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
edit
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
,
⋯
,
x
n
)
=
x
0
{\displaystyle f(x,x_{2},\cdots ,x_{n})=x_{0}}
.
Here
f
(
x
,
x
2
,
⋯
,
x
n
)
{\displaystyle f(x,x_{2},\cdots ,x_{n})}
is a multivariate function built by several binary operations or binary functions such as addition
f
a
{\displaystyle f_{a}}
,subtraction
f
s
{\displaystyle f_{s}}
, multiplication
f
m
{\displaystyle f_{m}}
,division
f
d
{\displaystyle f_{d}}
,power
f
p
{\displaystyle f_{p}}
,root
f
s
{\displaystyle f_{s}}
,logarithm
f
l
{\displaystyle 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
{\displaystyle x^{x}=a}
f
p
(
x
,
x
)
=
a
{\displaystyle f_{p}(x,x)=a}
C
i
,
j
(
f
p
)
(
x
)
=
a
{\displaystyle C_{i,j}(f_{p})(x)=a}
x
=
[
C
i
,
j
(
f
p
)
]
−
1
(
a
)
{\displaystyle x=[C_{i,j}(f_{p})]^{-1}(a)}
Example 2:
x
a
+
x
b
+
x
c
=
d
,
(
a
,
b
,
c
,
d
≥
0
)
{\displaystyle x^{a}+x^{b}+x^{c}=d,(a,b,c,d\geq 0)}
f
a
2
{
f
a
1
[
f
p
1
(
x
,
a
)
,
f
p
2
(
x
,
b
)
]
+
f
p
3
(
x
,
c
)
}
=
d
,
{\displaystyle 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
p
1
(
x
,
a
)
=
P
1
,
2
4
(
f
p
)
(
x
,
a
,
b
,
c
)
{\displaystyle f_{p1}(x,a)=P_{1,2}^{4}(f_{p})(x,a,b,c)}
,
f
p
2
(
x
,
b
)
=
P
1
,
3
4
(
f
p
)
(
x
,
a
,
b
,
c
)
{\displaystyle f_{p2}(x,b)=P_{1,3}^{4}(f_{p})(x,a,b,c)}
,
f
p
3
(
x
,
c
)
=
P
1
,
4
4
(
f
p
)
(
x
,
a
,
b
,
c
)
{\displaystyle f_{p3}(x,c)=P_{1,4}^{4}(f_{p})(x,a,b,c)}
,
f
a
1
(
x
1
,
x
3
)
=
P
1
,
3
4
(
f
a
)
(
x
1
,
x
2
,
x
3
,
x
4
)
{\displaystyle f_{a1}(x_{1},x_{3})=P_{1,3}^{4}(f_{a})(x_{1},x_{2},x_{3},x_{4})}
,
f
a
2
(
x
3
,
x
4
)
=
P
3
,
4
4
(
f
a
)
(
x
1
,
x
2
,
x
3
,
x
4
)
{\displaystyle 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
)
{\displaystyle P_{1,2}^{4}(f_{p})}
to
x
1
{\displaystyle x_{1}}
and
P
1
,
3
4
(
f
p
)
{\displaystyle P_{1,3}^{4}(f_{p})}
to
x
3
{\displaystyle x_{3}}
of
P
1
,
3
4
(
f
a
)
=
x
1
+
x
3
{\displaystyle 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
)
]
{\displaystyle 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
)
}
{\displaystyle 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
)
}
{\displaystyle 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
{\displaystyle x_{3}}
and
P
1
,
4
4
(
f
p
)
{\displaystyle P_{1,4}^{4}(f_{p})}
to
x
4
{\displaystyle x_{4}}
of
P
3
,
4
4
(
f
a
)
=
x
3
+
x
4
{\displaystyle P_{3,4}^{4}(f_{a})=x_{3}+x_{4}}
respectively,
C
3
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
)
}
{\displaystyle 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
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
)
]
{\displaystyle 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
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
)
{\displaystyle 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
{\displaystyle C_{i}}
and unary operators such as promotion
P
i
,
j
n
{\displaystyle P_{i,j}^{n}}
or oblique projection
C
i
,
j
{\displaystyle C_{i,j}}
or inverses
I
i
{\displaystyle 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
edit
For multivariate function composition:
f
(
x
1
,
…
,
x
i
−
1
,
g
(
x
1
,
x
2
,
…
,
x
n
)
,
x
i
+
1
,
…
,
x
n
)
.
