Finite differences¶
Much more can be said about interpolation. But now we turn to one of the most common and important applications of interpolants: finding derivatives. For the moment, we will continue to use \(x\) as the independent variable name and \(a=t_0,\ldots,t_n=b\) as the interpolation nodes. We also continue to use an equispaced grid, so that \(t_i=a+i h\) for \(i=0,\ldots,n\), where \(h=(ba)/n\).
Considering the most common definition of a derivative,
it stands to reason that an approximation to \(f'\) at a node \(t_i\) should depend on the values of \(f\) at the nodes closest to \(t_i\). Also, because differentiation is a linear operation, we will constrain ourselves to formulas that are linear in these nodal values. Consequently, the sort of formula we seek is the finite difference formula
where \(p\), \(q\) are integers, and the \(a_k\)’s are constants known as the weights of the formula. Crucially, the finite difference weights are independent of \(f\), although they do depend on the nodes. The factor of \(h^{1}\) is present to make the expression more convenient in what follows.
Before deriving some finite difference formulas, we make an important observation about them. Define the new variable \(s=xt_i\) and let \(\tilde{f}(s) = f(xt_i)\). Then it is elementary that
Applying the change of variables to (130) yields
with the same constants as before. These manipulations express a property that is simpler than it may appear: we can always derive and write the finite difference formula (130) with \(t_i=0\) without losing generality. In fact, \(t_i\) is just a “dummy” variable that we can replace by \(x\), as in
This property is translation invariance. The formula combines values of the function at points always placed the same way relative to \(x\).
An obvious candidate for a finite difference formula is based on the limit definition above:
which is (131) with \(p=0\), \(q=1\), \(a_0=1\), and \(a_1=1\). This is referred to as a forward difference formula, characterized by \(p=0\), because \(f\) is evaluated only at points “forward” from \(x\). Analogously, we could use the backward difference formula
in which \(q=0\).
Both the forward and backward difference formulas surely become equalities in the limit \(h\to 0\), provided \(f\) is differentiable at \(x\). However, they are not the only possibilities. One aesthetic objection is the lack of symmetry about the point \(x\). In response, we will derive a formula that uses \(p=q=1\), i.e., in which \(f(h)\), \(f(0)\), and \(f(h)\) are all available.
The formula (132) (with \(x=0\)) is simply the slope of the line through the points \(\bigl(0,f(0)\bigr)\) and \(\bigl(h,f(h)\bigr)\). A similar observation holds for the backward difference formula. Thus one route to using three function values is to differentiate the quadratic polynomial that interpolates them (see this exercise):
We now use \(f'(0)\approx Q'(0)\) to get the centered finite difference formula
This result is equivalent to (131) with \(p=q=1\) and weights \(a_{1}=1/2\), \(a_0=0\), and \(a_1=1/2\). Observe that while the value of \(f(0)\) was available during the derivation, its weight ends up being zero.
We can verify using L’Hôpital’s Rule that the approximation in (135) becomes an equality as \(h\to 0\). Such an analysis does not, however, reveal a significant accuracy advantage of the centered variant, one that we will take up in Convergence of finite differences.
We can in principle derive any finite difference formula from the same process: Interpolate the given function values, then differentiate the interpolant exactly. Some results are given here for two important special cases. Table 2 is for \(p=q\), or centered differences, while Table 3 is for \(p=0\), or forward differences. Both show the weights for estimating the derivative at zero, but they can be uniformly translated to any other point.
order 
\(4h\) 
\(3h\) 
\(2h\) 
\(h\) 
\(0\) 
\(h\) 
\(2h\) 
\(3h\) 
\(4h\) 

2 
\(\frac{1}{2}\) 
\(0\) 
\(\frac{1}{2}\) 

4 
\(\frac{1}{12}\) 
\(\frac{2}{3}\) 
\(0\) 
\(\frac{2}{3}\) 
\(\frac{1}{12}\) 

6 
\(\frac{1}{60}\) 
\(\frac{3}{20}\) 
\(\frac{3}{4}\) 
\(0\) 
\(\frac{3}{4}\) 
\(\frac{3}{20}\) 
\(\frac{1}{60}\) 

8 
\(\frac{1}{280}\) 
\(\frac{4}{105}\) 
\(\frac{1}{5}\) 
\(\frac{4}{5}\) 
\(0\) 
\(\frac{4}{5}\) 
\(\frac{1}{5}\) 
\(\frac{4}{105}\) 
\(\frac{1}{280}\) 
order 
\(0\) 
\(h\) 
\(2h\) 
\(3h\) 
\(4h\) 

1 
\(1\) 
\(1\) 

2 
\(\frac{3}{2}\) 
2 
\(\frac{1}{2}\) 

3 
\(\frac{11}{6}\) 
3 
\(\frac{3}{2}\) 
\(\frac{1}{3}\) 

