Calculus of one variable

Lens

MotivationH

How does a calculator know sin 0.3?

Press sin 0.3 on a calculator and $0.29552020666\ldots$ appears instantly. The chip did not draw a triangle. It added up a handful of terms of

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

(real libraries also use range reduction and carefully tuned polynomials, but the idea is the same). Two questions turn this trick into mathematics. Why does the infinite sum equal $\sin x$ ? And how many terms are enough for ten correct digits? The answer to both is Taylor's theorem with remainder, and its proof rests on a chain of results — limits, derivatives, the mean value theorem — that is the backbone of this lesson.

The same chain answers questions an engineer in Lagos, a pharmacologist in São Paulo or a climate modeller in Pune meets daily: if a quantity changes at a known rate, how much has it changed in total? (Integrate.) If a function has derivative zero, must it be constant? (Yes — but only on an interval, and the proof needs the mean value theorem.) If an infinite sum's terms shrink to zero, does the sum settle down? (Not necessarily: $1 + \tfrac12 + \tfrac13 + \cdots$ grows without bound.)

This lesson is a university-level treatment of calculus of one variable. It assumes the informal ideas of introductory calculus, exponentials and logarithms, sequences and geometric series and trigonometry, and makes them precise.

IntuitionH

Zooming in and adding up

Zoom in on the graph of a differentiable function and it looks more and more like a straight line. The slope of that line is the derivative: the derivative is the best linear approximation,

f(a + h) \approx f(a) + f'(a)\,h, \qquad \text{with an error that is small compared with } h.

Chop an interval into thin strips and add up "height × width": that is a Riemann sum, and as the strips get thinner the sums approach the integral. The fundamental theorem of calculus says these two processes undo each other.

Both ideas rest on the word approach. Newton and Leibniz worked with "vanishing quantities" and "infinitesimals" that were sometimes zero and sometimes not, and Bishop Berkeley mocked them as "the ghosts of departed quantities" (1734). The repair, completed in the nineteenth century by Cauchy, Bolzano and Weierstrass, replaced motion and infinitesimals by a game with inequalities: you name a tolerance $\varepsilon$ ; I must find a closeness $\delta$ that guarantees it.

The ε–δ game for x² at a = 3

A limit is a promise: for every tolerance ε there is a δ. Choose δ and see whether the graph stays inside the band.

Claim: lim_x→3 (x²) = 9Your move: find δ > 0 so that every x with 0 < |x − a| < δ satisfies |f(x) − L| < ε.

f(x) = x²
ε-band L ± ε
δ-window a ± δ
graph outside the band

The challenge ε

Your answer δδ = 0.5δ =

Claimed limit LL = 9

Not yet: near x ≈ 3.083 the graph leaves the band (|f(x) − L| ≈ 0.502). Try a smaller δ.

Checked at 2,118 sample points: evidence, not a proof.

Try shrinking $\varepsilon$ : you can always find a working $\delta$ , but it must shrink too. Then move the claimed limit $L$ away from $9$ and watch every $\delta$ fail once $\varepsilon$ is small. (The widget samples thousands of points: that is evidence, not proof. The proof is in the worked examples below.)

DefinitionsH

Definitions

Throughout, functions are real-valued and defined on subsets of $\R$ .

Definition— limit

Let $f$ be defined on an open interval around $a$ , except possibly at $a$ itself. We write $\lim_{x \to a} f(x) = L$ if for every $\varepsilon > 0$ there is $\delta > 0$ such that

0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.

One-sided limits $x \to a^+$ and $x \to a^-$ use only $a < x < a + \delta$ or $a - \delta < x < a$ .

Definition— continuity

$f$ is continuous at $a$ (a point of its domain $D$ ) if for every $\varepsilon > 0$ there is $\delta > 0$ such that $|f(x) - f(a)| < \varepsilon$ for all $x \in D$ with $|x - a| < \delta$ . It is continuous on $D$ if it is continuous at every point of $D$ .

The order of the quantifiers matters: $\delta$ may depend on $\varepsilon$ and on $a$ . If one $\delta$ works for all $a \in D$ at once, $f$ is uniformly continuous on $D$ .

