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Compute derivatives with finite-difference methods

Project description

FDM: Finite Difference Methods

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FDM estimates derivatives with finite differences.

See also FDM.jl.

See the docs.

Installation

pip install fdm

Multivariate Derivatives

from fdm import gradient, jacobian, jvp, hvp

For the purpose of illustration, let us consider a quadratic function:

>>> a = np.random.randn(3, 3); a = a @ a.T
>>> a
array([[ 3.57224794,  0.22646662, -1.80432262],
       [ 0.22646662,  4.72596213,  3.46435663],
       [-1.80432262,  3.46435663,  3.70938152]])
       
>>> def f(x):
...     return 0.5 * x @ a @ x

Consider the following input value:

>>> x = np.array([1.0, 2.0, 3.0])

Gradients

>>> grad = gradient(f)
>>> grad(x)
array([-1.38778668, 20.07146076, 16.25253519])

>>> a @ x
array([-1.38778668, 20.07146076, 16.25253519])

Jacobians

>>> jac = jacobian(f)
>>> jac(x)
array([[-1.38778668, 20.07146076, 16.25253519]])

>>> a @ x
array([-1.38778668, 20.07146076, 16.25253519])

But jacobian also works for multi-valued functions.

>>> def f2(x):
...     return a @ x

>>> jac2 = jacobian(f2)
>>> jac2(x)
array([[ 3.57224794,  0.22646662, -1.80432262],
       [ 0.22646662,  4.72596213,  3.46435663],
       [-1.80432262,  3.46435663,  3.70938152]])
       
>>> a
array([[ 3.57224794,  0.22646662, -1.80432262],
       [ 0.22646662,  4.72596213,  3.46435663],
       [-1.80432262,  3.46435663,  3.70938152]])

Jacobian-Vector Products (Directional Derivatives)

In the scalar case, jvp computes directional derivatives:

>>> v = np.array([0.5, 0.6, 0.7])  # A direction

>>> dir_deriv = jvp(f, v) 
>>> dir_deriv(x)
22.725757753354657

>>> np.sum(grad(x) * v)
22.72575775335481

In the multivariate case, jvp generalises to Jacobian-vector products:

>>> prod = jvp(f2, v)
>>> prod(x)
array([0.65897811, 5.37386023, 3.77301973])

>>> a @ v
array([0.65897811, 5.37386023, 3.77301973])

Hessian-Vector Products

>>> prod = hvp(f, v)
>>> prod(x)
array([[0.6589781 , 5.37386023, 3.77301973]])

>>> 0.5 * (a + a.T) @ v
array([0.65897811, 5.37386023, 3.77301973])

Scalar Derivatives

>>> from fdm import central_fdm

Let's try to estimate the first derivative of np.sin at 1 with a second-order method, where we know that np.sin is well conditioned.

>>> central_fdm(order=2, deriv=1, condition=1)(np.sin, 1) - np.cos(1)  
4.307577627926662e-10

And let's try to estimate the second derivative of np.sin at 1 with a third-order method.

>>> central_fdm(order=3, deriv=2, condition=1)(np.sin, 1) + np.sin(1)  
-1.263876664436836e-07

Hm. Let's check the accuracy of this third-order method. The step size and accuracy of the method are computed upon calling FDM.estimate().

>>> central_fdm(order=3, deriv=2, condition=1).estimate().acc
8.733476581980376e-06

We might want a little more accuracy. Let's check the accuracy of a fifth-order method.

>>> central_fdm(order=5, deriv=2, condition=1).estimate().acc
7.343652562575155e-10

And let's estimate the second derivative of np.sin at 1 with a fifth-order method.

>>> central_fdm(order=5, deriv=2, condition=1)(np.sin, 1) + np.sin(1)   
-9.145184609593571e-11

Hooray!

Finally, let us verify that increasing the order indeed reliably increases the accuracy.

>>> for i in range(3, 11):
...      print(central_fdm(order=i, deriv=2, condition=1)(np.sin, 1) + np.sin(1))
-1.263876664436836e-07
6.341286606925678e-09
-9.145184609593571e-11
2.7335911312320604e-12
6.588063428125679e-13
2.142730437526552e-13
2.057243264630415e-13
8.570921750106208e-14

Testing Sensitivities in a Reverse-Mode Automatic Differentation Framework

Consider the function

def mul(a, b):
    return a * b

and its sensitivity

def s_mul(s_y, y, a, b):
    return s_y * b, a * s_y

The sensitivity s_mul takes in the sensitivity s_y of the output y, the output y, and the arguments of the function mul; and returns a tuple containing the sensitivities with respect to a and b. Then function check_sensitivity can be used to assert that the implementation of s_mul is correct:

>>> from fdm import check_sensitivity

>>> check_sensitivity(mul, s_mul, (2, 3))  # Test at arguments `2` and `3`.

Suppose that the implementation were wrong, for example

def s_mul_wrong(s_y, y, a, b):
    return s_y * b, b * s_y  # Used `b` instead of `a` for the second sensitivity!

Then check_sensitivity should throw an AssertionError:

>>> check_sensitivity(mul, s_mul, (2, 3)) 
AssertionError: Sensitivity of argument 2 of function "mul" did not match numerical estimate.

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