Newton iteration for formal power series exponential
= Newton iteration for formal power series exponential
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To solve $\log G=A$ with $A(0)=0$, start with $G=1$ and update $G\gets G(1+A-\log G)$ at successively doubled truncation precision. The correction removes the current logarithmic error to first order. This is Newton iteration applied to the <formal power series exponential>, using a <formal power series logarithm>.