Convergence of finite differences - Fundamentals of Numerical Computation

All of the finite-difference formulas in the previous section based on equally spaced nodes converge as the node spacing $h$ decreases to zero. However, note that to discretize a function over an interval $[a,b]$ , we use $h=(b-a)/n$ , which implies $n=(b-a)/h=O(h^{-1})$ . As $h\to 0$ , the total number of nodes needed grows without bound. So we would like to make $h$ as large as possible while still achieving some acceptable accuracy.

Although we are measuring the truncation error only at $x=0$ , it could be defined for other $x$ as well. The definition adjusts naturally to use $f''(0)$ for difference formulas targeting the second derivative.

All of the finite-difference formulas given in Finite differences are convergent.

5.5.1Order of accuracy¶

Of major interest is the rate at which $\tau_f\to 0$ in a convergent formula.

Hence the forward-difference formula in Example 5.5.1 has order of accuracy equal to 1; i.e., it is first-order accurate. All else being equal, a higher order of accuracy is preferred, since $O(h^m)$ vanishes more quickly for larger values of $m$ . As a rule, including more function values in a finite-difference formula (i.e., increasing the number of weights in (5.4.1)) increases the order of accuracy, as can be seen in Table 5.4.1 and Table 5.4.2.

Order of accuracy is calculated by expanding $\tau_f$ in a Taylor series about $h=0$ and ignoring all but the leading term.^[1]

Example 5.5.2

We compute the truncation error of the centered difference formula (5.4.8):

\begin{split} \tau_f(h) &= f'(0) - \frac{ f(h)-f(-h)}{2h}\\ &= f'(0) - (2h)^{-1} \left[ \bigl( f(0) + h f'(0) + \tfrac{1}{2}h^2f''(0)+ \tfrac{1}{6}h^3f'''(0)+ O(h^4) \bigr) \right.\\ &\qquad - \left. \bigl( f(0) - h f'(0) + \tfrac{1}{2}h^2f''(0) - \tfrac{1}{6}h^3f'''(0)+O(h^4) \bigr) \right] \\ &= -(2h)^{-1} \left[ \tfrac{1}{3}h^3f'''(0) + O(h^4) \right] = O(h^2). \end{split}

(5.5.3)

Thus, this method has order of accuracy equal to 2.

Example 5.5.3 (Convergence of finite differences)

Julia

MATLAB

Python

Example 5.5.3

Let’s observe the convergence of the formulas in Example 5.5.1 and Example 5.5.2, applied to the function $\sin(e^{x+1})$ at $x=0$ .

f = x -> sin(exp(x + 1))
exact_value = exp(1) * cos(exp(1))

-2.478349732955235

We’ll compute the formulas in parallel for a sequence of $h$ values.

h = [5 / 10^n for n in 1:6]
FD = zeros(length(h), 2)
for (k, h) in enumerate(h)
    FD[k, 1] = (f(h) - f(0)) / h
    FD[k, 2] = (f(h) - f(-h)) / 2h
end
@pt :header=["h", "FD1", "FD2"] [h FD]

All that’s easy to see from this table is that FD2 appears to converge to the same result as FD1, but more rapidly. A table of errors is more informative.

error_FD = @. exact_value - FD
@pt :header=["h", "error in FD1", "error in FD2"] [h error_FD]

In each row, $h$ is decreased by a factor of 10, so that the error is reduced by a factor of 10 in the first-order method and 100 in the second-order method.

A graphical comparison can be useful as well. On a log-log scale, the error should (as $h\to 0$ ) be a straight line whose slope is the order of accuracy. However, it’s conventional in convergence plots to show $h$ decreasing from left to right, which negates the slopes.

using Plots
plot(h, abs.(error_FD); 
    m=:o,  label=["FD1" "FD2"], leg=:bottomleft,
    xflip=true,  xaxis=(:log10, L"h"),  yaxis=(:log10, "error"),
    title="Convergence of finite differences")

