Differential-Equations · Unit 2 · Video 6 · Interactive Practice

Pay Per Slope: RK4, Fourth-Order Error, and Why Nothing Is Free

IKey Formulas

Error	Method	Evaluations of $f$ per step
$e \approx C_1 h$	Euler	$1$ — the slope where you stand
$e \approx C_2 h^2$	Improved Euler (RK2)	$2$ — $A_n$ and $B_n$
$e \approx C_4 h^4$	RK4	$4$ — $A_n, B_n, C_n, D_n$
$y_{n+1} = y_n + h \cdot \dfrac{A_n + 2B_n + 2C_n + D_n}{6}$	The RK4 step	Weights $1, 2, 2, 1$ sum to $6$

Key Insight: Roughly, the number of evaluations per step is the power of $h$ in the error — and the trade is even. Cutting the error to $\tfrac{1}{16}$ costs $8$ evaluations per stretch of width $h$ whether you halve $h$ in RK4 or quarter it in RK2, where RK2 at step $h$ spent $2$ .

IIVisualization 1 — Counting the Readings

Each step reads the direction field once, twice, or four times, and the accuracy follows the count.

Method Steps 3

The exact curve here is itself an RK4 computation, run at a step so small that its own error sits below the width of the drawn line.

IIIVisualization 2 — The Super Slope

RK4 spends its four readings on a single weighted average, with weights $1, 2, 2, 1$ .

IVVisualization 3 — What a Sixteenth Costs

Every method can buy a sixteenth of its own error; the question is what each one charges.

Error target 1/2^d 4

The constants $C_1, C_2, C_4$ are different numbers, so each ray measures a method against its own error at step $h$ ; the three heights are not comparable with one another.

VQuiz Questions

Problem 1 · Halving the Step in RK4

Given: RK4's error over a fixed interval is $e \approx C_4 h^4$ . You replace $h$ by $\tfrac{h}{2}$ — find the new error estimate.

✅ Correct!

C_4\left(\tfrac{h}{2}\right)^4 = \tfrac{1}{16}C_4h^4

— fourth order means the error drops by a factor of sixteen.

❌ That is second-order behaviour.

\tfrac{1}{4}

is what the improved method gives, since

\left(\tfrac{h}{2}\right)^2 = \tfrac{h^2}{4}

. RK4 carries a fourth power.

❌ Check the exponent.

\tfrac{1}{8} = \left(\tfrac{1}{2}\right)^3

— you cubed rather than raised to the fourth power.

❌ Not quite. The whole step size sits inside the fourth power: compute

\left(\tfrac{h}{2}\right)^4

, not

\tfrac{h^4}{2}

Show solution

Substitute $\tfrac{h}{2}$ for $h$ in the error estimate:

e \approx C_4\left(\frac{h}{2}\right)^4 = C_4 \cdot \frac{h^4}{2^4} = \frac{1}{16}\, C_4 h^4

Every halving of the step divides a $p$ -th order error by $2^p$ : by $2$ for Euler, by $4$ for the improved method, by 16 for RK4.

Problem 2 · Counting the Bill

Given: one RK4 step of width $h$ covers a stretch of the $x$ -axis using $4$ evaluations of $f$ . You halve the step, so two RK4 steps now cover that same stretch.

How many evaluations of $f$ does the stretch now cost?

The improved method at step $h$ covered that stretch with $2$ evaluations. The count has risen by a factor of:

✅ Correct!

8 = 4 \times 2

: a sixteenth of the error, bought with four times the evaluations.

❌ That is the improved method's bill.

2

evaluations is what RK2 spends on the stretch at step

h

; RK4 reads the field four times in every step.

❌ That is one step's worth. Halving

h

puts two RK4 steps on the stretch, each costing

4

❌ Too many steps.

16 = 4 \times 4

would be four steps; halving

h

produces two.

❌ Not quite. Count steps first, then multiply by the evaluations each step costs.

❌ That is RK4 against itself.

8 \div 4 = 2

compares with RK4 at step

h

; here the comparison is with the improved method's

2

❌ That is the count, not the ratio.

8

is the new bill; divide it by the

2

the improved method spent.

❌ That is the error factor. The error fell by

16

; the work rose by much less.

❌ Not quite. Divide the new evaluation count by the

2

the improved method spent on the same stretch.

