Woodschain175

My name is Zi qian Wu, an engineer from China. I have studied 'Solution of general transcendental equations' for many years. I participated international congress of mathematicians 2010 in Hederabad.

Following is some ideas and results of my study.

Use a multivariate function to express the solution of a general transcendental equation

Please read 'Use a multivariate function to express the solution of a general transcendental equation', Is it easy to be understand? Can you accept it?You are welcome to improve it.

Note: Multivariate function composition and inverse multivariate function see below.

A general transcendental equation is like:

$f(x,x_{2},\cdots ,x_{n})=x_{0}$.

Here $f(x,x_{2},\cdots ,x_{n})$ is a multivariate function built by several binary operations or binary functions such as addition $f_{a}$ ,subtraction $f_{s}$, multiplication $f_{m}$ ,division $f_{d}$,power $f_{p}$ ,root $f_{s}$,logarithm $f_{l}$.

By the concepts of multivariate function compositions and its additional concept, function promotion, and multivariate inverse function we can give an expression to the solution of this general transcendental equation.

1 Obtain the number of all parameters including unknown variable 'x' in the left of the equation and change all binary operations to multivariate functions by promotion.

2 By multivariate function composition describe the structure of the left of the equation.

3 By multivariate inverse function give the expression to the solution of the equation.

Example 1: $x^{x}=a$

$f_{p}(x,x)=a$

$C_{i,j}(f_{p})(x)=a$

$x=[C_{i,j}(f_{p})]^{-1}(a)$

Example 2:

$x^{a}+x^{b}+x^{c}=d,(a,b,c,d\geq 0)$

$f_{a2}{\{}f_{a1}[f_{p1}(x,a),f_{p2}(x,b)]+f_{p3}(x,c){\}}=d,$

There are more than one additions or powers so we differ them in subscript.

First,there are four parameters,x,a,b,c. So we obtain:

$f_{p1}(x,a)=P_{1,2}^{4}(f_{p})(x,a,b,c)$,

$f_{p2}(x,b)=P_{1,3}^{4}(f_{p})(x,a,b,c)$,

$f_{p3}(x,c)=P_{1,4}^{4}(f_{p})(x,a,b,c)$,

$f_{a1}(x_{1},x_{3})=P_{1,3}^{4}(f_{a})(x_{1},x_{2},x_{3},x_{4})$,

$f_{a2}(x_{3},x_{4})=P_{3,4}^{4}(f_{a})(x_{1},x_{2},x_{3},x_{4})$,

Substituting $P_{1,2}^{4}(f_{p})$ to $x_{1}$ and $P_{1,3}^{4}(f_{p})$ to $x_{3}$ of $P_{1,3}^{4}(f_{a})=x_{1}+x_{3}$ respectively,

$C_{1}[P_{1,3}^{4}(f_{a}),P_{1,2}^{4}(f_{p})]$.

$C_{3}{\{}C_{1}[P_{1,3}^{4}(f_{a}),P_{1,2}^{4}(f_{p})],P_{1,3}^{4}(f_{p}){\}}$.

Substituting $C_{3}{\{}C_{1}[P_{1,3}^{4}(f_{a}),P_{1,2}^{4}(f_{p})],P_{1,3}^{4}(f_{p}){\}}$ to $x_{3}$ and $P_{1,4}^{4}(f_{p})$ to $x_{4}$ of $P_{3,4}^{4}(f_{a})=x_{3}+x_{4}$ respectively,

$C_{3}{\frac {P_{3,4}^{4}(f_{a})}{C_{3}{\{}C_{1}[P_{1,3}^{4}(f_{a}),P_{1,2}^{4}(f_{p})],P_{1,3}^{4}(f_{p}){\}}}}$.

$C_{4}[C_{3}{\frac {P_{3,4}^{4}(f_{a})}{C_{3}{\{}C_{1}[P_{1,3}^{4}(f_{a}),P_{1,2}^{4}(f_{p})],P_{1,3}^{4}(f_{p}){\}}}},P_{1,4}^{4}(f_{p})]$.

This is the structure of the left of the equation described by multivariate function composition .

