What is the Lagrange error bound? Basically, it’s a theoretical limit that measures how bad a Taylor polynomial estimate could be. Read on to find out more!

## Taylor Series and Taylor Polynomials

The whole point in developing Taylor series is that they replace more complicated functions with polynomial-like expressions. The properties of Taylor series make them especially useful when doing calculus.

Remember, a Taylor series for a function *f*, with center *c*, is:

Taylor series are wonderful tools. However, the biggest drawback associated with them is the fact that they typically involve *infinitely many* terms.

So in practice, we tend to use only the first few terms. Maybe 3 terms, maybe 30, but at least a finite number of terms is more reasonable than all infinitely many of them, right?

But there is a trade-off. You lose *accuracy*. A **Taylor polynomial** (that is, finitely many terms of a Taylor series) can provide a very good approximation for a function, but it can’t model the function exactly.

That’s where the *error* comes in.

By the way, now would be a great time to review: AP Calculus BC Review: Taylor Polynomials and AP Calculus BC Review: Taylor and Maclaurin Series.

## The Lagrange Error Bound

Let *T*(*x*) be the *n*th order Taylor polynomial for a given function *f*, with center at *c*.

Then the error between *T*(*x*) and *f*(*x*) is no greater than the **Lagrange error bound** (also called the **remainder term**),

Here, M stands for the maximum absolute value of the (n+1)-order derivative on the interval between *c* and *x*. In other words, *M* is found by plugging in the *z*-value between *x* and *c* that maximizes the following expression:

That may sound complicated, but in practice, there’s usually a quick way to decide what *M* should be.

## Example

Use the Lagrange error bound to estimate the error in using a 4th degree Maclaurin polynomial to approximate cos(π/4).

#### Solution

First, you need to find the 4th degree Maclaurin polynomial for cos *x*. A Maclaurin polynomial is simply a Taylor polynomial centered at *c* = 0.

Now, for the error bound, we’ll need to know what the 5th-derivative of *f*(*x*) = cos *x* is. (You probably would have computed all of the derivatives up to the 4th order when you constructed the Maclaurin polynomial for the function, anyway.)

Now, the largest that |-sin *x*| could possibly be is 1. (Actually, we could do even better than that if we realize that |-sin(π/4)| = 0.707 maximizes the quantity on the interval [0, π/4], but we’ll stick with our first estimate of 1. After all, this is *only* an estimate!)

So, we have *M* = 1. Plugging this into the error bound formula with *n* = 4, we get:

The error is roughly 0.0025

#### Follow-Up: How Good Was Our Estimate?

Because the computed error bound was so tiny, we can be sure that *T*(*x*) approximates the values of cos *x* incredibly well, at least when the input is within the interval from 0 to π/4.

Let’s compare values to see just how close the approximation really is. Computing each of *T*(π/4) and cos(π/4) to eight digits of accuracy:

The actual error is: 0.70742921 – 0.70710678 = 0.00032243. This is much better than our estimated error of 0.0025. Really, the Lagrange error bound is just a worst-case scenario in terms of estimating error; the actual error is often much less.

Why do you use the fifth derivative equation when it is supposed to be n=4?

because you need the n+1 term.