{\displaystyle 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 :
(
f
C
i
g
)
(
x
1
,
⋯
,
x
n
)
=
f
[
x
1
,
x
2
,
…
,
x
i
−
1
,
g
(
x
1
,
…
,
x
n
)
,
x
i
+
1
,
⋯
,
x
n
]
.
{\displaystyle (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
,
…
,
x
n
)
=
f
[
x
1
,
x
2
,
⋯
,
x
i
−
1
,
g
(
x
1
,
…
,
x
n
)
,
x
i
+
1
,
⋯
,
x
n
]
.
{\displaystyle [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
f
g
]
(
x
1
,
…
,
x
n
)
=
f
[
x
1
,
x
2
,
⋯
,
x
i
−
1
,
g
(
x
1
,
…
,
x
n
)
,
x
i
+
1
,
⋯
,
x
n
]
.
{\displaystyle [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
{\displaystyle x^{a}+x^{b}}
,first we denote it as
f
a
[
f
p
(
x
,
a
)
,
f
p
(
x
,
b
)
]
{\displaystyle f_{a}[f_{p}(x,a),f_{p}(x,b)]}
, in which
f
a
(
x
1
,
x
2
)
=
x
1
+
x
2
{\displaystyle f_{a}(x_{1},x_{2})=x_{1}+x_{2}}
and
f
p
(
x
1
,
x
2
)
=
x
1
x
2
{\displaystyle f_{p}(x_{1},x_{2})=x_{1}^{x_{2}}}
. In addition, we denote subtraction as
f
s
{\displaystyle f_{s}}
,multiplication as
f
m
{\displaystyle f_{m}}
, division as
f
d
{\displaystyle f_{d}}
, root as
f
r
{\displaystyle f_{r}}
and logarithm as
f
l
{\displaystyle f_{l}}
respectively. We want give an expression like
[
W
(
f
a
,
f
p
)
(
x
,
a
,
b
)
]
{\displaystyle [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
)
]
{\displaystyle f_{a}[f_{p}(x,a),f_{p}(x,b)]}
is an expression of a function of three variables . We consider
f
a
{\displaystyle f_{a}}
and
f
p
{\displaystyle f_{p}}
as especial functions of three variables too and introduce unary operator
P
i
,
j
n
{\displaystyle 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
)
{\displaystyle [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
{\displaystyle x_{1}}
or
x
3
{\displaystyle x_{3}}
is transitional variable and
O
(
x
)
=
0
{\displaystyle O(x)=0}
..
[
P
1
,
2
3
(
f
p
)
]
(
x
,
a
,
b
)
=
x
a
+
O
(
b
)
=
f
p
(
x
,
a
)
+
O
(
b
)
{\displaystyle [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
)
{\displaystyle [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
{\displaystyle 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
)
]
{\displaystyle f_{a}[f_{p}(x,a),f_{p}(x,b)]}
by substituting
x
1
{\displaystyle x_{1}}
and
x
3
{\displaystyle x_{3}}
in
f
a
(
x
1
,
x
3
)
{\displaystyle f_{a}(x_{1},x_{3})}
by
f
p
(
x
,
a
)
{\displaystyle f_{p}(x,a)}
and
f
p
(
x
,
b
)
{\displaystyle f_{p}(x,b)}
respectively. So
f
a
[
f
p
(
x
,
a
)
,
f
p
(
x
,
b
)
]
{\displaystyle 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
)
{\displaystyle {\{}[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
)
{\displaystyle 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
C
1
[
P
1
,
3
3
(
f
a
)
,
P
1
,
2
3
(
f
p
)
]
P
1
,
3
3
(
f
p
)
(
x
,
a
,
b
)
{\displaystyle 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
{\displaystyle 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
{\displaystyle x^{a}+x^{b}=c}
.
