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To solve $\log G=A$ with $A(0)=0$, start with $G=1$ and update $G\leftarrow G(1+A-\log G)$ 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.

The derivative fixed-point equation is unsuitable as an executable definition: requesting its first coefficient asks for the second coefficient of the same stream, then the third, indefinitely. It is not productive. As an algebraic equation it is also missing the constant coefficient: every scalar multiple of the formal power series exponential, including zero, satisfies it.

The integral equation gives the useful definition:

expx :: Fractional a => [a]
expx = plus one (integral expx)

The initial coefficient is available as 1 + 0; each subsequent coefficient needs only a previously determined coefficient. The recurrence relation is $a_0=1$ and $a_{n+1}=a_n/(n+1)$, so $a_n=1/n!$. This proves both uniqueness of the formal solution and productivity of every finite coefficient list prefix.