Topic 10.10: Alternating Series Error Bound

The lamp that hunts

Let’s find our first promise somewhere completely different, a lighthouse on a cliff, twenty minutes after sunset, right after a storm. Up in that lantern is a glass lens taller than you are, and to send a tight, nearly parallel beam out to ships twenty nautical miles away, the lamp has to sit at one exact spot called the focus. Now picture a keeper who focuses it the way you’d focus binoculars, sliding past the sharp spot and then back, with every correction smaller than the one before.

First she pushes it forward twenty-four millimeters, then back twelve, then forward four, and then back one. Look at her pattern. Each move is the last one divided by two, then by three, then by four, so her next move will be only a fifth of a millimeter. What I love is that after just two moves the focus is already trapped, because she went forward past it and came back, so it has to be somewhere between those two pins.

If she kept going forever, she’d end up right on the focus. Every move after this one is shorter than the last, and it flips direction, so she can never get back out past either pin, and neither can the focus. The same thing works for any two neighbors, so the fence keeps shrinking, from twelve millimeters wide to four and then to one.

She’s made four moves and she’s sitting at fifteen millimeters. Drag the two brass clamps to fence in the focus as tightly as you can promise, then lock it.

The first term you leave out

Let’s put the keeper’s moves on paper. What trapped her works for a whole family of series. Each place she lands is a partial sum, and the focus is the sum of the whole infinite series, twenty-four minus twelve plus four minus one, forever.

Her fence gives us a rule. When you stop after n terms, your error is at most the size of the first term you leave out. It only works when the signs take turns, the sizes keep shrinking, and the sizes head to zero, and those are exactly the things that kept the keeper trapped.

The first term you leave out also tells you which side you’re on. If it’s positive, your partial sum is on the low side, and if it’s negative, your partial sum is on the high side.

Let’s try one the way the exam asks it. We’ll use the first four terms of this series to approximate its sum, and then say how far off we could be. Over a common denominator of a hundred forty-four, the first four terms add up to a hundred fifteen over a hundred forty-four, about point seven nine eight six.

The first term we left out is one twenty-fifth, but before we use it, we check the conditions. They all hold, since the signs alternate and one over n squared keeps shrinking, all the way to zero.

That means our error is at most one twenty-fifth, which is point zero four. The term we left out is positive, so our partial sum is too small, and the true sum has to sit between about point seven nine eight six and point eight three eight seven. This series happens to add up to pi squared over twelve, about point eight two two five, and there it is, right inside our fence.

On the exam, full credit needs the reason too, so write it in one sentence, the way it’s written here, naming the conditions and the theorem. The most common slip is reaching for the last term you kept, one sixteenth, instead of the first term you left out. That’s still true, but it’s a looser promise than the theorem gives you, and on the exam it won’t earn the point.

Now let’s go back to the lighthouse and rewind her to her second move. This time you’re the one running the hunt. Tap the lamp to make each next correction, and pull the promise lever the moment you can promise she’s within a tenth of a millimeter of the focus. If she keeps going, her moves shrink to thousandths of a millimeter, and every single one of them hops over the focus.

Not every series closes in that fast. Try this one, where the terms only shrink like one over the square root of n. Using the first term you leave out as your bound, what’s the smallest number of terms that guarantees the error is no more than one tenth?

Terms that don’t take turns

Back at the rocket, let’s start with the simplest forecast, the tangent line, which only knows the rocket’s height and how fast it’s climbing. Last lesson, each extra derivative added a correction, about three hundred twenty-four meters and then about twenty-two, and every one of them pushed the forecast up, so nothing takes turns and nothing traps the answer here.

Let’s ask a different question instead. The tangent line acts as if the rocket weren’t accelerating at all, so how big can the acceleration actually get in these ten seconds? In the launch model it starts near six and a half meters per second squared and keeps climbing, but it never gets above eight.

Last lesson’s second correction, three hundred twenty-four meters, was one half times the acceleration, six point four eight, times ten squared. For a promise, we put the ceiling, eight, where the six point four eight was.

If the acceleration never goes above eight meters per second squared, what’s the most the tangent-line forecast could be off at seventy seconds, in meters? Now it’s your turn to set the ceiling. Drag the dashed line, fly the rocket, and find a ceiling that’s a promise you’d really stand behind.

