I haven't thought about this at all, but for what it's worth, sympy
gives the same answer (minus the tiny imaginary part):
from sympy import nroots
from sympy.abc import x
g = x**11 + 1000*x**10 + 500 *x**9 + 1
nroots(g)[0]
-999.499749749687
The documentation for
https://docs.sympy.org/latest/guides/solving/find-roots-polynomial.html
mentions that
nroots() can fail sometimes for polynomials that are numerically ill
conditioned, for example Wilkinson’s polynomial. Using RootOf() and
evalf() as described in Numerically Evaluate CRootOf Roots can avoid
both ill-conditioning and returning spurious complex parts because it
uses a more exact, but much slower, numerical algorithm based on
isolating intervals.
FWIW the slow/more exact method described here still gives the large
root.
from sympy import solve
g_solved = solve(g, x, dict=True);
for root in g_solved:
print(root[x].n(10))
Given the convergence of different methods, this is very unlikely to
be a bug in R or the underlying algorithm. Either there's a thinko and
-999.499... is actually a root, or this is a badly conditioned problem
for which it's extremely hard to get accurate numerical solutions ...
On 2025-10-02 12:14 p.m., tgs77m--- via R-help wrote:
Colleagues,
g <- function(x) ( x^11 + 1000*x^10 + 500 *x^9 + 1 )
coeffs <- c(1, rep(0, 8), 500, 1000, 1)
roots <- polyroot(coeffs)
Output
[1] 0.25770068+3.958197e-01i
[2] -0.34615184+3.782848e-01i
[3] -0.04089779-4.838134e-01i
[4] 0.44124314-1.517731e-01i
[5] -0.04089779+4.838134e-01i
[6] -0.56201931-1.282822e-01i
[7] -0.34615184-3.782848e-01i
[8] 0.44124314+1.517731e-01i
[9] -0.56201931+1.282822e-01i
[10] 0.25770068-3.958197e-01i
[11] -999.49974975+1.110223e-16i
[11] -999.49974975+1.110223e-16i makes no sense since f>0 for all x
Why does polyroot do this?
Thomas Subia
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