can
introduce a new binary function $\psi_{1}$ by $\psi$ and $\beta$, $\psi_{1}$
will meet the relation: $a_{1}\psi_{1}a_{2}=a_{0}$, that is say,
$a_{1}\psi_{1}a_{2}=\beta(a_{1})\psi a_{2}$. Introduce a special operator
$T_{1}$ to express the relation between $\psi_{1}$and $\psi$,$\beta$ .

(4.3a) $\psi_{1}=\psi T_{1}\beta$ In the same way if
$a_{1}\psi\beta(a_{2})=a_{0}$, we can introduce a new binary function
$\psi_{2}$ by $\psi$ and $\beta$, $\psi_{2}$ will meet the relation:
$a_{1}\psi_{2}a_{2}=a_{0}$, that is say,
$a_{1}\psi_{2}a_{2}=a_{1}\psi\beta(a_{2})$. Introduce a special operator
$T_{2}$ to express the relation between $\psi_{2}$and $\psi$,$\beta$. (4.3b)
$\psi_{2}=\psi T_{2}\beta$ If $a_{1}\psi a_{2}=\beta(a_{0})$,that is say
$\beta^{-1}[a_{1}\psi a_{2}]=a_{0}$,we can introduce a new binary function
$\psi_{0}$ by $\psi$ and $\beta$, $\psi_{0}$ will meet the relation:
$a_{1}\psi_{0}a_{2}=a_{0}$,that is say
$a_{1}\psi_{0}a_{2}=\beta^{-1}[a_{1}\psi a_{2}]$, and there is $T_{0}$: (4.3c)
$\psi_{0}=\psi T_{0}\beta$

For example, (1,-1,0) is an function of one variable and written in $\gamma$
then $\Omega$$T_{1}$$\gamma$ and $\Omega$$T_{2}$$\gamma$ and
$\Omega$$T_{0}$$\gamma$ will be

| -1 | 0 | 1
---|---|---|---
-1 | 1 | 0 | -1
0 | -1 | 1 | 0
1 | 0 | -1 | 1
| -1 | 0 | 1
---|---|---|---
-1 | 0 | -1 | 1
0 | 1 | 0 | -1
1 | -1 | 1 | 0
| -1 | 0 | 1
---|---|---|---
-1 | 0 | -1 | 1
0 | 1 | 0 | -1
1 | -1 | 1 | 0

respectively.

It’s occasional that $\Omega$$T_{2}$$\gamma$ is equal to
$\Omega$$T_{0}$$\gamma$. Only $T_{0}$ will be used in solving equation.

For an unary function we have only T and $T_{0}$:

(4.4a) $\beta_{1}T\beta_{2}=\beta_{1}\beta_{2}$ (4.4b)
$\beta_{1}T_{0}\beta_{2}=\beta_{2}^{-1}\beta_{1}$

Note,$\beta_{1}$$\beta_{2}$ means applying$\beta_{2}$ first and then
applying$\beta_{1}$.That is say

(4.5) $\\\ \beta_{1}\beta_{2}(x)=\beta_{1}\Big{[}\beta_{2}(x)\Big{]}$

A discrete point for $\beta_{1}$$\beta_{2}$ will be no-valued if it for any of
$\beta_{1}$ or $\beta_{2}$ is no-valued. $\beta_{1}$ and $\beta_{2}$ will be
each other inverse function if$\beta_{1}$$\beta_{2}$ =e . There are $8^{3}$
three numbers functions of one variable in which there is always inverse
function for any three numbers function of one variable.

This rule is right for many-valued functions of two variables because tension-
compression operators for functions of two variables involves actually only
composition of two functions of one variable.

Definition 4.3Superposition operator. Assume there are P functions of many
variables$\psi_{k}$(k=1,p),their superposition function $\psi$ will be:

(4.6) $\psi=\sum_{k=1}^{P}\psi_{k}$

Value of $\psi$ will be the sum of value of $\psi_{k}$(k=1,p). This is a kind
of operator by it we can get a new function by several known functions with
same variables. $\psi$ will be no-valued in a point if any of $\psi_{k}$ is
no-valued in this point. $\psi_{1}+\psi_{2}$ will be many-valued in a point if
$\psi_{1}$ is single-valued and $\psi_{2}$ is many-valued in this point.

Definition 4.4Decomposition operator.

(4.7)
$\psi_{3}=\sum_{i=1}^{27}f_{i}\Bigg{\\{}g_{i4}\Big{[}g_{i1}(\widetilde{\alpha_{1}})+g_{i2}(\widetilde{\alpha_{2}})\Big{]}+g_{i3}(\widetilde{\alpha_{3}})\Bigg{\\}}$

We can express the relations between$f_{i}$ or $g_{ij}$ and $\psi_{3}$with
special operators $V_{3}$ and $P_{ij}$ and actually $g_{ij}$ is not change
with $\psi_{3}$.

(4.8a) $f_{i}=V_{i}(\psi_{3})\qquad\qquad\qquad\qquad(i=1,27)$ (4.8b)
$g_{ij}=P_{ij}(\psi_{3})\qquad\qquad(i=1,27,j=1,4)$

Otherwise there are more than one decomposition for any function of 3
variables but we select only one of them. Correspondence between $\psi_{3}$
and$f_{i}$ , $g_{ij}$ is clear and easy to be gotten. So decomposition
operator is not occult at all.

Please note commutation operator or tension-compression operator or
decomposition operator or superposition operator will be close within all
three numbers functions if they contain ones being many-valued and no-valued.
This is very important and is the sufficient reason for existing of many-
valued functions and no-valued functions.

So four special operators are very clear and not perplexed at all.

Definition 4.5False function of M+K variables. We can change an function of M
variables to a false one of (M+K ) variables by adding
$o\widetilde{\alpha_{k}}$ in which o is a zero function and function $\psi$
will not change with K variables.