# Add lines for perfect 1st and 2nd order.
plot!(h, [h h .^ 2], l=:dash, label=[L"O(h)" L"O(h^2)"])

Example 5.5.3

Let’s observe the convergence of the formulas in Example 5.5.1 and Example 5.5.2, applied to the function $\sin(e^{x+1})$ at $x=0$ .

f = @(x) sin(exp(x + 1));
exact_value = exp(1) * cos(exp(1))

We’ll compute the formulas in parallel for a sequence of $h$ values.

h = 5 ./ 10.^(1:6)';
FD1 = zeros(size(h));
FD2 = zeros(size(h));
for i = 1:length(h)
    h_i = h(i);
    FD1(i) = (f(h_i) - f(0)    ) / h_i;
    FD2(i) = (f(h_i) - f(-h_i)) / (2*h_i);
end
table(h, FD1, FD2)

All that’s easy to see from this table is that FD2 appears to converge to the same result as FD1, but more rapidly. A table of errors is more informative.

err1 = abs(exact_value - FD1);
err2 = abs(exact_value - FD2);
table(h, err1, err2, variableNames=["h", "error in FD1", "error in FD2"])

In each row, $h$ is decreased by a factor of 10, so that the error is reduced by a factor of 10 in the first-order method and 100 in the second-order method.

clf
loglog(h, abs([err1 err2]), "o-")
set(gca, "xdir", "reverse")
order1 = 0.1 * err1(end) * (h / h(end)) .^ 1;
order2 = 0.1 * err2(end) * (h / h(end)) .^ 2;
hold on
loglog(h, order1, "--", h, order2, "--")
xlabel("h");  ylabel("error")
title("Convergence of finite differences")
legend("FD1", "FD2", "O(h)", "O(h^2)");

Example 5.5.3

Let’s observe the convergence of the formulas in Example 5.5.1 and Example 5.5.2, applied to the function $\sin(e^{x+1})$ at $x=0$ .

f = lambda x: sin(exp(x + 1))
exact_value = exp(1) * cos(exp(1))

We’ll compute the formulas in parallel for a sequence of $h$ values.

h_ = array([5 / 10**(n+1) for n in range(6)])
FD = zeros((len(h_), 2))
for (i, h) in enumerate(h_):
    FD[i, 0] = (f(h) - f(0)) / h 
    FD[i, 1] = (f(h) - f(-h)) / (2*h)
results = PrettyTable()
results.add_column("h", h_)
results.add_column("FD1", FD[:, 0])
results.add_column("FD2", FD[:, 1])
print(results)

+--------+---------------------+---------------------+
|   h    |         FD1         |         FD2         |
+--------+---------------------+---------------------+
|  0.5   |  -2.768575766550465 | -1.9704719803862911 |
|  0.05  | -2.6127952856136947 |  -2.475520256824717 |
| 0.005  | -2.4921052189334048 | -2.4783216951179465 |
| 0.0005 | -2.4797278609705042 | -2.4783494526022243 |
| 5e-05  | -2.4784875710526233 |  -2.478349730152263 |
| 5e-06  |  -2.478363517077753 |  -2.478349732970564 |
+--------+---------------------+---------------------+

All that’s easy to see from this table is that FD2 appears to converge to the same result as FD1, but more rapidly. A table of errors is more informative.

errors = FD - exact_value
results = PrettyTable()
results.add_column("h", h_)
results.add_column("error in FD1", errors[:, 0])
results.add_column("error in FD2", errors[:, 1])
print(results)

+--------+-------------------------+------------------------+
|   h    |       error in FD1      |      error in FD2      |
+--------+-------------------------+------------------------+
|  0.5   |   -0.29022603359523025  |   0.5078777525689437   |
|  0.05  |   -0.1344455526584598   | 0.0028294761305178717  |
| 0.005  |   -0.01375548597816989  | 2.8037837288330536e-05 |
| 0.0005 |  -0.0013781280152693753 | 2.8035301058437767e-07 |
| 5e-05  |  -0.0001378380973884319 | 2.8029720766653554e-09 |
| 5e-06  | -1.3784122518067932e-05 | -1.532907134560446e-11 |
+--------+-------------------------+------------------------+

