Chapter 8 · Section 8.4

8.4 Numerical Methods I: Euler's Method

Graphical methods give pictures, and series give good approximations near the initial point. Engineers usually need something else: reliable numbers for the solution over a whole interval, for any reasonable $f$ . Numerical methods provide exactly that. They do not produce a formula; they produce a table of approximate values at chosen points.

This section develops the simplest numerical method, due to Euler. It is too crude for serious work, but every better method (Sections 8.5 to 8.7) is built on the same ideas, and Euler's method is where those ideas, especially the different kinds of error, are easiest to see.

The setting

We want to approximate the solution $y = φ (x)$ of

y^{'} = f (x, y), y (x_{0}) = y_{0}

on an interval $x_{0} \leq x \leq X$ . We assume that this problem has a unique solution there (Section 8.3 says when). Choose a step size $h > 0$ and the equally spaced grid points

x_{1} = x_{0} + h, x_{2} = x_{0} + 2 h, \dots, x_{n} = x_{0} + nh .

A numerical method produces numbers $y_{1}, y_{2}, \dots$ with $y_{n} \approx φ (x_{n})$ . We keep the two symbols apart: $φ (x_{n})$ is the true value, which we usually do not know, and $y_{n}$ is what we compute.

Deriving Euler's method

We give three short derivations. They lead to the same formula, and each one teaches something different.

1. Follow the tangent line

At the starting point we know the slope of the solution exactly: $φ^{'} (x_{0}) = f (x_{0}, y_{0})$ . Over a short step, the curve stays close to its tangent line, so we walk along the tangent:

y_{1} = y_{0} + h f (x_{0}, y_{0}) .

At $(x_{1}, y_{1})$ we do not know the true solution any more, but the direction field still gives the slope $f (x_{1}, y_{1})$ of the solution curve through this point. So we take another tangent step from there, and so on. The result is a broken line, the Euler polygon, built from line elements of the direction field (Figure 8.4.1).

2. Cut off a Taylor series

If $φ^{''}$ is continuous, Taylor's theorem with remainder gives, for some $ξ_{n}$ between $x_{n}$ and $x_{n + 1}$ ,

φ (x_{n + 1}) = φ (x_{n}) + h φ^{'} (x_{n}) + \frac{h ^{2}}{2} φ^{''} (ξ_{n}) .

Since $φ^{'} (x_{n}) = f (x_{n}, φ (x_{n}))$ , dropping the last term and replacing $φ (x_{n})$ by $y_{n}$ gives Euler's formula. This view tells us the size of what we threw away: a term proportional to $h^{2}$ . We will use it below.

3. Approximate an integral

Integrating the equation over one step, as in Section 8.3,

φ (x_{n + 1}) = φ (x_{n}) + \int_{x_{n}}^{x_{n + 1}} f (t, φ (t)) d t .

Approximating the integral by a rectangle of width $h$ and height equal to the integrand at the left end, $f (x_{n}, φ (x_{n}))$ , gives Euler's formula once more. Better ways of approximating this integral (a trapezoid, Simpson's rule) lead to the better methods of the next sections.

Definition 8.4.1 (Euler's method)

With step size $h$ and $x_{n} = x_{0} + nh$ , Euler's method for $y^{'} = f (x, y)$ , $y (x_{0}) = y_{0}$ computes $y_{n + 1} = y_{n} + h f (x_{n}, y_{n}), n = 0, 1, 2, \dots$

Figure 8.4.1. Two Euler steps (drawn with a large

h

to make the errors visible). After the first step we are on a different solution curve (dashed). The local error is what one step adds, measured from the solution through the current point; the global error is the total distance from the true solution.

Method in a nutshell: Euler's method

Choose $h$ and the number of steps $N$ (so that $x_{0} + N h$ is the end of the interval).
Start with $x = x_{0}$ , $y = y_{0}$ .
Compute the slope $k = f (x, y)$ .
Update $y \leftarrow y + h k$ and $x \leftarrow x + h$ . Record $(x, y)$ .
Repeat steps 3 and 4 $N$ times. Keep 6 or more significant digits in intermediate values.

Euler's method at work

We use the equation whose direction field we sketched in Section 8.1. It has the exact solution $φ (x) = x^{2} - 2 x + 2 - e^{- x}$ , so we can measure the errors.

Example 8.4.1 (A first table)

Use Euler's method with $h = 0.2$ to approximate the solution of $y^{'} = x^{2} - y$ , $y (0) = 1$ on $[0, 1]$ . Compare with the exact values.