(4.9)
$\psi=\sum_{i=1}^{L}fi\sum_{j=1}^{M}g_{ij}\widetilde{\alpha_{i}}=\sum_{i=1}^{L}fi\Bigg{\\{}\sum_{j=1}^{M}g_{ij}\widetilde{\alpha_{i}}+\sum_{k=M+1}^{M+K}o\widetilde{\alpha_{k}}\Bigg{\\}}$

We can also get false function of (M+K ) variables from one of M variables by
$T_{k}o$ (k=M,M+K). For example:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 0 | 1 | -1 |  | 0,1 | N | -1*0*1 |  | 1 | 0 | N
0 | 0 | 1 | -1 |  | 0,1 | N | -1*0*1 |  | 1 | 0 | N
1 | 0 | 1 | -1 |  | 0,1 | N | -1*0*1 |  | 1 | 0 | N

This is a false function of three variables and value of it will not change
with the first variable. Below table is a false function of two variables.

| -1 | 0 | 1
---|---|---|---
-1 | 1 | 1 | 1
0 | -1 | -1 | -1
1 | 0 | 0 | 0

False function of many variables will be used in solving equations with
parameterized functions.

##  5\. Formula solution for equations with parameterized functions

What’s an analytic solution or formula solution for an equation?Formula
solution can only contain known parameters or constants and known
parameterized functions or known numerical functions and four kinds of special
operators and we call them valid symbols and all others invalid ones. This is
the standard to verify a formula solution of an equation. Commutation
operators and tension-compression operators and superposition operators and
decomposition operators are the sufficient condition but not the necessary
condition to give formula solutions of equations. There may be another
equivalence set of operators that can express formula solutions of equations.
We will solve two branches equation as a example below. At the same time we
will solve an equation with digital functions below then we can understand the
procedure more clearly. We must believe that it’s not complex to solve this
equation because we have known already the solution exists surely and only
four special operators will be deal with to get it. We will take any new
function met in procedure of solving the equation as a normal one and will
never be puzzled by its appearance.

Step 1: Decomposing function $\psi_{3}$ as:

$\psi_{3}=\sum_{i=1}^{9}f_{i}(g_{i1}\widetilde{\alpha_{1}}+g_{i2}\widetilde{\alpha_{2}})$

$\sum_{i=1}^{9}f_{i}\Big{[}g_{i1}(x\psi_{1}a)+g_{i2}(x\psi_{2}b)\Big{]}=c$

$\Omega_{3}=~{}(0,0,1)\Big{[}(1,0,0)\widetilde{\alpha_{1}}+(0,-1,-1)\widetilde{\alpha_{2}}\Big{]}+(0,0,-1)\Big{[}(1,0,0)\widetilde{\alpha_{1}}+(-1,0,-1)\widetilde{\alpha_{2}}\Big{]}$
$+(0,01)\Big{[}(0,1,0)\widetilde{\alpha_{1}}+(-1,0,-1)\widetilde{\alpha_{2}}\Big{]}+(0,0,-1)\Big{[}(0,1,0)\widetilde{\alpha_{1}}+(-1,-1,0)\widetilde{\alpha_{2}}\Big{]}$
$+(0,0,-1)\Big{[}(0,0,1)\widetilde{\alpha_{1}}+(0,-1,-1)\widetilde{\alpha_{2}}\Big{]}+(0,0,1)\Big{[}(0,0,1)\widetilde{\alpha_{1}}+(-1,-1,0)\widetilde{\alpha_{2}}\Big{]}$

$(x\Omega_{1}a)\Omega_{3}(x\Omega_{2}b)=$
$~{}(0,0,1)\Big{[}(1,0,0)(x\Omega_{1}a)+(0,-1,-1)(x\Omega_{2}b)\Big{]}+(0,0,-1)\Big{[}(1,0,0)(x\Omega_{1}a)+(-1,0,-1)(x\Omega_{2}b)\Big{]}$
$+(0,01)\Big{[}(0,1,0)(x\Omega_{1}a)+(-1,0,-1)(x\Omega_{2}b)\Big{]}+(0,0,-1)\Big{[}(0,1,0)(x\Omega_{1}a)+(-1,-1,0)(x\Omega_{2}b)\Big{]}$
$+(0,0,-1)\Big{[}(0,0,1)(x\Omega_{1}a)+(0,-1,-1)(x\Omega_{2}b)\Big{]}+(0,0,1)\Big{[}(0,0,1)(x\Omega_{1}a)+(-1,-1,0)(x\Omega_{2}b)\Big{]}$
=c

Step 2: By tension-compression of$g_{i1},g_{i2}$ we have:

$\sum_{i=1}^{9}f_{i}\Big{[}x(\psi_{1}T_{0}g_{i1}^{-1})a+x(\psi_{2}T_{0}g_{i2}^{-1})b\Big{]}=c$

Note,$(\psi_{1}T_{0}g_{i1}^{-1})$ in $x(\psi_{1}T_{0}g_{i1}^{-1})a$ and
$(\psi_{2}T_{0}g_{i2}^{-1})$ in $x(\psi_{2}T_{0}g_{i2}^{-1})b$ are two
functions of two variables.

$(x\Omega_{1}a)\Omega_{3}(x\Omega_{2}b)=~{}$
$(0,0,1)\Bigg{\\{}x\Big{[}\Omega_{1}T_{0}(1,0,0)^{-1}\Big{]}a+x\Big{[}\Omega_{2}T_{0}(0,-1,-1)^{-1}\Big{]}b\Bigg{\\}}+$

$(0,0,-1)\Bigg{\\{}x\Big{[}\Omega_{1}T_{0}(1,0,0)^{-1}\Big{]}a+x\Big{[}\Omega_{2}T_{0}(-1,0,-1)^{-1}\Big{]}b\Bigg{\\}}+$
$(0,01)\Bigg{\\{}x\Big{[}\Omega_{1}T_{0}(0,1,0)^{-1}\Big{]}a$

$+x\Big{[}\Omega_{2}T_{0}(-1,0,-1)^{-1}\Big{]}b\Bigg{\\}}+$
$(0,0,-1)\Bigg{\\{}x\Big{[}\Omega_{1}T_{0}(0,1,0)^{-1}\Big{]}a+x\Big{[}\Omega_{2}T_{0}(-1,-1,0)^{-1}\Big{]}b\Bigg{\\}}+$

$(0,0,-1)\Bigg{\\{}x\Big{[}\Omega_{1}T_{0}(0,0,1)^{-1}\Big{]}a+x\Big{[}\Omega_{2}T_{0}(0,-1,-1)^{-1}\Big{]}b\Bigg{\\}}+$
$(0,0,1)\Bigg{\\{}x\Big{[}\Omega_{1}T_{0}(0,0,1)^{-1}\Big{]}a+$

