OFFSET
0,1
COMMENTS
Let p(n,x) denote the n-th Maclaurin polynomial of e^x, and let p'(n,x) denote its derivative. Then p'(n+1,x) = p(n,x), so that the real zero of p(n,x), for odd n, is also the value of x that minimizes p(n+1,x). See A117605 for the (negated) real zero p(3,x).
EXAMPLE
Minimum value = 0.2703947652051846079624596133831109119614602128142...
MATHEMATICA
z = 150; p[n_, x_] := Normal[Series[E^x, {x, 0, n}]];
t = x /. NSolve[p[3, x] == 0, x, z][[1]]
RealDigits[t][[1]]
Plot[Evaluate[p[4, x]], {x, -3, 1}, PlotRange -> {-1, 3}]
CROSSREFS
KEYWORD
AUTHOR
Clark Kimberling, Feb 11 2020
STATUS
approved