Approximating Using Taylor’s Formula, calculus lesson

Before computers, tables of $\sin x$ , $e^x$ and $\ln x$ were filled in by hand, and the people who computed them needed to know every digit was right. Taylor’s formula gives both halves of that job. The polynomial gives the number, and the remainder term says how far off it can be. The same idea, integrated term by term, handles integrals that have no elementary antiderivative.

Key terms

Taylor’s formula with remainder

Definition Let $n \ge 0$ be an integer and let $f$ be $n + 1$ times differentiable on an open interval $I$ containing $a$ . For each $x$ in $I$ ,

f(x) = P_n(x) + R_n(x), \qquad P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x - a)^k,

and there is a number $\xi$ strictly between $a$ and $x$ (or $\xi = a$ when $x = a$ ) with

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

Consequently, if $\lvert f^{(n+1)}(t) \rvert \le M$ for every $t$ between $a$ and $x$ , then

\lvert R_n(x) \rvert \le \frac{M \lvert x - a \rvert^{n+1}}{(n+1)!}.

This is Taylor’s inequality.

In plain words The Taylor polynomial is the best polynomial match at $a$ , and the remainder is the exact error. You rarely know $\xi$ , but bounding the derivative bounds the error.

Example For $f(x) = e^x$ at $a = 0$ , every derivative is $e^x$ , so

P_2(x) = 1 + x + \frac{x^2}{2}, \qquad R_2(x) = \frac{e^{\xi}}{6} x^3

for some $\xi$ between $0$ and $x$ .

Evaluating functions with Taylor polynomials

Definition To evaluate $f(x)$ with a Taylor polynomial, compute $P_n(x)$ and report it together with a bound $B$ on $\lvert R_n(x) \rvert$ from Taylor’s inequality, so that

P_n(x) - B \le f(x) \le P_n(x) + B.

In plain words You get the approximate value and a guarantee in one step. The answer is an interval that must contain the true value.

Example Approximate $e^{0.1}$ with $n = 3$ and $a = 0$ . The polynomial gives

P_3(0.1) = 1 + 0.1 + 0.005 + \frac{0.001}{6} \approx 1.1051667.

On $[0, 0.1]$ the fourth derivative satisfies $f^{(4)}(t) = e^t \le e^{0.1} < 1.2$ , so

\lvert R_3(0.1) \rvert \le \frac{1.2 \cdot 0.1^4}{24} = 5 \times 10^{-6}.

The true value is $e^{0.1} \approx 1.1051709$ , and the actual error $4.3 \times 10^{-6}$ is within the bound.

Taylor-based quadrature

Definition To approximate $\int_c^d f(x) \, dx$ , expand $f$ in a Taylor polynomial $P_n$ about a point $a$ , where $[c, d]$ lies inside an open interval $I$ containing $a$ on which $f$ is $n + 1$ times differentiable, and integrate the polynomial exactly. Since $f = P_n + R_n$ on $I$ ,

\int_c^d f(x) \, dx = \int_c^d P_n(x) \, dx + \int_c^d R_n(x) \, dx.

The first integral is the approximation. The second is its error. When $f$ is the sum of a power series about $a$ and $[c, d]$ lies inside the open interval of convergence, the series may be integrated term by term over $[c, d]$ instead.

In plain words Polynomials are easy to integrate. Replace the hard integrand by its Taylor polynomial and integrate that.

Example Approximate $\int_0^{1/2} e^{-x^2} \, dx$ , which has no elementary antiderivative. Put $u = -x^2$ in $e^{u} = 1 + u + \frac{u^2}{2} + \dots$ and keep three terms.

\int_0^{1/2} \left( 1 - x^2 + \frac{x^4}{2} \right) dx = \frac{1}{2} - \frac{1}{24} + \frac{1}{320} \approx 0.4614583.

The full series integrates to an alternating series whose terms decrease to $0$ , so the error is at most the first omitted term,

\frac{(1/2)^7}{6 \cdot 7} \approx 1.86 \times 10^{-4}.