Definition— derivative

Let $f$ be defined on an open interval containing $a$ . Then $f$ is differentiable at $a$ , with derivative $f'(a)$ , if the limit

f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

exists (as a real number). Higher derivatives $f'', f''', \dots, f^{(n)}$ are defined by repeating. $f$ is of class $C^n$ on an interval if $f^{(n)}$ exists and is continuous there, and smooth ( $C^\infty$ ) if it is $C^n$ for every $n$ .

Definition— Riemann integral

Let $f$ be bounded on $[a, b]$ . A tagged partition is a choice of points $a = x_0 < x_1 < \cdots < x_n = b$ together with tags $t_i \in [x_{i-1}, x_i]$ ; its mesh is $\max_i (x_i - x_{i-1})$ . The Riemann sum is $S = \sum_{i=1}^n f(t_i)\,(x_i - x_{i-1})$ . We say $f$ is Riemann integrable with integral $I = \int_a^b f(x)\,dx$ if for every $\varepsilon > 0$ there is $\delta > 0$ such that $|S - I| < \varepsilon$ for every tagged partition of mesh less than $\delta$ .

Every continuous function on $[a, b]$ is Riemann integrable, and so is every monotone function; both facts are proved in real analysis (the first using uniform continuity). We use freely the elementary properties: linearity, additivity over adjacent intervals, and the estimate $\left|\int_a^b g\right| \le (b - a)\sup_{[a, b]} |g|$ .

Definition— improper integral

If $f$ is integrable on $[a, b]$ for every $b > a$ , then $\int_a^\infty f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx$ when this limit exists (the integral converges). Integrals of functions that are unbounded near an endpoint are defined by the same kind of limit.

Definition— convergent series

The series $\sum_{n=0}^\infty a_n$ converges to $S$ if its partial sums $s_N = \sum_{n=0}^N a_n$ satisfy $s_N \to S$ . It converges absolutely if $\sum |a_n|$ converges, and conditionally if it converges but not absolutely.

Definition— Taylor polynomial and Taylor series

If $f$ is $n$ times differentiable at $a$ , its Taylor polynomial of degree $n$ at $a$ is

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

and $R_n(x) = f(x) - T_n(x)$ is the remainder. If $f$ is smooth, its Taylor series at $a$ is $\sum_{k=0}^\infty f^{(k)}(a)(x - a)^k/k!$ (a Maclaurin series when $a = 0$ ). $f$ is (real) analytic at $a$ if its Taylor series converges to $f$ on some interval around $a$ .

Worked examplesH

Worked examples and non-examples

Example— An ε–δ proof: x² is continuous

Fix $a$ and $\varepsilon > 0$ . Since $|x^2 - a^2| = |x - a|\,|x + a|$ , we must control $|x + a|$ . If $|x - a| < 1$ then $|x + a| \le |x - a| + 2|a| < 2|a| + 1$ . So choose

\delta = \min\left(1, \frac{\varepsilon}{2|a| + 1}\right).

Then $|x - a| < \delta$ gives $|x^2 - a^2| < \delta\,(2|a| + 1) \le \varepsilon$ . Notice that $\delta$ depends on $a$ : near $a = 1000$ you must stay much closer. In fact $x^2$ is not uniformly continuous on $\R$ . (This exact argument is checked by Lean below.)

Example— A derivative from the definition, and a corner

For $f(x) = x^n$ ( $n \ge 1$ ), the binomial theorem gives $\frac{(a + h)^n - a^n}{h} = n a^{n-1} + \binom n2 a^{n-2}h + \cdots + h^{n-1} \to n a^{n-1}$ .

For $f(x) = |x|$ at $0$ : the difference quotient $|h|/h$ is $1$ for $h > 0$ and $-1$ for $h < 0$ , so the limit does not exist. $|x|$ is continuous but not differentiable at $0$ . Continuity does not imply differentiability — Weierstrass (1872) even built a function continuous everywhere and differentiable nowhere.

