Is there a general way to solve transcendental equations?

Question

To make things definite, let's narrow them and call transcendental equation of the form

$$f(x) = 0$$

where $f$ is a real elementary function in the usual sense. For example

$$\cos(\pi x) + x^2 = 0$$ or $$a = x \tan x$$

Is there a general way to solve such equations except for numerics? That is to produce an expression$$x_0 = \text{RHS}$$ where $\text{RHS}$ does not depend on $x_0$ and can be actually computed? Maybe in a form of series or something similar, not in "finite terms".

There is of course a question about the existance of solutions, it would be nice if the form of $\text{RHS}$ incorporatted the answer to it.

In the formulation above the problem seems to be equivalent to finding local inverse function.

Answer 1

Under certain assumptions you can use inversion of power series to obtain a solution. If $f(z)$ be analytic at $z_0$ where $f'(z_0)\ne 0$, then $w=f(z)$ has an analytic inverse $z=g(w)$ in some neighbourhood of $w_0=f(z_0)$, hence $$z-z_0=\sum_{k=1}^{\infty}a_k(w-w_0)^k$$ where $$a_k=\frac{1}{n!}\left[\frac{(z-z_0)^n}{(f(z)-w_0)^n}\right]^{(n-1)}_{z-z_0}$$

If the equation is of the form $$z=a+wf(z)$$ where $a$ is inside the domain of analyticity of $f(z)$ and $f(a)\ne 0$, then $$z=a+\sum_{k=1}^{\infty}\frac{w^k}{k!}\left[f^{k}(a)\right]^{(k)}$$ For example, a solution to the Kepler's equation $$z=m+E\sin z$$ can be expressed as follows $$z=m+E\sin m+\frac{E^2}{2!}(\sin m)' +...$$

Your second example can be written as $$z=w\cot z$$ So if you work your way through the formulae, you should be able to obtain some form of a solution.

Answer 2

If your hope is to conclude with a statement like $x=\mathrm{RHS}$, where $\mathrm{RHS}$ is itself an expression using elementary functions, composition, rational numbers, and well-known constants, then you cannot even do this for general polynomial equations of degree $5$ or higher without using some kind of infinite process, like a series or a recursion. See here, for example.

This is not a full answer to your question though - you are also asking if solutions can be found through series.

Answer 3

In general, the elementary functions are analytic except for some isolated singularities and branch cuts, so they and their local inverses will have convergent Taylor series expansions in suitable disks. Look up "reversion of series".

For example, with $f(x) = \cos(\pi x) + x^2$, a convenient starting point might be $x=1/2$ for which $f(x) = 1/4$. We have $$f(x) = {\frac {1}{4}}+ \left( 1-\pi \right) \left( x-{\frac {1}{2}} \right) + \left( x-{\frac {1}{2}} \right) ^{2}+\frac16\,{\pi }^{3} \left( x-{\frac {1}{2}} \right) ^{3}-{ \frac {1}{120}}\,{\pi }^{5} \left( x-{\frac {1}{2}} \right) ^{5}+\ldots $$ and then a solution of $f(x) = t$ is $$ x = {\frac {1}{2}}- \left( -1+\pi \right) ^{-1} \left( t-{\frac {1}{4}} \right) + \left( -1+\pi \right) ^{-3} \left( t-{\frac {1}{4}} \right) ^{2}-1/6\,{\frac {{\pi }^{4}-{\pi }^{3}+12}{ \left( -1+\pi \right) ^{5}}} \left( t-{\frac {1}{4}} \right) ^{3}+5/6\,{\frac {-{ \pi }^{3}+6+{\pi }^{4}}{ \left( -1+\pi \right) ^{7}}} \left( t-{ \frac {1}{4}} \right) ^{4}-{\frac {1}{120}}\,{\frac {{\pi }^{5}+9\,{ \pi }^{8}-17\,{\pi }^{7}+1680+7\,{\pi }^{6}-420\,{\pi }^{3}+420\,{\pi }^{4}}{ \left( -1+\pi \right) ^{9}}} \left( t-{\frac {1}{4}} \right) ^{5}+O \left( \left( t-{\frac {1}{4}} \right) ^{6} \right) $$