Improve approximation of $exp(x)$ Given some crude approximation of $exp(x)$, how to improve its precision?
I could compute $exp(x)$ with, for example, Taylor series from scratch but it is not clear to me, how to take advantage of the approximation I already have.
 A: You can use the exponential properties
$$
\exp(x)=\exp(x/n)^n
$$
esp. for dyadic powers in $n$, and
$$
\exp(x)=2^n·\exp(x-n\ln(2))
$$
to reduce the size of the argument and thus hopefully get better results.
A: This can be done if for some $r>0$ you have an approximation to $\exp(x)$ in some functional form that is valid for all $x\in [0,r]$. You can then use that $y(x) = \exp(x)$ is a solution to the differential equation:
$$y' = y$$
with $y(0) = 1$. It then follows that:
$$y(x) = 1 + \int_0^{x} y(t) dt$$
Suppose that instead of $y(x) = \exp(x)$ we substitute some arbitrary function $f(x)$ that satisfies the boundary condition $f(0) = 1$ on the right hand side of the equation. Then the left hand side won't be equal to $f(x)$, it will be some other function $g(x)$. It can be shown that in some sense $g(x)$ will be a better approximation to $\exp(x)$ than $f(x)$, this means that iterating this integral equation will lead to a sequence of functions that converges to $\exp(x)$. This is true for general linear differential equations and it is used  to prove that solutions that satisfy the boundary conditions are unique.