Example— Three integration techniques

Substitution ( $u = x^2$ , $du = 2x\,dx$ ): $\displaystyle\int_0^{\sqrt\pi} x \sin(x^2)\,dx = \tfrac12\int_0^{\pi} \sin u\,du = 1.$
Integration by parts ( $u = x$ , $dv = e^x dx$ ): $\displaystyle\int_0^1 x e^x\,dx = \big[x e^x\big]_0^1 - \int_0^1 e^x\,dx = e - (e - 1) = 1.$
Partial fractions: $\dfrac{1}{x(x + 1)} = \dfrac1x - \dfrac1{x+1}$ , so $\displaystyle\int_1^N \frac{dx}{x(x+1)} = \ln\frac{2N}{N + 1} \to \ln 2$ as $N \to \infty$ : a convergent improper integral.

Example— Series: convergent, divergent, and conditionally convergent

$\sum 1/n^2$ converges (integral test, $p = 2 > 1$ ); Euler found its sum, $\pi^2/6$ , in the 1730s.
$\sum 1/n$ diverges although $1/n \to 0$ : group the terms as $1 + \tfrac12 + (\tfrac13 + \tfrac14) + (\tfrac15 + \cdots + \tfrac18) + \cdots$ , each bracket at least $\tfrac12$ .
$\sum (-1)^{n+1}/n = 1 - \tfrac12 + \tfrac13 - \cdots$ converges (to $\ln 2$ ) by the alternating series test, but not absolutely. By Riemann's rearrangement theorem its terms can be reordered to converge to any real number you like: conditional convergence depends on the order of the terms.

ProofH

Derivatives and the mean value theorem

Theorem— Proposition (differentiable implies continuous)

If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ .

Proof

For $h \ne 0$ , $f(a + h) - f(a) = h \cdot \dfrac{f(a + h) - f(a)}{h}$ , and as $h \to 0$ this tends to $0 \cdot f'(a) = 0$ .

∎

To prove the chain rule cleanly we use a reformulation of the derivative due to Carathéodory, which avoids dividing by quantities that might be zero.

Theorem— Lemma (Carathéodory's criterion)

$f$ is differentiable at $a$ if and only if there is a function $\varphi$ , continuous at $a$ , with $f(x) - f(a) = \varphi(x)(x - a)$ for all $x$ near $a$ . In that case $f'(a) = \varphi(a)$ .

Proof

If $f'(a)$ exists, put $\varphi(x) = \frac{f(x) - f(a)}{x - a}$ for $x \ne a$ and $\varphi(a) = f'(a)$ ; continuity of $\varphi$ at $a$ is exactly the definition of $f'(a)$ . Conversely, given $\varphi$ , the difference quotient equals $\varphi(x)$ for $x \ne a$ , which tends to $\varphi(a)$ .

∎

Theorem— Theorem (chain rule)

If $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$ , then $g \circ f$ is differentiable at $a$ and $(g \circ f)'(a) = g'(f(a))\,f'(a)$ .

Proof

Write $f(x) - f(a) = \varphi(x)(x - a)$ and $g(y) - g(f(a)) = \psi(y)(y - f(a))$ with $\varphi$ continuous at $a$ , $\psi$ continuous at $f(a)$ , $\varphi(a) = f'(a)$ , $\psi(f(a)) = g'(f(a))$ . Then

g(f(x)) - g(f(a)) = \psi(f(x))\,\big(f(x) - f(a)\big) = \psi(f(x))\,\varphi(x)\,(x - a).

The function $x \mapsto \psi(f(x))\varphi(x)$ is continuous at $a$ , because $f$ is continuous at $a$ (previous proposition) and $\psi$ at $f(a)$ . By Carathéodory's criterion, $(g \circ f)'(a) = \psi(f(a))\varphi(a) = g'(f(a))f'(a)$ .

∎

The product, quotient and inverse-function rules are proved in the same spirit. The theorems that make derivatives useful rest on one deep fact about the real line, proved in real analysis from the completeness of $\R$ (see the topology viewpoint below):

Theorem— Theorem (extreme value theorem)

A continuous function on a closed bounded interval $[a, b]$ attains a maximum and a minimum value.

Theorem— Lemma (Fermat's interior extremum theorem)

If $f$ has a local maximum or minimum at an interior point $c$ and is differentiable at $c$ , then $f'(c) = 0$ .