The expression to the solution of the equation is:

$x=I_{3}{\{}C_{4}[C_{3}{\frac {P_{3,4}^{4}(f_{a})}{C_{3}{\{}C_{1}[P_{1,3}^{4}(f_{a}),P_{1,2}^{4}(f_{p})],P_{1,3}^{4}(f_{p}){\}}}},P_{1,4}^{4}(f_{p})]{\}}(d,a,b,c)$

It is enough for you to know how to obtain the expression of the solution for a given equation and are clear the structure of the multivariate function consisted of some binary functions and binary operators being composition $C_{i}$ and unary operators such as promotion $P_{i,j}^{n}$ or oblique projection $C_{i,j}$ or inverses $I_{i}$.

Solving an equation is reducing several X to one then putting the X on one side of '=' and putting all the known things on the other side. The expression shown here meets this requirement.Is the expression a real solution? We obtain this expression by three steps,function promotion,multivariate function composition and multivariate inverse function. Which step can not be accepted by us? Function promotion? It is just changing binary operations or binary functions as special ones of n variables. Multivariatefunction composition? There is unary function composition. Why is no there multivariate function composition? In the same reason, there is unary inverse function, there must be multivariate inverse function! So I can not find any reason to reject such an expression.

Multivariate function composition

For multivariate function composition:

$f(x_{1},\ldots ,x_{i-1},g(x_{1},x_{2},\ldots ,x_{n}),x_{i+1},\ldots ,x_{n}).$

Here we give it three expressions like (f.g) for unary function composition. In the expression of (f.g), '.' can be considered as a binary operation taking f and g as its operands or a binary function taking f and g as its variables.

For multivariate function, the first expression is like an operation:

$(fC_{i}g)(x_{1},\cdots ,x_{n})=f[x_{1},x_{2},\ldots ,x_{i-1},g(x_{1},\ldots ,x_{n}),x_{i+1},\cdots ,x_{n}].$

The second one is like a function:

$[C_{i}(f,g)](x_{1},\ldots ,x_{n})=f[x_{1},x_{2},\cdots ,x_{i-1},g(x_{1},\ldots ,x_{n}),x_{i+1},\cdots ,x_{n}].$

The third one is like a fraction:

$[C_{i}{\frac {f}{g}}](x_{1},\ldots ,x_{n})=f[x_{1},x_{2},\cdots ,x_{i-1},g(x_{1},\ldots ,x_{n}),x_{i+1},\cdots ,x_{n}].$

Why do we use these forms? We can describe any expression in a fire-new way. For example,$x^{a}+x^{b}$,first we denote it as $f_{a}[f_{p}(x,a),f_{p}(x,b)]$, in which $f_{a}(x_{1},x_{2})=x_{1}+x_{2}$ and $f_{p}(x_{1},x_{2})=x_{1}^{x_{2}}$. In addition, we denote subtraction as $f_{s}$,multiplication as $f_{m}$, division as $f_{d}$, root as $f_{r}$ and logarithm as $f_{l}$ respectively. We want give an expression like $[W(f_{a},f_{p})(x,a,b)]$ in which the left part is called bare function containing only symbolics of function and the right part contains only variables.

$f_{a}[f_{p}(x,a),f_{p}(x,b)]$ is an expression of a function of three variables. We consider $f_{a}$ and $f_{p}$ as especial functions of three variables too and introduce unary operator $P_{i,j}^{n}$ to express these especial functions of three variables.

$[A_{1,3}^{3}(f_{a})](x_{1},x_{2},x_{3})=x_{1}+x_{3}+O(x_{2})=f_{a}(x_{1},x_{3})+O(x_{2})$

Here $x_{1}$ or $x_{3}$ is transitional variable and $O(x)=0$..

$[P_{1,2}^{3}(f_{p})](x,a,b)=x^{a}+O(b)=f_{p}(x,a)+O(b)$

$[P_{1,3}^{3}(f_{p})](x,a,b)=x^{b}+O(a)=f_{p}(x,b)+O(a)$

By these examples we know the meaning of superscript and subscript of $P_{i,j}^{n}$ and we call it function promotion.