The true value is about $0.4612810$ , an error of $1.77 \times 10^{-4}$ .

Integrated remainder bound

Definition Suppose Taylor’s formula holds on an interval containing $[c, d]$ . If $\lvert R_n(x) \rvert \le \rho(x)$ for all $x$ in $[c, d]$ and $\rho$ is integrable on $[c, d]$ , then

\left\lvert \int_c^d f(x) \, dx - \int_c^d P_n(x) \, dx \right\rvert = \left\lvert \int_c^d R_n(x) \, dx \right\rvert \le \int_c^d \rho(x) \, dx.

With Taylor’s inequality one can take

\rho(x) = \frac{M \lvert x - a \rvert^{n+1}}{(n+1)!},

where $M$ bounds $\lvert f^{(n+1)} \rvert$ on the smallest closed interval containing both $a$ and $[c, d]$ .

In plain words Bound the error at each point, then integrate the bound. This works whether or not the series alternates.

Example Approximate $\int_0^1 \frac{\sin x}{x} \, dx$ , where the integrand is defined as $1$ at $x = 0$ . For $\sin x$ about $0$ the polynomials $P_5$ and $P_6$ are both $x - \frac{x^3}{6} + \frac{x^5}{120}$ , and $\lvert \sin^{(7)} \rvert \le 1$ , so $\lvert R_6(x) \rvert \le \frac{\lvert x \rvert^7}{5040}$ . Dividing by $x > 0$ gives

\left\lvert \frac{\sin x}{x} - \left( 1 - \frac{x^2}{6} + \frac{x^4}{120} \right) \right\rvert \le \frac{x^6}{5040},

which also holds at $x = 0$ . Integrating over $[0, 1]$ gives the approximation

1 - \frac{1}{18} + \frac{1}{600} \approx 0.946111

with error at most

\int_0^1 \frac{x^6}{5040} \, dx = \frac{1}{35280} \approx 2.83 \times 10^{-5}.

The true value is about $0.946083$ , an error of $2.80 \times 10^{-5}$ .

Expansion center

Definition The expansion center is the point $a$ about which $P_n$ is built. It must lie in an open interval on which $f$ has the required derivatives, and the values $f^{(k)}(a)$ must be known exactly or cheaply. Because the remainder bound contains the factor $\lvert x - a \rvert^{n+1}$ , a center close to $x$ gives a smaller error for the same degree.

In plain words Expand around a nearby point where you already know the function exactly.

Example To approximate $\sqrt{4.1}$ , a center at $0$ is useless because $\sqrt{x}$ is not differentiable there. Center at $a = 4$ , where $\sqrt{4} = 2$ . With $f'(x) = \frac{1}{2}x^{-1/2}$ and $f''(x) = -\frac{1}{4}x^{-3/2}$ ,

P_2(x) = 2 + \frac{x - 4}{4} - \frac{(x - 4)^2}{64}, \qquad P_2(4.1) = 2.02484375.

On $[4, 4.1]$ the third derivative satisfies

f'''(t) = \frac{3}{8} t^{-5/2} \le \frac{3}{8} \cdot 4^{-5/2} = \frac{3}{256},

so the error is at most $\frac{3}{256} \cdot \frac{0.1^3}{6} \approx 1.95 \times 10^{-6}$ . In fact $\sqrt{4.1} \approx 2.02484567$ , off by $1.92 \times 10^{-6}$ .

Polynomial degree

Definition For a fixed center $a$ and point $x$ , the degree $n$ of $P_n$ is the highest power of $(x - a)$ allowed in it. To meet a target error, choose the smallest $n$ whose remainder bound

\frac{M_{n+1} \lvert x - a \rvert^{n+1}}{(n+1)!}

is small enough, where $M_{n+1}$ bounds $\lvert f^{(n+1)} \rvert$ between $a$ and $x$ . The guarantee comes from the bound, so a degree that happens to give a small actual error without a bound to show it is not guaranteed.

In plain words More terms usually means a better answer, and the factorial in the bound tells you how many terms are enough.