In each row, $h$ is decreased by a factor of 10, so that the error is reduced by a factor of 10 in the first-order method and 100 in the second-order method.

plot(h_, abs(errors), "o-", label=["FD1", "FD2"])
gca().invert_xaxis()
# Add lines for perfect 1st and 2nd order.
loglog(h_, h_, "--", label="$O(h)$")
loglog(h_, h_**2, "--", label="$O(h^2)$")
xlabel("$h$")
ylabel("error")
legend();

5.5.2Stability¶

The truncation error $\tau_f(h)$ of a finite-difference formula is dominated by a leading term $O(h^m)$ for an integer $m$ . This error decreases as $h\to 0$ . However, we have not yet accounted for the effects of roundoff error. To keep matters as simple as possible, let’s consider the forward difference

\delta(h) = \frac{f(x+h)-f(x)}{h}.

(5.5.4)

As $h\to 0$ , the numerator approaches zero even though the values $f(x+h)$ and $f(x)$ are not necessarily near zero. This is the recipe for subtractive cancellation error! In fact, finite-difference formulas are inherently ill-conditioned as $h\to 0$ . To be precise, recall that the condition number for the problem of computing $f(x+h)-f(x)$ is

\kappa(h) = \frac{ \max\{\,|f(x+h)|,|f(x)|\,\} }{ |f(x+h)-f(x) | },

(5.5.5)

implying a relative error of size $\kappa(h) \epsilon_\text{mach}$ in its computation. Hence the numerical value we actually compute for δ is

\begin{split} \tilde{\delta}(h) &= \frac{f(x+h)-f(x)}{h}\, (1+\kappa(h)\epsilon_\text{mach}) \\ &= \delta(h) + \frac{ \max\{\,|f(x+h)|,|f(x)|\,\} }{ |f(x+h)-f(x) | }\cdot \frac{f(x+h)-f(x)}{h} \cdot \epsilon_\text{mach}. \end{split}

(5.5.6)

Hence as $h\to 0$ ,

\bigl| \tilde{\delta}(h) - \delta(h) \bigr| = \frac{ \max\{\,|f(x+h)|,|f(x)|\,\} }{ h}\,\epsilon_\text{mach} \sim |f(x)|\, \epsilon_\text{mach}\cdot h^{-1}.

(5.5.7)

Combining the truncation error and the roundoff error leads to

\bigl| f'(x) - \tilde{\delta}(h) \bigr| \le \bigl| \tau_f(h) \bigr| + \bigl|f(x) \bigr|\, \epsilon_\text{mach} \, h^{-1}.

(5.5.8)

Equation (5.5.8) indicates that while the truncation error τ vanishes as $h$ decreases, the roundoff error actually increases thanks to the subtractive cancellation. At some value of $h$ the two error contributions will be of roughly equal size. This occurs when

\bigl|f(x)\bigr|\, \epsilon_\text{mach}\, h^{-1} \approx C h, \quad \text{or} \quad h \approx K \sqrt{\rule[0.05em]{0mm}{0.4em}\epsilon_\text{mach}},

(5.5.9)

for a constant $K$ that depends on $x$ and $f$ , but not $h$ . In summary, for a first-order finite-difference method, the optimum spacing between nodes is proportional to $\epsilon_\text{mach}^{\,\,1/2}$ . (This observation explains the choice of δ in Function 4.6.1.)

For a method of truncation order $m$ , the details of the subtractive cancellation are a bit different, but the conclusion generalizes.

A different statement of the conclusion is that for a first-order formula, at most we can expect accuracy in only about half of the available machine digits. As $m$ increases, we get ever closer to using the full accuracy available. Higher-order finite-difference methods are both more efficient and less vulnerable to roundoff than low-order methods.

