# Taylor expansion 1 / [1+ln(e+x) ]Watch

#1
Taylor expansion 1 / [1+ln(e+x) ] about a point a.

No idea what to do. If it were just 1 / [ln(e+x)] it would be easy...
0
#2
Actually I think I get 1/2 - x/4e when a =0

f(x) = [1+ln(e+x) ] ^-1
f'(x) = -(e+x)^-1.[1+ln(x+e)]^-2

f(x) --> f(a) + (x-a).f'(a)/(1!) + (x-a)^2.f''(a)/(2!) + ...

WHen x = a = 0
We get f(x) = 1/2 - x/(4e) +... (the first two terms of the power series)
0
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