Transcendental equation#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.

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

• applying the natural logarithm to both sides may yield an algebraic equation,Bronstein et al., Sect.1.6.4.2.a, p.46 e.g.

: $4^x\; =\; 3^\{x^2-1\}\; \backslash cdot\; 2^\{5x\}$ transforms to $x\; \backslash ln\; 4\; =\; (x^2-1)\; \backslash ln\; 3\; +\; 5x\; \backslash ln\; 2$, which simplifies to $x^2\; \backslash ln\; 3\; +\; x(5\; \backslash ln\; 2\; -\; \backslash ln\; 4)\; -\backslash ln\; 3\; =\; 0$, which has the solutions $x\; =\; \backslash frac\{\; -3\; \backslash ln\; 2\; \backslash pm\; \backslash sqrt\{9(\backslash ln\; 2)^2\; -\; 4\; (\backslash ln\; 3)^2\}\; \}\{\; 2\; \backslash ln\; 3\; \}\; .$

: 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=q^x may succeed, e.g.

: $2^\{x-1\}\; +\; 4^\{x-2\}\; -\; 8^\{x-2\}\; =\; 0$ transforms, using y=2^x, to $\backslash frac\{1\}\{2\}\; y\; +\; \backslash frac\{1\}\{16\}\; y^2\; -\; \backslash frac\{1\}\{64\}\; y^3\; =\; 0$ which has the solutions $y\; \backslash in\; \backslash \{\; 0,\; -4,\; 8\backslash \}$, hence $x=\; \backslash log\_2\; 8\; =\; 3$ is the only real solution.Bronstein et al., Sect.1.6.4.2.b, p.46

: 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=xe^x 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|date=January 2022}} e.g.:

: $x^2e^\{2x\}\; +\; 2\; =\; 3x\; e^x$ transforms to $y^2\; +\; 2\; =\; 3y,$ which has the solutions $y\; \backslash in\; \backslash \{1,2\backslash \},$ hence $x\; \backslash in\; \backslash \{\; W\_0(1),\; W\_0(2),\; W\_\{-1\}(1),\; W\_\{-1\}(2)\; \backslash \}$, where $W\_0$ and $W\_\{-1\}$ denote the real-valued branches of the multivalued $W$ function.

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

• applying exponentiation to both sides may yield an algebraic equation, e.g.

:$2\; \backslash log\_5\; (3x-1)\; -\; \backslash log\_5\; (12x+1)\; =\; 0$ transforms, using exponentiation to base $5.$ to $\backslash frac\{\; (3x-1)^2\; \}\{\; 12x+1\; \}\; =\; 1,$ which has the solutions $x\; \backslash in\; \backslash \{\; 0,\; 2\backslash \}\; .$ If only real numbers are considered, $x\; =\; 0$ is not a solution, as it leads to a non-real subexpression $\backslash log\_5(-1)$ 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.Bronstein et al., Sect.1.6.4.3.b, p.46

• if all "logarithm calls" have a unique base $b$ and a unique argument expression $f(x),$ then substituting $y\; =\; \backslash log\_b\; (f(x))$ may lead to a simpler equation,Bronstein et al., Sect.1.6.4.3.a, p.46 e.g.

: $5\; \backslash ln(\backslash sin\; x^2)\; +\; 6\; =\; 7\; \backslash sqrt\{\; \backslash ln(\backslash sin\; x^2)\; +\; 8\; \}$ transforms, using $y\; =\; \backslash ln(\backslash sin\; x^2)\; ,$ to $5\; y\; +\; 6\; =\; 7\; \backslash sqrt\{\; y\; +\; 8\; \},$ which is algebraic and has the single solution $y=\backslash frac\{89\}\{25\}$.{{efn|Squaring both sides obtains $25y^2+60y+36\; =\; 49\; \backslash cdot\; (y+8)$ which has the additional solution $y=-4$; however, the latter does not solve the unsquared equation.}} After that, applying inverse operations to the substitution equation yields $x\; =\; \backslash sqrt\{\; \backslash arcsin\; \backslash exp\; y\; \}\; =\; \backslash sqrt\{\; \backslash arcsin\; \backslash exp\; \backslash frac\{89\}\{25\}\; \}.$

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

• applying Pythagorean identities and trigonometric sum and multiple formulas, arguments of the forms $\backslash sin(nx+a),\; \backslash cos(mx+b),\; \backslash tan(lx+c),\; ...$ with integer $n,m,l,...$ might all be transformed to arguments of the form, say, $\backslash sin\; x$. After that, substituting $y\; =\; \backslash sin(x)$ yields an algebraic equation,Bronstein et al., Sect.1.6.4.4, p.46-47 e.g.