Solution

The formula. Here $f (x, y) = x^{2} - y$ , so $y_{n + 1} = y_{n} + 0.2 (x_{n}^{2} - y_{n}) .$
First step. $f (0, 1) = 0 - 1 = - 1$ , so $y_{1} = 1 + 0.2 (- 1) = 0.8$ at $x_{1} = 0.2$ .
Second step. $f (0.2, 0.8) = 0.04 - 0.8 = - 0.76$ , so $y_{2} = 0.8 + 0.2 (- 0.76) = 0.648$ at $x_{2} = 0.4$ .

Continue. The full table (6 decimals; error = exact − Euler):

$n$	$x_{n}$	$y_{n}$	$f (x_{n}, y_{n})$	exact $φ (x_{n})$	error
0	0.0	1.000000	−1.000000	1.000000	0.000000
1	0.2	0.800000	−0.760000	0.821269	0.021269
2	0.4	0.648000	−0.488000	0.689680	0.041680
3	0.6	0.550400	−0.190400	0.611188	0.060788
4	0.8	0.512320	0.127680	0.590671	0.078351
5	1.0	0.537856	0.462144	0.632121	0.094265

Observations. The error grows from step to step. At $x = 1$ it is $0.094$ , which is $14.9%$ of the true value. Every Euler value lies below the exact one. The reason is concavity: all the points involved lie above the inflection curve $y = x^{2} - 2 x$ of Example 8.1.3, where solution curves are concave up, and a tangent line of a concave-up curve lies below the curve. Each step therefore lands a little below the solution curve it started on.

Example 8.4.2 (Halving the step)

Repeat Example 8.4.1 with $h = 0.1$ . How does the error at $x = 1$ change?

Solution

The formula. $y_{n + 1} = y_{n} + 0.1 (x_{n}^{2} - y_{n})$ , now with 10 steps. The first step gives $y_{1} = 1 + 0.1 (- 1) = 0.9$ .

Table.

$x_{n}$	$y_{n}$	exact	error
0.1	0.900000	0.905163	0.005163
0.2	0.811000	0.821269	0.010269
0.3	0.733900	0.749182	0.015282
0.4	0.669510	0.689680	0.020170
0.5	0.618559	0.643469	0.024910
0.6	0.581703	0.611188	0.029485
0.7	0.559533	0.593415	0.033882
0.8	0.552580	0.590671	0.038092
0.9	0.561322	0.603430	0.042109
1.0	0.586189	0.632121	0.045931

Compare. At $x = 1$ the error dropped from $0.094265$ to $0.045931$ , a factor of $2.05$ ; at $x = 0.4$ , from $0.041680$ to $0.020170$ . Halving $h$ roughly halves the error, at the price of twice as many steps. Figure 8.4.2 shows both polygons.

Figure 8.4.2. Euler polygons for

y^{'} = x^{2} - y

y (0) = 1

with

h = 0.2

and

h = 0.1

, continued to

x = 2

. Both lag below the concave-up exact solution; the gap is roughly halved when

h

is halved.

Use the calculator below to run Euler's method yourself. It lists every intermediate quantity and, when you give the exact solution, the error at each step. Start with the preset (Example 8.4.2), then try $h = 0.05$ with $n = 20$ and check that the error at $x = 1$ is about halved again.

Nothing in the method requires linearity. Here is a nonlinear problem from engineering.

Example 8.4.3 (A draining tank)

By Torricelli's law, the water depth $y$ (in m) in a tank emptying through a hole in its bottom satisfies, for a certain tank, $y^{'} = - 0.6 y$ with $y (0) = 4$ ( $t$ in minutes). Use Euler's method with $h = 0.5$ to estimate the depth for $0 \leq t \leq 2$ . Compare with the exact solution $y = (2 - 0.3 t)^{2}$ .

Solution

Exact solution (check). Separating, $y^{- 1/2} d y = - 0.6 d t$ , so $2 y = 4 - 0.6 t$ , i.e. $y = 2 - 0.3 t$ and $y = (2 - 0.3 t)^{2}$ . It is valid until the tank is empty at $t = 20/3 \approx 6.67$ min.
Euler steps. $y_{n + 1} = y_{n} - 0.5 \cdot 0.6 y_{n} = y_{n} - 0.3 y_{n}$ . For example, $y_{1} = 4 - 0.3 \cdot 2 = 3.4$ and $y_{2} = 3.4 - 0.3 3.4 = 3.4 - 0.553173 = 2.846827$ .

