Vinzent Steinberg

bc.. r722 **Status:** Fixed p. Original comment: "http://code.google.com/p/mpmath/issues/detail?id=103#c3":http://code.google.com/p/mpmath/issues/detail?id=103#c3 p. Original author: "https://code.google.com/u/[email protected]/":https://code.google.com/u/[email protected]/

bc.. The problem is that the method to evaluate f(*x0) does not work here. It should however work with multidimensional=True, which it does not. Is there a way to get...

bc.. I thinks that it's a one dimensional function and thus tries to use the secant method. p. Original comment: "http://code.google.com/p/mpmath/issues/detail?id=103#c7":http://code.google.com/p/mpmath/issues/detail?id=103#c7 p. Original author: "https://code.google.com/u/[email protected]/":https://code.google.com/u/[email protected]/

bc.. For the secant method x0 can be twodimensional, for other onedimesional solvers even threedimensional. So this works only for len(x0) > 3. p. Original comment: "http://code.google.com/p/mpmath/issues/detail?id=103#c9":http://code.google.com/p/mpmath/issues/detail?id=103#c9 p. Original author:...

bc.. Function using *args are usually multidimensional, because you don't want to use *args for one argument. So it should be implemented this way. BTW, findroot(lambda x, y: x+y, (1,...

bc.. Being underdetermined means that the solution is a function. You can transform x+y==0 to x==-y and you got a solution depending on another unknown. However you have turn some...

bc.. > Yes, i think it would be useful to return an arbitrary solution. So let's assign random values to random variables until we get a determined system? Shouldn't this...

bc.. I fixed findroot() for f(x0) == 0. > If it's easy for the rootfinder to > pick a zero, then it should do it. How? Let's say we have...

bc.. Well, this should work: > > > findroot([lambda x,y:(x-1)**2+y**2, lambda x,y:0], (2, 2)) > > > Traceback (most recent call last): > > > File "", line 1, in...