ODE solvers — Euler's method (derivation, global error): Visual walkthrough

We are chasing a curve we cannot write down. We only know its slope at every point. Watch how that single fact becomes a working method, step by step.

Step 1 — What we are given: a slope field, not a curve

WHAT. We have an Initial Value Problem: $slope of the curve \frac{d y}{d x} = a rule that gives the slope f (x, y), value at the start y (x_{0}) = a known number y_{0}$

Read every symbol out loud:

$x$ — the horizontal position (how far along we've walked).
$y$ — the height of the curve at that position (what we want to know).
$\frac{d y}{d x}$ — "how steep is the curve here", the rise over run of its tangent line.
$f (x, y)$ — a machine: feed it a point $(x, y)$ , it hands back a slope. We do have this.
$(x_{0}, y_{0})$ — the one point we know for certain. Our starting pin.

WHY this framing. We cannot integrate directly because $f$ depends on $y$ , and $y$ is exactly the unknown. So instead of a formula, think of $f$ as painting a tiny arrow at every point of the plane — a slope field. The true solution is the curve that always lines up with the local arrow.

PICTURE. Below, the pastel arrows are the slope field for $y^{'} = x + y$ . The single black dot is our known start $(0, 1)$ . There is no curve drawn yet — that's the whole problem.

Figure — ODE solvers — Euler's method (derivation, global error)

Step 2 — The one honest move: trust the tangent for a short distance

WHAT. At our known point $(x_{0}, y_{0})$ the arrow points with slope $m_{0} = f (x_{0}, y_{0})$ . We build the straight line through the point with exactly that slope — the tangent line — and follow it a short horizontal distance $h$ .

$our guess for the height at x_{1} y_{1} = height now y_{0} + how far right we step h \cdot slope here f (x_{0}, y_{0})$

Term by term: $h \cdot f (x_{0}, y_{0})$ is $run \times slope = rise$ . Add that rise to the current height $y_{0}$ and you land at the new height $y_{1}$ .

WHY. Right at the start the tangent line is the curve (they touch and share a slope). For a short run they stay close. We can't follow the curve — we have no formula — but we can follow its tangent, because the tangent needs only the slope, which $f$ gives us for free.

PICTURE. The right triangle makes the arithmetic visible: horizontal leg $h$ , vertical leg $h \cdot m_{0}$ , and the hypotenuse is the tangent we walk along.

Step 3 — Where does the step formula come from? Taylor's theorem

WHAT. Taylor's theorem writes the true height a distance $h$ ahead exactly: $y (x_{n} + h) = start height y (x_{n}) + straight-line part h y^{'} (x_{n}) + the curving correction \frac{h ^{2}}{2} y^{''} (ξ)$

$y^{'} (x_{n})$ is the slope — which the ODE says equals $f (x_{n}, y_{n})$ .
$y^{''} (ξ)$ is the curvature: how fast the slope itself changes. $ξ$ is some hidden point between $x_{n}$ and $x_{n} + h$ .
The $h^{2}$ term is small when $h$ is small, because squaring a small number makes it much smaller.

WHY this tool and not another. Why Taylor? Because it is the one expansion that separates a curve into "a straight line plus a labelled, honest remainder". That remainder tells us exactly what we are throwing away when we keep only the straight-line part. No other tool hands us the error term on a plate.

Drop the $h^{2}$ term and you get the boxed Euler step of Step 2. The discarded $\frac{h ^{2}}{2} y^{''} (ξ)$ is the local truncation error — the error of a single step.

PICTURE. The true curve rises above the tangent line by exactly the gap $\frac{h ^{2}}{2} y^{''} (ξ)$ . That gap is the correction we chose to ignore.

Step 4 — Repeat: chain the steps into a staircase

WHAT. We only reached $x_{1}$ . To go further, treat the landing point $(x_{1}, y_{1})$ as a new start, read the slope there, and step again: $y_{2} = y_{1} + h f (x_{1}, y_{1}), y_{3} = y_{2} + h f (x_{2}, y_{2}), \dots$

WHY. This is the "self-bootstrapping" idea: each answer becomes the input for the next question. We never needed the whole curve at once — only the slope wherever we currently stand.

Notice the danger being born: at $(x_{1}, y_{1})$ we are already slightly below the true curve, so the slope we read is the slope of the wrong point. Small mistakes feed the next step.