$x\Big{[}\Omega_{2}T_{0}(-1,-1,0)^{-1}\Big{]}b\Bigg{\\}}=c$

$\Omega_{1}T_{0}(1,0,0)^{-1}$, $\Omega_{1}T_{0}(0,1,0)^{-1}$,
$\Omega_{1}T_{0}(0,0,1)^{-1}$ is

| -1 | 0 | 1
---|---|---|---
-1 | 1 | 0 | 0
0 | 0 | 1 | 0
1 | 0 | 0 | 1
| -1 | 0 | 1
---|---|---|---
-1 | 0 | 0 | 1
0 | 1 | 0 | 0
1 | 0 | 1 | 0
| -1 | 0 | 1
---|---|---|---
-1 | 0 | 1 | 0
0 | 0 | 0 | 1
1 | 1 | 0 | 0

$\Omega_{2}T_{0}(0,-1,-1)^{-1}$, $\Omega_{2}T_{0}(-1,0,-1)^{-1}$,
$\Omega_{2}T_{0}(-1,-1,0)^{-1}$ is

| -1 | 0 | 1
---|---|---|---
-1 | -1 | 0 | -1
0 | 0 | -1 | 0
1 | -1 | -1 | -1
| -1 | 0 | 1
---|---|---|---
-1 | 0 | -1 | -1
0 | -1 | 0 | -1
1 | -1 | -1 | 0
| -1 | 0 | 1
---|---|---|---
-1 | -1 | -1 | 0
0 | -1 | -1 | -1
1 | 0 | 0 | -1

respectively.

Step 3: Changing $\psi_{1}T_{0}g_{i1}^{-1}$ by $T_{3}o$ to get a false
function of three variables $\psi_{1}T_{3}oT_{0}g_{i1}^{-1}$ in which variable
c is a false one and Changing $\psi_{2}T_{0}g_{i2}^{-1}$ by $T_{2}$o to get a
false function of three variables $\psi_{2}T_{2}oT_{0}g_{i2}^{-1}$ in which
variable b is a false one ,respectively. Adding them to get a real function of
three variables $\psi_{i3}$. This is the application of tension-compression
operator in solving equation.

$\psi_{i3}=\psi_{1}T_{3}oT_{0}g_{i1}^{-1}+\psi_{2}T_{2}oT_{0}g_{i2}^{-1}\qquad(i=1,9)$

$\theta_{1}=\Omega_{1}T_{3}oT_{0}(1,0,0)^{-1}+\Omega_{2}T_{2}oT_{0}(0,-1,-1)^{-1}$
is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 0 | -1 | -1 |  | 1 | 0 | 0 |  | 0 | -1 | -1
0 | 0 | 1 | 0 |  | -1 | 0 | -1 |  | 0 | 1 | 0
1 | -1 | -1 | 0 |  | -1 | -1 | 0 |  | -1 | -1 | 0

$\theta_{2}=\Omega_{1}T_{3}oT_{0}(1,0,0)^{-1}+\Omega_{2}T_{2}oT_{0}(-1,0,-1)^{-1}$
is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 1 | 0 | 0 |  | 0 | -1 | -1 |  | 0 | -1 | -1
0 | -1 | 0 | -1 |  | 0 | 1 | 0 |  | -1 | 0 | -1
1 | -1 | -1 | 0 |  | -1 | -1 | 0 |  | 0 | 0 | 1

$\theta_{3}=\Omega_{1}T_{3}oT_{0}(0,1,0)^{-1}+\Omega_{2}T_{2}oT_{0}(-1,0,-1)^{-1}$
is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 0 | 0 | 1 |  | -1 | -1 | 0 |  | -1 | -1 | 0
0 | 0 | -1 | -1 |  | 1 | 0 | 0 |  | 0 | -1 | -1
1 | -1 | 0 | -1 |  | -1 | 0 | -1 |  | 0 | 1 | 0

$\theta_{4}=\Omega_{1}T_{3}oT_{0}(0,1,0)^{-1}+\Omega_{2}T_{2}oT_{0}(-1,-1,0)^{-1}$
is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | -1 | -1 | 0 |  | -1 | -1 | 0 |  | 0 | 0 | 1
0 | 0 | -1 | -1 |  | 0 | -1 | -1 |  | 0 | -1 | -1
1 | 0 | 1 | 0 |  | 0 | 1 | 0 |  | -1 | 0 | -1

$\theta_{5}=\Omega_{1}T_{3}oT_{0}(0,0,1)^{-1}+\Omega_{2}T_{2}oT_{0}(0,-1,-1)^{-1}$
is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | -1 | 0 | -1 |  | 0 | 1 | 0 |  | -1 | 0 | -1
0 | 0 | 0 | 1 |  | -1 | -1 | 0 |  | 0 | 0 | 1
1 | 0 | -1 | -1 |  | 0 | -1 | -1 |  | 0 | -1 | -1

$\theta_{6}=\Omega_{1}T_{3}oT_{0}(0,0,1)^{-1}+\Omega_{2}T_{2}oT_{0}(-1,-1,0)^{-1}$
is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | -1 | 0 | -1 |  | -1 | 0 | -1 |  | 0 | 1 | 0
0 | -1 | -1 | 0 |  | -1 | -1 | 0 |  | -1 | -1 | 0
1 | 1 | 0 | 0 |  | 1 | 0 | 0 |  | 0 | -1 | -1

Step 4: Changing $\psi_{i3}$ by $T_{0}f_{i}^{-1}$we get:

$\psi_{i4}=\psi_{i3}T_{0}f_{i}^{-1}\qquad(i=1,9)$

$\theta_{1}T_{0}(0,0,1)^{-1}$ is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 0 | 0 | 0 |  | 1 | 0 | 0 |  | 0 | 0 | 0
0 | 0 | 1 | 0 |  | 0 | 0 | 0 |  | 0 | 1 | 0
1 | 0 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 0 | 0

$\theta_{2}T_{0}(0,0,-1)^{-1}$ is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | -1 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 0 | 0
0 | 0 | 0 | 0 |  | 0 | -1 | 0 |  | 0 | 0 | 0
1 | 0 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 0 | -1

