[CALCULUS] Taylor's theorem

This page gives an exact formula for the error of a Taylor polynomial $P_n$ , called the remainder $R_{n+1}$ :

f(x)=P_n(x)+R_{n+1}(x).

Bounding the remainder turns an approximation into a guaranteed one and its leading term classifies stationary points.

The remainder measures what the polynomial misses

[DEFINITION]

Taylor polynomial

If $f$ has derivatives through order $n$ at $a$ , its Taylor polynomial of degree at most $n$ about $a$ is

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

[1]:Taylor polynomials and a Lagrange remainder
[2]:Seen Questions
[3]:Sequences and limits
[4]:Maclaurin polynomial
4 more

[THEOREM]

Taylor's theorem with remainder

Suppose $f$ has $n+1$ continuous derivatives on an interval containing $a$ and $x$ . If

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

then

f(x)=P_n(x)+R_{n+1}(x),

where the integral remainder is

R_{n+1}(x) =\frac1{n!}\int_a^x f^{(n+1)}(t)(x-t)^n\,dt.

For $x\ne a$ , there is also a point $c$ between $a$ and $x$ for which the Lagrange remainder is

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

The integral form accumulates the first omitted derivative. The Lagrange form replaces the integral by one unknown derivative value. If

|f^{(n+1)}(t)|\le M

throughout the interval between $a$ and $x$ , then

|f(x)-P_n(x)| =|R_{n+1}(x)| \le \frac{M|x-a|^{n+1}}{(n+1)!}.

The bound depends on the distance $|x-a|$ from the centre and the size of the first omitted derivative. The factorial often dominates the numerator for large $n$ , but this must be checked for the interval and degree in use.

[1]:Taylor polynomial
[2]:When a Taylor series represents its function
[3]:Taylor series
[4]:Proof
[5]:The Mean Value Theorem as the first case
3 more

[PROOF]

We prove the integral formula by induction. When $n=0$ , the Fundamental Theorem of Calculus gives

R_1(x)=\int_a^x f'(t)\,dt=f(x)-f(a).

Assume that

R_k(x) =\frac1{(k-1)!}\int_a^x f^{(k)}(t)(x-t)^{k-1}\,dt.

For the next remainder, take $u=(x-t)^k$ and $dv=f^{(k+1)}(t)\,dt$ , so $du=-k(x-t)^{k-1}\,dt$ and $v=f^{(k)}(t)$ . We obtain

\begin{aligned} R_{k+1}(x) &=\frac1{k!}\int_a^x f^{(k+1)}(t)(x-t)^k\,dt\\ &=\left[\frac{f^{(k)}(t)}{k!}(x-t)^k\right]_{t=a}^{t=x} +\frac1{(k-1)!}\int_a^x f^{(k)}(t)(x-t)^{k-1}\,dt\\ &=-\frac{f^{(k)}(a)}{k!}(x-a)^k+R_k(x)\\ &=f(x)-P_{k-1}(x)-\frac{f^{(k)}(a)}{k!}(x-a)^k\\ &=f(x)-P_k(x), \end{aligned}

The $t=x$ endpoint vanishes because $(x-x)^k=0$ and the penultimate line uses the induction hypothesis $R_k(x)=f(x)-P_{k-1}(x)$ . Each induction step removes one Taylor term, which proves the integral form for every $n$ .

To obtain the Lagrange form, apply the mean value theorem for integrals to the continuous function $f^{(n+1)}$ with the non-negative weight $|x-t|^n$ between $a$ and $x$ . Accounting for the orientation when $x<a$ gives $\int_a^x f^{(n+1)}(t)(x-t)^n\,dt =f^{(n+1)}(c)(x-a)^{n+1}/(n+1)$ for some intermediate $c$ .

[1]:Taylor's theorem with remainder

Taylor's theorem extends the Mean Value Theorem

When $n=0$ , Taylor's theorem reads

f(x)=f(a)+f'(c)(x-a).

[COROLLARY]

The Mean Value Theorem as the first case

For some $c$ between $a$ and $x$ ,

\frac{f(x)-f(a)}{x-a}=f'(c)

whenever $x\ne a$ and the hypotheses of the Mean Value Theorem hold.

[1]:Taylor's theorem with remainder

Turn the remainder into an error bound

[METHOD]

Bound a Taylor approximation error

To prove that $P_n(x)$ approximates $f(x)$ within a tolerance $\varepsilon$ :

identify $f^{(n+1)}$ ;
find a bound $M$ for $|f^{(n+1)}(t)|$ between $a$ and the relevant values of $x$ ;
use $|f(x)-P_n(x)| \le \frac{M|x-a|^{n+1}}{(n+1)!};$
choose $n$ or the interval so that the right-hand side is at most $\varepsilon$ .

The estimate is uniform on an interval when one $M$ and one distance bound work throughout that interval. It bounds the error from above; it is not the actual error.

[EXAMPLE]

Approximating $e^x$ by a quadratic

Take

P_2(x)=1+x+\frac{x^2}{2}

for $f(x)=e^x$ about zero. The third derivative is $e^x$ . On $[-1/2,1/2]$ , an intermediate point satisfies $e^c\le e^{1/2}$ , so

|R_3(x)| \le \frac{e^{1/2}|x|^3}{3!} \le \frac{e^{1/2}}{48} \approx0.03435.

The bound holds for every $x$ in the interval.

[routine]

[1]:Bound a Taylor approximation error

[EXAMPLE]

Computing $e$ to within $10^{-6}$

At $x=1$ , the Maclaurin polynomial gives

P_n(1)=\sum_{k=0}^{n}\frac1{k!}.