Example Approximate $e = e^1$ with center $0$ within $10^{-6}$ . Here $\lvert f^{(n+1)}(t) \rvert = e^t \le e < 3$ on $[0, 1]$ , so you need

\frac{3}{(n+1)!} \le 10^{-6}, \quad \text{that is} \quad (n+1)! \ge 3 \times 10^6.

Since $9! = 362880$ and $10! = 3628800$ , take $n = 9$ . Then $P_9(1) \approx 2.7182815$ with actual error $3.0 \times 10^{-7}$ . With $n = 8$ the error is $3.1 \times 10^{-6}$ , too large.

Error tolerance

Definition An error tolerance is a number $\tau > 0$ that the error must not exceed. An approximation $A$ of a quantity $Q$ meets an absolute tolerance $\tau$ if $\lvert Q - A \rvert \le \tau$ , and a relative tolerance $\tau$ if $\lvert Q - A \rvert \le \tau \lvert Q \rvert$ . To certify a Taylor approximation, show that the remainder bound is at most $\tau$ .

In plain words The tolerance is the accuracy you were asked for. You are done when your error bound, not your hope, is below it.

Example Find $\ln(1.1)$ within $\tau = 10^{-5}$ using $f(x) = \ln(1 + x)$ at $a = 0$ . Here

f^{(n+1)}(t) = \frac{(-1)^n n!}{(1 + t)^{n+1}},

so for $0 \le t \le x$ we get

\lvert R_n(x) \rvert \le \frac{n!}{(n+1)!} x^{n+1} = \frac{x^{n+1}}{n+1}.

At $x = 0.1$ the choice $n = 3$ gives $\frac{10^{-4}}{4} = 2.5 \times 10^{-5}$ , too big, while $n = 4$ gives $\frac{10^{-5}}{5} = 2 \times 10^{-6}$ . So

\ln(1.1) \approx 0.1 - \frac{0.01}{2} + \frac{0.001}{3} - \frac{0.0001}{4} \approx 0.0953083,

and the true value $0.0953102$ is indeed within $10^{-5}$ .

Check yourself

Approximate $\cos(0.2)$ by $1 - \frac{x^2}{2}$ and bound the error.

Show the answer

The approximation is $1 - 0.02 = 0.98$ . For cosine $P_2 = P_3$ , and $\lvert \cos^{(4)} \rvert \le 1$ , so the error is at most

\frac{0.2^4}{24} \approx 6.7 \times 10^{-5}.

In fact $\cos 0.2 \approx 0.980067$ .

What degree $n$ of the Taylor polynomial for $\sin x$ about $0$ guarantees $\sin 1$ within $10^{-4}$ ?

Show the answer

Every derivative of $\sin$ is bounded by $1$ , so $\lvert R_n(1) \rvert \le \frac{1}{(n+1)!}$ . You need $(n+1)! \ge 10^4$ . Since $7! = 5040$ and $8! = 40320$ , take $n = 7$ , giving

1 - \frac{1}{6} + \frac{1}{120} - \frac{1}{5040} \approx 0.841468.

Use $\cos u \approx 1 - \frac{u^2}{2}$ to approximate $\int_0^{1/2} \cos(x^2) \, dx$ and bound the error.

Show the answer

The approximation is

\int_0^{1/2} \left( 1 - \frac{x^4}{2} \right) dx = \frac{1}{2} - \frac{(1/2)^5}{10} = 0.496875.

Since $\lvert \cos u - (1 - \frac{u^2}{2}) \rvert \le \frac{u^4}{24}$ , the integrand error is at most $\frac{x^8}{24}$ , and

\int_0^{1/2} \frac{x^8}{24} \, dx = \frac{(1/2)^9}{216} \approx 9.0 \times 10^{-6}.

Approximating Using Taylor’s Formula

Key terms

Taylor’s formula with remainder

Evaluating functions with Taylor polynomials

Taylor-based quadrature

Integrated remainder bound

Expansion center

Polynomial degree

Error tolerance

Check yourself

How this lesson connects

Builds on

Resources for this lesson

Taylor polynomial explorer