$\theta_{3}T_{0}(0,0,1)^{-1}$ is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 0 | 0 | 1 |  | 0 | 0 | 0 |  | 0 | 0 | 0
0 | 0 | 0 | 0 |  | 1 | 0 | 0 |  | 0 | 0 | 0
1 | 0 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 1 | 0

$\theta_{4}T_{0}(0,0,-1)^{-1}$ is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 0 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 0 | -1
0 | 0 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 0 | 0
1 | 0 | -1 | 0 |  | 0 | -1 | 0 |  | 0 | 0 | 0

$\theta_{5}T_{0}(0,0,-1)^{-1}$ is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 0 | 0 | 0 |  | 0 | -1 | 0 |  | 0 | 0 | 0
0 | 0 | 0 | -1 |  | 0 | 0 | 0 |  | 0 | 0 | -1
1 | 0 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 0 | 0

$\theta_{6}T_{0}(0,0,1)^{-1}$ is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | 0 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 1 | 0
0 | 0 | 0 | 0 |  | 0 | 0 | 0 |  | 0 | 0 | 0
1 | 1 | 0 | 0 |  | 1 | 0 | 0 |  | 0 | 0 | 0

Step 5: To sum $\psi_{i4}$ we get:

$\psi_{5}=\sum_{i=1}^{9}\psi_{i4}$

Original equation will be:

$\psi_{5}(x,a,b)=c$

$\theta_{7}=\theta_{1}T_{0}(0,0,1)^{-1}+\theta_{2}T_{0}(0,0,-1)^{-1}+\theta_{3}T_{0}(0,0,1)^{-1}+\theta_{4}T_{0}(0,0,-1)^{-1}$

+$\theta_{5}T_{0}(0,0,-1)^{-1}+\theta_{6}T_{0}(0,0,1)^{-1}$ is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | -1 | 0 | 1 |  | 1 | -1 | 0 |  | 0 | 1 | -1
0 | 0 | 1 | -1 |  | 1 | -1 | 0 |  | 0 | 1 | -1
1 | 1 | -1 | 0 |  | 1 | -1 | 0 |  | 0 | 1 | -1

Step6: By commutation operator we get:

$x=\Big{[}C(2,3,0,1)\psi_{5}\Big{]}(a,b,c)=W(a,b,c)$
$C(2,3,0,1)\theta_{7}$is:

c=-1 | -1 | 0 | 1 | c=0 | -1 | 0 | 1 | c=1 | -1 | 0 | 1
---|---|---|---|---|---|---|---|---|---|---|---
-1 | -1 | N | N |  | 0 | N | -1*0*1 |  | 1 | -1*0*1 | N
0 | 1 | -1*0*1 | N |  | -1 | N | N |  | 0 | N | -1*0*1
1 | 0 | N | -1*0*1 |  | 1 | -1*0*1 | N |  | -1 | N | N

It’s not oddball there are many-valued discrete points or no-valued discrete
points. Not all commutation operators are used in solving equation.

Step 7: by decomposition operator we get:

$x=\sum_{k=1}^{27}u_{k}\Bigg{\\{}v_{k4}\Big{[}v_{k1}(a)+v_{k2}(b)\Big{]}+v_{k3}(c)\Bigg{\\}}$
$=\sum_{k=1}^{27}(V_{k}W)\Bigg{\\{}(P_{k4}W)\Big{[}(P_{k1}W)(a)+(P_{k2}W)(b)\Big{]}+(P_{k3}W)(c)\Bigg{\\}}$

we replace logogram symbols by complete ones.

$x=\sum_{k=1}^{27}V_{k}\Big{[}C(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\psi_{1}T_{3}oT_{0}(P_{i1}\psi_{3})^{-1}+\psi_{2}T_{2}oT_{0}(P_{i2}\psi_{3})^{-1}\big{]}T_{0}(V_{i}\psi_{3})^{-1}\big{\\}}\Big{)}\Big{]}$

$\Bigg{[}P_{k4}\Big{[}C(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\psi_{1}T_{3}oT_{0}(P_{i1}\psi_{3})^{-1}+\psi_{2}T_{2}oT_{0}(P_{i2}\psi_{3})^{-1}\big{]}T_{0}(V_{i}\psi_{3})^{-1}\big{\\}}\Big{)}\Big{]}$

$\Bigg{(}P_{k1}\Big{[}C(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\psi_{1}T_{3}oT_{0}(P_{i1}\psi_{3})^{-1}+\psi_{2}T_{2}oT_{0}(P_{i2}\psi_{3})^{-1}\big{]}T_{0}(V_{i}\psi_{3})^{-1}\big{\\}}\Big{)}\Big{]}(a)$

$+P_{k2}\Big{[}C(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\psi_{1}T_{3}oT_{0}(P_{i1}\psi_{3})^{-1}+\psi_{2}T_{2}oT_{0}(P_{i2}\psi_{3})^{-1}\big{]}T_{0}(V_{i}\psi_{3})^{-1}\big{\\}}\Big{)}\Big{]}(b)\Bigg{)}$

$+P_{k3}\Big{[}C(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\psi_{1}T_{3}oT_{0}(P_{i1}\psi_{3})^{-1}+\psi_{2}T_{2}oT_{0}(P_{i2}\psi_{3})^{-1}\big{]}T_{0}(V_{i}\psi_{3})^{-1}\big{\\}}\Big{)}\Big{]}(c)\Bigg{]}$

Actually location functions $P_{ij}\psi_{k}$ do not change with $\psi_{k}$ and
can be called constant functions. Solution of equation with function
$\Omega_{1}$,$\Omega_{2}$ and $\Omega_{3}$ has been given already above by
getting a function of many variables . Giving the procedure of it is just only
make the method clearer. We deal with the function of three variables in
solving this equation. Can we avoid to use it in the procedure? Never! In
history one reason to introduce complex number is that we have to deal with
complex number even if three roots of a cubic equation are all real number.
It’s the most important that we have gotten the solution expressed by function
of one variable however.