Proof

Say $c$ is a local maximum: $f(c + h) \le f(c)$ for $|h|$ small. For small $h > 0$ the quotient $\frac{f(c + h) - f(c)}{h}$ is $\le 0$ , so its limit $f'(c) \le 0$ ; for small $h < 0$ it is $\ge 0$ , so $f'(c) \ge 0$ . Hence $f'(c) = 0$ . A minimum is a maximum of $-f$ .

∎

Theorem— Theorem (Rolle)

If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $f(a) = f(b)$ , then $f'(c) = 0$ for some $c \in (a, b)$ .

Proof

By the extreme value theorem $f$ attains a maximum $M$ and a minimum $m$ on $[a, b]$ . If both are attained only at the endpoints, then $M = m = f(a)$ , so $f$ is constant and every $c \in (a, b)$ works. Otherwise one of them is attained at some $c \in (a, b)$ , and $f'(c) = 0$ by Fermat's lemma.

∎

Theorem— Theorem (mean value theorem)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , where $a < b$ , then

f'(c) = \frac{f(b) - f(a)}{b - a} \quad \text{for some } c \in (a, b).

Proof

Let $m = \frac{f(b) - f(a)}{b - a}$ and tilt the graph: $g(x) = f(x) - m(x - a)$ . Then $g$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $g(a) = f(a) = g(b)$ (check: $g(b) = f(b) - (f(b) - f(a)) = f(a)$ ). Rolle's theorem gives $c$ with $0 = g'(c) = f'(c) - m$ .

∎

Theorem— Corollary (what the sign of f′ tells you)

Let $f$ be differentiable on an interval $I$ .

If $f' = 0$ on $I$ , then $f$ is constant on $I$ .
If $f' > 0$ on $I$ , then $f$ is strictly increasing on $I$ (and if $f' \ge 0$ , increasing).

Proof

For $x < y$ in $I$ , the mean value theorem on $[x, y]$ gives $f(y) - f(x) = f'(c)(y - x)$ for some $c \in (x, y) \subseteq I$ . If $f' = 0$ the right side is $0$ ; if $f' > 0$ it is positive.

∎

The interval matters: $f(x) = x/|x|$ has derivative $0$ on $\R \setminus \{0\}$ but is not constant. This corollary is what makes antiderivatives unique up to a constant on an interval.

Theorem— Theorem (Cauchy's mean value theorem)

If $f, g$ are continuous on $[a, b]$ and differentiable on $(a, b)$ , then for some $c \in (a, b)$ ,

\big(f(b) - f(a)\big)\,g'(c) = \big(g(b) - g(a)\big)\,f'(c).

Proof

Apply Rolle's theorem to $h(x) = \big(f(b) - f(a)\big)g(x) - \big(g(b) - g(a)\big)f(x)$ ; a short computation gives $h(a) = f(b)g(a) - f(a)g(b) = h(b)$ .

∎

Theorem— Theorem (L'Hôpital's rule, 0/0 form)

Let $f, g$ be differentiable on $(a, a + r)$ with $g' \ne 0$ there, and suppose $f(x) \to 0$ and $g(x) \to 0$ as $x \to a^+$ . If $\dfrac{f'(x)}{g'(x)} \to L$ as $x \to a^+$ , then $\dfrac{f(x)}{g(x)} \to L$ as $x \to a^+$ . The same holds for left-hand and two-sided limits.

Proof

Set $f(a) = g(a) = 0$ ; then $f, g$ are continuous on $[a, x]$ for every $x \in (a, a + r)$ . First, $g(x) \ne 0$ : otherwise Rolle's theorem on $[a, x]$ would give a zero of $g'$ . Cauchy's theorem on $[a, x]$ gives $c_x \in (a, x)$ with $f(x)\,g'(c_x) = g(x)\,f'(c_x)$ , that is,

\frac{f(x)}{g(x)} = \frac{f'(c_x)}{g'(c_x)}.

Given $\varepsilon > 0$ , choose $\delta \in (0, r)$ with $\left|\frac{f'(c)}{g'(c)} - L\right| < \varepsilon$ for $a < c < a + \delta$ . For $a < x < a + \delta$ we have $a < c_x < x < a + \delta$ , so $\left|\frac{f(x)}{g(x)} - L\right| < \varepsilon$ .

