I have the following Mathematica code:
Gdd = {{-((E^(2 AA[z]) L^2 g[z, B])/z^2), 0, 0, 0, 0}, {0, (
E^(2 AA[z]) L^2)/(z^2 g[z, B]), 0, 0, 0}, {0, 0, (
E^(2 AA[z]) L^2)/z^2, 0, 0}, {0, 0, 0, (
E^(B^2 z^2 + 2 AA[z]) L^2)/z^2, B/bb}, {0, 0, 0, -(B/bb), (
E^(B^2 z^2 + 2 AA[z]) L^2)/z^2}};
Gddinverse = {{-((E^(-2 AA[z]) z^2)/(L^2 g[z, B])), 0, 0, 0, 0}, {0, (
E^(-2 AA[z]) z^2 g[z, B])/L^2, 0, 0, 0}, {0, 0, (
E^(-2 AA[z]) z^2)/L^2, 0, 0}, {0, 0, 0, (
bb^2 E^(B^2 z^2 + 2 AA[z]) L^2 z^2)/(
bb^2 E^(2 B^2 z^2 + 4 AA[z]) L^4 + B^2 z^4), -((B bb z^4)/(
bb^2 E^(2 B^2 z^2 + 4 AA[z]) L^4 + B^2 z^4))}, {0, 0, 0, (
B bb z^4)/(bb^2 E^(2 B^2 z^2 + 4 AA[z]) L^4 + B^2 z^4), (
bb^2 E^(B^2 z^2 + 2 AA[z]) L^2 z^2)/(
bb^2 E^(2 B^2 z^2 + 4 AA[z]) L^4 + B^2 z^4)}};
AA[z_] = -a z^2;
g[z_, B_] =
1 - Integrate[z^3 Exp[-B^2 z^2 - 3 AA[z]], {z, 0, z}]/
Integrate[z^3 Exp[-B^2 z^2 - 3 AA[z]], {z, 0, zh}];
ff[z_, B_] = -(1/z^2) 2 E^(2 (B^2 z^2 + AA[z])) L^2 Sqrt[
1 + (B^2 E^(-2 B^2 z^2 - 4 AA[z]) z^4)/(
bb^2 L^4)] (g[z, B] (-2 + 2 B^2 z^2 + 3 z Derivative[1][AA][z]) +
z D[g[z, B], z]);
term[z_, B_] =
Sqrt[-Det[Gdd]] Gddinverse[[2, 2]] Gddinverse[[3, 3]] ff[z, B] //
PowerExpand // Simplify;
V2[z_] = c1 Log[z] (z^4 (1 + a2 z^2)) + (1 + b2 z^2 + b4 z^4);
eq2[z_] =
D[V2[z], {z, 2}] +
D[Log[term[z, B] ], z] D[V2[z],
z] - \[Omega]^2 ((Gddinverse[[1, 1]]/Gddinverse[[2, 2]])) V2[z];
seriesexpneq12 = Series[eq2[z], {z, 0.0, 1}] // Normal
It gives the answer as:
-4 b2 + \[Omega]^2
But when I use 0.0000000000000001 instead of 0.0, I get the answer:
2 b2 + (1. + 1.*10^-32 b2 + 1.*10^-64 b4 - 3.68414*10^-63 c1 -
3.68414*10^-95 a2 c1) \[Omega]^2 + (2.*10^-16 b2 +
4.*10^-48 b4 + (-1.46365*10^-46 -
2.20048*10^-78 a2) c1) (-1.*10^-16 + z) \[Omega]^2
Why does Mathematica flip the sign of the leading order term in this case? Which answer should be considered the correct one?
Series
, and that the first is the incorrect one. $\endgroup$