##  6\. Composition of special operators

There are 10 compositions for commutation operators and tension-compression
operators and superposition operators and decomposition operators as bellow
tables:
table1
commutation tension-compression superposition decomposition commutation 1
tension-compression 2 3 superposition 4 5 6 decomposition 7 8 9 10

We will mention them below. Here we give only results of binary function and
they can be extended to functions of many variables easily.
Composition 1 commutation and commutation:

see table 2
Composition 2 tension-compression and commutation:

see table 3
Composition 3 tension-compression and tension-compression:

see table4
Composition 4 superposition and commutation:

is equal to commutation and superposition for commutation C(2,1,0):

(6.1)
$C(2,1,0)(\sum_{k=1}^{H}\psi_{k})=\sum_{k=1}^{H}\Big{[}C(2,1,0)(\psi_{k})\Big{]}\qquad$

It will be complex for commutation C(0,2,1) and commutation C(1,0,2).
Composition 5 superposition and tension-compression:

is equal to tension-compression and superposition for tension-compression
$T_{1}$ and $T_{2}$:

(6.2a)
$(\sum_{k=1}^{H}\psi_{k})T_{1}\beta=\sum_{k=1}^{H}(\psi_{k}T_{1}\beta)\qquad\qquad\qquad$
(6.2b)
$(\sum_{k=1}^{H}\psi_{k})T_{2}\beta=\sum_{k=1}^{H}(\psi_{k}T_{2}\beta)\qquad\qquad\qquad$

is complex for tension-compression $T_{0}$.
Composition 6 superposition and superposition:

It is very simple.
Composition 7 decomposition - commutation:

Value functions will hold the line and location functions will exchange for
commutation C(2,1,0)

(6.3a) $V_{i}\Big{[}C(2,1,0)(\psi)\Big{]}=V_{i}(\psi)\qquad(i=1,L)$ (6.3b)
$P_{i1}\Big{[}C(2,1,0)(\psi)\Big{]}=P_{i2}(\psi)\qquad(i=1,L)$ (6.3c)
$P_{i2}\Big{[}C(2,1,0)(\psi)\Big{]}=P_{i1}(\psi)\qquad(i=1,L)$

is complex for commutation C(0,2,1) and commutation C(1,0,2).
Composition 8 decomposition and tension-compression:

Value functions will hold the line and location functions will be acted by
$T_{1}\beta$ or $T_{2}\beta$ for tension-compression $T_{1}$ and $T_{2}$.

(6.4a) $V_{i}(\psi T_{j}\beta)=V_{i}\psi\qquad(i=1,L\qquad j=1,2)$ (6.4b)
$P_{ij}(\psi T_{j}\beta)=(P_{ij}\psi)T_{j}\beta\qquad(i=1,L\qquad j=1,2)$

is complex for $T_{0}$. But there is relation between value functions of
$\psi$ and of $\psi$ acted by $T_{0}$ if it’s a trivial decomposition.

(6.5) $V_{i}(\psi T_{0}\beta)=(V_{i}\psi)T_{0}\beta\qquad(i=1,L)$

This relation is very important.
Composition 9 decomposition and superposition:

Value functions will be composition of value functions and location functions
will be any location functions.

(6.6a)
$V_{i}(\sum_{k=1}^{H}\psi_{k})=\sum_{k=1}^{H}V_{i}(\psi_{k})\qquad(i=1,L)$
(6.6b) $P_{ij}(\sum_{k=1}^{H}\psi_{i})=P_{ij}(\psi_{k})\qquad(i=1,L\qquad
j=1,2)$

Composition 10 decomposition and decomposition:

None.

Law and composition of special operators can be extended to high degree
operators in form.
Table2 commutation and commutation

| C(1,2,0) | C(1,0,2) | C(0,2,1) | C(2,1,0) | C(2,0,1) | C(0,1,2)
---|---|---|---|---|---|---
C(1,2,0) | C(1,2,0) | C(1,0,2) | C(0,2,1) | C(2,1,0) | C(2,0,1) | C(0,1,2)
C(1,0,2) | C(1,0,2) | C(1,2,0) | C(2,0,1) | C(0,1,2) | C(0,2,1) | C(2,1,0)
C(0,2,1) | C(0,2,1) | C(0,1,2) | C(1,2,0) | C(2,0,1) | C(2,1,0) | C(1,0,2)
C(2,1,0) | C(2,1,0) | C(2,0,1) | C(0,1,2) | C(1,2,0) | C(1,0,2) | C(0,2,1)
C(2,0,1) | C(2,0,1) | C(2,1,0) | C(1,0,2) | C(0,2,1) | C(0,1,2) | C(1,2,0)
C(0,1,2) | C(0,1,2) | C(0,2,1) | C(2,1,0) | C(1,0,2) | C(1,2,0) | C(2,0,1)

Table3 tension-compression and commutation

| C(1,2,0) | C(1,0,2) | C(0,2,1) | C(2,1,0) | C(2,0,1) | C(0,1,2)
---|---|---|---|---|---|---
$T_{1}\beta$ | $T_{1}\beta$ | $C(1,0,2)T_{1}\beta$ | $C(0,2,1)T_{0}\beta$ | $C(2,1,0)T_{2}\beta$ | $C(2,0,1)T_{0}\beta$ | $C(0,1,2)T_{2}\beta$
$T_{2}\beta$ | $T_{2}\beta$ | $C(1,0,2)T_{0}\beta$ | $C(0,2,1)T_{2}\beta$ | $C(2,1,0)T_{1}\beta$ | $C(2,0,1)T_{1}\beta$ | $C(0,1,2)T_{0}\beta$
$T_{0}\beta$ | $T_{0}\beta$ | $C(1,0,2)T_{2}\beta$ | $C(0,2,1)T_{1}\beta$ | $C(2,1,0)T_{0}\beta$ | $C(2,0,1)T_{2}\beta$ | $C(0,1,2)T_{1}\beta$

table4 tension-compression and tension-compression

| $T_{1}\beta_{2}$ | $T_{2}\beta_{2}$ | $T_{0}\beta_{2}$
---|---|---|---
$T_{1}\beta_{1}$ | $T_{1}(\beta_{1}\beta_{2})$ | $(T_{2}\beta_{2})T_{1}\beta_{1}$ | ($T_{0}\beta_{2})T_{1}\beta_{1}$
$T_{2}\beta_{1}$ | $(T_{1}\beta_{2})T_{2}\beta_{1}$ | $T_{2}(\beta_{1}\beta_{2})$ | $(T_{0}\beta_{2})T_{2}\beta_{1}$
$T_{0}\beta_{1}$ | $(T_{1}\beta_{2})T_{0}\beta_{1}$ | $(T_{2}\beta_{2})T_{0}\beta_{1}$ | $T_{0}(\beta_{1}\beta_{2})$

All of them are easy to be validated by readers.

