Transcendent equation

In mathematics , a transcendent equation is an equation in an unknown in which the unknown occurs in the argument of at least one transcendent function . examples are

(1) : (2) : (3) : ${\displaystyle \ e^{x}-1=x\ ,\quad }$ ${\displaystyle \ \sin(2x)+\ln x=0\ ,\quad }$ ${\displaystyle \ \sinh(x^{2})\cdot \arccos x=e^{x}-1\ .}$

While one can estimate the maximum number and position of the zeros when determining the zeros of a polynomial and reduce the problem with a known solution by polynomial division , this is not possible with transcendent equations. For example, the equation has an infinite number of solutions. In practice, however, the area in which one suspects solutions can usually be limited by the respective problem. ${\displaystyle \sin x=0}$

The variety of transcendent functions is very large. With practical problems, however, one usually comes across equations that contain one or more functions of the following types:

${\displaystyle \sin ,\cos ,\tan ,\exp ,\sinh ,\cosh \ }$and their inverse functions .

As they are provided on pocket calculators, they are called TTR functions here for the sake of simplicity . Transcendent functions that differ from these standard functions can often be described with the help of a power series , e.g. B. the Si function .

Note: Root functions are not transcendent functions.

Simple cases

(1) , where is a TTR function (see above) ${\displaystyle \ f(x)=c\ }$ ${\displaystyle f}$

In this case, the solution is calculated with the help of the respective inverse function (on the calculator or computer with math software). For example:

${\displaystyle \ln x^{2}=3\ \to \ x=e^{3/2}}$

(2) , where is a polynomial and a TTR function. ${\displaystyle \ p(f(x))=0\ }$ ${\displaystyle p(z)}$ ${\displaystyle z=f(x)}$

Here one first calculates the zeros of the polynomial (possibly numerically) and then solutions of the equations using the inverse function of . For example: ${\displaystyle z_{1},z_{2},...}$ ${\displaystyle f(x)=z_{i}}$ ${\displaystyle f}$

${\displaystyle \ (\tan x)^{2}+\tan x-2=0\ \to \ p(z)=z^{2}+z-2=0\ \to \ z_{1}=1,\;z_{2}=-2}$

${\displaystyle \ \to \ x_{1}=\tan ^{-1}(1)\;,\quad x_{2}=\tan ^{-1}(-2)\ .}$

General case

The general case

${\displaystyle F_{1}(x)=F_{2}(x)}$

can always be on the form

${\displaystyle G(x)=F_{1}(x)-F_{2}(x)=0}$

bring. The zeros of are solutions of the given equation. ${\displaystyle G}$

If the given equation is not of a simple kind, there is usually only the numerical way left, i. H. one looks for solutions with a numerical approximation method . The simplest methods are bisection , regula falsi and the Newton method . Good initial approximations are important in these procedures. These can usually be recognized from the given problem or through a graphic representation.

literature

• IN Bronstein, KA Semendjajew: Taschenbuch der Mathematik , Harri Deutsch Verlag, Frankfurt, 1979, ISBN 3 871444928 , p. 190.

Individual evidence

1. ↑ Small Encyclopedia of Mathematics. Harry Deutsch Verlag, 1977, ISBN 3 87144 323 9 , p. 88.
2. ^ Arnfried Kemnitz: Mathematics at the beginning of the course. Springer-Verlag, 2010, ISBN 3834812935 , p. 73.