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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.

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Would you call $x=0$ a transcendental equation? :) – anon Jun 6 '12 at 17:35
Iterative methods abound. – user31373 Jun 6 '12 at 17:36
That depends on what you mean by "solve." – Qiaochu Yuan Jun 6 '12 at 18:11
up vote 3 down vote accepted

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.

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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 $$ x = {\frac {1}{2}}- \left( -1+\pi \right) ^{-1} \left( t-{\frac {1}{4}} \right) + \left( -1+\pi \right) ^{-3} \left( t-{\frac {1}{4}} \right) ^{2}-1/6\,{\frac {{\pi }^{4}-{\pi }^{3}+12}{ \left( -1+\pi \right) ^{5}}} \left( t-{\frac {1}{4}} \right) ^{3}+5/6\,{\frac {-{ \pi }^{3}+6+{\pi }^{4}}{ \left( -1+\pi \right) ^{7}}} \left( t-{ \frac {1}{4}} \right) ^{4}-{\frac {1}{120}}\,{\frac {{\pi }^{5}+9\,{ \pi }^{8}-17\,{\pi }^{7}+1680+7\,{\pi }^{6}-420\,{\pi }^{3}+420\,{\pi }^{4}}{ \left( -1+\pi \right) ^{9}}} \left( t-{\frac {1}{4}} \right) ^{5}+O \left( \left( t-{\frac {1}{4}} \right) ^{6} \right) $$

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Maybe I am reading too much into OP's question, but I don't see this kind of answer as what OP wants since the series coefficients (post inversion) are not expressed as $f(n)$ for some elementary $f$. For each $n$, $f(n)$ seems to require hands-on computation. And if say, we are willing to settle for some inaccuracy and stop at $n=5$ as in this example, then why not use an iterative method (like say Newton's method) computed out to it's 5th iterate instead? – alex.jordan Jun 6 '12 at 18:28
@alex.jordan I don't think Robert did it by hand, though I don't know what algorithm was used. Maybe the one in Mathematica ( or other CAS. – Yrogirg Jun 6 '12 at 18:32
@alex.jordan Oh,sorry, I've missed your point, you were talking about the general expression for n'th term, should study that in detail. – Yrogirg Jun 6 '12 at 18:35
True, I didn't do it by hand, I used Maple. The general formula for the series coefficients is the one given by Valentin in his response. True, this is in general not an elementary function of $n$. – Robert Israel Jun 6 '12 at 20:54
@RobertIsrael, can one reverse Puiseux Series? – Yrogirg Jun 7 '12 at 12:34

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.

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