##  7\. Extend results to discrete operators

Now we extend results about discrete functions to discrete operators. We limit
the field of definition and range of operators within three discrete functions
-e=(1,0,-1),o=(0,0,0),e=(-1,0,1) for simplicity.

Definition 7.1Assume there are three numbers functions -e=(1,0,-1),o=(0,0,0)
and e=(-1,0,1)we let A=$\\{$-e,0,e$\\}$ and define three functions operator of
one variable $S_{1}$ as

$S_{1}$:A$\longrightarrow$A

define three functions operator of two variables $S_{2}$ as

$S_{2}$:$A^{2}$$\longrightarrow$A

define three functions operator of three variables $S_{3}$ as

$S_{3}$:$A^{3}$$\longrightarrow$A

There are $3^{3}$ single-valued three functions operators of one variable and
$8^{3}$ ones if they contain many-valued or no-valued.There are $3^{9}$
single-valued three functions operators of two variables and $8^{9}$ ones if
they contain many-valued or no-valued.There are $3^{27}$ single-valued three
functions operators of three variable and $8^{27}$ ones if they contain many-
valued or no-valued.

Functions will be partitioned by the symbol ’*’for many-valued point and no-
valued point will be indicated by ’N’.

’+’operation will be expressed as:

| -1 | 0 | 1
---|---|---|---
-1 | 1 | -1 | 0
0 | -1 | 0 | 1
1 | 0 | 1 | -1

’+’operator will be expressed as:

| -e | o | e
---|---|---|---
-e | e | -e | o
o | -e | o | e
e | o | e | -e

Compare two tables we know -e,o,e in discrete operators system is like -1,0,1
in discrete functions system ,respectively. We can also introduce concepts of
singular three functions operator and standard singular three functions
operator.

A standard singular three functions operator of two variables can be expressed
by table:

| -e | o | e
---|---|---|---
-e | o | o | o
o | o | o | o
e | o | o | e

It can be represented as a superposition of three functions operators of one
variable:

$G=(o,o,e)\Big{[}(o,o,e)\widetilde{\beta_{1}}+(-e,-e,o)\widetilde{\beta_{2}}\Big{]}$

By the same reason for three numbers function we know a standard singular
binary three functions operator can be represented as a superposition of unary
three functions operators and so does a general singular binary three
functions operator. A general binary three functions operator can be expressed
to sum of 9 singular binary three functions operators so we have

Theorem 7.1 Every binary three functions operator can be represented as a
superposition of three functions operators of one variable.

A standard singular three functions operator of three variables $\phi_{3}$ can
be represented as:

$\phi_{3}=(o,o,e)\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)(\widetilde{\beta_{1}})+(-e,-e,o)(\widetilde{\beta_{2}})\Big{]}+(-e,-e,o)(\widetilde{\beta_{3}})\Bigg{\\}}$

Theorem 7.2 Every three functions operator of two or of three variables can be
represented as a superposition of three functions operators of one variable.

All conclusions here are not suit to discrete 2 operator.

There are great number of operator equations constructed by $8^{9}$ operators
of two variables.

Theorem 7.3 Every operator equation constructed by three functions operators
of two variables can be give formula solution represented as a superposition
of three functions operators of one variable.

Although there are many operator equations we give formula solution for only
double branches operator equation with digital operators and with
parameterized operators.

$(y\phi_{1}f)\phi_{3}(y\phi_{2}g)=h$

Assume $\phi_{1}$,$\phi_{2}$,$\phi_{3}$ is $\Theta_{1},\Theta_{2},\Theta_{3}$
as below,respectively:

| -e | o | e
---|---|---|---
-e | -e | e | o
o | o | -e | e
e | e | o | -e
| -e | o | e
---|---|---|---
-e | o | -e | e
o | -e | o | -e
e | e | e | o
| -e | o | e
---|---|---|---
-e | e | -e | o
o | o | e | -e
e | -e | o | e

Solution expressed by superposition of operators of one variable will be:

$y=(o,o,-e)\Bigg{\\{}(o,o,e)\Big{[}(e,o,o)f+(o,-e,-e)g\Big{]}+(o,-e,-e)h\Bigg{\\}}$

+$(o,o,e)\Bigg{\\{}(o,o,e)\Big{[}(e,o,o)f+(o,-e,-e)g\Big{]}+(-e,-e,o)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(e,o,o)f+(-e,o,-e)g\Big{]}+(o,-e,-e)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(e,o,o)f+(-e,o,-e)g\Big{]}+(-e,o,-e)h\Bigg{\\}}$

+$(o,o,$-e*o*e$)\Bigg{\\{}(o,o,e)\Big{[}(e,o,o)f+(-e,o,-e)g\Big{]}+(-e,-e,o)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(e,o,o)f+(-e,-e,o)g\Big{]}+(o,-e,-e)h\Bigg{\\}}$

+$(o,o,$-e*o*e$)\Bigg{\\{}(o,o,e)\Big{[}(e,o,o)f+(-e,-e,o)g\Big{]}+(-e,o,-e)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(e,o,o)f+(-e,-e,o)g\Big{]}+(-e,-e,o)h\Bigg{\\}}$

+$(o,o,e)\Bigg{\\{}(o,o,e)\Big{[}(o,e,o)f+(o,-e,-e)g\Big{]}+(o,-e,-e)h\Bigg{\\}}$

+$(o,o,-e)\Big{]}\Bigg{\\{}(o,o,e)\Big{[}(o,e,o)f+(o,-e,-e)g\Big{]}+(-e,o,-e)h\Bigg{\\}}$

+$(o,o,$-e*o*e$)\Bigg{\\{}(o,o,e)\Big{[}(o,e,o)f+(-e,o,-e)g\Big{]}+(o,-e,-e)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(o,e,o)f+(-e,o,-e)g\Big{]}+(-e,o,-e)h\Bigg{\\}}$

