Take the group of all Translation in $R_{1}$.

The way to think about this is:

- the translation group operates on the argument of a function $f(x)$
- the generator is an operator that operates on $f$ itself

So let's take the exponential map:
and we notice that this is exactly the Taylor series of $f(x)$ around the identity element of the translation group, which is 0! Therefore, if $f(x)$ behaves nicely enough, within some radius of convergence around the origin we have for finite $x_{0}$:

$e_{x_{0}∂x∂}f(x)=(1+x_{0}∂x∂ +x_{0}∂x_{2}∂_{2} +…)f(x)$

$e_{x_{0}∂x∂}f(x)=f(x+x_{0})$

This example shows clearly how the exponential map applied to a (differential) operator can generate finite (non-infinitesimal) Translation!