Talk:Transcendental equation

Very bad

This is a terrible article. I reads like an answer to "Explain a transcendental equation off the top of your head, right now!" -98.228.254.203 (talk) 05:13, 9 May 2011 (UTC)

I totally agree. But currently we have few ways to do with transcendent cases. --IkamusumeFan (talk) 05:59, 3 November 2014 (UTC)

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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.

• @Woodchain175:  Declined No reliable sources to support change. -KAP03(^{Talk • Contributions • Email}) 20:19, 26 June 2017 (UTC)

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.

Definition

@Jochen Burghardt: Defining a transcendental equation as one where one or both sides are transcendental functions is problematic. That would mean that $\sin x=\sin x$ and $\sin ^{2}x=1-cos^{2}x$ are transcendental equations (both sides are transcendental functions), but the equivalent $0=0$ and $\sin ^{2}x+cos^{2}x=1$ are not (neither side is). Determining whether an expression represents a transcendental function is non-trivial in general. Is $\sin(a\arcsin(x))={\frac {1}{2}}$ only a transcendental equation for some values of $a$?

Other sources agree that a transcendental equation is one which involves transcendental functions.[1] --Macrakis (talk) 20:29, 6 January 2022 (UTC)

@Macrakis and D.Lazard: I just checked Bronstein et al. again; they say "Eine Gleichung F(x) = f(x) ist transzendent, wenn wenigstens eine der Funktionen F(x) oder f(x) nicht algebraisch ist." "An equation F(x) = f(x) is transcendental if at least one of the functions F(x) or f(x) is not algebraic." So there are different definitions around (I'm not sure that Bronstein et al. are aware of the difference).

Maybe "involves" is the best choice for the definition, per your above arguments. The equation $\exp({\frac {\log(x)}{2}})$ would then be considered a transcendental one, which could trivially be transformed into an algebraic one. This is similar to $x^{3}+x^{2}-1=x^{3}$ which is not a quadratic equation in the strict sense, but can be trivially transformed into one. There are many different degrees of transformation difficulty, while a definition should provide a sharp yes/no criterion. So, if there are no objections, I'd change the definition to that of Macrakis/mathworld, and add a remark about Bronstein's deviation. - Jochen Burghardt (talk) 21:39, 6 January 2022 (UTC)

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