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Martin R
11:01
This is perhaps known:
e^{-x}
and
\exp(-x)
are treated differently by Approach0.
playing with Approach0 I found another duplicate:
math.stackexchange.com/questions/933462/…
. However, the search engine is quite sensitive : searching with $exp(-x)$ doesn't pick proper results while $e^{-x}$ does! —
Pierre H.
1 hour ago
I can confirm that
$e^{-x} \geq \frac{2-x}{2+x} $
and
$\exp(-x) \geq \frac{2-x}{2+x} $
return quite different results.
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