Now let the forecast use the acceleration too. The only thing that can still surprise it is the acceleration changing, which is the jerk, and over these ten seconds the jerk stays under point one five meters per second cubed. That one half was really one over two factorial. One step up, it’s the next term again with the jerk’s ceiling in place, point one five times ten cubed over three factorial, twenty-five meters. The rocket comes through twenty-two meters off, inside again.

Now you’re the flight engineer. Choose the forecast that lets you promise one meter at seventy seconds, certify it, and only then let it fly.

Physicists really do call that fourth derivative of position the snap, by the way, and the next two the crackle and the pop. Let’s take this graph down to the board. Those three promises are really one formula.

Lagrange’s promise

Here are our three promises side by side. Each one is the next Taylor term, with a ceiling on that derivative in place of the derivative itself. That’s the Lagrange error bound. For a Taylor polynomial of degree n centered at a, the error at x is at most M times the distance from a to x to the power n plus one, over n plus one factorial.

M is a ceiling on the next derivative, the n plus first. It has to hold for every t between a and x, and for the rocket, with a at sixty and x at seventy, that meant every moment of those ten seconds. We raise the distance to the n plus first power, so the guaranteed interval widens fast as we move away from the center.

Let’s try an exam-style problem, with the second-degree Taylor polynomial for a function centered at two. Its third derivative stays between negative three and three from two all the way to two point four, and we’ll need that two point four in a minute. To bound the error at two point two, M is three, the distance from the center is point two, and n plus one is three. That makes the bound three times point two cubed over three factorial, which is point zero zero four.

Same function and same ceiling, but now we look at two point four, twice as far from the center. What happens to the bound?

Now let’s do one with a real function. The third-degree Maclaurin polynomial for e to the x says e to the one half is about one point six four five eight. Every term here is positive, so nothing alternates, and Lagrange is the only fence we’ve got. We need a ceiling on the fourth derivative, and for e to the x that’s just e to the t, for every t between zero and one half.

Which value of M can you use? We can trust two without a calculator, because e is less than four, so its square root, e to the one half, is less than two. Two isn’t the smallest ceiling, and that’s fine, because any true ceiling gives a true promise. That means the true value of e to the one half has to sit inside this interval, a little over five thousandths on either side of our estimate.

Two fences, one sine

Your calculator shows a long string of digits for the sine of point one. Before we look at them, how many could we certify ourselves? The third-degree Taylor polynomial for sine is x minus x cubed over six, so the sine of point one is about point one minus one six-thousandth, which is point zero nine nine eight three three three, repeating.

Plug in point one and the series takes turns, and its terms shrink fast, so the keeper’s fence works on a Taylor series too. The first term we left out is point one to the fifth over five factorial, one twelve-millionth, and it’s positive, so the polynomial is a hair too small.

That traps the sine of point one in a tiny window. Every number in that window starts point zero nine nine eight three three, so those six decimal places are guaranteed, while the seventh could still be a three or a four. Now we can peek at the calculator, and sure enough, it starts point zero nine nine eight three three.

Lagrange works here too, with M equal to one, since sine’s derivatives never go above one, but with n equal to three its promise comes out fifty times looser, so here the first term we left out gives the tighter guarantee. Which of these approximations can’t use the alternating series error bound?

On the exam

Let’s finish with a question built like the last part of a BC free-response question, about a function whose value and first three derivatives we know at x equals one. Building its third-degree Taylor polynomial about one, the way we did last lesson, gives about two point six six four for f of one point four.

The part we care about asks you to use the Lagrange error bound to show this approximation is within two hundredths of the true value. What’s the Lagrange error bound for P three of one point four?

For full credit, show the bound with its numbers in place. Then tie it to the question, because an error of at most point zero one two eight is less than point zero two. Graders also watch one small detail. If your bound had come out exactly point zero two, you could only say the error is at most point zero two, never that it’s less.

Promises kept

Let’s go back to where we started. Now you can make this promise yourself. When a series takes turns and its terms shrink to zero, the first term you leave out is your promise, and its sign tells you which side you’re on.

When it doesn’t, you find the most the next derivative can be anywhere between the center and your x, and Lagrange turns that ceiling into a promise. Either way, you know how wrong you could possibly be before you ever see the answer, and that’s what lets anyone trust a forecast, a lamp, or the digits on a calculator.