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Let $\Phi (z,t)$ be a polynomial given by $$ \Phi(z,t) := z^n + A_{n-1}(t) z^{n-1} + \ldots + A_1(t) z + A_0(t).$$ Assume that $\Phi(0,0) =0$. It is a fact that a solution $z(t)$ of the equation $$ \Phi(z(t), t) =0 $$ that is close to zero, can be expressed as a formal power series in $t^{1/r}$ for some positive integer $r$. Moreover, this formal power series has a non zero radius of convergence. This follows from the "Newton Pusieux Theorem".

My question is the following: Is there some procedure/algorithm to find out what this $r$ is? Of course it will depend on the $A_i$.

As an example, suppose we want solutions for $$ \Phi(z,t) = z^2 + K z + t =0$$ where $K$ is some constant. Then a solution $z(t)$ is a power series in $t$ if $K \neq 0$. It is a power series in $t^{1/2}$ if $K=0$.

In general is there some algorithm to find this $r$?

Everything is over the field of complex numbers $\mathbb{C}$

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up vote 8 down vote accepted

This algorithm is called the Newton Polygon, and it was really invented and carefuly described by Newton, with examples, see for example,

MR1836037 Fischer, Gerd Plane algebraic curves. Translated from the 1994 German original by Leslie Kay. Student Mathematical Library, 15. American Mathematical Society, Providence, RI, 2001.

or any other book with a similar title, or Mathematical Papers of Newton himself.

In XIX century this algorithm was a part of high school curriculum, at least in England, see, for exapmple

JFM 31.0103.05 Chrystal, G. Algebra. Elementary textbook. 2. edition. Part II. (English) London. 632 S. Published: 1900

I think, for this reason, those who wrote university algebra textbooks in XX century did not care to include it (Lang, Van der Waerden,...).

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I see. Is there any computer algebra system that implements this algorithm? The input should be a polynomial $\Phi(z,t)$ and the output should give me $r$, and ideally the Pusieux series $z(t)$. –  Ritwik Jul 16 '13 at 2:39
I am sure there is. But let some person more knowledgeable in computer algebra answer this. –  Alexandre Eremenko Jul 16 '13 at 10:29
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