For $0<c<1$ , the Lagrange remainder satisfies $e^c<3$ , hence

|R_{n+1}(1)|<\frac3{(n+1)!}.

With $n=9$ ,

\frac3{10!}<10^{-6}.

Therefore

e\approx P_9(1) =\sum_{k=0}^{9}\frac1{k!} =2.7182815256

with the requested accuracy. The bound guarantees the accuracy; the actual error is about $3.03\times10^{-7}$ .

[exam]

[1]:Bound a Taylor approximation error

[EXAMPLE]

Error for a sine approximation

Every derivative of $\sin x$ is one of $\pm\sin x$ or $\pm\cos x$ , so its absolute value is at most $1$ . For a degree- $n$ Taylor polynomial about $a$ ,

|R_{n+1}(x)| \le \frac{|x-a|^{n+1}}{(n+1)!}.

Near the expansion centre, the factorial makes the bound decrease rapidly.

[routine]

[1]:Bound a Taylor approximation error

The first nonzero term controls local behaviour

At a stationary point the linear term vanishes, so the first nonzero higher derivative decides whether values rise, fall or change sign on each side. The remainder is of strictly higher order, so close to the centre it is too small to reverse that sign.

[THEOREM]

Classify a stationary point with the first nonzero derivative

Suppose $k$ is an integer with $k\ge2$ and $f$ has $k+1$ continuous derivatives on a neighbourhood of $a$ . If

f^{(j)}(a)=0\quad(1\le j<k), \qquad f^{(k)}(a)\ne0.

then Taylor's theorem gives

f(x)-f(a) =\frac{f^{(k)}(a)}{k!}(x-a)^k+R_{k+1}(x), \qquad R_{k+1}(x)=O\bigl(|x-a|^{k+1}\bigr),\qquad R_{k+1}(x)=o\bigl(|x-a|^k\bigr).

Thus, locally:

if $k$ is even and $f^{(k)}(a)>0$ , $a$ is a local minimum;
if $k$ is even and $f^{(k)}(a)<0$ , $a$ is a local maximum;
if $k$ is odd, the leading term changes sign across $a$ , giving a horizontal point of inflection.

Here $O(|x-a|^{k+1})$ means an error bounded by a constant times $|x-a|^{k+1}$ near $a$ and $o(|x-a|^k)$ means an error whose ratio to $|x-a|^k$ tends to zero.

The test requires $k\ge2$ , a genuinely first nonzero derivative and enough continuity to ensure that the remainder is $o(|x-a|^k)$ . If all derivatives examined vanish, those terms give no classification.

[PROOF]

Taylor's theorem gives

f(x)-f(a) =\frac{f^{(k)}(a)}{k!}(x-a)^k+R_{k+1}(x),

because the lower-order terms vanish. Since $f^{(k+1)}$ is continuous near $a$ , it is bounded on a smaller neighbourhood, so the remainder estimate gives $R_{k+1}(x)=O(|x-a|^{k+1})$ and hence $R_{k+1}(x)=o(|x-a|^k)$ . Therefore the first nonzero term determines the sign of $f(x)-f(a)$ sufficiently close to $a$ .

When $k$ is even, $(x-a)^k$ is positive on both sides of $a$ , so the sign of $f^{(k)}(a)$ determines a minimum or maximum. When $k$ is odd, then $k\ge3$ and the same expansion applied to $f''$ gives

f''(x)=\frac{f^{(k)}(a)}{(k-2)!}(x-a)^{k-2}+o\bigl(|x-a|^{k-2}\bigr).

Since $k-2$ is odd, $f''$ changes sign across $a$ ; together with $f'(a)=0$ , this gives a horizontal point of inflection.

[1]:Classify a stationary point with the first nonzero derivative

[EXAMPLE]

A fifth-order stationary point

Let

f(x)=x^7-10x^6+40x^5-80x^4+80x^3-32x^2-10.

At $x=2$ ,

f'(2)=f''(2)=f^{(3)}(2)=f^{(4)}(2)=0, \qquad f^{(5)}(2)=480>0.

The first nonzero derivative has odd order, so $x=2$ is a horizontal point of inflection. Since $f(2)=-10$ , the point is $(2,-10)$ .

[challenge]

[1]:Classify a stationary point with the first nonzero derivative

[PROPOSITION]

Comparing the first nonzero Taylor terms

Suppose $m\ge1$ , $f$ and $g$ are $m$ times differentiable near $a$ , $f^{(j)}(a)=g^{(j)}(a)=0$ for $0\le j<m$ and $f^{(m)}(a),g^{(m)}(a)\ne0$ . Then

\lim_{x\to a}\frac{f(x)}{g(x)} =\frac{f^{(m)}(a)}{g^{(m)}(a)}.

The leading expansions are

f(x)=\frac{f^{(m)}(a)}{m!}(x-a)^m+o((x-a)^m),

and

g(x)=\frac{g^{(m)}(a)}{m!}(x-a)^m+o((x-a)^m).

The common power $(x-a)^m$ cancels in the quotient. This explains why repeated L'Hôpital differentiation can work, but the hypotheses of that rule must still be checked.

[1]:Taylor's theorem with remainder

From one polynomial to a sequence

For a fixed $x$ , the Taylor polynomials of degree $0,1,2,\ldots$ form a sequence of approximations. Two questions decide its limit: the convergence tests settle whether the series converges and the remainder decides whether the sum is $f(x)$ . The Taylor series page combines the two and power series adds termwise calculus inside the radius.

Taylor's theorem

Taylor polynomial

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