Topic 7.5: Approximating Solutions Using Euler’s Method

5:05 a.m.

It’s five past five on a school morning in January, it’s been snowing all night, and in this whole town full of sleeping houses, exactly one kid is awake. They look outside, see a foot of fresh snow, and text a friend, “school’s canceled, my mom works for the district,” and nobody has any idea whether that’s true. Watch what that one text does between now and seven o’clock.

For the first half hour it barely spreads at all, and then suddenly it’s everywhere at once. After that it slows right back down, until the last few kids hear it when they’re already standing at the bus stop.

If you were listening to those pings, you caught something strange. There’s one exact moment when this rumor spreads faster than at any other time, and it isn’t the moment most people would guess. By the end of this lesson, you’ll be able to forecast this rumor five minutes at a time without solving for its curve. You’ll also prove how many kids knew it when it spread fastest.

Let’s rewind to six in the morning. In our model of this thousand-kid school, two hundred kids have heard it by then, and the rumor is spreading at sixteen kids a minute.

Five minutes from now

If it keeps spreading at exactly sixteen kids a minute, how many kids will have heard the rumor by six oh five? Now let’s see what really happens. At six oh five our model gives about two hundred ninety-two kids, so our forecast came up twelve kids short.

We came up short because we pretended the rate stayed at sixteen. Every kid who hears the rumor starts texting too, so it kept speeding up the whole time we weren’t looking. The fix is to take a shorter step, check the rate again from wherever you land, and keep going. That only works if you have a rule that tells you the rate at any moment.

A rule like that is a differential equation, and we’ll find the rumor’s own rule soon. First let’s practice on one you’ve already seen, right here on this river. The current follows d y d x equals x minus y, so at every point the water tells a lantern which way to go. This lantern just rides it, straight through the gate between those two posts.

Now meet Euler’s boat. It can only read the current at these buoys, one unit apart, and between buoys it locks its rudder and goes dead straight. At each buoy it reads the current, turns to match it, and heads straight for the next buoy. There it checks again and turns again, so its whole path is a chain of short straight lines.

It misses the gate, coming in low by more than a quarter of a unit, because the current kept bending while the boat was busy going straight. We call that chain of straight steps Euler’s method. We keep repeating what we did with the rumor at six, checking the slope before each new step.

On paper the river becomes a slope field. We start at the point we know, zero, one. The equation tells us the slope right there, and x minus y is zero minus one, which is negative one. We’ll take a step of one half to the right, and we’ll call that step h. A slope is the rise for every one unit of run, so over a run of one half, the rise is one half times negative one, negative one half.

That means the new y is the old y plus h times the slope, where the slope always comes from the point where the step starts. That one line is the whole of Euler’s method. We land at one half, one half, and now we have to decide which way to go next. For the second step, which slope should we use?

Let’s write it the way you’d write it on the exam, as a table. The first row starts at zero, one, where the slope is negative one, so the change in y is one half times negative one, negative one half. That puts us at one half, one half, where the slope is zero, so the second step is flat and we land at one, one half.

After two steps, Euler’s method says y of one is about one half, and I’m writing approximately equal, because this is an estimate. The real solution through zero, one is at about zero point seven four when x is one, so we’re low, just like the boat. If we’d kept the first slope the whole way we’d have landed at zero, which is way worse.

Two steps, AP style

Let’s try an AP-style question. We have d y d x equal to two x minus y, and f of one is three. We need two equal Euler steps to estimate f of two.

The first thing I do is find the step size. Two equal steps from one to two means each step is one half. A mistake AP graders have written about is using one, the whole distance, as a single step.

The first step starts at one, three. The slope there is two times one minus three, negative one, so f of one point five is about three plus one half times negative one, which is two point five. You can write exactly this line on the exam and leave the arithmetic unsimplified.

Now you take the second step. What does Euler’s method give for f of two? The exact value turns out to be about three point one, so we’re low again, and in a few minutes you’ll see why that keeps happening.

Now let’s see you do both steps. The slope is x plus two y, and we start at zero, one. Use two equal steps to estimate y of one.

Smaller steps

Let’s go back to the river, because we still owe Euler’s boat a better trip through that gate. Find the widest buoy spacing that still gets Euler’s boat through the gate, so it does as little work as possible.

Let’s write down every trip. With buoys a whole unit apart the boat misses by about zero point two seven, and the table shows what happens as the spacing shrinks. Look at the miss column, which is the error. Every time we halve the step, the error roughly halves too, which is great news, but it also means doing twice as many steps.

You might have noticed the boat always comes in low. This route is concave up the whole way, and a tangent line to a concave-up curve sits below the curve, so the very first step lands low.

From there the boat never catches back up, and every estimate in the table stays below the true value. Our exam problem came in low too, two point seven five against about three point one.