“The art of numbering and measuring exactly a thing whose existence cannot be conceived.” Voltaire on the calculus of infinitesimals, Letters on the English (1733)
Follow one question through four thousand years, from Babylonian clay tablets to the limits of modern analysis, with labs and step-by-step proofs along the way.
Start the journey arrow_downwardThe question behind calculus
Curved areas, changing speeds and growing quantities share one difficulty. To measure them exactly, it seems you must cut them into pieces that are infinitely many and infinitely small. These pieces came to be called infinitesimals.
For two thousand years mathematicians used them to get correct answers without being able to say what they were. The timeline below follows that single question through twenty smaller ones, each answer, objection and repair. Pick a question to jump to it.
Discussed for two thousand years, defined only in the 1800s. Zeno asked the question; Eudoxus and Archimedes had the logic; Newton had the word; d’Alembert said calculus should rest on it; Cauchy and Weierstrass finally said exactly what it means. Every card on this thread shows how far along the road it is.
A strange idea, argued over since Zeno. Aristotle allowed only endless processes, never completed infinities; Galileo found that infinite collections defy “more” and “less”; Cantor measured them, and found some infinities larger than others.
For centuries mathematicians got right answers from methods they could not justify. Archimedes kept his slicing method as a private tool for discovery, Fermat rarely published, and Newton sat on his fluxions for decades. Newton and Leibniz’s foundations had the same gaps as everyone’s before them; what was new was a general method, and publishing it, so others could build on it.
The answer
Cauchy and Weierstrass answered the question by changing it. Calculus no longer calculates with infinitely small quantities. It works with ordinary, finite quantities and asks what value they approach. Zeno’s race, Archimedes’ polygons, Newton’s fluxions and Riemann’s sums can all be restated as limits.
The story has one more twist. In 1961 Abraham Robinson used mathematical logic to show that infinitesimals can be made rigorous after all. His non-standard analysis vindicated Leibniz’s intuition almost three centuries later.
Four beads leave A together and slide without friction to B. The straight line is shortest, but it is not fastest. Move B and race again.
Times for a 1 metre drop.
If two solids have the same height and equal cross-sectional areas at every level, they have the same volume.
Analogy: Think of a stack of coins. Whether you stack them straight up or push them to the side (shear), the total amount of metal (volume) remains exactly the same.
Cut a circle into sectors and lay them top to tail. As the pieces get thinner, the shape becomes a rectangle of width $\pi r$ and height $r$.
Treating each sector as a triangle gives $\tfrac{n}{2}\sin\!\left(\tfrac{2\pi}{n}\right)r^2$, which tends to $\pi r^2$.
A function is continuous at $c$ if, for every target $\varepsilon$, some window $\delta$ keeps $f(x)$ within $\varepsilon$ of $f(c)$. Unlike a limit, the value at $c$ itself must fit.
Each short line shows the rate of change at that point. A solution follows the lines. Choose a problem, then click anywhere on the graph to start a new solution from that point.
Drag point Q towards P to visualize the secant line becoming the tangent.
Definition:
$\displaystyle f'(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}$
Observe: When $h$ is large, the secant line (gold) cuts through the curve. As $h$ approaches 0, the secant line becomes indistinguishable from the tangent line (cyan).
The "Limit Game": For a given target range ($\varepsilon$), can you find a window ($\delta$) where the function stays inside?
"How close do we need to be?"
"How close must $x$ be?"
Goal: Adjust $\delta$ so that, inside the gold window, the cyan curve stays within the blue $\varepsilon$ band. Any escape is marked in pink.
Top: the area under $f$ up to $x$, plus a thin strip of width $\Delta x$. Bottom: the area function $A(x)$. Shrink $\Delta x$ and the strip becomes a rectangle $f(x)\,\Delta x$, so the gradient of $A$ is exactly $f(x)$.
As $\Delta x$ shrinks, $\Delta A / \Delta x$ closes in on $f(x)$: the slope of the area graph is the height of the curve.
Compare numerical methods to approximate the area under a curve.
