Taylor's Theorem

The mean value theorem approximates $f$ near $x_{0}$ by its tangent line and measures the error with one derivative at an unknown interior point:

f (x) = f (x_{0}) + f^{'} (c) (x - x_{0}) .

Taylor's theorem repeats this to every order. It replaces the tangent line by a polynomial that matches $f$ and its first $n$ derivatives at $x_{0}$ , and it writes the error through the $(n + 1)$ st derivative in exactly the same shape. It underpins quantitative approximation throughout analysis.

Higher-order derivatives

Differentiating $f$ yields $f^{'}$ ; if $f^{'}$ is itself differentiable, its derivative is the second derivative $f^{''}$ . Iterating gives $f^{'''}, f^{''''}, \dots$ , and past the third or fourth order the prime notation becomes unwieldy.

Each order requires the previous one to exist on a neighborhood, not just at a point, so higher differentiability is a genuinely stronger hypothesis than plain differentiability.

The Taylor polynomial

The unique polynomial of degree $\leq n$ agreeing with $f$ in value and in its first $n$ derivatives at $x_{0}$ is the Taylor polynomial.

Differentiating the sum term by term and evaluating at $x_{0}$ collapses every term but one, giving $(P_{n}^{x_{0}})^{(k)} (x_{0}) = f^{(k)} (x_{0})$ for $0 \leq k \leq n$ : the polynomial's derivatives at $x_{0}$ match those of $f$ up to order $n$ . That matching is the entire reason it approximates well nearby.

Algorithm:constructing the

n

th Taylor polynomial of

f

x_0

1
input: a function $f$ that is $n$ times differentiable near $x_0$
2
$P \gets 0$
3
for $k = 0$ to $n$ do
4
compute the derivative $f^{(k)}(x_0)$
5
$P \gets P + \dfrac{f^{(k)}(x_0)}{k!}\,(x - x_0)^k$
6
return $P$

For $f = sin$ at $x_{0} = 0$ the derivatives cycle through $sin, cos, - sin, - cos$ , so the even-order terms all vanish and only odd powers remain.

$k$	$f^{(k)} (0)$	contribution to $P_{n}^{0}$
$0$	$0$	$0$
$1$	$1$	$x$
$2$	$0$	$0$
$3$	$- 1$	$- \frac{x ^{3}}{6}$
$4$	$0$	$0$
$5$	$1$	$\frac{x ^{5}}{120}$

Successive odd-degree polynomials $P_{1}, P_{3}, P_{5}, \dots$ track $sin$ over ever wider intervals around the origin.

A consistency check confirms the construction: a polynomial $p$ of degree $d$ is its own $d$ th Taylor polynomial at every point. Matching value and first $d$ derivatives at $x_{0}$ determines a degree- $d$ polynomial uniquely, and $p$ matches itself, so $P_{d}^{x_{0}} = p$ . Taylor approximation reproduces polynomials exactly and only genuinely approximates functions that are not polynomials.¹

Taylor polynomials of the sine at the origin. The degree-1 tangent

P_{1} = x

leaves the curve almost immediately; the degree-3 polynomial

P_{3} = x - x^{3} /6

peaks just short of the crest, tracks the sine to about

x = 2

, then dives through its zero at

6 \approx 2.45

. Each higher odd degree hugs the curve over a wider range.

Taylor's theorem with Lagrange remainder

The approximation is exact once the error term is added, and the error has the same form as the mean value theorem's, one derivative higher.

Taking $n = 0$ reduces the statement to $f (x) = f (x_{0}) + f^{'} (c) (x - x_{0})$ , which is the mean value theorem; Taylor's theorem is its exact generalization.

Proof. Fix $x$ and choose the number $M$ so that $f (x) = P_{n}^{x_{0}} (x) + M (x - x_{0})^{n + 1}$ ; the claim is that $M = \frac{f ^{(n + 1)} ( c )}{( n + 1 )!}$ for some interior $c$ . Define

g (s) = f (s) - P_{n}^{x_{0}} (s) - M (s - x_{0})^{n + 1} .

