From: "Will Self" <wself@msubillings.edu>
Subject: Re: is a Taylo polynomial approximant of exp(z) Hurwitz ?
Date: Mon, 12 Feb 2001 15:45:49 -0700
Newsgroups: sci.math
Summary:  Lucas' theorem (roots of derivatives of polynomials)

<rge11@my-deja.com> wrote in message news:968rfq$1c4$1@nnrp1.deja.com...
> exp(z) has no roots in the finite z plane but expand exp(z) in to its
> Taylor series. The first three approximants 1+z, 1+z+z^2/2! and
> 1+z+z^2/2!+z^3/3! are Hurwitz, that is the roots fall on the Re(z) < 0
> half plane. The next few approximants all have roots on Re(z) > 0, and
> therefore are not Hurwitz.
>
> My question: is there an n>3 for which the polynomial
>
> 1+z+z^2/2!+z^3/3!+....+z^n/n!
>
> has all its roots on Re(z)<0?
>
> In general, what can be said about the location of the roots of the
> Taylor polynomial approximants of an entire function that has no roots
> in the finite plane?
>
By Lucas' theorem, the roots of the derivative of a polynomial must lie
inside the convex hull of the roots of the polynomial.  In this interesting
situation, the derivative of the nth Taylor polynomial is the (n-1)st Taylor
polynomial.  Thus, if any of the Taylor polynomials has a root with positive
real part, so do all the later ones.

I did some crunching and found that the real part of the root with largest
real part (could I have phrased that better?) seems to go up roughly
linearly with n, approximately r = .6n.  This is certainly fascinating.

Will Self