Numerical solution to a simple exponential equation
Subject: Numerical solution to a simple exponential equation
From: William Schelter
Date: Wed, 4 Jul 2001 05:37:28 -0500
You could use 'newton'
For example to get a root to 100 decimal places do
(C1) load("newton")$
(C2) fpprec:100$
(C3) NEWTON(x*exp(-x)-bfloat(1/4),x,bfloat(0),1.0b-100);
(D3) 3.5740295618138890306881110405590475331659055507601204362762044858967140#
25961457962896168513444411851B-1