PICTURE. A polyline (bent line) of tangent segments — the Euler path — climbing beside the smooth true curve. See the gap open up.

Step 5 — Local error: one step's mistake is $O (h^{2})$

WHAT. The mistake made in one step (assuming we started exactly on the curve) is the term we dropped: $LTE = \frac{h ^{2}}{2} y^{''} (ξ) = O (h^{2})$

The notation $O (h^{2})$ means: "behaves like a constant times $h^{2}$ for small $h$ ." Halve $h$ , and this single-step mistake drops to a quarter.

WHY it matters. This is the tempting trap. $O (h^{2})$ looks great. But it is the cost of one step, and we take many steps to cross a fixed distance.

PICTURE. Zoom into one step: the vertical bracket between "where the tangent lands" and "where the curve actually is" is the LTE — a tiny sliver proportional to $h^{2}$ .

Step 6 — Global error: why we lose one power of $h$

WHAT. To reach a fixed endpoint $x = b$ we need $N = \frac{b - x _{0}}{h} steps.$ As $h$ shrinks, $N$ grows (smaller steps → more of them). Add up the mistakes: $\propto 1/ h N \times per step O (h^{2}) = \frac{b - x _{0}}{h} \cdot O (h^{2}) = O (h)$

$N \propto 1/ h$ : this factor of $1/ h$ eats one power of $h$ .
Result: the global error is $O (h)$ , not $O (h^{2})$ .

WHY. This is the whole reason Euler is called first order. Each step is cheap ( $h^{2}$ ), but you pay for it $1/ h$ times. One power of $h$ survives.

PICTURE. Two staircases across the same interval $[x_{0}, b]$ : a coarse one (big $h$ , few steps, wide final gap) and a fine one (small $h$ , many steps, narrow gap). The final gap shrinks in proportion to $h$ — linearly, not quadratically.

Step 7 — The degenerate case: when a big step overshoots

WHAT. Take $y^{'} = - 2 y$ , $y (0) = 1$ , $h = 0.5$ . The true solution decays smoothly: $y = e^{- 2 x}$ . But Euler gives $y_{1} = 1 + 0.5 \cdot (- 2) (1) = 1 - 1 = 0, y_{2} = 0 + 0.5 \cdot (- 2) (0) = 0.$ It slams to zero and stays there. True $y (1) = e^{- 2} \approx 0.135$ .

WHY. The tangent at the start is steeper than the curve for the whole step, so following it straight overshoots the gentle decay entirely. The governing quantity is $∣1 + hλ ∣$ with $λ = - 2$ : here $∣1 + 0.5 (- 2) ∣ = ∣0∣ = 0$ . If that number exceeds $1$ , the numbers would blow up. This is the doorway to Numerical Stability and Backward Euler & Implicit Methods.

PICTURE. The smooth decaying curve versus the Euler point crashing to (and sticking at) zero — the tangent line diving straight through the true curve.

The one-picture summary

Everything on one canvas: the slope field, the known start, the tangent staircase we walk, the growing gap to the true curve, and the label that a coarse step gives a wide gap ( $O (h)$ ) while a fine step gives a narrow one.

Recall Feynman retelling of the whole walkthrough

You are lost in fog and can only see, at your feet, which way the path points. You cannot see the path ahead. So you face that direction and take one small step (Step 2). Taylor's theorem is just the honest bookkeeper who whispers, "By walking straight you missed the curve by about $\frac{h ^{2}}{2} \times (how sharply it bends)$ " (Step 3). At the new spot you look down again, face the new direction, step again — that's the staircase (Step 4). One step's error is tiny, like $h^{2}$ (Step 5). But to cross the whole valley you take about $1/ h$ steps, so the total drift is $1/ h$ times $h^{2}$ , which is just $h$ (Step 6). That's why halving your step size only halves your final error — Euler is first order. And if you take a wildly big step on a curve that's turning fast, you can leap right past it and land somewhere silly (Step 7) — the warning that step size also controls stability.

Connections

Taylor Series Expansion — the theorem behind Step 3's exact remainder
Finite Difference Approximations — the tangent step read as a forward difference
Runge-Kutta Methods — sample several slopes per step to kill more of the error
Backward Euler & Implicit Methods — cures the Step 7 overshoot
Numerical Stability — the $∣1 + hλ ∣$ condition from Step 7
Lipschitz Continuity — supplies the $L$ in the Step 6 bound