Example 5.5.4 (Roundoff error in finite differences)

Julia

MATLAB

Python

Example 5.5.4

Let $f(x)=e^{-1.3x}$ . We apply finite-difference formulas of first, second, and fourth order to estimate $f'(0)=-1.3$ .

f = x -> exp(-1.3 * x);
exact = -1.3

h = [1 / 10^n for n in 1:12]
FD = zeros(length(h), 3)
for (k, h) in enumerate(h)
    nodes = h * (-2:2)
    vals = @. f(nodes)
    FD[k, 1] = dot([0 0 -1 1 0] / h, vals)
    FD[k, 2] = dot([0 -1 / 2 0 1 / 2 0] / h, vals)
    FD[k, 3] = dot([1 / 12 -2 / 3 0 2 / 3 -1 / 12] / h, vals)
end
@pt :header=["h", "FD1", "FD2", "FD4"] [h FD]

They all seem to be converging to -1.3. The convergence plot reveals some interesting structure to the errors, though.

err = @. abs(FD - exact)

plot(h, err;
    m=:o, label=["FD1" "FD2" "FD4"],  legend=:bottomright,
    xaxis=(:log10, L"h"),  xflip=true,  yaxis=(:log10, "error"),
    title="FD error with roundoff")

# Add line for perfect 1st order.
plot!(h, 0.1 * eps() ./ h, l=:dash, color=:black, label=L"O(h^{-1})")

Again the graph is made so that $h$ decreases from left to right. The errors are dominated at first by truncation error, which decreases most rapidly for the fourth-order formula. However, increasing roundoff error eventually equals and then dominates the truncation error as $h$ continues to decrease. As the order of accuracy increases, the crossover point moves to the left (greater efficiency) and down (greater accuracy).

Example 5.5.4

Let $f(x)=e^{-1.3x}$ . We apply finite-difference formulas of first, second, and fourth order to estimate $f'(0)=-1.3$ .

f = @(x) exp(-1.3 * x);
exact = -1.3;

h = 10 .^ (-(1:12))';
FD = zeros(length(h), 3);
for i = 1:length(h)
    h_i = h(i);
    nodes = h_i * (-2:2);
    vals = f(nodes);
    FD(i, 1) = dot([0      0 -1   1    0] / h_i, vals);
    FD(i, 2) = dot([0    -1/2 0 1/2    0] / h_i, vals);
    FD(i, 3) = dot([1/12 -2/3 0 2/3 -1/12] / h_i, vals);
end
format long
table(h, FD(:, 1), FD(:, 2), FD(:, 3), variableNames=["h", "FD1", "FD2", "FD4"])

They all seem to be converging to -1.3. The convergence plot reveals some interesting structure to the errors, though.

err = abs(FD - exact);
clf
loglog(h, err, "o-")
set(gca, "xdir", "reverse")
order1 = 0.1 * err(end, 1) * (h / h(end)) .^ (-1);
hold on
loglog(h, order1, "k--")
xlabel("h");  ylabel("error")
title("FD error with roundoff")
legend("FD1", "FD2", "FD4", "O(1/h)", "location", "northeast");

Example 5.5.4

Let $f(x)=e^{-1.3x}$ . We apply finite-difference formulas of first, second, and fourth order to estimate $f'(0)=-1.3$ .

f = lambda x: exp(-1.3 * x)
exact = -1.3

h_ = array([1 / 10**(n+1) for n in range(12)])
FD = zeros((len(h_), 3))
for (i, h) in enumerate(h_):
    nodes = h * linspace(-2, 2, 5)
    vals = f(nodes)
    FD[i, 0] = dot(array([0, 0, -1, 1, 0]) / h, vals)
    FD[i, 1] = dot(array([0, -1/2, 0, 1/2, 0]) / h, vals)
    FD[i, 2] = dot(array([1/12, -2/3, 0, 2/3, -1/12]) / h, vals)

results = PrettyTable()
results.add_column("h", h_)
results.add_column("FD1", FD[:, 0])
results.add_column("FD2", FD[:, 1])
results.add_column("FD4", FD[:, 2])
print(results)

