The Remainder Term

If the Taylor series for a function $f(x)$ is truncated at the $n^{\rm th}$ term, what is the difference between $f(x)$ and the value given by the $n^{\rm th}$ Taylor polynomial? That is, what is the error involved in using the Taylor polynomial to approximate the function?

Theorem. (Taylor's Theorem with Remainder) Suppose you expand f around c, and that f is $(n + 1)$ -times continuously differentiable on an open interval containing c. If x is another point in this interval, then for some z in the open interval between x and c,

$$f(x) = \sum_{k = 0}^n \dfrac{f^{(k)}(c)}{k!} (x - c)^k + \dfrac{f^{(n+1)}(z)}{(n + 1)!} (x - c)^{n+1}.$$

$\displaystyle p_n(x; c) = \sum_{k=0}^n \dfrac{f^{(k)}(c)}{k!} (x - c)^k$ is the $n^{\rm th}$ degree Taylor polynomial at c. The other term on the right is called the Lagrange remainder term:

$$R_n(x; c) = \dfrac{f^{(n+1)}(z)}{(n + 1)!} (x - c)^{n+1}.$$

The appearance of z, a point between x and c, and the fact that it's being plugged into a derivative suggest that there is a connection between this result and the Mean Value Theorem. In fact, for $n = 0$ the theorem says that there is a number z between c and x such that

$$f(x) = f(c) + f'(z)\cdot (x - c).$$

This is the Mean Value Theorem. Thus, Taylor's Theorem with Remainder is a generalization of the Mean Value Theorem.

The connection goes further. I'll give a proof of the theorem at the end, since it's rather technical and isn't needed to do the error approximation examples which are the focus of this section. The proof uses Rolle's Theorem, which you may recall is a special case of the Mean Value Theorem.

Example. Compute the Remainder Term $R_3(x;1)$ for $f(x) = \sin 2 x$ .

For the third remainder term, I need the fourth derivative:

$$f'(x) = 2 \cos 2 x, \quad f''(x) = -4 \sin 2 x, \quad f'''(x) = -8 \cos 2 x, \quad f^{(4)}(x) = 16 \sin 2 x.$$

The Remainder Term is

$$R_3(x; 1) = \dfrac{16 \sin 2 z}{4!}(x - 1)^4.$$

z is a number between x and 1.

Example. Compute the Remainder Term $R_n(x; 3)$ for $f(x) = e^{4 x}$ .

Since I want the $n^{\rm th}$ Remainder Term, I need to find an expression for the $(n + 1)^{\rm st}$ derivative. I'll compute derivative until I see a pattern:

$$f'(x) = 4 e^{4 x}, \quad f''(x) = 4^2 e^{4 x}, \quad f'''(x) = 4^3 e^{4 x}.$$

Notice that it's easier to see the pattern if you don't multiply out the power of 4.

Thus,

$$f^{(n)}(x) = 4^n e^{4 x}, \quad\hbox{so}\quad f^{(n+1)}(x) = 4^{n+1} e^{4 x}.$$

The Remainder Term is

$$R_n(x;3) = \dfrac{4^{n+1} e^{4 z}}{(n + 1)!} (x - 3)^{n+1}.$$

z is a number between x and 3.

There are several things you might do with the Remainder Term:

1. Estimate the error in using $p_n(x; c)$ to estimate $f(x)$ on a given interval $(c - r, c + r)$ . (The interval and the degree n are fixed; you want to find the error.)

2. Find the smallest value of n for which $p_n(x; c)$ approximates $f(x)$ to within a given error ("tolerance") on a given interval $(c - r, c + r)$ . (The interval and the error are fixed; you want to find the degree.)

3. Find the largest interval $(c - r, c + r)$ on which $p_n(x; c)$ approximates $f(x)$ to within a given error ("tolerance"). (The degree and the error are fixed; you want to find the interval.)

Example. The Maclaurin series for $\ln (1 + x)$ is

$$\ln (1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots.$$

What is the largest error which might result from using the first three terms of the series to approximate $\ln (1 + x)$ , if $0 \le x \le 1$ ?

