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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 ...


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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 ...


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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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