Taylor Polynomials and Taylor’s Theorem, calculus lesson

A calculator cannot compute $\sqrt{4.1}$ or $e^{0.1}$ directly. It can only add and multiply, so it evaluates a polynomial. The tangent line is the first such polynomial, matching the value and the slope at one point. Matching the second, third and later derivatives as well gives better and better polynomials, and Taylor’s theorem tells you exactly how far off each one can be.

Key terms

Higher derivatives

Definition The derivatives of $f$ of higher order are defined one after another by $f^{(0)} = f$ and $f^{(n+1)} = \bigl(f^{(n)}\bigr)'$ , wherever that derivative exists. So $f^{(1)} = f'$ , $f^{(2)} = f''$ , $f^{(3)} = f'''$ . A function is $n$ times differentiable at $a$ if $f^{(n)}(a)$ exists, and it is of class $C^n$ on an interval if $f^{(n)}$ exists and is continuous there.

In plain words Keep differentiating. Each new derivative describes a finer feature of the graph near a point.

Example For $f(x) = \ln x$ on $(0, \infty)$ ,

f'(x) = \frac{1}{x}, \quad f''(x) = -\frac{1}{x^2}, \quad f'''(x) = \frac{2}{x^3}, \quad f^{(4)}(x) = -\frac{6}{x^4},

and in general

f^{(n)}(x) = (-1)^{n-1} \frac{(n - 1)!}{x^n} \qquad \text{for } n \ge 1.

Expansion center

Definition The expansion center is the point $a$ at which a polynomial approximation is built. It must be a point where $f$ has as many derivatives as the polynomial uses, so $f^{(n)}(a)$ must exist for a polynomial of degree $n$ .

In plain words Choose the center where you know everything about $f$ and close to the $x$ you care about. The approximation is best near the center.

Example To approximate $\sqrt{4.1}$ , center at $a = 4$ , where $\sqrt{4} = 2$ and the derivatives are easy. Centering at $a = 0$ would fail, since $\sqrt{x}$ is not differentiable at $0$ .

Taylor polynomial

Definition Let $f$ be $n$ times differentiable at $a$ . The Taylor polynomial of degree $n$ of $f$ at $a$ is

T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x - a)^k = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n.

It is the unique polynomial of degree at most $n$ with $T_n^{(k)}(a) = f^{(k)}(a)$ for $k = 0, 1, \ldots, n$ . For $n = 1$ it is the tangent line.

In plain words It is the polynomial whose value, slope, bending and so on up to order $n$ agree with those of $f$ at $a$ .

Example Take $f(x) = \sqrt{x}$ at $a = 4$ . Then $f(4) = 2$ and

f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}, \qquad f''(4) = -\frac{1}{4} \cdot 4^{-3/2} = -\frac{1}{32}.

T_2(x) = 2 + \frac{x - 4}{4} - \frac{(x - 4)^2}{64}, \qquad T_2(4.1) = 2.02484375,

while $\sqrt{4.1} = 2.0248456\ldots$ .

Maclaurin polynomial

Definition A Maclaurin polynomial is a Taylor polynomial with center $a = 0$ ,

T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^k.

In plain words It is the special case centered at the origin, which is the most common choice when $f$ is well behaved at $0$ .

Example Every derivative of $e^x$ is $e^x$ , which equals $1$ at $0$ . So the Maclaurin polynomial of degree $3$ is

T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}.

At $x = 0.1$ it gives $1.1051666\ldots$ , and $e^{0.1} = 1.1051709\ldots$ . You can reproduce this in the Taylor polynomial explorer by choosing $e^x$ with degree $3$ and $x = 0.1$ , and its error readout shows the gap of about $4.25 \times 10^{-6}$ .

Lagrange remainder

Definition The remainder is $R_n(x) = f(x) - T_n(x)$ . Taylor’s theorem with Lagrange remainder says the following. Let $f$ be $n + 1$ times differentiable on an open interval $I$ containing $a$ . Then for each $x \in I$ with $x \neq a$ there is a number $z$ strictly between $a$ and $x$ such that

R_n(x) = \frac{f^{(n+1)}(z)}{(n + 1)!} (x - a)^{n+1}.

For $n = 0$ this is the mean value theorem.

In plain words The error looks exactly like the next term of the polynomial, except that the derivative is taken at some unknown point between $a$ and $x$ .

Example For $e^x$ with $n = 3$ , $a = 0$ and $x = 0.1$ , there is a $z$ in $(0, 0.1)$ with

R_3(0.1) = \frac{e^z}{24}(0.1)^4.

