On Sat, 24 Sept 2022 at 09:24, Peter Stahlecker <peter.stahlec...@gmail.com> wrote: > > Thanks! I did not know about nsolve. > Would nsolve be ‚better‘ than converting the equations to numpy, using > lambdify, and the use scipy.optimize.fsolve ?
When you use nsolve more or less the same thing happens: the equations are lambdified but converted to mpmath rather than numpy and then solved with mpath's findroot function rather than scipy's fsolve. The tradeoffs between using mpmath vs numpy are the usual ones for variable precision vs fixed precision: accuracy vs speed. Since numpy/scipy use fixed precision hardware level floating point (i.e. 64 bit floats) they can be faster than mpmath which uses software level arbitrary precision floating point. Likewise since numpy/scipy use lots of C and Fortran code they can also be faster than mpmath which is pure Python. On the other hand there may not be much of a speed difference for relatively "small" systems that do not benefit significantly from vectorisation and where the overheads of the Python wrappers in numpy/scipy dominate over the routines that are implemented in the compiled parts. Since mpmath uses variable precision arithmetic it can calculate answers that are more accurate than is possible when using numpy/scipy or any other fixed precision libraries e.g.: In [5]: nsolve(x**2 - 2, 1) Out[5]: 1.41421356237310 In [6]: nsolve(x**2 - 2, 1, prec=50) Out[6]: 1.4142135623730950488016887242096980785696718753769 The default precision used by nsolve is 15 decimal digits i.e. basically the same precision as you would expect for 64 bit floats (53 binary digits, ~15 decimal digits) so it should return a result with the same precision as fsolve. That does not mean that there is no accuracy gain by default though. Solvers like fsolve that work *within* the same precision as the output format that they return will typically return a result that is less accurate than implied by the precision. In other words the fact that fsolve uses 64 bit floats does not imply that it returns the most accurate possible 64 bit floats: In [7]: from scipy.optimize import fsolve In [8]: fsolve(lambda x: x**2 - 2, 1) # the digits shown are correct Out[8]: array([1.41421356]) In [9]: fsolve(lambda x: x**2 - 2, 1)[0] # last digits not quite right Out[9]: 1.4142135623730947 With mpmath's findroot even if you only want the accuracy of 64 bit floats it will use a higher precision internally in order to round the result to the desired precision at the end. The example shown above of x**2 - 2 is a relatively well behaved one where both nsolve and fsolve can easily find an accurate answer although the answer from nsolve is slightly more accurate. If you try something more numerically difficult then you can see bigger output errors from both functions as well as cases where nsolve throws an error because it doesn't think that it has enough precision to be able to converge. Lastly, nsolve is more convenient just because you don't have to use lambdify explicitly. Ideally it would be possible to have nsolve call fsolve so that it would be easier to switch between findroot and fsolve. -- Oscar -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/CAHVvXxRqaf2xOiL%2BuM6zDMLpk8UYgwX44ay%2BGb%2Bbi-8KwqiPUA%40mail.gmail.com.
