Differential-Equations · Unit 2 · Video 4 · Interactive Practice

Halve the Step, Halve the Error: The First-Order Price of Euler's Method

IKey Formulas

Formula	Name	What you need
$e = y(x_n) - y_n$	The error — true value minus Euler value	The exact solution at $x_n$ , and the computed $y_n$
$\text{convex} \Rightarrow e > 0$ $\text{concave} \Rightarrow e < 0$	Systematic error: the sign is fixed in advance	The sign of $y''$ along the solution, not only at the start
$e \approx C h$	First-order error law (asymptotic, small $h$ )	Nothing computable — $C$ depends on the equation and on how far you go
$e\!\left(\tfrac{h}{2}\right) \approx \tfrac{1}{2}\, e(h)$	Halve the step, halve the error	One error at one step size

Key Insight: First order counts the power of $h$ in the error, not the order of the differential equation. And it buys one factor of two per round: $2^{10} = 1024$ , so a thousandfold cut in the error costs ten halvings and about a thousandfold increase in the number of steps.

IIBigger Step, Bigger Error

Every Euler polygon for $y' = x^2 - y^2$ , $y(0) = 1$ runs below the convex solution; how far below depends on $h$ .

x reached 0.80 Step size

💡 The polygons stay low because $y'' > 0$ along the whole of $[0, 0.8]$ , checked on the curve itself — the sign of $y''(0)$ alone guarantees nothing this far out.

IIIHalving the Step, Round by Round

Each halving of $h$ lifts $y(0.2)$ by about half of the previous lift.

Halvings 0

$h$	steps	$y(0.2)$	change
$0.2$	$1$	$0.80000$	—
$0.1$	$2$	$0.82000$	$0.02000$
$0.05$	$4$	$0.82830$	$0.00830$
$0.025$	$8$	$0.83213$	$0.00383$
$0.0125$	$16$	$0.83398$	$0.00185$
$h \to 0$	—	$0.835785$	—

IVPredicting the Next Error

If $e \approx Ch$ , the errors sit on a straight line through the origin.

Halving Your e(h/2) 0.01184

💡 Theory never supplies $C$ — you read it off one computed error. In practice you keep halving until the digits stop moving, and call that the solution.

VQuiz Questions

Problem 1 · The Sign of the Error

Given: $y' = x^2 - y^2$ , $y(0) = 1$ . Euler's method with $h = 0.1$ returns $y_2 = 0.82$ at $x = 0.2$ , and the solution is convex on $[0, 0.2]$ — what can you say about $e = y(0.2) - 0.82$ ?

✅ Correct! Convexity fixes the sign before anything is computed: Euler runs low, so the true value is the larger one and

e > 0

❌ Not quite. A convex curve bends up away from every line element, so the broken line falls short. Overshooting is the concave case.

❌ The size of $e$ needs the exact solution — the sign does not. That asymmetry is exactly what makes the error systematic.

❌ Not quite.

e = y(0.2) - 0.82

, and on a convex solution the curve lies above the polygon at every

x_n

Show solution

By definition the error is the true value minus the computed one:

e = y(x_n) - y_n

On $[0, 0.2]$ the solution is convex, so at each step the curve bends up away from the line element the step follows. Every vertex of the broken line therefore lands below the curve:

y(0.2) > 0.82 \quad \Longrightarrow \quad e > 0

Numerically $y(0.2) = 0.835785\ldots$ , so $e = 0.015785\ldots$ — but that number required solving the equation. Convexity gave the sign for free.

That is the complaint against Euler: the miss is guaranteed and its direction is known in advance. A method whose errors scattered above and below would at least sometimes be right.

Problem 2 · Two Halvings

Given: on the same problem Euler's method has error $e \approx 0.0158$ at $h = 0.1$ — using $e \approx Ch$ , estimate the error at $h = 0.025$ .

✅ Correct!

0.1 \to 0.05 \to 0.025

is two halvings, so the error is quartered:

0.0158/4 \approx 0.0040

❌ Close — that is one halving. Going from

h = 0.1

h = 0.025

takes two:

0.1 \to 0.05 \to 0.025

❌ That is three halvings.