Since $1 < e^z < e^{0.1}$ , the error lies between about $4.167 \times 10^{-6}$ and $4.605 \times 10^{-6}$ . The true error is about $4.25 \times 10^{-6}$ .

Integral remainder

Definition Let $f^{(n+1)}$ be continuous on an interval containing $a$ and $x$ . Then

R_n(x) = \int_a^x \frac{f^{(n+1)}(t)}{n!} (x - t)^n \, dt.

For $n = 0$ this is the fundamental theorem of calculus,

f(x) = f(a) + \int_a^x f'(t) \, dt,

and the general case follows by integrating by parts $n$ times.

In plain words This form gives the error exactly as an integral, with no unknown point in it. It needs a little more smoothness than the Lagrange form.

Example For $e^x$ with $n = 1$ and $a = 0$ ,

R_1(x) = \int_0^x e^t (x - t) \, dt = \Bigl[e^t (x - t)\Bigr]_0^x + \int_0^x e^t \, dt = -x + e^x - 1,

which is exactly $e^x - (1 + x)$ . At $x = 1$ the error is $e - 2 \approx 0.718$ .

Derivative bounds on the interval

Definition To estimate $R_n(x)$ from either form you need a number $M$ with

\bigl\lvert f^{(n+1)}(t) \bigr\rvert \le M \quad \text{for every } t \text{ between } a \text{ and } x,

or for every $t$ in a fixed interval $\lvert t - a \rvert \le d$ that contains $x$ . The bound must hold on the whole interval, not only at $a$ , because the point $z$ in the Lagrange form is unknown.

In plain words You do not know where $z$ is, so bound the next derivative everywhere it could be.

Example For $\ln x$ on $[1, 1.5]$ we have $f''(t) = -\frac{1}{t^2}$ , so $\lvert f''(t) \rvert \le 1$ with the largest value at $t = 1$ . For $\sin x$ every derivative is $\pm \sin x$ or $\pm \cos x$ , so $M = 1$ works on any interval for every $n$ .

Taylor’s inequality

Definition Let $f$ be $n + 1$ times differentiable on $[a - d, a + d]$ , where $d > 0$ , and suppose $\bigl\lvert f^{(n+1)}(t) \bigr\rvert \le M$ for all $t$ in that interval. Then

\lvert R_n(x) \rvert \le \frac{M}{(n + 1)!} \lvert x - a \rvert^{n+1} \qquad \text{for all } x \text{ with } \lvert x - a \rvert \le d.

In plain words The error is at most the bound on the next derivative, times the distance to the center to the power $n + 1$ , divided by $(n + 1)!$ .

Example For $\sin x$ at $a = 0$ the polynomials $T_3$ and $T_4$ are both $x - \frac{x^3}{6}$ , since $\sin^{(4)}(0) = 0$ . Using $n = 4$ and $M = 1$ ,

\left\lvert \sin 0.5 - \left(0.5 - \frac{0.5^3}{6}\right) \right\rvert \le \frac{0.5^5}{120} \approx 0.000260.

The actual error is $0.4794255 - 0.4791667 \approx 0.000259$ , just under the bound.

Check yourself

Find the Taylor polynomial of degree $2$ of $\ln x$ at $a = 1$ , and use it to estimate $\ln 1.2$ .

Show the answer

The values are $f(1) = 0$ , $f'(1) = 1$ and $f''(1) = -1$ , so

T_2(x) = (x - 1) - \frac{(x - 1)^2}{2}.

Then $T_2(1.2) = 0.2 - 0.02 = 0.18$ , while $\ln 1.2 \approx 0.1823$ .

Bound the error of $e^x \approx 1 + x + \frac{x^2}{2}$ for $0 \le x \le 0.5$ .

Show the answer

Here $n = 2$ and $f'''(t) = e^t \le e^{0.5} < 2$ on $[0, 0.5]$ . Taylor’s inequality gives

\lvert R_2(x) \rvert \le \frac{2}{3!}(0.5)^3 = \frac{1}{24} \approx 0.042.

What is the Maclaurin polynomial of degree $3$ of $\cos x$ ?

Show the answer

The derivatives at $0$ are $1, 0, -1, 0$ , so $T_3(x) = 1 - \frac{x^2}{2}$ . The cubic coefficient is $0$ .

Taylor Polynomials and Taylor’s Theorem

Key terms

Higher derivatives

Expansion center

Taylor polynomial

Maclaurin polynomial

Lagrange remainder

Integral remainder

Derivative bounds on the interval

Taylor’s inequality

Check yourself

How this lesson connects

Builds on

Used later in

Resources for this lesson

Taylor polynomial explorer