4 
\(\frac{25}{12}\) 
\(4\) 
\(3\) 
\(\frac{4}{3}\) 
\(\frac{1}{4}\) 
The main motivation for using more function values in a formula is to improve the accuracy. This is measured by order of accuracy, which is show in the tables and explained in Convergence of finite differences. To get backward differences with \(q=0\), you can use the change of variable \(\hat{f}(x)=f(x)\), which changes the sign and reverses the order of the coefficients in Table 3; see this exercise.
According to the tables, here are two finite difference formulas:
Higher derivatives¶
Many applications require the second derivative of a function. It’s tempting to use the finite difference of a finite difference. For example, applying (135) twice leads to
This is a valid formula, but it uses values at \(\pm 2h\) rather than the closer values at \(\pm h\). A better and more generalizable tactic is to return to the quadratic \(Q(x)\) in (134) and use \(Q''(0)\) to approximate \(f''(0)\). Doing so yields
which is the simplest centered seconddifference formula. As with the first derivative, we can choose larger values of \(p\) and \(q\) in (130) to get new formulas, such as
and
Arbitrary nodes¶
Although function values at equally spaced nodes are a common and convenient situation, the node locations may be arbitrary. The general form of a finite difference formula is
We no longer assume equally spaced nodes, so there is no “\(h\)” to be used in the formula. As before, the weights may be applied after any translation of the independent variable. The weights again follow from the interpolate/differentiate recipe, but the algebra becomes complicated. Fortunately there is an elegant recursion known as Fornberg’s algorithm that can calculate these weights for any desired formula. We present it without derivation as fdweights.
Fornberg’s algorithm for finite difference weights.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36  """
fdweights(t,m)
Return weights for the `m`th derivative of a function at zero using
values at the nodes in vector `t`.
"""
function fdweights(t,m)
# This is a compact implementation, not an efficient one.
function weight(t,m,r,k)
# Recursion for one weight. Input: t nodes (vector) m
# order of derivative sought r number of nodes to use from
# t (<= length(t)) k index of node whose weight is found
if (m<0)  (m>r) # undefined coeffs must be zero
c = 0
elseif (m==0) && (r==0) # base case of onepoint interpolation
c = 1
else # generic recursion
if k<r
c = (t[r+1]*weight(t,m,r1,k) 
m*weight(t,m1,r1,k))/(t[r+1]t[k+1])
else
numer = r > 1 ? prod(t[r]x for x in t[1:r1]) : 1.0
denom = r > 0 ? prod(t[r+1]x for x in t[1:r]) : 1.0
β = numer/denom
c = β*(m*weight(t,m1,r1,r1)  t[r]*weight(t,m,r1,r1))
end
end
return c
end
r = length(t)1
w = zeros(size(t))
return [ weight(t,m,r,k) for k=0:r ]
end

Exercises¶
✍ This problem refers to \(Q(x)\) defined by (134).
(a) Show that \(Q(x)\) interpolates the three values of \(f\) at \(x=h\), \(x=0\), and \(x=h\).
(b) Show that \(Q'(0)\) gives the finite difference formula defined by (135).
(a) ✍ Weights for forward finite difference formulas. lists forwarddifference formulas in which \(p=0\) in (131). Show that the change of variable \(g(x) = f(x)\) transforms these formulas into backward difference formulas with \(q=0\), and write out the table analogous to Weights for forward finite difference formulas. for backward differences.
(b) ⌨ Suppose you are given the nodes \(t_0=0.9\), \(t_1=1\), and \(t_2=1.1\), and \(f(x) = \sin(2x)\). Using formulas from Table 2 and Table 3, compute secondorder accurate approximations to \(f'\) at each of the three nodes.
⌨ Using fdweights to get the necessary weights, find finitedifference approximations to the first, second, third, and fourth derivatives of \(f(x)=e^{x}\) at \(x=0.5\). In each case use a centered stencil of minimum possible width. Make a table showing the values and the errors in each case.
⌨ Use fdweights to write out a table analogous to Weights for centered finite difference formulas. that lists centered finite difference weights for the second derivative \(f''(0)\). (Hint: The “rat” command will let you express the results as exact rational numbers.)
⌨ For this problem, let \(f(x)=\tan(2x)\).
(a) ⌨ Apply fdweights to find a finite difference approximation to \(f''(0.3)\) using the five nodes \(t_j=0.3+jh\) for \(j=2,\ldots,2\) and \(h=0.05\). Compare to the exact value of \(f''(0.3)\).
(b) ⌨ Repeat part~(a) for \(f''(0.75)\) and \(t_j=0.75+jh\). Why is the finite difference result so inaccurate?
(a) ✍ Derive (136) by applying applying (135) twice (i.e., apply once to get \(f'\) values and then apply again to those values).
(b) ✍ Find the formula for \(f''(0)\) that results from applying (132) and then (133).
(a) ✍ Show using L’Hôpital’s Rule that the centered formula approximation (135) converges to an equality as \(h\to 0\).
(b) ✍ Derive two conditions on the finite difference weights in (131) that arise from requiring convergence as \(h\to 0\).