Transcendental equation

Short description: Equation whose side(s) describe a transcendental function

File:Herschel - Description of a machine for resolving by inspection certain important forms of transcendental equations, 1832 - 687143.tiff

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.[1] Examples include:

[math]\displaystyle{ \begin{align} x &= e^{-x} \\\ x &= \cos x \\\ 2^x &= x^2 \end{align} }[/math]

A transcendental equation need not be an equation between elementary functions, although most published examples are.

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.[2]

In general, however, only approximate solutions can be found.[3]

## Contents

* 1 Transformation into an algebraic equation
* 1.1 Exponential equations
* 1.2 Logarithmic equations
* 1.3 Trigonometric equations
* 1.4 Hyperbolic equations
* 2 Approximate solutions
* 3 Other solutions
* 4 See also
* 5 References

## 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,[4] e.g.

[math]\displaystyle{ 4^x = 3^{x^2-1} \cdot 2^{5x} }[/math] transforms to [math]\displaystyle{ x \ln 4 = (x^2-1) \ln 3 + 5x \ln 2 }[/math], which simplifies to [math]\displaystyle{ x^2 \ln 3 + x(5 \ln 2 - \ln 4) -\ln 3 = 0 }[/math], which has the solutions [math]\displaystyle{ x = \frac{ -3 \ln 2 \pm \sqrt{9(\ln 2)^2 - 4 (\ln 3)^2} }{ 2 \ln 3 } . }[/math]
This will not work if addition occurs "at the base line", as in [math]\displaystyle{ 4^x = 3^{x^2-1} + 2^{5x} . }[/math]

* if all "base constants" can be written as integer or rational powers of some number q, then substituting y=qx may succeed, e.g.

[math]\displaystyle{ 2^{x-1} + 4^{x-2} - 8^{x-2} = 0 }[/math] transforms, using y=2x, to [math]\displaystyle{ \frac{1}{2} y + \frac{1}{16} y^2 - \frac{1}{64} y^3 = 0 }[/math] which has the solutions [math]\displaystyle{ y \in \\{ 0, -4, 8\\} }[/math], hence [math]\displaystyle{ x= \log_2 8 = 3 }[/math] is the only real solution.[5]
This will not work if squares or 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,[citation needed] e.g.:

[math]\displaystyle{ x^2e^{2x} + 2 = 3x e^x }[/math] transforms to [math]\displaystyle{ y^2 + 2 = 3y, }[/math] which has the solutions [math]\displaystyle{ y \in \\{1,2\\}, }[/math] hence [math]\displaystyle{ x \in \\{ W_0(1), W_0(2), W_{-1}(1), W_{-1}(2) \\} }[/math], where [math]\displaystyle{ W_0 }[/math] and [math]\displaystyle{ W_{-1} }[/math] the denote the real-valued branches of the multivalued [math]\displaystyle{ W }[/math] 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.

[math]\displaystyle{ 2 \log_5 (3x-1) - \log_5 (12x+1) = 0 }[/math] transforms, using exponentiation to base [math]\displaystyle{ 5. }[/math] to [math]\displaystyle{ \frac{ (3x-1)^2 }{ 12x+1 } = 1, }[/math] which has the solutions [math]\displaystyle{ x \in \\{ 0, 2\\} . }[/math] If only real numbers are considered, [math]\displaystyle{ x = 0 }[/math] is not a solution, as it leads to a non-real subexpression [math]\displaystyle{ \log_5(-1) }[/math] 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.[6]

* if all "logarithm calls" have a unique base [math]\displaystyle{ b }[/math] and a unique argument expression [math]\displaystyle{ f(x), }[/math] then substituting [math]\displaystyle{ y = \log_b (f(x)) }[/math] may lead to a simpler equation,[7] e.g.

[math]\displaystyle{ 5 \ln(\sin x^2) + 6 = 7 \sqrt{ \ln(\sin x^2) + 8 } }[/math] transforms, using [math]\displaystyle{ y = \ln(\sin x^2) , }[/math] to [math]\displaystyle{ 5 y + 6 = 7 \sqrt{ y + 8 }, }[/math] which is algebraic and can be solved.[clarification needed] After that, applying inverse operations to the substitution equation yields [math]\displaystyle{ \sqrt{ \arcsin \exp y } = x. }[/math]

### Trigonometric equations

If the unknown x occurs only as argument of trigonometric functions:

* applying Pythagorean identities and trigonometric sum and multiple formulas, arguments of the forms [math]\displaystyle{ \sin(nx+a), \cos(mx+b), \tan(lx+c), ... }[/math] with integer [math]\displaystyle{ n,m,l,... }[/math] might all be transformed to arguments of the form, say, [math]\displaystyle{ \sin x }[/math]. After that, substituting [math]\displaystyle{ y = \sin(x) }[/math] yields an algebraic equation,[8] e.g.

[math]\displaystyle{ \sin(x+a) = (\cos^2 x) - 1 }[/math] transforms to [math]\displaystyle{ (\sin x)(\cos a) + \sqrt{ 1 - \sin^2 x }(\sin a) = 1 - (\sin^2 x) - 1 }[/math], and, after substitution, to [math]\displaystyle{ y (\cos a) + \sqrt{ 1 - y^2 }(\sin a) = - y^2 }[/math] which is algebraic[9] and can be solved. After that, applying [math]\displaystyle{ x = 2k\pi + \arcsin y }[/math] obtains the solutions.

