Python fsolve fprime 的形状是错误的吗？

Question

The following could not be much simpler, yet I cannot find how to make it work...

from scipy.optimize import fsolve

def a(x):
    return x*x-x

def ap(x):
    return 2*x-1

#works
print(fsolve(a, 0.3))

#works
print(fsolve(a, 0.3, fprime=ap))

#works
print(fsolve(a, [0.3], fprime=ap))

#works
print(fsolve(a, [0.3, 0.7]))

#crashes
print(fsolve(a, [0.3, 0.7], fprime=ap))

When it crashes is gives the error

TypeError: fsolve: there is a mismatch between the input and output shape of the 'fprime' argument 'ap'.Shape should be (2, 2) but it is (2,).

The output dimension of ap seems like it should definitely be the same as the input. How could this possibly be going wrong (and how do I fix it)?

I think some downvoters are missing the subtlety of the question so here is a more in depth explanation of why I am confused:

It seems that scipy interprets a as a function of one variable with [0.3,0.7] starting estimates of two roots in fsolve(a, [0.3, 0.7]) , but in fsolve(a, [0.3, 0.7], fprime=ap) it interprets a as function of two variables with [.3,.7] being the estimate of a single root. According to the documentation ( https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fsolve.html ), the second argument

x0 : ndarray
    The starting estimate for the roots of func(x) = 0.

says that it is looking for estimate of the roots ( plural ). But its behavior in the case where fprime is given makes it sound like x0 is interpreted as an estimate of a single root.

Answer 1

I suspect you expected fsolve(a, [0.3, 0.7], fprime=ap) to return two solutions to the scalar problem. fsolve does not work that way.

When you call fsolve(a, x0, fprime=ap) , the fsolve function infers the dimensions of the problem from the shape of x0 . If x0 is a scalar, it expects a to accept a scalar, and fprime must accept a scalar and return a scalar (or a 1x1 array). If x0 is a sequence of length 2 (as in your example that didn't work), fsolve expects a to accept an array of length 2 as its x argument and return a sequence of length 2. It also expects fprime to accept a array of length 2 and return an array (the Jacobian matrix) with shape 2x2.

Answer 2

Thanks to everyone who explained why I was getting the behavior I was getting. No one actually told me how to fix the issue though, so I'm writing this answer since I figured it out.

The confusion stems from the fact that the scipy documentation is somewhere between wrong and misleading.

The documentation says:

Return the roots of the (non-linear) equations defined by func(x) = 0 given a starting estimate.

suggesting that for a scalar function it may return an array of roots. The actual behavior is closer to

Returns a root of the (non-linear) equation defined by func(x) = 0 given a starting estimate.

This would have made what to do obvious. To find multiple roots of the equation, we can let scipy (as it always does) interpret a(x) as a function of n variables, where n is the length of x . To find multiple roots let a be vectorized, ie

a([x1, x2, ..., xn]) = [a(x1), a(x2), ..., a(xn)] .

The jacobian of this function is a diagonal matrix

diag([ap(x1), ap(x2), ..., ap(xn)]) .

Thus the answer to my question is :

Change

print(fsolve(a, [0.3, 0.7], fprime=ap))

to

print(fsolve(a, [0.3, 0.7], fprime=lambda x: np.diag(ap(x))))
# correctly outputs
# [ -3.69321143e-16   1.00000000e+00]

or simply loop over [0.3, 0.7] calling fsolve once per initial guess.