∎

Example— L'Hôpital twice, and the hypotheses that matter

$\displaystyle\lim_{x \to 0} \frac{e^{x} - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac12.$ Each step is legitimate because the previous quotient was of the form $0/0$ and the next limit exists. By contrast, $\lim_{x \to 0} \frac{x + 1}{x + 2} = \frac12$ by continuity, while blindly differentiating gives $1$ : the rule only applies to indeterminate forms.

Secants become a tangent

Shrink h and watch the secant line turn into the tangent: its slope approaches the derivative.

f(x) = sin(x)
secant through P and Q
tangent line (numerical slope)

Point x₀x₀ = 0.79Shrink h toward 0h = 1

Slope of the secant PQ

(f(x₀ + h) − f(x₀)) / h = (0.9771 − 0.7071) / 1 = 0.27

Slope of the tangent at P

f′(0.785) ≈ 0.7071

Numerical estimate (central difference with a tiny step).

Difference quotients as h shrinks

h	(f(x₀ + h) − f(x₀)) / h
1	0.26995
0.1	0.6706
0.01	0.70356
0.001	0.70675

As h shrinks, Q slides towards P, the secant turns into the tangent, and its slope settles on the derivative f′(x₀).

ProofH

Integration and the fundamental theorem

Theorem— Theorem (fundamental theorem of calculus, part 1)

Let $f$ be integrable on $[a, b]$ and $F(x) = \int_a^x f(t)\,dt$ . If $f$ is continuous at $x_0 \in (a, b)$ , then $F$ is differentiable at $x_0$ and $F'(x_0) = f(x_0)$ .

Proof

For $h \ne 0$ small, additivity gives

\frac{F(x_0 + h) - F(x_0)}{h} - f(x_0) = \frac1h \int_{x_0}^{x_0 + h} \big(f(t) - f(x_0)\big)\,dt.

∎

In particular every continuous function on an interval has an antiderivative.

Theorem— Theorem (fundamental theorem of calculus, part 2)

Let $F$ be continuous on $[a, b]$ and differentiable on $(a, b)$ with $F' = f$ there, and suppose $f$ is integrable on $[a, b]$ . Then

\int_a^b f(x)\,dx = F(b) - F(a).

Proof

Take any partition $a = x_0 < \cdots < x_n = b$ . By the mean value theorem on each $[x_{i-1}, x_i]$ there is $t_i \in (x_{i-1}, x_i)$ with $F(x_i) - F(x_{i-1}) = f(t_i)(x_i - x_{i-1})$ . Summing, the left side telescopes:

F(b) - F(a) = \sum_{i=1}^n f(t_i)\,(x_i - x_{i-1}).

So $F(b) - F(a)$ is a Riemann sum, for partitions of every mesh. As the mesh tends to $0$ these sums tend to $\int_a^b f$ ; hence the two numbers are equal.

∎

Theorem— Corollary (integration by parts and substitution)

If $u, v$ are $C^1$ on $[a, b]$ , then $\int_a^b u\,v'\,dx = \big[u v\big]_a^b - \int_a^b u'\,v\,dx$ . If $\varphi$ is $C^1$ on $[a, b]$ and $f$ is continuous on an interval containing $\varphi([a, b])$ , then $\int_a^b f(\varphi(t))\,\varphi'(t)\,dt = \int_{\varphi(a)}^{\varphi(b)} f(x)\,dx$ .

Proof

For parts, $(uv)' = u'v + uv'$ is continuous, so part 2 gives $\int_a^b (u'v + uv') = [uv]_a^b$ ; rearrange. For substitution, let $F$ be an antiderivative of $f$ (part 1). By the chain rule $(F \circ \varphi)' = f(\varphi)\,\varphi'$ , so both sides equal $F(\varphi(b)) - F(\varphi(a))$ by part 2.

∎

Theorem— Proposition (p-integrals)

$\displaystyle\int_1^\infty \frac{dx}{x^p}$ converges if and only if $p > 1$ , and then equals $\dfrac{1}{p - 1}$ .

Proof

For $p \ne 1$ , $\int_1^b x^{-p}\,dx = \frac{b^{1-p} - 1}{1 - p}$ , which tends to $\frac{1}{p - 1}$ if $p > 1$ and to $\infty$ if $p < 1$ . For $p = 1$ the integral is $\ln b \to \infty$ .

