超越方程

举例

以下的方程分别因为有指数函数、三角函数等超越函数，因此均为超越方程。 e^(x)x=1

x=sin x

在天文学中，有关轨道偏近点角 E 的开普勒方程也是超越方程： M=E-esin E

其中： M 为轨道的平近点角。 e 为轨道的离心率。

求解方法

超越方程的求解可以利用绘图法及数值方法求解。若利用绘图法，可以分别令等式二边的式子等于另一变量（例如 y），然后在二个图绘制在一起，二个图的交点即为超越方程的解。数值方法也是以此想法往下延伸，利用数学公式求得二个图交点的位置。

若是数值很小，或是已知解在某一数值附近，也可以用泰勒级数的方式来用多项式近似超越函数，因此超越方程可用代数方程近似，再针对代数方程求解。用牛顿法也可以求超越方程的数值解。

有些特殊的函数可用来表示超越方程的解。例如复变函数朗伯W函数就可以表示一些超越方程的解。以下的超越方程 xe^(x)=1

其解为 W(1)，近似值为 0.56714329… （欧米加常数）。

参照

• 代数方程
• 超越函数
• 超越数
• 朗伯W函数

In applied mathematics, a transcendental equation is an equation over the real (or complex) numbers that is not algebraic, that is, if at least one of its sides describes a transcendental function.

Examples include: （公式略）

A transcendental equation may involve also non-elementary functions, although most published examples do not.

In some cases, a transcendental equation can be solved by transforming it into an equivalent algebraic equation.

Some such transformations are sketched below; computer algebra systems may provide more elaborated transformations.

In general, however, only approximate solutions can be found.

Transformation into an algebraic equation

Ad hoc methods exist for some classes of transcendental equations in one variable to transform them into algebraic equations which then might be solved.

Exponential equations

If the unknown, say x, occurs only in exponents:

applying the natural logarithm to both sides may yield an algebraic equation, e.g. 4^(x)=3^(x)²-1· 2^(5x) transforms to xln 4=(x²-1)ln 3+5xln 2, which simplifies to x²ln 3+x(5ln 2-ln 4)-ln 3=0, which has the solutions （公式略）

This will not work if addition occurs "at the base line", as in 4^(x)=3^(x)²-1+2^(5x).

if all "base constants" can be written as integer or rational powers of some number q, then substituting y=qx may succeed, e.g. 2^(x-1)+4^(x-2)-8^(x-2)=0 transforms, using y=2x, to 1/2y+1/16y²-1/64y³=0 which has the solutions （公式略）, hence x=log ₂8=3 is the only real solution.

This will not work if a square or a higher power of x occurs in an exponent, or if the "base constants" do not "share" a common q.

sometimes, substituting y=xex may obtain an algebraic equation; after the solutions for y are known, those for x can be obtained by applying the Lambert W function, e.g.: x²e^(2x)+2=3xe^(x) transforms to y²+2=3y, which has the solutions （公式略） hence （公式略）, where W₀ and W₋₁ denote the real-valued branches of the multivalued W function.

Logarithmic equations

If the unknown x occurs only in arguments of a logarithm function:

applying exponentiation to both sides may yield an algebraic equation, e.g. 2log ₅(3x-1)-log ₅(12x+1)=0 transforms, using exponentiation to base 5, to （公式略） which has the solutions （公式略） If only real numbers are considered, x=0 is not a solution, as it leads to a non-real subexpression log ₅(-1) in the given equation.

This requires the original equation to consist of integer-coefficient linear combinations of logarithms w.r.t. a unique base, and the logarithm arguments to be polynomials in x.

Approximate solutions

Approximate numerical solutions to transcendental equations can be found using numerical, analytical approximations, or graphical methods.

These equations can be solved by direct iteration by reordering the equation into the form x=f(x) and

making an initial guess x₀, computing f(x) which becomes x₁ and substituting it back into f(x), etc. Convergence may be very slow. Some reorderings may diverge, so some other reordering that converges must be found. f(x) must be continuous and "sufficiently smooth" or the method may fail.

Numerical methods for solving arbitrary equations are called root-finding algorithms. By rearranging the equation into the form f(x)=0, if f(x) is continuous and differentiable, Newton's method involving taking the derivative of f(x), is a common iterative method of approximating a root; an initial guess x₀ must be "sufficiently close" to the root of interest to converge to it.

In some cases, the equation can be well approximated using Taylor series near the zero. For example, for k≈ 1, the solutions of sin x=kx are approximately those of (1-k)x-x³/6=0, namely x=0 and x=± √6√(1-k).

For a graphical solution, one method is to set each side of a single-variable transcendental equation equal to a dependent variable and plot the two graphs, using their intersecting points to find solutions (see picture).

Other solutions

Some transcendental systems of high-order equations can be solved by “separation” of the unknowns, reducing them to algebraic equations.

The following can also be used when solving transcendental equations/inequalities: If x₀ is a solution to the equation f(x)=g(x) and f(x)≤ c≤ g(x), then this solution must satisfy f(x₀)=g(x₀)=c. For example, we want to solve （公式略）. The given equation is defined for -1<x<3. Let f(x)=log ₂(3+2x-x²) and （公式略）. It is easy to show that f(x)≤ 2 and g(x)≥ 2 so if there is a solution to the equation, it must satisfy f(x)=g(x)=2. From f(x)=2 we get x=1∈ (-1,3). Indeed, f(1)=g(1)=2 and so x=1 is the only real solution to the equation.