+--------+---------------------+---------------------+---------------------+
|   h    |         FD1         |         FD2         |         FD4         |
+--------+---------------------+---------------------+---------------------+
|  0.1   | -1.2190456907943865 | -1.3036647620203026 |  -1.299987598641898 |
|  0.01  | -1.2915864979712381 | -1.3000366169760815 | -1.2999999987623418 |
| 0.001  | -1.2991553660477848 | -1.3000003661667074 | -1.2999999999999687 |
| 0.0001 | -1.2999155036613956 |  -1.300000003662075 | -1.3000000000002956 |
| 1e-05  | -1.2999915500404313 | -1.3000000000396366 | -1.3000000000138243 |
| 1e-06  |  -1.299999154987745 | -1.2999999999900496 | -1.3000000000320142 |
| 1e-07  | -1.2999999150633812 | -1.3000000000289944 |  -1.300000001094304 |
| 1e-08  | -1.2999999970197678 | -1.3000000042305544 | -1.3000000128522515 |
| 1e-09  | -1.2999999523162842 | -1.3000000340328768 | -1.3000001162290573 |
| 1e-10  | -1.2999992370605469 | -1.2999996723115146 | -1.3000001907348633 |
| 1e-11  | -1.3000030517578125 | -1.3000038001061966 | -1.3000049591064453 |
| 1e-12  |   -1.2999267578125  |  -1.300004483829298 | -1.3000030517578125 |
+--------+---------------------+---------------------+---------------------+

They all seem to be converging to -1.3. The convergence plot reveals some interesting structure to the errors, though.

loglog(h_, abs(FD[:, 0] + 1.3), "-o", label="FD1")
loglog(h_, abs(FD[:, 1] + 1.3), "-o", label="FD2")
loglog(h_, abs(FD[:, 2] + 1.3), "-o", label="FD4")
gca().invert_xaxis()
plot(h_, 0.1 * 2 ** (-52) / h_, "--", color="k", label="$O(h^{-1})$")
xlabel("$h$")
ylabel("total error")
title("FD error with roundoff")
legend();

5.5.3Exercises¶

⌨ Evaluate the centered second-order finite-difference approximation to $f'(4\pi/5)$ for $f(x)=\cos(x^3)$ and $h=2^{-1},2^{-2},\ldots,2^{-8}$ . On a log-log graph, plot the error as a function of $h$ and compare it graphically to second-order convergence.
✍ Derive the first two nonzero terms of the Taylor series at $h=0$ of the truncation error $\tau_{f}(h)$ for the formula (5.4.3).
✍ Calculate the first nonzero term in the Taylor series of the truncation error $\tau_{f}(h)$ for the finite-difference formula defined by the second row of Table 5.4.2.
✍ Calculate the first nonzero term in the Taylor series of the truncation error $\tau_{f}(h)$ for the finite-difference formula defined by the third row of Table 5.4.2.
✍ Show that the formula (5.4.12) is second-order accurate.

✍ A different way to derive finite-difference formulas is the method of undetermined coefficients. Starting from (5.4.1),
$f'(x) \approx \frac{1}{h}\sum_{k=-p}^q a_k f(x+kh),$
(5.5.11)
let each $f(x+k h)$ be expanded in a series around $h=0$ . When the coefficients of powers of $h$ are collected, one obtains
$\frac{1}{h} \sum_{k=-p}^q a_k f(x+kh) = \frac{b_0}{h} + b_1 f'(x) + b_2 f''(x)h + \cdots,$
(5.5.12)
where
$b_i = \sum_{k=-p}^q k^i a_k.$
(5.5.13)
In order to make the result as close as possible to $f'(x)$ , we impose the conditions
$b_0 = 0,\, b_1=1,\, b_2=0,\, b_3=0,\,\ldots,\,b_{p+q}=0.$
(5.5.14)
This provides a system of linear equations for the weights.
(a) For $p=q=2$ , write out the system of equations for $a_{-2}$ , $a_{-1}$ , $a_0$ , $a_1$ , $a_2$ .
(b) Verify that the coefficients from the appropriate row of Table 5.4.1 satisfy the equations you wrote down in part (a).
(c) Derive the finite-difference formula for $p=1$ , $q=2$ using the method of undetermined coefficients.

Footnotes¶

The term truncation error is derived from the idea that the finite-difference formula, being finite, has to truncate the series representation and thus cannot be exactly correct for all functions.
↩

Preface

Finite differences

Preface

Numerical integration