Show solution

Step 1 — the new bill. Where one step of width $h$ stood there are now two of width $\tfrac{h}{2}$ , and each evaluates $f$ four times:

2 \text{ steps} \times 4 \text{ evaluations} = 8

Step 2 — the comparison. The improved method crossed the same stretch in one step of width $h$ , for $2$ evaluations:

\frac{8}{2} = 4

So the ledger reads $\text{error} \div 16$ , $\ \text{evaluations} \times 4$ . The accuracy is bought, not given.

Problem 3 · One RK4 Step

Given: an RK4 step starting at $y_n = 3.0$ with $h = 0.5$ produces the four slopes $A_n = 0.6$ , $B_n = 1.2$ , $C_n = 1.4$ , $D_n = 1.4$ — find the super slope and then $y_{n+1}$ .

What is the super slope?

What is $y_{n+1}$ ?

✅ Correct!

\tfrac{7.2}{6} = 1.2

, and

3.0 + 0.5(1.2) = 3.6

❌ That is the unweighted average.

\tfrac{0.6 + 1.2 + 1.4 + 1.4}{4} = 1.15

, but

B_n

and

C_n

are each counted twice.

❌ Wrong denominator.

\tfrac{7.2}{4} = 1.8

divides by the number of slopes; a weighted average divides by the sum of the weights,

1 + 2 + 2 + 1 = 6

❌ That is the weighted sum.

7.2

still has to be divided by

6

to become an average.

❌ The step size is missing.

3.0 + 1.2 = 4.2

adds the slope itself; the update adds

h

times the slope.

❌ Not quite. Weight the slopes

1, 2, 2, 1

, add them, and divide by

6

❌ Not quite. Use

y_{n+1} = y_n + h \cdot (\text{super slope})

with

h = 0.5

Show solution

Step 1 — the weighted sum.

A_n + 2B_n + 2C_n + D_n = 0.6 + 2(1.2) + 2(1.4) + 1.4 = 0.6 + 2.4 + 2.8 + 1.4 = 7.2

Step 2 — divide by the sum of the weights. The weights are $1, 2, 2, 1$ and $1 + 2 + 2 + 1 = 6$ :

\text{super slope} = \frac{7.2}{6} = 1.2

Step 3 — take the step.

y_{n+1} = y_n + h \cdot \frac{A_n + 2B_n + 2C_n + D_n}{6} = 3.0 + 0.5(1.2) = 3.6

Why $6$ ? If all four slopes were equal to some $m$ , the sum would be $m + 2m + 2m + m = 6m$ , and $\tfrac{6m}{6} = m$ — the average of four equal slopes has to be that slope.

Problem 4 · A Third-Order Method

Given: a method reads the direction field $3$ times per step and has error $e \approx C_3 h^3$ . You halve its step — find the new error factor and the new cost of one stretch of width $h$ (which used to cost $3$ evaluations).

The error is now what fraction of what it was?

How many evaluations of $f$ does the stretch now cost?

✅ Correct!

\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}

of the error for

2 \times 3 = 6

evaluations — twice the work. The same kind of bargain, in a third-order currency.

❌ The order is an exponent, not a multiplier.

\tfrac{1}{6}

multiplies

2

3

; the factor is

\left(\tfrac{1}{2}\right)^3

❌ That is second-order behaviour.

\tfrac{1}{4} = \left(\tfrac{1}{2}\right)^2

; this method carries a cube.

❌ Not a division by the order. Halving

h

multiplies

h^3

\left(\tfrac{1}{2}\right)^3

❌ Not quite. Put

\tfrac{h}{2}

into

C_3h^3

and read off the numerical factor.

❌ That is the old bill.

3

evaluations covered the stretch in one step; halving

h

needs two steps.

❌ That is RK4's bill.

8 = 2 \times 4

belongs to a method reading four times per step; this one reads three.

❌ Too many steps.

12 = 4 \times 3

would be four steps; halving

h

produces two.

❌ Not quite. Two steps now cover the stretch, and each one reads the field three times.

Show solution

Step 1 — the error.

C_3\left(\frac{h}{2}\right)^3 = C_3 \cdot \frac{h^3}{8} = \frac{1}{8}\, C_3 h^3

Step 2 — the bill. Two steps of width $\tfrac{h}{2}$ replace one of width $h$ , at $3$ evaluations each:

2 \text{ steps} \times 3 \text{ evaluations} = 6

So $\tfrac{1}{8}$ of the error for $2\times$ the work. Compare RK4, which halves $h$ for $\tfrac{1}{16}$ of the error at $2\times$ its own work, and Euler, which halves $h$ for only $\tfrac{1}{2}$ of the error at the same $2\times$ . Higher order buys a steeper discount for the same doubling — but a discount, never a gift.

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