It is clear that we obtain $f_{a}[f_{p}(x,a),f_{p}(x,b)]$ by substituting $x_{1}$ and $x_{3}$ in $f_{a}(x_{1},x_{3})$ by $f_{p}(x,a)$ and $f_{p}(x,b)$ respectively. So $f_{a}[f_{p}(x,a),f_{p}(x,b)]$ can be written in:

${\{}[P_{1,3}^{3}(f_{a})]C_{1}[P_{1,2}^{3}(f_{p})]{\}}C_{3}[P_{1,3}^{3}(f_{p})](x,a,b)$

or

$C_{3}{\{}C_{1}[P_{1,3}^{3}(f_{a}),P_{1,2}^{3}(f_{p})],P_{1,3}^{3}(f_{p}){\}}(x,a,b)$

or

$C_{3}{\frac {C_{1}[P_{1,3}^{3}(f_{a}),P_{1,2}^{3}(f_{p})]}{P_{1,3}^{3}(f_{p})}}(x,a,b)$

We never mind how complex they are. We consider them as multivariate functions being composition results of two other multivariate functions being composition results and or promotion results. These new expressions are different from $x^{a}+x^{b}$. Actually we had departed bare function from variables in these new expressions and there is only one "x" in them. This is what we want to do when we solve transcendental equations like $x^{a}+x^{b}=c$.

For an unary function promotion, $P_{j}^{n}(u)=u(x_{j})$. In special,$u=x_{j}$,$P_{j}^{n}(e)=x_{j}$ in which 'e' is the identity function.

In $C_{i}(f,g)$ if $g=P_{j}^{n}(e)=x_{j}$ and $i\neq j,$

$C_{i}[f,P_{j}^{n}(e)](x_{1},\ldots ,x_{n})=f[x_{1},x_{2},\cdots ,x_{i-1},x_{i+1},\cdots ,x_{j-1},x_{j},x_{j+1},\cdots ,x_{n}].$

Note,there is no $x_{i}$ in the expression.

$C_{i}[f,P_{j}^{n}(e)]$ is called oblique projection of f. Actually it is a function of n-1 variables and is dependent on only f and i,j so we denote it as $C_{i,j}(f)$. For example,$C_{i,j}(f_{p})(x)=f_{p}(x,x)=x^{x}$

Inverse multivariate function

For multivariate function $x_{0}=f(x_{1},x_{2},\cdots ,x_{i},\cdots ,x_{n})$,$f_{i}:x_{i}\mapsto f(x_{1},x_{2},\cdots ,x_{i},\cdots ,x_{n})$.

If $f_{i}$ is bijection for any $x_{j}(j=1,2,\cdots ,n,j\neq i)$ we call $f_{i}^{-1}$ an multivariate inverse function about $x_{i}$. Introduce unary operator $I_{i}$ and denote :

$f_{i}^{-1}=I_{i}(f)$.

For example,$f(x_{1},x_{2})=x_{1}^{3}+x_{2}^{2}$ is invertible about variable $x_{1}$ and is not invertible about variable $x_{2}$.

Partial inverses can be extend to multivariate functions too. We can define multivariate inverse function for an irreversible function if we can divide it into r partial functions $f_{(k)},k=1,\cdots ,r$ and denote its inverses as :

$f_{i,(k)}^{-1}=I_{i}(f_{(k)}),k=1,\cdots ,r$.

For example,$f(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4}$

$x_{0}=f_{(1)}(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4},x_{1}\geq 0,x_{3}\geq 0$

$x_{0}=f_{(2)}(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4},x_{1}\geq 0,x_{3}<0$

$x_{0}=f_{(3)}(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4},x_{1}<0,x_{3}\geq 0$

$x_{0}=f_{(4)}(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4},x_{1}<0,x_{3}<0$

$x_{1}=f_{1,(1)}^{-1}(x_{0},x_{2},x_{3})=I_{i}(f_{(1)})(x_{0},x_{2},x_{3})=[x_{0}-x_{2}^{3}-x_{3}^{4}]^{1/2},x_{0}\geq 0,x_{3}\geq 0$

$x_{3}=f_{3,(3)}^{-1}(x_{0},x_{1},x_{2})=I_{3}(f_{(3)})(x_{0},x_{1},x_{2})=-[x_{0}-x_{1}^{2}-x_{2}^{3}]^{1/4},x_{0}\geq 0,x_{0}\geq 0$

The concept of multivariate inverse function is useful to express the solution of a transcendental equation.