Because $P_{n}^{x_{0}}$ matches $f$ through order $n$ at $x_{0}$ and the last term and its first $n$ derivatives vanish at $x_{0}$ ,

g (x_{0}) = g^{'} (x_{0}) = \dots = g^{(n)} (x_{0}) = 0,

and by the choice of $M$ also $g (x) = 0$ . Apply Rolle's theorem to $g$ on the interval between $x_{0}$ and $x$ : since $g (x_{0}) = g (x) = 0$ there is an $x_{1}$ with $g^{'} (x_{1}) = 0$ . Now $g^{'} (x_{0}) = 0 = g^{'} (x_{1})$ , so Rolle applied to $g^{'}$ gives $x_{2}$ with $g^{''} (x_{2}) = 0$ . Repeating $n + 1$ times produces a point $c$ with $g^{(n + 1)} (c) = 0$ . Differentiating $g$ exactly $n + 1$ times annihilates the polynomial $P_{n}^{x_{0}}$ and leaves

g^{(n + 1)} (s) = f^{(n + 1)} (s) - (n + 1)! M,

so $0 = g^{(n + 1)} (c) = f^{(n + 1)} (c) - (n + 1)! M$ gives the claimed $M$ . $■$

The proof iterates the mean value theorem through the derivatives: each application of Rolle lowers the order of $g$ by one while producing a new zero.

Error estimates

The remainder is useful because $f^{(n + 1)}$ is often bounded on the interval. If $∣ f^{(n + 1)} ∣ \leq K$ between $x_{0}$ and $x$ , then

f (x) - P_{n}^{x_{0}} (x) \leq \frac{K}{( n + 1 )!} ∣ x - x_{0} ∣^{n + 1} .

Two features drive the bound down: the factorial $(n + 1)!$ in the denominator and the power $∣ x - x_{0} ∣^{n + 1}$ , which is small when $x$ is close to $x_{0}$ . Together they mean the error goes to zero faster than $(x - x_{0})^{n}$ as $x \to x_{0}$ , which is the precise sense in which $P_{n}^{x_{0}}$ is a degree- $n$ approximation.

The remainder formula converts a bound on a derivative into a certified numerical error, with no appeal to the unknown exact value. The exponential shows the same mechanism where the bounding derivative is not constant.

The remainder is the vertical gap between the graph and its Taylor polynomial at x; it is controlled by the next derivative and shrinks faster than the nth power of the distance from x0.

The second-derivative test falls out of the case $n = 1$ . If $f$ is twice continuously differentiable, $f^{'} (x_{0}) = 0$ , and $f^{''} (x_{0}) > 0$ , then near $x_{0}$

f (x) = f (x_{0}) + \frac{f ^{''} ( c )}{2} (x - x_{0})^{2},

with $f^{''} (c) > 0$ for $x$ close enough (continuity of $f^{''}$ ), so $f (x) > f (x_{0})$ : the point is a strict relative minimum.³ The sign of $f^{''} (x_{0})$ classifies the critical point, and the same idea iterated gives the $n$ th-derivative test.

At a critical point with positive second derivative the order-2 Taylor expansion is an upward parabola, so f sits above f(x0) on both sides and the point is a strict minimum.

Taylor series and analytic functions

When $f$ is infinitely differentiable, the polynomials extend to an infinite series.

Two independent failures separate a function from its Taylor series:

The series may diverge. Expanding $\frac{x}{1 - x}$ at $0$ gives Taylor polynomials $P_{n}^{0} (x) = x + x^{2} + \dots + x^{n}$ , the partial sums of a geometric series. These converge to the function only on $(- 1, 1)$ ; for $x \leq - 1$ the approximations grow visibly worse as the degree rises, no matter that $\frac{x}{1 - x}$ is perfectly finite there. The Taylor series carries the convergence radius of a power series, a restriction the function itself need not respect.
The series may converge to the wrong function. Even where the Taylor series converges, its sum need not be $f$ .

The second failure is the sharper one. The Taylor series can converge everywhere and still equal $f$ at only one point.

The smooth bump e^{-1/x^2} is zero at the origin together with every derivative, so its Taylor series at 0 is identically zero and never recovers the function away from 0.

The function

f (x) = {e^{- 1/ x^{2}}, 0, x \neq = 0, x = 0,

is infinitely differentiable, and every derivative at the origin is zero. Its Taylor series at $0$ is therefore the zero series, which converges everywhere to the zero function, yet $f (x) > 0$ for all $x \neq = 0$ . Smoothness does not imply analyticity. The power-series methods distinguish the two, and the same non-analytic bump is what makes smooth partitions of unity possible in the calculus of several variables.

Lebl, Basic Analysis I, §4.3, Exercise 4.3.2 — a degree- $d$ polynomial equals its own $d$ th Taylor polynomial. ↩
Lebl, Basic Analysis I, §4.3, Definition 4.3.1 and Theorem 4.3.2 — the Taylor polynomial and Taylor's theorem with the Lagrange form of the remainder, proved by iterating the mean value theorem. See also Rosenlicht, Ch. 5. ↩
Lebl, Basic Analysis I, §4.3, Proposition 4.3.3 — the second-derivative test as the order-1 case of Taylor's theorem. ↩