∎

Riemann sums for ∫₀¹ eˣ dx = e − 1

Approximate the area under a curve with thin strips, and watch the error shrink as the strips get thinner.

f(x) = exp(x)

Number of strips nn = 8

Midpoint sum with 8 strips

M₈ = 1.717164

Exact value of the integral

∫₀¹ f(x) dx = 1.718282

Error (sum − integral)

−0.001118

All four rules with n = 8

Rule	Sum	Error
Left (L₈)	1.613126	−0.105156
Right (R₈)	1.827911	0.109629
Midpoint (M₈)	1.717164	−0.001118
Trapezoid (T₈)	1.720519	0.002237

Double n and watch the error: left and right sums roughly halve it, midpoint and trapezoid sums cut it to about a quarter.

Compare the four rules as you double the number of strips: the left-rule error roughly halves, while the midpoint and trapezoid errors drop by a factor of about four. The experiment below measures these rates.

ProofH

Series and convergence tests

All the tests rest on the monotone convergence theorem (a bounded increasing sequence of reals converges), which is one form of the completeness of $\R$ . For a series with non-negative terms, the partial sums increase, so such a series converges exactly when its partial sums are bounded.

Theorem— Theorem (comparison test)

If $0 \le a_n \le b_n$ for all $n$ and $\sum b_n$ converges, then $\sum a_n$ converges.

Proof

The partial sums of $\sum a_n$ increase and are bounded by $\sum_{n=0}^\infty b_n$ .

∎

Theorem— Theorem (absolute convergence implies convergence)

If $\sum |a_n|$ converges, then $\sum a_n$ converges.

Proof

Since $-|a_n| \le a_n \le |a_n|$ , we have $0 \le a_n + |a_n| \le 2|a_n|$ , so $\sum (a_n + |a_n|)$ converges by comparison. Then $\sum a_n = \sum (a_n + |a_n|) - \sum |a_n|$ is a difference of two convergent series.

∎

Theorem— Theorem (ratio test)

Suppose $a_n \ne 0$ for large $n$ and $|a_{n+1}/a_n| \to \rho$ . If $\rho < 1$ , then $\sum a_n$ converges absolutely; if $\rho > 1$ , it diverges. If $\rho = 1$ the test says nothing.

Proof

If $\rho < 1$ , pick $r$ with $\rho < r < 1$ . For $n \ge N$ we have $|a_{n+1}| \le r|a_n|$ , so $|a_n| \le |a_N|\,r^{n - N}$ , and $\sum |a_n|$ converges by comparison with a geometric series. If $\rho > 1$ then $|a_{n+1}| > |a_n|$ for large $n$ , so $a_n \not\to 0$ and the series diverges (its partial sums cannot settle if the terms do not tend to $0$ ). For $\rho = 1$ : $\sum 1/n$ diverges while $\sum 1/n^2$ converges, and both have $\rho = 1$ .

∎

Theorem— Theorem (integral test)

Let $f$ be positive, continuous and decreasing on $[1, \infty)$ . Then $\sum_{n \ge 1} f(n)$ converges if and only if $\int_1^\infty f(x)\,dx$ converges.

Proof

Because $f$ decreases, $f(k + 1) \le \int_k^{k+1} f \le f(k)$ for each $k \ge 1$ . Summing over $k = 1, \dots, N - 1$ :

\sum_{n=2}^{N} f(n) \;\le\; \int_1^N f(x)\,dx \;\le\; \sum_{n=1}^{N-1} f(n).

All three quantities increase with $N$ . If the integral converges, the left inequality bounds the partial sums; if the series converges, the right inequality bounds $\int_1^N f$ , which then converges as $N \to \infty$ (and $\int_1^b f$ increases in $b$ ).

∎

With $f(x) = x^{-p}$ and the $p$ -integrals: $\sum 1/n^p$ converges if and only if $p > 1$ .

Theorem— Theorem (alternating series test)

If $b_n$ decreases to $0$ , then $\sum_{n \ge 1} (-1)^{n+1} b_n$ converges to some $S$ , and $|S - s_N| \le b_{N+1}$ .