For an unary function promotion,
P
j
n
(
u
)
=
u
(
x
j
)
{\displaystyle P_{j}^{n}(u)=u(x_{j})}
. In special,
u
=
x
j
{\displaystyle u=x_{j}}
,
P
j
n
(
e
)
=
x
j
{\displaystyle P_{j}^{n}(e)=x_{j}}
in which 'e' is the identity function .
In
C
i
(
f
,
g
)
{\displaystyle C_{i}(f,g)}
if
g
=
P
j
n
(
e
)
=
x
j
{\displaystyle g=P_{j}^{n}(e)=x_{j}}
and
i
≠
j
,
{\displaystyle i\neq j,}
C
i
[
f
,
P
j
n
(
e
)
]
(
x
1
,
…
,
x
n
)
=
f
[
x
1
,
x
2
,
⋯
,
x
i
−
1
,
x
i
+
1
,
⋯
,
x
j
−
1
,
x
j
,
x
j
+
1
,
⋯
,
x
n
]
.
{\displaystyle 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
{\displaystyle x_{i}}
in the expression.
C
i
[
f
,
P
j
n
(
e
)
]
{\displaystyle 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
)
{\displaystyle C_{i,j}(f)}
.
For example,
C
i
,
j
(
f
p
)
(
x
)
=
f
p
(
x
,
x
)
=
x
x
{\displaystyle C_{i,j}(f_{p})(x)=f_{p}(x,x)=x^{x}}
Inverse multivariate function
edit
For multivariate function
x
0
=
f
(
x
1
,
x
2
,
⋯
,
x
i
,
⋯
,
x
n
)
{\displaystyle x_{0}=f(x_{1},x_{2},\cdots ,x_{i},\cdots ,x_{n})}
,
f
i
:
x
i
↦
f
(
x
1
,
x
2
,
⋯
,
x
i
,
⋯
,
x
n
)
{\displaystyle f_{i}:x_{i}\mapsto f(x_{1},x_{2},\cdots ,x_{i},\cdots ,x_{n})}
.
If
f
i
{\displaystyle f_{i}}
is bijection for any
x
j
(
j
=
1
,
2
,
⋯
,
n
,
j
≠
i
)
{\displaystyle x_{j}(j=1,2,\cdots ,n,j\neq i)}
we call
f
i
−
1
{\displaystyle f_{i}^{-1}}
an multivariate inverse function about
x
i
{\displaystyle x_{i}}
. Introduce unary operator
I
i
{\displaystyle I_{i}}
and denote :
f
i
−
1
=
I
i
(
f
)
{\displaystyle f_{i}^{-1}=I_{i}(f)}
.
For example,
f
(
x
1
,
x
2
)
=
x
1
3
+
x
2
2
{\displaystyle f(x_{1},x_{2})=x_{1}^{3}+x_{2}^{2}}
is invertible about variable
x
1
{\displaystyle x_{1}}
and is not invertible about variable
x
2
{\displaystyle 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
,
⋯
,
r
{\displaystyle f_{(k)},k=1,\cdots ,r}
and denote its inverses as :
f
i
,
(
k
)
−
1
=
I
i
(
f
(
k
)
)
,
k
=
1
,
⋯
,
r
{\displaystyle 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
{\displaystyle 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
≥
0
,
x
3
≥
0
{\displaystyle 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
≥
0
,
x
3
<
0
{\displaystyle 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
≥
0
{\displaystyle 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
{\displaystyle 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
≥
0
,
x
3
≥
0
{\displaystyle 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
≥
0
,
x
0
≥
0
{\displaystyle 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 .