Table.

$t_{n}$	$y_{n}$	$f (t_{n}, y_{n})$	exact	error
0.0	4.000000	−1.200000	4.000000	0.000000
0.5	3.400000	−1.106345	3.422500	0.022500
1.0	2.846827	−1.012353	2.890000	0.043173
1.5	2.340651	−0.917951	2.402500	0.061849
2.0	1.881675	−0.823045	1.960000	0.078325

Check by halving. With $h = 0.25$ (8 steps) we get $y (2) \approx 1.921709$ , error $0.038291$ : again about half of $0.078325$ .

How large is the error?

A numerical answer without an idea of its error is of little use. Three kinds of error are at work.

Truncation error: the error caused by the method itself, because it replaces the curve by straight pieces (the Taylor terms we dropped). It is present even with perfect arithmetic.
Round-off error: the error caused by doing arithmetic with finitely many digits.
Errors in the data ( $y_{0}$ or coefficients of $f$ known only approximately). We do not study these here.

Sizes of errors are reported in two ways. The error at $x_{n}$ is $φ (x_{n}) - y_{n}$ (its absolute value is the absolute error). The percentage relative error is

100 \cdot \frac{∣ φ ( x _{n} ) - y _{n} ∣}{∣ φ ( x _{n} ) ∣} %,

which compares the error with the size of the quantity itself. (It is meaningless when $φ (x_{n})$ is zero or nearly so.) In Example 8.4.1 the percentage relative error at $x = 1$ is $14.9%$ ; with $h = 0.1$ it is $7.3%$ .

Local truncation error

Definition 8.4.2 (Local truncation error)

The local truncation error of Euler's method at step $n + 1$ is the error made in a single step that starts from the exact value: $τ_{n + 1} = φ (x_{n + 1}) - [φ (x_{n}) + h f (x_{n}, φ (x_{n}))] .$

By the Taylor formula of the second derivation, $τ_{n + 1} = \frac{h ^{2}}{2} φ^{''} (ξ_{n})$ . So if $∣ φ^{''} ∣ \leq M_{2}$ on the interval,

∣ τ_{n + 1} ∣ \leq \frac{M _{2}}{2} h^{2} .

We say the local truncation error is of order $h^{2}$ , written $O (h^{2})$ : halving $h$ divides it by about $4$ .

Example 8.4.4 (Measuring the local error)

For $y^{'} = x^{2} - y$ , $y (0) = 1$ , compute the error of the very first Euler step for $h = 0.1$ and $h = 0.05$ , and compare with $\frac{h ^{2}}{2} φ^{''} (0)$ .

Solution

Why the first step? The first step starts from the exact value $y_{0} = φ (0)$ , so its error is exactly the local truncation error $τ_{1}$ .
$h = 0.1$ . $y_{1} = 0.9$ and $φ (0.1) = 0.905163$ , so $τ_{1} = 0.005163$ .
$h = 0.05$ . $y_{1} = 1 + 0.05 (- 1) = 0.95$ and $φ (0.05) = 0.951271$ , so $τ_{1} = 0.001271$ .
Compare. The ratio is $0.005163/0.001271 \approx 4.06$ : halving $h$ divided the local error by about $4$ . With $φ^{''} (x) = 2 - e^{- x}$ we have $φ^{''} (0) = 1$ , and $\frac{h ^{2}}{2} φ^{''} (0)$ equals $0.005$ and $0.00125$ , very close to the measured values.

Global truncation error

The error we actually care about is the global error $e_{n} = φ (x_{n}) - y_{n}$ , the total distance from the truth after $n$ steps. It is built from two ingredients: each step adds a new local error, and each step also carries forward the error accumulated so far.

A rough count already shows the answer. To cross an interval of length $L = X - x_{0}$ we take $N = L / h$ steps, each adding an error of size about $h^{2}$ . Adding them up gives about $N h^{2} = L h$ . So the global error should be of order $h$ , one power lower than the local error. The following theorem makes this precise.

Theorem 8.4.1 (Global error of Euler's method)

Suppose $f$ is continuous on the strip $x_{0} \leq x \leq X$ , $- \infty < y < \infty$ , and satisfies there a Lipschitz condition $∣ f (x, y_{1}) - f (x, y_{2}) ∣ \leq K ∣ y_{1} - y_{2} ∣$ with $K > 0$ . Suppose the solution $φ$ of $y^{'} = f (x, y)$ , $y (x_{0}) = y_{0}$ has a continuous second derivative on $[x_{0}, X]$ with $∣ φ^{''} (x) ∣ \leq M_{2}$ . Then the Euler approximations with step $h$ satisfy $∣ φ (x_{n}) - y_{n} ∣ \leq \frac{M _{2} h}{2 K} (e^{K (x_{n} - x_{0})} - 1) for all x_{n} \leq X .$ In particular, the global error is $O (h)$ : Euler's method is a first-order method.

