In general in a correctly implemented fixed-step ODE solver method you have 3 sources for numerical errors: the theoretical method truncation error, the floating point error from evaluating the ODE function and composing the method step and the error from accumulating the single updates of size $O(h)$ to the integration result of size $O(1)$. So the total ...


I find an 8 times overhead a bit surprising, the most intense operation i have in mind is 4 times overhead, namely multiplication. The rest could be due to an inefficient library though. I am using JuliaIntervals and never has such a slow down. About your question, Interval Arithmetics computes the error along the path of computation you took you can then ...


In terms of Python, use from cmath import log, exp def clog(x): if x == 0: return -float('inf') else: return log(x) For a negative value of $x$, this gives $\log(x) = \log(-x) + i \pi$, and for $\log(0)$, this gives $\infty$.

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