3 ms·
>since it requires drawing two real numbers within some absurdly small epsilon of each other. Not small, zero.
by thrown_22 4y ago
>since it requires drawing two real numbers within some absurdly small epsilon of each other.
Not small, zero.
- MontyCarloHall 4y agoDefinitely, if we care about the outcome of the trajectories as t\to\infty. For certain systems, at smaller timescales, merely an absurd amount of precision will suffice [0,1]. [0] https://en.m.wikipedia.org/wiki/Lyapunov_exponent https://en.m.wikipedia.org/wiki/Lyapunov_exponent [1] https://en.m.wikipedia.org/wiki/Lyapunov_time https://en.m.wikipedia.org/wiki/Lyapunov_time