Why it works

Subtract the Euler formula $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ from the exact relation $φ (x_{n + 1}) = φ (x_{n}) + h f (x_{n}, φ (x_{n})) + τ_{n + 1}$ : $e_{n + 1} = e_{n} + h [f (x_{n}, φ (x_{n})) - f (x_{n}, y_{n})] + τ_{n + 1} .$ By the Lipschitz condition the bracket is at most $K ∣ e_{n} ∣$ in size, and $∣ τ_{n + 1} ∣ \leq \frac{M _{2}}{2} h^{2}$ . Hence $∣ e_{n + 1} ∣ \leq (1 + h K) ∣ e_{n} ∣ + \frac{M _{2}}{2} h^{2} .$ The old error is amplified by at most $1 + h K$ per step, and a new piece of size $h^{2}$ is added. Starting from $e_{0} = 0$ and unrolling this inequality gives a geometric sum: $∣ e_{n} ∣ \leq \frac{M _{2} h ^{2}}{2} \frac{( 1 + h K ) ^{n} - 1}{h K} \leq \frac{M _{2} h}{2 K} (e^{nh K} - 1),$ using $1 + h K \leq e^{h K}$ and $nh = x_{n} - x_{0}$ .

Example 8.4.5 (The error bound and the halving table)

For $y^{'} = x^{2} - y$ , $y (0) = 1$ : (a) evaluate the bound of Theorem 8.4.1 at $x = 1$ for $h = 0.1$ ; (b) tabulate the actual error at $x = 1$ as $h$ is halved repeatedly, starting from $h = 0.2$ .

Solution

Constants. $f_{y} = - 1$ , so $K = 1$ on the whole strip. Also $φ^{''} (x) = 2 - e^{- x}$ is positive and increasing on $[0, 1]$ , so $M_{2} = 2 - e^{- 1} = 1.632121$ .
(a) Bound. $∣ e ∣ \leq \frac{1.632121 \times 0.1}{2 \times 1} (e^{1} - 1) \approx 0.1402.$ The actual error is $0.045931$ . The bound is about three times too large, but it is guaranteed, and it shows the right dependence on $h$ .

(b) Halving table. (Exact value

φ (1) = 0.632121

$h$	steps	$y_{N} \approx φ (1)$	error	ratio to previous error
0.2	5	0.537856	0.094265	–
0.1	10	0.586189	0.045931	2.05
0.05	20	0.609438	0.022682	2.02
0.025	40	0.620848	0.011272	2.01
0.0125	80	0.626501	0.005619	2.01
0.00625	160	0.629315	0.002805	2.00
0.003125	320	0.630719	0.001402	2.00
0.0015625	640	0.631420	0.000701	2.00

Reading the table. The ratios approach $2$ , and error $/ h$ settles near $0.45$ (the limit is $1 - 1.5/ e \approx 0.448$ ). So error $\approx 0.45 h$ at $x = 1$ . To guarantee three correct decimals (error below $0.001$ ) we need $h \approx 0.002$ , that is, 500 steps. Accuracy is expensive with a first-order method: each extra correct digit costs ten times more work.

Round-off error

Every arithmetic operation on a computer or calculator rounds its result to a fixed number of digits. Each Euler step therefore adds a small rounding error, say of size up to $ε$ . Over $N = L / h$ steps these can add up to roughly $Nε = L ε / h$ , which grows as $h$ shrinks. A simple model of the total error is

E (h) \approx C h + \frac{L ε}{h},

a decreasing truncation part plus an increasing round-off part. Such a function has a minimum at $h^{*} = L ε / C$ : making $h$ smaller than $h^{*}$ makes the answer worse. The model is a worst case (rounding errors of different steps partly cancel), but its shape is real.