The remainder term is

$$R_n(x;0) = \dfrac{f^{(n+1)}(z)}{(n + 1)!} x^{n+1},$$

I have $0 < z < x$ . I want to estimate the maximum size of $|R_3(x;0)|$ . I take absolute values, because I don't care whether the error is positive or negative, only how large it is.

$f(x) = \ln (1 + x)$ , and you can check by taking derivatives that $f^{(4)}(x) = \dfrac{-6}{(1 + x)^4}$ . Thus, $f^{(4)}(z) = \dfrac{-6}{(1 + z)^4}$ . So

$$|R_3(x;0)| = \left|\dfrac{\dfrac{-6}{(1 + z)^4}}{4!} (x - 0)^4\right| = \dfrac{1}{4} \dfrac{1}{(1 + z)^4} |x|^4.$$

Since I want the largest possible error, I want to see how large the terms $\dfrac{1}{(1 + z)^4}$ and $|x|^4$ could be.

Remember that z is between 0 and x, and $0 \le x \le 1$ . So

$$0 < z < x \le 1.$$

First, $0 \le x \le 1$ means that

$$|x|^4 \le 1^4 = 1.$$

How large can $\dfrac{1}{(1 + z)^4}$ be, given that $0 < z < 1$ ? As z goes from 0 to 1, $\dfrac{1}{(1 + z)^4}$ decreases, so it is largest if $z = 0$ . This means that

$$\dfrac{1}{(1 + z)^4} \le 1.$$

You can also see this by doing the algebra:

$$\eqalign{ 0 <\ & z < 1 \cr 1 <\ & z + 1 < 2 \cr 1 <\ & (z + 1)^4 < 16 \cr \noalign{\vskip2pt} 1 >\ & \dfrac{1}{(1 + z)^4} > \dfrac{1}{16} \cr}$$

In general, to estimate the z-term you'd have to find the absolute max on the interval for z. If you know that the z-term is either increasing or decreasing, you can check its value at the interval endpoints, and take the largest.

Using the estimates for $\dfrac{1}{(1 + z)^4}$ and $|x|^4$ , I have

$$|R_3(x;0)| \le \dfrac{1}{4} \cdot 1 \cdot 1 = \dfrac{1}{4}.$$

The error is no greater than $\dfrac{1}{4}$ .

I can check this by plotting the difference between the $3^{\rm rd}$ degree Taylor polynomial and $\ln (1 + x)$ .

$$\hbox{\epsfysize=2in \epsffile{remainder-term-1.eps}}$$

From the picture, it looks as though the maximum error is around 0.15 (in absolute value). The estimated error was pretty conservative.

Example. (a) Compute $R_3(x;0)$ for $f(x) = \dfrac{1}{2 + x}$ , and express $f(x)$ using $p_3(x)$ and the remainder term.

(b) Use $R_3(x;0)$ to approximate the largest error that occurs in using $p_3(x)$ to approximate $\dfrac{1}{2 + x}$ for $0 \le x \le 1$ .

(a) Since I want $R_3(x;0)$ , I need the fourth derivative:

$$f'(x) = \dfrac{-1}{(2 + x)^2}, \quad f''(x) = \dfrac{2}{(2 + x)^3}, \quad f'''(x) = \dfrac{-6}{(x + x)^4}, \quad f^{(4)}(x) = \dfrac{24}{(2 + x)^5}.$$

Thus,

$$R_3(x;0) = \dfrac{24}{(2 + z)^5} \cdot \dfrac{1}{4!} x^4 = \dfrac{x^4}{(2 + z)^5}.$$

Now

$$\dfrac{1}{2 + x} = \dfrac{1}{2} \cdot \dfrac{1}{1 - \left(-\dfrac{x}{2}\right)} = \dfrac{1}{2} \cdot\left(1 - \dfrac{x}{2} + \dfrac{x^2}{4} - \dfrac{x^3}{8} + \cdots\right).$$

Therefore,

$$\dfrac{1}{2 + x} = \dfrac{1}{2} \cdot\left(1 - \dfrac{x}{2} + \dfrac{x^2}{4} - \dfrac{x^3}{8}\right) + \dfrac{x^4}{(2 + z)^5}.$$

Here z is between 0 and x.