+$(o,o,N)\Big{]}\Bigg{\\{}(o,o,e)\Big{[}(o,e,o)f+(-e,o,-e)g\Big{]}+(-e,-e,o)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(o,e,o)f+(-e,-e,o)g\Big{]}+(o,-e,-e)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(o,e,o)f+(-e,-e,o)g\Big{]}+(-e,o,-e)h\Bigg{\\}}$

+$(o,o,$-e*o*e$)\Bigg{\\{}(o,o,e)\Big{[}(o,e,o)f+(-e,-e,o)g\Big{]}+(-e,-e,o)h\Bigg{\\}}$

+$(o,o,e)\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)f+(o,-e,-e)g\Big{]}+(-e,o,-e)h\Bigg{\\}}$

+$(o,o,-e)\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)f+(o,-e,-e)g\Big{]}+(-e,-e,o)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)f+(-e,o,-e)g\Big{]}+(o,-e,-e)h\Bigg{\\}}$

+$(o,o,$-e*o*e$)\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)f+(-e,o,-e)g\Big{]}+(-e,o,-e)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)f+(-e,o,-e)g\Big{]}+(-e,-e,o)h\Bigg{\\}}$

+$(o,o,$-e*o*e$)\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)f+(-e,-e,o)g\Big{]}+(o,-e,-e)h\Bigg{\\}}$

+$(o,o,N)\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)f+(-e,-e,o)g\Big{]}+(-e,o,-e)h\Bigg{\\}}$

+$(o,o,N)\Big{]}\Bigg{\\{}(o,o,e)\Big{[}(o,o,e)f+(-e,-e,o)g\Big{]}+(-e,-e,o)h\Bigg{\\}}$

Definition 7.2 High Commutation Operators. Assume there is an operator of two
variables $\phi$, $y_{1}\phi y_{2}=y_{0}$, its commutation operators
$\phi$(1,2,0),$\phi$(1,0,2),$\phi$(0,
2,1),$\phi$(2,1,0),$\phi$(2,0,1),$\phi$(0,1,2) will be defined by following
formulas and we introduce high commutation operators $\overline{C}$[1,2,0],
$\overline{C}$[1,0,2], $\overline{C}$[0,2,1], $\overline{C}$[2,1,0],
$\overline{C}$[2,0,1],$\overline{C}$[0,1,2] then new operators can be
expressed by $\phi$ and high commutation operators.

(7.1a)
$y_{1}\phi[1,2,0]y_{2}=y_{0}\qquad\qquad\phi[1,2,0]=\overline{C}[1,2,0](\phi)$
(7.1b)
$y_{1}\phi[1,0,2]y_{0}=y_{2}\qquad\qquad\phi[1,0,2]=\overline{C}[1,0,2](\phi)$
(7.1c)
$y_{0}\phi[0,2,1]y_{2}=y_{1}\qquad\qquad\phi[0,2,1]=\overline{C}[0,2,1](\phi)$
(7.1d)
$y_{2}\phi[2,1,0]y_{1}=y_{0}\qquad\qquad\phi[2,1,0]=\overline{C}[2,1,0](\phi)$
(7.1e)
$y_{2}\phi[2,0,1]y_{0}=y_{1}\qquad\qquad\phi[2,0,1]=\overline{C}[2,0,1](\phi)$
(7.1f)
$y_{0}\phi[0,1,2]y_{1}=y_{2}\qquad\qquad\phi[0,1,2]=\overline{C}[0,1,2](\phi)$

There are only two high commutation functions for a unary operator:

(7.2a) $\zeta_{e}(y)=y_{0}\qquad\qquad\qquad\zeta_{e}=\zeta$

(7.2b) $\zeta_{t}(y_{0})=y\qquad\qquad\qquad\zeta_{t}=\overline{C}(\zeta)$

Definition 7.3 High Tension-compression Operator.Assume there is a binary
operator $\phi$ and an unary operator $\zeta$, $\zeta(y_{1})\phi y_{2}=y_{0}$,
we can introduce a new binary operator $\phi_{1}$ by $\phi$ and $\zeta$,
$\phi_{1}$ will meet the relation: $y_{1}\phi_{1}y_{2}=y_{0}$, that is say,
$y_{1}\phi_{1}y_{2}=\zeta(y_{1})\phi y_{2}$. Introduce a special high operator
$\overline{T}_{1}$ to express the relation between $\phi_{1}$and
$\phi$,$\zeta$ .

(7.3a) $\phi_{1}=\phi\overline{T}_{1}\zeta$ In the same way if
$y_{1}\phi\zeta(y_{2})=y_{0}$, we can introduce a new binary operator
$\phi_{2}$ by $\phi$ and $\zeta$, $\phi_{2}$ will meet the relation:
$y_{1}\phi_{2}y_{2}=y_{0}$, that is say,
$y_{1}\phi_{2}y_{2}=y_{1}\phi\zeta(y_{2})$. Introduce a special high operator
$\overline{T}_{2}$ to express the relation between $\phi_{2}$and
$\phi$,$\zeta$. (7.3b) $\phi_{2}=\phi\overline{T}_{2}\zeta$ If $y_{1}\phi
y_{2}=\zeta(y_{0})$,that is say $\zeta^{-1}[y_{1}\phi y_{2}]=y_{0}$, we can
introduce a new binary operator $\phi_{0}$ by $\phi$ and $\zeta$, $\phi_{0}$
will meet the relation: $y_{1}\phi_{0}y_{2}=y_{0}$, that is say
$y_{1}\phi_{0}y_{2}=\zeta^{-1}[y_{1}\phi y_{2}]$, and there is
$\overline{T}_{0}$: (7.3c) $\phi_{0}=\phi\overline{T}_{0}\zeta$

For an unary operator we have only $\overline{T}$ and $\overline{T}_{0}$:

(7.4a) $\zeta_{1}\overline{T}\zeta_{2}=\zeta_{1}\zeta_{2}$ (7.4b)
$\zeta_{1}\overline{T}_{0}\zeta_{2}=\zeta_{2}^{-1}\zeta_{1}$

Definition 7.4High Superposition Operator. Assume there are P operators of
many variables $\phi_{k}$ (k=1,p), its superposition operator $\phi$ will be:

(7.5) $\phi=\sum_{k=1}^{P}\phi_{k}$

Function of $\phi$ will be the sum of function of $\phi_{k}$(k=1,p). $\phi$
will be no-valued in a point if any of $\phi_{k}$ is no-valued in this point.
$\phi_{1}+\phi_{2}$ will be many-valued in a point if $\phi_{1}$ is single-
valued and $\phi_{2}$ is many-valued in this point.