0.1/0.025 = 4

, so divide the error by

4

, not by

8

❌ Wrong direction. A smaller step gives a smaller error —

e \approx Ch

is a proportionality, so shrinking

h

shrinks

e

❌ Not quite.

e \approx Ch

means

e

falls in the same proportion as

h

Show solution

The first-order law makes $e$ proportional to $h$ , so the ratio of the step sizes is the ratio of the errors:

\frac{e(0.025)}{e(0.1)} \approx \frac{0.025}{0.1} = \frac{1}{4}

e(0.025) \approx \frac{0.0158}{4} \approx 0.0040

Equivalently, two rounds of halving, each one buying a single factor of $2$ .

Check against the real computation: eight Euler steps of $h = 0.025$ give $y(0.2) = 0.83213$ , so $e = 0.835785 - 0.83213 = 0.00365$ . The estimate is a little high, because the successive ratios approach $2$ from above — the law is asymptotic, exact only in the limit $h \to 0$ .

Problem 3 · The Price of Three More Digits

Given: you keep halving $h$ until Euler's error is about one thousandth of what it was — how many halvings does that take, and what happens to the number of steps?

✅ Correct!

2^{10} = 1024 \approx 1000

: ten rounds, and since each round doubles the step count, the work grows by the same

2^{10}

❌ The count of halvings is right, the work is not. Each round doubles the number of steps, so ten rounds multiply it by

2^{10} \approx 1000

❌ That is the second-order answer. With

e \approx Ch^2

you would need

h/\sqrt{1000} \approx h/32

. Euler's error is only

\approx Ch

❌ The thousandfold is the error reduction, not the number of rounds. Each round supplies one factor of

2

, and

2^{10}

already reaches

1000

❌ Not quite. One halving buys one factor of

2

in the error and costs one doubling of the steps.

Show solution

Since $e \approx Ch$ , dividing the error by $1000$ means dividing $h$ by about $1000$ . Halving is the tool, and each halving supplies exactly one factor of $2$ :

2^{10} = 1024 \approx 1000

so ten rounds. Reaching a fixed $x$ with a step size $1024$ times smaller needs $1024$ times as many steps, so the work grows by the same factor:

h \to \frac{h}{1024}, \qquad n \to 1024\,n

Why this is the whole complaint. A second-order method, $e \approx Ch^2$ , would need only

\frac{h}{\sqrt{1000}} \approx \frac{h}{32},

about five halvings and $32$ times the work for the same three digits. Euler creeps; that is why the next step is to replace its slope with a better one.

Problem 4 · A Different Equation

Given: a first-order method on some other initial value problem has error $|e| = 0.048$ at $h = 0.08$ — find the step size that brings the error to about $0.006$ .

✅ Correct!

C \approx 0.048/0.08 = 0.6

, so

h \approx 0.006/0.6 = 0.01

— three halvings of

0.08

❌ One halving only takes the error to $0.024$ . You need a factor of

8

, which is three halvings.

❌ That is the target error, not a step size. The two are different quantities; divide by

C

to convert one into the other.

❌ $0.08/\sqrt{8} \approx 0.028$ is what a second-order method would need. A first-order error falls only in direct proportion to

h

❌ Not quite. Read

C \approx e/h

off the one computation you have, then solve

Ch \approx 0.006

Show solution

Step 1 — read off the constant. Theory never supplies $C$ ; one computed error does:

C \approx \frac{e}{h} = \frac{0.048}{0.08} = 0.6

Step 2 — solve for the step size.

Ch \approx 0.006 \quad \Longrightarrow \quad h \approx \frac{0.006}{0.6} = 0.01

Step 3 — read it as halvings. The error must fall by $0.048/0.006 = 8$ , so $h$ must fall by $8$ :

0.08 \to 0.04 \to 0.02 \to 0.01

three rounds, and eight times as many steps to reach the same point. Note that $C$ is only approximately constant — it drifts with $h$ and with how far you compute — so $h = 0.01$ is an estimate, not a guarantee.

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