(b) I have

$$|R_3(x;0)| = \dfrac{1}{(2 + z)^5} \cdot |x|^4.$$

I'll estimate the z and x-terms one at a time.

Since $0 \le x \le 1$ , I have

$$|x|^4 \le 1^4 = 1.$$

Since $0 \le x \le 1$ and z is between 0 and x, it follows that $0 \le z \le 1$ . On this interval, $\dfrac{1}{(2 + z)^5}$ decreases, so it attains its largest value at $z = 0$ . Therefore,

$$\dfrac{1}{(2 + z)^5} \le \dfrac{1}{(2 + 0)^5} = \dfrac{1}{32}.$$

Alternatively,

$$\eqalign{ 0 <\ & z < 1 \cr 2 <\ & 2 + z < 3 \cr 32 <\ & (2 + z)^5 < 243 \cr \noalign{\vskip2pt} \dfrac{1}{32} >\ & \dfrac{1}{(2 + z)^5} > \dfrac{1}{243} \cr}$$

Thus,

$$|R_3(x;0)| \le \dfrac{1}{32} \cdot 1 = \dfrac{1}{32}.$$

The error is no greater than $\dfrac{1}{32}$ .

Example. Find the smallest value of n for which the $n^{\rm th}$ degree Taylor series for $f(x) = e^{2 x}$ at $c = 0$ approximates $e^{2 x}$ on the interval $0 \le x \le 0.3$ with an error no greater than $10^{-6}$ .

Notice that

$$f'(x) = 2 e^{2 x}, \quad f''(x) = 2^2 e^{2 x}, \quad f^{(3)}(x) = 2^3 e^{2 x}, \quad \ldots, \quad f^{(n)}(x) = 2^n e^{2 x}.$$

So

$$|R_n(x; 0)| = \left|\dfrac{2^{n+1} e^{2 z}}{(n + 1)!} x^{n+1}\right| = \dfrac{2^{n+1} e^{2 z}}{(n + 1)!} |x|^{n+1} \quad\hbox{for}\quad 0 \le z \le x \le 0.3.$$

First, I'll estimate how large the z and x-terms can be. Since $0 \le x \le 0.3$ and $x^n$ is an increasing function, I have

$$|x|^{n+1} \le 0.3^{n+1}.$$

Since $0 \le z \le 0.3$ and since $e^{2 z}$ is an increasing function, I have

$$e^{2 z} \le e^{0.6}.$$

Thus,

$$|R_n(x; 0)| \le \dfrac{2^{n+1} e^{0.6}}{(n + 1)!} \cdot 0.3^{n + 1} = e^{0.6} \dfrac{0.6^{n + 1}}{(n + 1)!}.$$

Therefore, I want the smallest n for which

$$e^{0.6} \dfrac{0.6^{n + 1}}{(n + 1)!} < 10^{-6}.$$

I can't solve this inequality algebraically, so I'll have to use trial-and-error:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & n & & $e^{0.6} \dfrac{0.6^{n + 1}}{(n + 1)!}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 1 & & $0.32798 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 2 & & $0.06559 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 3 & & $0.00983 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 4 & & $0.00118 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 5 & & $1.18073 \ldots \cdot 10^{-4}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 6 & & $1.01205 \ldots \cdot 10^{-5}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 7 & & $7.59042 \ldots \cdot 10^{-7}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

The smallest value of n is $n = 7$ .

You can also use the Remainder Term to estimate the error in using a Taylor polynomial to approximate an integral.

Example. Calvin Butterball wants to impress Phoebe Small by using the MacLaurin series for $e^{2 x}$ to approximate $\displaystyle \int_0^{0.5} x e^{2 x}\,dx$ to within 0.0001. How many terms of the series should he use?

The Maclaurin series for $e^{2 x}$ is

$$e^{2 x} = \sum_{n=0}^{\infty} \dfrac{2^n x^n}{n!}.$$

(Substitute $u = 2 x$ in the standard series for $e^u$ .) I want to know how many terms of the series to use to approximate the integral.