Proof

The even partial sums increase, since $s_{2k+2} = s_{2k} + (b_{2k+1} - b_{2k+2})$ , and the odd ones decrease, since $s_{2k+1} = s_{2k-1} - (b_{2k} - b_{2k+1})$ ; also $s_{2k} \le s_{2k+1} \le s_1$ . Both subsequences are monotone and bounded, so they converge, and to the same limit because $s_{2k+1} - s_{2k} = b_{2k+1} \to 0$ . The limit $S$ lies between any two consecutive partial sums, so $|S - s_N| \le |s_{N+1} - s_N| = b_{N+1}$ .

∎

The p-series: slide p across 1

Add the terms one by one. Do the partial sums Sₙ settle down, or keep growing?

Examples

Term aₙ of the series

a_n =

partial sums Sₙ
terms aₙ

Partial sums up to NN = 300

pp = 2

S₃₀₀ = a₁ + … + a₃₀₀ = 1.641606

Partial sums on a screen are evidence, never proof: a series can look settled and still diverge, as the harmonic series does.

ProofH

Taylor's theorem

Theorem— Theorem (Taylor's theorem with Lagrange remainder)

Let $f^{(n)}$ be continuous on the closed interval between $a$ and $x \ne a$ , and let $f^{(n+1)}$ exist on the open interval between them. Then for some $\xi$ strictly between $a$ and $x$ ,

f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k + \frac{f^{(n+1)}(\xi)}{(n + 1)!}(x - a)^{n+1}.

Proof

Keep $x$ fixed and let the base point vary: for $t$ between $a$ and $x$ put

F(t) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(t)}{k!}(x - t)^k, \qquad G(t) = (x - t)^{n+1}.

Differentiating term by term, the sum telescopes (the derivative of the $k$ -th term is $\frac{f^{(k+1)}(t)}{k!}(x - t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x - t)^{k-1}$ for $k \ge 1$ ), leaving

F'(t) = -\frac{f^{(n+1)}(t)}{n!}(x - t)^n, \qquad G'(t) = -(n + 1)(x - t)^n.

Now $F(x) = 0$ , $F(a) = R_n(x)$ , $G(x) = 0$ and $G(a) = (x - a)^{n+1}$ . Cauchy's mean value theorem on the interval between $a$ and $x$ gives $\xi$ strictly between them with $\big(F(x) - F(a)\big)G'(\xi) = \big(G(x) - G(a)\big)F'(\xi)$ , that is, $R_n(x)\,(n + 1)(x - \xi)^n = (x - a)^{n+1}\,\frac{f^{(n+1)}(\xi)}{n!}(x - \xi)^n$ . Since $\xi \ne x$ we may cancel $(x - \xi)^n$ , which gives the formula.

∎

For $n = 0$ this is the mean value theorem. The remainder formula is a certificate of accuracy: if $|f^{(n+1)}| \le M$ between $a$ and $x$ , then $|R_n(x)| \le M|x - a|^{n+1}/(n+1)!$ .

Theorem— Corollary (the exponential series)

For every real $x$ , $\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ . Similarly $\sin x = \sum (-1)^n \frac{x^{2n+1}}{(2n+1)!}$ and $\cos x = \sum (-1)^n \frac{x^{2n}}{(2n)!}$ for all $x$ .

Proof

Every derivative of $e^t$ is $e^t$ , so Taylor's theorem at $0$ gives $|R_n(x)| = e^{\xi}|x|^{n+1}/(n+1)! \le e^{|x|}|x|^{n+1}/(n+1)!$ . For fixed $x$ , the numbers $c_n = |x|^n/n!$ satisfy $c_{n+1}/c_n = |x|/(n + 1) \to 0$ , so $c_n \to 0$ (by the ratio argument above). Hence $R_n(x) \to 0$ . For $\sin$ and $\cos$ use $|f^{(n+1)}| \le 1$ .

∎

Example— How many terms for sin 0.3?

The Maclaurin polynomial $x - x^3/6 + x^5/120$ of $\sin$ is also $T_6$ , since the $x^6$ coefficient is $0$ . With $|\sin^{(7)}| \le 1$ the error at $x = 0.3$ is at most $0.3^7/7! \approx 4.3 \times 10^{-8}$ . Three terms already give seven correct decimals; the experiment below compares such bounds with the actual errors.