: $\backslash sin(x+a)\; =\; (\backslash cos^2\; x)\; -\; 1$ transforms to $(\backslash sin\; x)(\backslash cos\; a)\; +\; \backslash sqrt\{\; 1\; -\; \backslash sin^2\; x\; \}(\backslash sin\; a)\; =\; 1\; -\; (\backslash sin^2\; x)\; -\; 1$, and, after substitution, to $y\; (\backslash cos\; a)\; +\; \backslash sqrt\{\; 1\; -\; y^2\; \}(\backslash sin\; a)\; =\; -\; y^2$ which is algebraic{{efn|over an appropriate field, containing $\backslash sin\; a$ and $\backslash cos\; a$}} and can be solved. After that, applying $x\; =\; 2k\backslash pi\; +\; \backslash arcsin\; y$ obtains the solutions.

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

• unfolding them by their defining exponential expressions and substituting $y\; =\; \backslash exp(x)$ yields an algebraic equation,Bronstein et al., Sect.1.6.4.5, p.47 e.g.

: $3\; \backslash cosh\; x\; =\; 4\; +\; \backslash sinh\; (2x-6)$ unfolds to $\backslash frac\{3\}\{2\}\; (e^x\; +\; \backslash frac\{1\}\{e^x\})\; =\; 4\; +\; \backslash frac\{1\}\{2\}\; \backslash left(\; \backslash frac\{(e^x)^2\}\{e^6\}\; -\; \backslash frac\{e^6\}\{(e^x)^2\}\; \backslash right)\; ,$ which transforms to the equation $\backslash frac\{3\}\{2\}\; (y\; +\; \backslash frac\{1\}\{y\})\; =\; 4\; +\; \backslash frac\{1\}\{2\}\; \backslash left(\; \backslash frac\{y^2\}\{e^6\}\; -\; \backslash frac\{e^6\}\{y^2\}\; \backslash right)\; ,$ which is algebraic{{efn|over an appropriate field, containing $e^6$}} and can be solved. Applying $x\; =\; \backslash ln\; y$ obtains the solutions of the original equation.

File:Sin x = ln x svg.svg

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 $x\_0$, computing $f(x)$ which becomes $x\_1$ 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 $x\_0$ 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 $k\; \backslash approx\; 1$, the solutions of $\backslash sin\; x\; =\; k\; x$ are approximately those of $(1-k)\; x\; -\; x^3/6=0$, namely $x=0$ and $x\; =\; \backslash plusmn\; \backslash sqrt\{6\}\; \backslash sqrt\{1-k\}$.

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).

• Some transcendental systems of high-order equations can be solved by “separation” of the unknowns, reducing them to algebraic equations.[http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=zvmmf&paperid=7414&option_lang=eng 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), 214–221V.A. Varyukhin, Fundamental Theory of Multichannel Analysis (VA PVO SV, Kyiv, 1993) [in Russian]
• The following can also be used when solving transcendental equations/inequalities: If $x\_0$ is a solution to the equation $f(x)=g(x)$ and $f(x)\backslash leq\; c\backslash leq\; g(x)$, then this solution must satisfy $f(x\_0)=g(x\_0)=c$. For example, we want to solve $\backslash log\_\{2\}\backslash left(3+2x-x^\{2\}\backslash right)=\backslash tan^\{2\}\backslash left(\backslash frac\{\backslash pi\; x\}\{4\}\backslash right)+\backslash cot^\{2\}\backslash left(\backslash frac\{\backslash pi\; x\}\{4\}\backslash right)$. The given equation is defined for

• {{annotated link|Mrs. Miniver's problem}}
• Goat grazing problem - another problem on areas of intersecting circles

{{notelist}}

{{reflist}}

• {{cite book | doi=10.1137/1.9781611973525 | isbn=978-1-61197-351-8 | author=John P. Boyd | editor= | title=Solving Transcendental Equations: The Chebyshev Polynomial Proxy and Other Numerical Rootfinders, Perturbation Series, and Oracles | location=Philadelphia | publisher=Society for Industrial and Applied Mathematics (SIAM) | series=Other Titles in Applied Mathematics | volume= | edition= | year=2014 }}

Category:Equations