### Hyperbolic equations

If the unknown x occurs only in linear expressions inside arguments of hyperbolic functions,

* unfolding them by their defining exponential expressions and substituting [math]\displaystyle{ y = exp(x) }[/math] yields an algebraic equation,[10] e.g.

[math]\displaystyle{ 3 \cosh x = 4 + \sinh (2x-6) }[/math] unfolds to [math]\displaystyle{ \frac{3}{2} (e^x + \frac{1}{e^x}) = 4 + \frac{1}{2} \left( \frac{(e^x)^2}{e^6} - \frac{e^6}{(e^x)^2} \right) , }[/math] which transforms to the equation [math]\displaystyle{ \frac{3}{2} (y + \frac{1}{y}) = 4 + \frac{1}{2} \left( \frac{y^2}{e^6} - \frac{e^6}{y^2} \right) , }[/math] which is algebraic[11] and can be solved. Applying [math]\displaystyle{ x = \ln y }[/math] obtains the solutions of the original equation.

## Approximate solutions

Graphical solution of sin(x)=ln(x)

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

Numerical methods for solving arbitrary equations are called root-finding algorithms.

In some cases, the equation can be well approximated using Taylor series near the zero. For example, for [math]\displaystyle{ k \approx 1 }[/math], the solutions of [math]\displaystyle{ \sin x = k x }[/math] are approximately those of [math]\displaystyle{ (1-k) x - x^3/6=0 }[/math], namely [math]\displaystyle{ x=0 }[/math] and [math]\displaystyle{ x = \plusmn \sqrt{6} \sqrt{1-k} }[/math].

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.[12][13]
* The following can also be used when solving transcendental equations/inequalities: If [math]\displaystyle{ x_0 }[/math] is a solution to the equation [math]\displaystyle{ f(x)=g(x) }[/math] and [math]\displaystyle{ f(x)\leq c\leq g(x) }[/math], then this solution must satisfy [math]\displaystyle{ f(x_0)=g(x_0)=c }[/math]. For example, we want to solve [math]\displaystyle{ \log_{2}\left(3+2x-x^{2}\right)=\tan^{2}\left(\frac{\pi x}{4}\right)+\cot^{2}\left(\frac{\pi x}{4}\right) }[/math]. The given equation is defined for [math]\displaystyle{ -1\lt x\lt 3 }[/math]. Let [math]\displaystyle{ f(x)=\log_{2}\left(3+2x-x^{2}\right) }[/math] and [math]\displaystyle{ g(x)=\tan^{2}\left(\frac{\pi x}{4}\right)+\cot^{2}\left(\frac{\pi x}{4}\right) }[/math]. It is easy to show that [math]\displaystyle{ f(x)\leq 2 }[/math] and [math]\displaystyle{ g(x)\geq 2 }[/math] so if there is a solution to the equation, it must satisfy [math]\displaystyle{ f(x)=g(x)=2 }[/math]. From [math]\displaystyle{ f(x)=2 }[/math] we get [math]\displaystyle{ x=1\in(-1,3) }[/math]. Indeed, [math]\displaystyle{ f(1)=g(1)=2 }[/math] and so [math]\displaystyle{ x=1 }[/math] is the only real solution to the equation.

## See also

* Mrs. Miniver's problem - Problem on areas of intersecting circles

## References

1. ↑ I.N. Bronstein and K.A. Semendjajew and G. Musiol and H. Mühlig (2005) (in German). Taschenbuch der Mathematik. Frankfurt/Main: Harri Deutsch. Here: Sect.1.6.4.1, p.45. The domain of equations is left implicit throughout the book.
2. ↑ For example, according to the Wolfram Mathematica tutorial page on equation solving, both [math]\displaystyle{ 2^x = x }[/math] and [math]\displaystyle{ e^x + x + 1 = 0 }[/math] can be solved by symbolic expressions, while [math]\displaystyle{ x = \cos x }[/math] can only be solved approximatively.
3. ↑ Bronstein et al., p.45-46
4. ↑ Bronstein et al., Sect.1.6.4.2.a, p.46
5. ↑ Bronstein et al., Sect.1.6.4.2.b, p.46
6. ↑ Bronstein et al., Sect.1.6.4.3.b, p.46
7. ↑ Bronstein et al., Sect.1.6.4.3.a, p.46
8. ↑ Bronstein et al., Sect.1.6.4.4, p.46-47
9. ↑ over an appropriate field, containing [math]\displaystyle{ \sin a }[/math] and [math]\displaystyle{ \cos a }[/math]
10. ↑ Bronstein et al., Sect.1.6.4.5, p.47
11. ↑ over an appropriate field, containing [math]\displaystyle{ e^6 }[/math]
12. ↑ V. A. Varyuhin, S. A. Kas'yanyuk, “On a certain method for solving nonlinear systems of a special type”, Zh. Vychisl. Mat. Mat. Fiz., 6:2 (1966), 347–352; U.S.S.R. Comput. Math. Math. Phys., 6:2 (1966), 214–221
13. ↑ V.A. Varyukhin, Fundamental Theory of Multichannel Analysis (VA PVO SV, Kyiv, 1993) [in Russian]

* John P. Boyd (2014). Solving Transcendental Equations: The Chebyshev Polynomial Proxy and Other Numerical Rootfinders, Perturbation Series, and Oracles. Other Titles in Applied Mathematics. Philadelphia: Society for Industrial and Applied Mathematics (SIAM). doi:10.1137/1.9781611973525. ISBN 978-1-61197-351-8.

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