Approximating derivatives of discrete functions

### 13.2.4 Approximating derivatives of discrete functions

The numdiff command finds numerical approximations to derivatives.

- numdiff takes three mandatory arguments and one optional argument.
  - X=[α0,α1,…,αn], Y=[β0,β1,…,βn], two lists of real numbers, where n≥ 1.
  - x0, a real number.
  - Optionally, m, an integer or a sequence of integers (by default 1).
  Note that X, Y and x0 can also be symbolic expressions.
- numdiff(X,Y,x0 ⟨,m⟩) returns an approximation of the mth derivative of a function f at x0, or a sequence of derivatives of order given by the sequence m, where f has values given by f(αk)=βk, k=0,1,…,n.
- numdiff uses Fornberg’s algorithm with complexity O(n2m) in both time and space.
- Note that α0,α1,…,αn do not have to be equally spaced, but they must be mutually different and input in ascending order. There are no restrictions on the choice of x0.
- The algorithm performs reasonably fast on inexact input data, with rounding errors usually not damaging for m≤ 4. For higher m, consider providing input data in an exact form. This will make the algorithm run considerably slower, but will also avoid numerical instabilities due to the unlimited precision arithmetic being used.

In case you need only the first or second derivative at a sequence of points, you can use the interp command (see Section 17.1.2) which uses cubic spline interpolation. Note that this method does not accept symbolic input.

#### Examples

Let f(x)=sin(x) e−x, x∈[0,1]. Sample this function at the points in

X=[0,0.1,0.2,0.4,0.5,0.7,0.8,1]

to approximate f′′(1/π).

f:=unapply(sin(x)*exp(-x),x):; X:=[0,0.1,0.2,0.4,0.5,0.7,0.8,1]:; Y:=apply(f,X):;

Now you can approximate the second derivative of f at the point x0=1/π.

x0:=1/pi:; d:=numdiff(X,Y,x0,2)

−1.38167652799 |  |  |  |  |  |  |  |  |  |

Finally, compute the relative error of the obtained approximation.

abs(d-f''(x0))/abs(f''(x0))*100

2.82975186496×10−5 |  |  |  |  |  |  |  |  |  |

The result is expressed in percentages.

Use a sequence of values for the parameter m to find a list of approximations of the respective derivatives at x0. This is faster than calling numdiff to approximate one derivative at a time. For example, approximate the first, second and third derivative of the function

f(x)=1− |
1
1+x2
| , x∈[0,1],

at the point x0=0.57 by sampling f at 21 equidistant points in the segment [0,1].

f:=unapply(1-1/(1+x^2),x):; X:=[(0.05*k)$(k=0..20)]:; Y:=apply(f,X):; numdiff(X,Y,0.57,1,2,3)

| ⎡ ⎣ | 0.649439427528,0.0217571104587,−2.99724196738 | ⎤ ⎦

Actually, f′(x0)=0.649439427528, f′′(x0)=0.0217571104558 and f′′′(x0)=−2.99724196764.

numdiff can be used for generating custom finite-difference stencils for approximation of derivatives. For example, let X=[−1,0,2,4], Y=[a,b,c,d] and x0=1. To obtain an approximation formula for the second derivative:

numdiff([-1,0,2,4],[a,b,c,d],1,2)

2
5
| a− |
b
2
| + |
d
10

The approximation is always a linear combination of elements in Y, regardless of X, x0 and m.

Given the lists X=[α0,α1,…,αn] and Y=[β0,β1,…,βn], the Lagrange polynomial passing through points (αk,βk) where k=0,1,…,n can be obtained by setting m=0 and entering a symbol for x0. Let X=[−2,0,1] and Y=[2,4,1]:

expand(numdiff([-2,0,1],[2,4,1],x,0))

− |
4
3
| x2− |
5
3
| x+4

Note that by entering

lagrange([-2,0,1],[2,4,1],x)

we obtain the same result.