Is every smooth function the sum of its Taylor series? Cauchy showed in 1823 that the answer is no.

Theorem— Theorem (a smooth function that is not analytic)

Let $f(x) = e^{-1/x^2}$ for $x \ne 0$ and $f(0) = 0$ . Then $f$ is smooth on $\R$ and $f^{(n)}(0) = 0$ for every $n$ . So its Taylor series at $0$ is identically zero, yet $f(x) > 0$ for every $x \ne 0$ .

Proof

Step 1. For each integer $m \ge 0$ , $\lim_{x \to 0} |x|^{-m} e^{-1/x^2} = 0$ . Indeed, with $y = 1/x^2 \to \infty$ , this is $y^{m/2} e^{-y} \to 0$ , because $e^y \ge y^{k}/k!$ for every $k$ (one term of its series) — take $k > m/2$ .

Step 2. For $x \ne 0$ , $f^{(n)}(x) = P_n(1/x)\,e^{-1/x^2}$ for some polynomial $P_n$ . This holds for $n = 0$ with $P_0 = 1$ , and if it holds for $n$ , then by the chain and product rules

f^{(n+1)}(x) = \Big(-\frac{1}{x^2}P_n'(1/x) + \frac{2}{x^3}P_n(1/x)\Big)e^{-1/x^2} = P_{n+1}(1/x)\,e^{-1/x^2}

with $P_{n+1}(u) = 2u^3 P_n(u) - u^2 P_n'(u)$ .

Step 3. By induction, $f^{(n)}(0)$ exists and equals $0$ . For $n = 0$ this is the definition. If $f^{(n)}(0) = 0$ , then

f^{(n+1)}(0) = \lim_{h \to 0} \frac{f^{(n)}(h) - 0}{h} = \lim_{h \to 0} \frac1h P_n(1/h)\,e^{-1/h^2} = 0,

because $\frac1h P_n(1/h)$ is a finite combination of powers $h^{-m}$ , and each $h^{-m}e^{-1/h^2} \to 0$ by Step 1. Steps 2 and 3, together with Step 1 again, also show each $f^{(n)}$ is continuous at $0$ , so $f$ is smooth.

∎

Flat at 0, yet not zero

Move the sliders and trace along the curves to read off values.

f(x) = exp(−1/x²)
its Taylor series at 0

Trace along xx = 1

f(x) = exp(−1/x²): y = 0.3679
its Taylor series at 0: y = 0

Near $0$ the graph is so flat that no polynomial can see it: every Taylor polynomial at $0$ is the zero polynomial. The lesson? Taylor's theorem says $f = T_n + R_n$ — always true — but whether $R_n \to 0$ must be proved, function by function, usually with the Lagrange bound.

Formal mathematicsF

Calculus in Lean and Mathlib

Mathlib has a large, general library of calculus: HasDerivAt f f' x says $f$ has derivative $f'$ at $x$ , ∫ x in a..b, f x is the interval integral (defined via Lebesgue integration, which agrees with the Riemann integral for Riemann-integrable functions), HasSum a s says a series converges (unconditionally) to $s$ , and taylorWithinEval gives Taylor polynomials. The lesson's module re-proves several results from more basic library facts — these are marked independent — and cites others directly.

First, continuity. Mathlib defines it with neighbourhood filters; for real functions this unfolds to exactly our ε–δ definition, and the example $x^2$ is proved with the same $\delta = \min(1, \varepsilon/(2|a| + 1))$ as on paper, each calc line matching a step of the human proof:

continuousAt_iff_eps_deltaLean 4Applies a library theoremFormally verified

/-- Mathlib defines continuity with neighbourhood filters; for real functions this is exactly
the ε–δ definition of the lesson. This applies Mathlib's `Metric.continuousAt_iff` and
`Real.dist_eq` (the distance on `ℝ` is `|x - y|`). -/
theorem continuousAt_iff_eps_delta (f : ℝ → ℝ) (a : ℝ) :
    ContinuousAt f a ↔ ∀ ε > 0, ∃ δ > 0, ∀ x, |x - a| < δ → |f x - f a| < ε := by
  simp only [Metric.continuousAt_iff, Real.dist_eq]