To see this, we repeated the computation of Example 8.4.5 while rounding $y_{n}$ to 4 decimal places after every step, as one might on a hand calculator:

steps $N$	$h$	absolute error, full precision	absolute error, 4 decimals kept
64	0.015625	0.007029	0.006721
256	0.003906	0.001752	0.001921
1024	0.000977	0.000438	0.000321
4096	0.000244	0.000109	0.013179
16384	0.000061	0.000027	0.197779

Down to about $h = 0.001$ both columns agree in size. Beyond that, the rounded computation falls apart: with $h = 1/16384$ , every increment $h f$ smaller than $0.00005$ in size is rounded away completely, so the computed $y$ stops changing wherever $∣ f ∣ < 0.8192$ . Figure 8.4.3 shows the two behaviours on log–log axes.

Figure 8.4.3. Error at

x = 1

against

h

(log–log;

h

decreases to the right). In full double precision the points lie on a line of slope

1

: error

\approx 0.448 h

, the signature of a first-order method. Keeping only 4 decimals per step, the error stops improving near

h \approx 1 0^{- 3}

and then grows as round-off takes over.

What this means in practice

Computers work with about 16 significant digits, so for Euler's method round-off becomes visible only for extremely small $h$ , long after the run time has become absurd. The practical lesson is different: reducing $h$ is an expensive and, eventually, futile way to buy accuracy. The efficient cure is a method whose error falls faster than $h$ , such as the improved Euler method ( $O (h^{2})$ , Section 8.5) or the Runge–Kutta method ( $O (h^{4})$ , Section 8.6).

A warning: when the step is too large

Error estimates like Theorem 8.4.1 describe what happens as $h \to 0$ . With a step that is too large, something worse than inaccuracy can happen.

Example 8.4.6 (Instability)

Apply Euler's method to $y^{'} = - 20 y$ , $y (0) = 1$ (exact solution $e^{- 20 x}$ , a fast decay) with $h = 0.02$ and with $h = 0.12$ . For which $h$ do the Euler values decay like the true solution?

Solution

The Euler formula. $y_{n + 1} = y_{n} + h (- 20 y_{n}) = (1 - 20 h) y_{n}$ , so $y_{n} = (1 - 20 h)^{n} .$
$h = 0.02$ . The factor is $0.6$ : $y_{n} = 1, 0.6, 0.36, 0.216, 0.1296, 0.0778, \dots$ , a decay like the true solution (which has $e^{- 2} = 0.1353$ at $x = 0.1$ ).
$h = 0.12$ . The factor is $- 1.4$ : $y_{n} = 1, - 1.4, 1.96, - 2.744, 3.8416, - 5.3782, \dots$ . The values oscillate and grow, while the true solution is practically zero (Figure 8.4.4).
Condition. The Euler values decay exactly when $∣1 - 20 h ∣ < 1$ , that is, $0 < h < 0.1$ . For $0.05 < h < 0.1$ they decay but alternate in sign; for $h \leq 0.05$ they decay without oscillation.

Figure 8.4.4.

y^{'} = - 20 y

y (0) = 1

. With

h = 0.02

Euler's method decays like the exact solution; with

h = 0.12

the computed values oscillate with growing amplitude. This is numerical instability, not just inaccuracy.

Equations containing fast-decaying components (large negative $f_{y}$ ) appear in circuit and chemical-reaction models; they are called stiff. For them, the step size of Euler's method is limited by stability rather than by accuracy. When a numerical solution oscillates wildly where the direction field says it should not, suspect instability and reduce $h$ .

Summary

Euler's method $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ follows tangent lines from grid point to grid point.
Its local truncation error is $O (h^{2})$ ; its global error is $O (h)$ . Halving $h$ halves the error and doubles the work.
Round-off error grows as $h$ shrinks, so there is a smallest useful step; and a step that is too large can make the computation unstable.

Euler's method uses the slope only at the beginning of each step. Section 8.5 shows how averaging two slopes gives a method whose error falls like $h^{2}$ , at almost no extra cost.

Exercises

Give numerical answers to 6 decimal places.