Definition 7.5High Decomposition Operator.

(7.6)
$\phi_{3}=\sum_{i=1}^{27}\zeta_{i}\Bigg{\\{}\eta_{i4}\Big{[}\eta_{i1}(\widetilde{\beta_{1}})+\eta_{i2}(\widetilde{\beta_{2}})\Big{]}+\eta_{i3}(\widetilde{\beta_{3}})\Bigg{\\}}$

We can express the relations between$\zeta_{i}$ or $\eta_{ij}$ and
$\phi_{3}$with special operators $\overline{V}_{i}$ and $\overline{P}_{ij}$
and actually $\eta_{ij}$ is not change with $\phi_{3}$.

(7.7a) $\zeta_{i}=\overline{V}_{i}(\phi_{3})\qquad\qquad\qquad\qquad(i=1,27)$

(7.7b) $\eta_{ij}=\overline{P}_{ij}(\phi_{3})\qquad\qquad(i=1,27,j=1,4)$

Please note high commutation operator or high tension-compression operator or
high decomposition operator or high superposition operator will be close
within all three numbers operators if they contain ones being many-valued and
no-valued.

Definition 7.6False operator of M+K variables. We can change an operator of M
variables to a false one of (M+K ) variables by adding
$\sigma\widetilde{\zeta_{k}}$ in which $\sigma$ is a zero operator and
operator $\phi$ will not change with K variables.

(7.8)
$\phi=\sum_{i=1}^{L}fi\sum_{j=1}^{M}g_{ij}\widetilde{\zeta_{i}}=\sum_{i=1}^{L}fi\Bigg{\\{}\sum_{j=1}^{M}g_{ij}\widetilde{\zeta_{i}}+\sum_{k=M+1}^{M+K}\sigma\widetilde{\zeta_{k}}\Bigg{\\}}$

We can also get false operator of (M+K ) variables from one of M variables by
$\overline{T}_{k}\sigma$ (k=M,M+K).

Formula solution of double branches operator equation with parameterized
operators will be:

$y=\sum_{k=1}^{27}\overline{V}_{k}\Big{[}\overline{C}(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\phi_{1}T_{3}o\overline{T}_{0}(\overline{P}_{i1}\phi_{3})^{-1}+\phi_{2}\overline{T}_{2}\sigma\overline{T}_{0}(\overline{P}_{i2}\phi_{3})^{-1}\big{]}\overline{T}_{0}(\overline{V}_{i}\phi_{3})^{-1}\big{\\}}\Big{)}\Big{]}$

$\Bigg{[}\overline{P}_{k4}\Big{[}\overline{C}(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\phi_{1}T_{3}o\overline{T}_{0}(\overline{P}_{i1}\phi_{3})^{-1}+\phi_{2}\overline{T}_{2}\sigma\overline{T}_{0}(\overline{P}_{i2}\phi_{3})^{-1}\big{]}\overline{T}_{0}(\overline{V}_{i}\phi_{3})^{-1}\big{\\}}\Big{)}\Big{]}$

$\Bigg{(}\overline{P}_{k1}\Big{[}\overline{C}(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\phi_{1}T_{3}o\overline{T}_{0}(\overline{P}_{i1}\phi_{3})^{-1}+\phi_{2}\overline{T}_{2}\sigma\overline{T}_{0}(\overline{P}_{i2}\phi_{3})^{-1}\big{]}\overline{T}_{0}(\overline{V}_{i}\phi_{3})^{-1}\big{\\}}\Big{)}\Big{]}(f)$

$+\overline{P}_{k2}\Big{[}\overline{C}(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\phi_{1}T_{3}o\overline{T}_{0}(\overline{P}_{i1}\phi_{3})^{-1}+\phi_{2}\overline{T}_{2}\sigma\overline{T}_{0}(\overline{P}_{i2}\phi_{3})^{-1}\big{]}\overline{T}_{0}(\overline{V}_{i}\phi_{3})^{-1}\big{\\}}\Big{)}\Big{]}(g)\Bigg{)}$

$+\overline{P}_{k3}\Big{[}\overline{C}(2,3,0,1)\Big{(}\sum_{i=1}^{9}\big{\\{}\big{[}\phi_{1}T_{3}o\overline{T}_{0}(\overline{P}_{i1}\phi_{3})^{-1}+\phi_{2}\overline{T}_{2}\sigma\overline{T}_{0}(\overline{P}_{i2}\phi_{3})^{-1}\big{]}\overline{T}_{0}(\overline{V}_{i}\phi_{3})^{-1}\big{\\}}\Big{)}\Big{]}(h)\Bigg{]}$

Please note solution for double branches operator equation has the same form
with one for double branches algebraic equation. Is it appropriate to class
algebraic equation and operator equation to different fields? But we have done
it! Mathematics has been parted to many alone islands. This situation is not
good and will be changed in future.

These results mean that there is a new accurate analytical route beside
approximate numerical method and topological way in study of operator
equations certainly including functional equations and function equations and
differential equations.We can extend results to N numbers operators of M
variables but there are many works to be done.

##  8\. Try to extend to continuous situation

We can extend results about discrete functions to continue functions if we
accept results about Hilbert’s 13th problem. We can express formula solution
of equation constructed by continue functions in the same form of equation
constructed by discrete functions even though we can’t give a procedure to
decompose a continue function of many variables to a superposition of
functions of one variable. But there are many tasks to be done if we want to
make results to be strict in logic.

We must prove that every continue operator of many variables can be
represented as a superposition of continue operators of one variable if we
want to extend results in this paper to continue operators and equations
constructed by them. I don’t know if there is such a result in current
literature. Please give it if there isn’t.

There are enough space for us to write our results so we are luckier than
Pierre de Fermat (1601-1665) who could not write the proof of his last
theorem. Now we have only poor results shown here but mathematicians will find
more and more good results because there is huge mineral deposit in this
direction. Please believe this point!

## References

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