Since $f(x) = e^{2 x}$ , I have

$$f'(x) = 2 e^{2 x}, \quad f''(x) = 2^2 e^{2 x}, \ldots f^n(x) = 2^n e^{2 x}.$$

Therefore,

$$R_n(x) = \dfrac{1}{(n + 1)!} f^{(n+1)}(z) (x - c)^{n+1} = \dfrac{1}{(n + 1)!} \cdot 2^n \cdot e^{2 z} \cdot x^{n+1}.$$

In the integral, x goes from 0 to 0.5, and z is a number between 0 (the expansion point) and x. Therefore, I know that z is a number between 0 and 0.5. Taking the worst possible case, the largest $e^{2 c}$ could be is $e^{2 \cdot 0.5} = e$ . Replace $e^{2 z}$ with e to obtain

$$R_n(x) \le \dfrac{1}{(n + 1)!} \cdot 2^n \cdot e \cdot x^{n+1}.$$

Insert this into the integral (remembering to multiply by x):

$$\hbox{error} \le \int_0^{0.5} \dfrac{1}{(n + 1)!} \cdot 2^n \cdot e \cdot x^{n+2}\,dx = \dfrac{1}{(n + 1)!} 2^{n+1} \cdot e \cdot \dfrac{1}{n + 3} \cdot (0.5)^{n+3}.$$

I want the smallest value of n for which this ugly mess is less than 0.0001. The easiest way to do this is by trial: Plug in successive values of n.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & n & & $\dfrac{2^{n + 1}}{(n + 1)!} \cdot \dfrac{e}{n + 3} \cdot 0.5^{n + 3}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 0 & & $0.226523485 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 1 & & $0.084946307 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 2 & & $0.022652348 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 3 & & $0.004719239 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 4 & & $0.000809012 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 5 & & $0.000117980 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 6 & & $0.000014981 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

$n = 6$ is the smallest value that works.

Now I'll give a proof of the Remainder Term Theorem. The proof is a little tricky, and as the previous examples show, it isn't necessary to know the proof in order to apply the Theorem. This version of the proof roughly follows that given by G. H. Hardy in his classic A Course of Pure Mathematics [1].

In what follows, I'll take $c < x$ . You can adapt everything to the case $x < c$ without too much trouble.

As I noted at the start, the theorem says roughly that there's a number z in an interval $(c, x)$ and z satisfies certain conditions. It's natural to think of other theorems with similar conclusions, with the idea of using those theorems to prove this one. The Mean Value Theorem has a similar conclusion, stating that there's a number z in $(c, x)$ such that

$$f(x) = f(c) + f'(z) (x - c).$$

I'll actually use Rolle's Theorem, which is a special case of the Mean Value Theorem. It says that, with appropriate conditions on f, if $f(c) = 0$ and $f(x) = 0$ , then $f'(z) = 0$ for some z in $(c, x)$ .

The tricky part is knowing what function to apply Rolle's Theorem to: It isn't just the "f" in the statement of the theorem. If you look at the proof, you can see how the function used might have been concocted by working backwards from the conclusion.

Theorem. Suppose f is $(n + 1)$ -times continuously differentiable on an open interval containing c. If x is another point in this interval, then for some z in the open interval between x and c,

$$f(x) = f(c) + (x - c) f'(c) + \dfrac{1}{2!} (x - c)^2 f''(c) + \cdots + \dfrac{1}{n!} (x - c)^n f^{(n)}(c) + \dfrac{1}{(n + 1)!} (x - c)^{n + 1} f^{(n + 1)}(z).$$

Proof. I'll write the proof for the case $x > c$ , but the same approach works if $x < c$ . To make the writing easier, I'll define two helper functions:

$$g(t) = f(x) - f(t) - (x - t) f'(t) - \dfrac{}1{2!} (x - t)^2 f''(t) - \cdots - \dfrac{1}{n!} (x - t)^n f^{(n)}(t).$$

$$h(t) = g(t) - \dfrac{(x - t)^{n + 1}}{(x - c)^{n + 1}} g(c).$$

(Note that the variable in these functions is t.) Observe that

$$g(x) = f(x) - f(x) - (x - x) f'(t) - \cdots - \dfrac{1}{n!} (x - x)^n f^{(n)}(x) = 0.$$