Use Euler's method with $h = 0.25$ to approximate $y (1)$ for $y^{'} = 1 - y$ , $y (0) = 0$ . Show that $y_{n} = 1 - 0.7 5^{n}$ and compare with the exact value $1 - e^{- 1}$ .
Answer
$y_{4} = 1 - 0.7 5^{4} = 0.683594$ ; exact $0.632121$ ; error $- 0.051473$ (Euler overestimates here).
For $y^{'} = 3 x - y$ , $y (0) = 2$ , take three Euler steps with $h = 0.1$ . Compare $y_{3}$ with the exact solution $y = 3 x - 3 + 5 e^{- x}$ at $x = 0.3$ .
Answer
$y_{1} = 1.8$ , $y_{2} = 1.65$ , $y_{3} = 1.545$ ; exact $y (0.3) = 1.604091$ ; error $0.059091$ .
For $y^{'} = - 2 x y$ , $y (0) = 1$ , use $h = 0.25$ to approximate $y$ at $x = 0.25, 0.5, 0.75, 1$ . Compare $y_{4}$ with the exact value $e^{- 1}$ .
Answer
$1, 0.875, 0.65625, 0.410156$ ; exact $y (1) = 0.367879$ .
For $y^{'} = y^{2}$ , $y (0) = 0.5$ , use $h = 0.2$ to approximate $y (1)$ . The exact solution is $y = 1/ (2 - x)$ .
Answer
$0.55, 0.6105, 0.685042, 0.778899, 0.900235$ ; exact $y (1) = 1$ ; error $0.099765$ .
For $y^{'} = e^{- x} - y$ , $y (0) = 0$ , use $h = 0.2$ to approximate $y (0.6)$ and compare with the exact solution $y = x e^{- x}$ .
Answer
$y_{1} = 0.2$ , $y_{2} = 0.323746$ , $y_{3} = 0.393061$ ; exact $0.329287$ ; Euler is too high by $0.063774$ .
A tank drains according to $y^{'} = - 0.4 y$ , $y (0) = 9$ (depth in m, $t$ in min). Take three Euler steps with $h = 1$ and compare with the exact solution $y = (3 - 0.2 t)^{2}$ .
Answer
$7.8, 6.682861, 5.648812$ ; exact $7.84, 6.76, 5.76$ .
For $y^{'} = 3 y$ , $y (0) = 1$ , show that the local truncation error of the first step is $e^{3 h} - 1 - 3 h$ , and that it is approximately $\frac{9}{2} h^{2}$ for small $h$ . Evaluate both for $h = 0.1$ .
Answer
$φ = e^{3 x}$ and $y_{1} = 1 + 3 h$ . Taylor: $e^{3 h} = 1 + 3 h + \frac{9}{2} h^{2} + \dots$ . For $h = 0.1$ : $0.049859$ versus $0.045$ .
Show that Euler's method for $y^{'} = λ y$ , $y (0) = y_{0}$ gives $y_{n} = (1 + λh)^{n} y_{0}$ . If we reach a fixed $x$ in $n$ steps ( $h = x / n$ ), show that $y_{n} \to y_{0} e^{λ x}$ as $n \to \infty$ .
Answer
$y_{n + 1} = y_{n} + hλ y_{n} = (1 + λh) y_{n}$ . Then $y_{n} = y_{0} (1 + \frac{λ x}{n})^{n} \to y_{0} e^{λ x}$ (a standard limit).
Euler's method with $h = 0.1$ gives an error of $0.08$ at some point. Predict the error at the same point with $h = 0.025$ . About what $h$ is needed for an error of $0.001$ ?
Answer
Error $\approx 0.8 h$ : about $0.02$ for $h = 0.025$ ; $h \approx 0.00125$ for $0.001$ .
If the error at $x = 1$ behaves like $C h$ , then $2 y^{(h /2)} - y^{(h)}$ should cancel most of it (here $y^{(h)}$ is the Euler value with step $h$ ). Test this with the unrounded values $y^{(0.2)} = 0.537856$ and $y^{(0.1)} = 0.5861894$ from Examples 8.4.1 and 8.4.2.
Answer
$2 (0.5861894) - 0.537856 = 0.634523$ ; error $- 0.002402$ (a slight overshoot), about 19 times smaller in size than the error $0.045931$ of $y^{(0.1)}$ . (This trick is called Richardson extrapolation.)
Suppose the total error of a computation is modelled by $E (h) = 0.5 h + 2 \times 1 0^{- 6} / h$ . Find the step size that minimizes $E$ and the minimum error.
Answer
$E^{'} (h) = 0.5 - 2 \times 1 0^{- 6} / h^{2} = 0$ gives $h^{*} = 0.002$ ; $E (h^{*}) = 0.001 + 0.001 = 0.002$ .
For $y^{'} = - 40 y$ , for which step sizes do the Euler values decay to zero? For which do they decay without changing sign?
Answer
$y_{n} = (1 - 40 h)^{n} y_{0}$ . Decay iff $∣1 - 40 h ∣ < 1$ , i.e. $0 < h < 0.05$ ; no sign change iff $1 - 40 h \geq 0$ , i.e. $h \leq 0.025$ .