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As stated above, can I solve things like $p(x) = e^{h(x)}$ without approximation? If not is there an explanation?

Lets say here that $h$ and $p$ are polynomials.

I can't say I've done much more than try the typical algebraic tricks on easy equations, but nothing normal worked. I also determined by graphing some basic equations that often solutions do not exist, so maybe there is something to be said there about when solutions exist.

So I am interested if there is some trick method here that works, either in specific cases, or in general. Or any theory behind these types of equations involving analytic functions with infinite series expansions such as $exp$.

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I looked briefly for duplicate questions but didn't find any immediately. – AnonymousCoward Mar 14 '12 at 2:58
They had to invent a special function just to deal with inverting $xe^x$; I don't think there's a method here. – anon Mar 14 '12 at 4:23
Relevant reading: Lambert's W. – user2468 Mar 14 '12 at 4:35
In what sense is there a non-numerical way to solve an equation of the form $x^2 = 2$? We "know" that the answer is $\sqrt{2}$ but at the end of the day all this means is "the unique positive real such that $x^2 = 2$" and this is really a tautology. What counts is that we know how to compute arbitrarily many digits of this number and so forth and this is a numerical thing. – Qiaochu Yuan Mar 14 '12 at 5:50
@Qiaochu I do not find your comment remotely helpful. If you have a disagreement with the question perhaps suggest a refinement? – AnonymousCoward Mar 14 '12 at 6:09
up vote 1 down vote accepted

The method I have seen is to differentiate the equation, so $p'(x) = h'(x) e^{h(x)} = h'(x) p(x)$. If you then expand $p(x)$ and $h(x)$ as power series, you can then iteratively get the coefficients of $h(x)$ in terms of $p(x)$.

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