g is continuous on $[c, x]$ and differentiable on $(c, x)$ , so the same is true for h. If I plug the endpoints c and x into h, I get

$$h(c) = g(c) - \dfrac{(x - c)^{n + 1}}{(x - c)^{n + 1}} g(c) = g(c) - g(c) = 0,$$

$$h(x) = g(x) - \dfrac{(x - x)^{n + 1}}{(x - c)^{n + 1}} g(c) = g(x) = 0.$$

Thus, the conditions for applying Rolle's Theorem to h on the interval $[c, x]$ are satisfied. I'll compute the derivative of h, first computing the derivative of g:

$$\hbox{\hskip-0.75in} g'(t) = 0 - f'(t) - [(-1) f'(t) + (x - t) f''(t)] - \dfrac{1}{2} [-2(x - t) f''(t) + (x - t)^2 f^{(3)}(t)] - \cdots$$

$$\hbox{\hskip2.5in} - \dfrac{1}{n!} [-n(x - t)^{n - 1} + (x - t)^n f^{(n + 1)}(t)]$$

$$= -\dfrac{1}{n!} (x - t)^n f^{(n + 1)}(t).$$

Note how adjacent terms cancel in pairs, except for the last one. Using this, the derivative of h is

$$h'(t) = g'(t) + (n + 1) \dfrac{(x - t)^n}{(x - c)^{n + 1}} g(c) = -\dfrac{1}{n!} (x - t)^n f^{(n + 1)}(t) + (n + 1) \dfrac{(x - t)^n}{(x - c)^{n + 1}} g(c) =$$

$$(n + 1) \dfrac{(x - t)^n}{(x - c)^{n + 1}} \left(g(c) - \dfrac{1}{(n + 1)!} (x - c)^{n + 1} f^{(n + 1)}(t)\right).$$

Rolle's Theorem implies that $h'(z) = 0$ for some z in $(c, x)$ . That is,

$$\eqalign{ (n + 1) \dfrac{(x - z)^n}{(x - c)^{n + 1}} \left(g(c) - \dfrac{1}{(n + 1)!} (x - c)^{n + 1} f^{(n + 1)}(z)\right) & = 0 \cr g(c) - \dfrac{1}{(n + 1)!} (x - c)^{n + 1} f^{(n + 1)}(z) & = 0 \cr}$$

Now

$$g(c) = f(x) - f(c) - (x - c) f'(c) - \dfrac{}1{2!} (x - c)^2 f''(c) - \cdots - \dfrac{1}{n!} (x - c)^n f^{(n)}(c).$$

So

$$f(x) - f(c) - (x - c) f'(c) - \dfrac{}1{2!} (x - c)^2 f''(c) - \cdots - \dfrac{1}{n!} (x - c)^n f^{(n)}(c) - \dfrac{1}{(n + 1)!} (x - c)^{n + 1} f^{(n + 1)}(z) = 0.$$

Hence,

$$f(x) = f(c) + (x - c) f'(c) + \dfrac{}1{2!} (x - c)^2 f''(c) + \cdots + \dfrac{1}{n!} (x - c)^n f^{(n)}(c) + \dfrac{1}{(n + 1)!} (x - c)^{n + 1} f^{(n + 1)}(z).\quad\halmos$$

[1] G. H. Hardy, A Course of Pure Mathematics. Cambridge, UK: Cambridge University Press, 1952 (tenth edition). [0-521-09227-2]

The proof above is essentially the one in Section 150, page 286.

Hardy's book was first published in 1909. Not all books from that era have aged well, but this one has. Hardy was an excellent expositor, and the writing is very clear. Most of the material would be found in our current first-term calculus or elementary functions courses. However, Hardy covers topics at substantial depth: For instance, the book begins with a construction of the real numbers using Dedekind cuts. Proofs are given for all major results.

Thus, this is a proof-oriented calculus book, somewhat like Tom Apostol's two-volume calculus text. It's a little bit below a typical real analysis text in some respects, though some of the problems would be challenging for real analysis students. I'd recommend it if you'd like to deepen your knowledge of calculus.

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