I was playing around with Vitaliy Kaurov's neat answer to this question, and found something a bit unusual:
In[1]:= formula = 0.006101617526907695` + E^-(-7.` + x)^2 +
E^(-1.0000502772595525` (4.899999190832501` + x)^2) -
0.002006784176061687` x^2;
In[2]:= formula // FullSimplify
Out[2]= 0.00610162 - 0.00200678 x^2
Wait, what?
In[3]:= FullSimplify[E^-(-7.` + x)^2 +
E^(-1.0000502772595525` (4.899999190832501` + x)^2)]
Out[3]= 0.
I have to say this is not the result I expected at all.
In[4]:= FullSimplify[E^-(-7.` + x)^2]
Out[4]= 0.
In[5]:= FullSimplify[E^(-1.0000502772595525` (4.899999190832501` + x)^2)]
Out[5]= 0.
Well, at least it gets points for consistency.
In[6]:= FullSimplify[E^-(-7.` + x)]
Out[6]= 1096.63 E^-x
In[7]:= FullSimplify[E^-(-7.` + x)^3]
Out[7]= E^(-1. (-7. + x)^3)
In[8]:= FullSimplify[E^-(-7.` + x)^2.]
Out[8]= E^-(-7. + x)^2.
In[9]:= FullSimplify[E^-(-1.` + x)^2]
Out[9]= E^(-1. (-1. + x)^2)
In[10]:= FullSimplify[E^-(5.` + x)^2]
Out[10]= 0.
I gotta say, the last one was sort of a relief. Just to be sure, I'll plot something I never expected to plot:
Plot[{Boole[FullSimplify[E^(-(y + x)^2)] === 0.]}, {y, -7, 7}]
FWIW, this is Mathematica 10.3.
FullSimplify[E^-(5 + x)^2]
$\endgroup$TrigExpand /@ ExpandAll /@ ExpToTrig[expr]
. $\endgroup$