Left Sum: Uses the height of the function at the left edge of each rectangle. Often inaccurate for curvy functions.
Watch Jupiter move in the 60 days after it first appears at dawn (the tablet gives its daily motion; placing it among the stars of the Lion is our illustration). Dots mark every fifth night: they bunch up as Jupiter slows, and the shaded area under the speed graph is the distance it has moved.
Guess first: slide to the day when Jupiter has covered half of the 60-day distance.
Choose a limit. Watch the ratio as $x$ approaches its target, then apply the rule step by step. Some limits exist, some are zero, and some do not exist at all.
Add terms one at a time. The first series zig-zags slowly around π; Madhava's faster series homes in within a handful of terms.
Pick a function, move the centre $a$ and add terms. The shaded band shows where the series converges; outside it, more terms make things worse.
Explore the "infinite" steps required to catch up.
Achilles runs only 2x faster than the Tortoise in this simulation to make the convergence easier to see. The Tortoise starts 100m ahead.
The Calculus Solution: Zeno assumed an infinite sum of times is infinite. Calculus shows that $\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}$, a finite value.
From the start point, follow a slope for one step of size $h$, then recalculate and repeat. Euler uses the slope at the start of each step; improved Euler (Heun) and Runge–Kutta (RK4) sample the slope at more points in the step and average. Compare each method with the true solution in white, and halve $h$ to see how fast each error shrinks.
Problem
Method
Results
Press Halve $h$ a few times. The ratio of errors settles near $2$ for Euler, $4$ for Heun and $16$ for RK4: the error is roughly proportional to $h$, $h^2$ and $h^4$, so the orders are $1$, $2$ and $4$.
Steps
Build a model for two interacting species, term by term, and see how it behaves. Top: the populations over time. Bottom: the phase plane, where each point is a pair of populations; a dot over a letter means its rate of change, $\dot{x} = \tfrac{dx}{dt}$. Click or drag in the phase plane to choose the starting populations. The white square marks the start; filled circles are stable equilibria and hollow circles unstable ones.
Model
Outcome
Parameters
Behaviour
Data: hares and lynx
Numbers of snowshoe hare and Canada lynx pelts (thousands) traded by the Hudson’s Bay Company, as tabulated in many textbooks (after Odum, 1953, and Leigh, 1968). Squared error is the sum of the squared gaps between the model and both data series at the 21 yearly records. Try to beat the fit by hand.
The best classic Lotka–Volterra fit captures the roughly ten-year cycle and the lag of lynx behind hares. But its cycles repeat exactly, with a fixed period and size, which the longer fur records do not. Pelts are not populations: trapping effort, prices and weather all changed the counts. Field experiments in the Yukon (Krebs and colleagues, 1995) found that the hare cycle depends on both food and predation, with several predators besides lynx; in parts of the longer record the lynx peaks even come first (Gilpin, 1973, “Do hares eat lynx?”). A good fit does not prove the model.
In Linz the wine gauger pushed one rod through the bung hole to the far bottom edge of the barrel and read the volume off its scale. Kepler asked whether one length $d$ could be fair to every barrel. Shape the barrel, cut it into thin discs to find its volume, then keep the rod fixed and change the proportions: near the best shape the volume hardly changes. Drag the barrel to turn it; drag on the graph to reshape it.
Indian astronomers found the Moon’s true place by taking its steadily moving mean place and subtracting a correction, the equation, worked out from the sine of the anomaly on an epicycle. The old daily motion was the difference between two positions a day apart. Bhāskara II (1150) wanted the motion at a single instant, and got it from the cosine (koṭi). Shrink the interval and watch the old answer close in on his.
Time
Old method: interval
Readouts
The model
$\lambda = \bar\lambda - \mu$
$R\sin\mu = \tfrac{32}{360}\,R\sin\kappa$
$\bar\lambda$ is the mean place, moving $13^\circ 10' 35''$ a day; $\kappa$ is the mean anomaly, the angle from the apogee (mandocca), growing $13^\circ 3' 54''$ a day. $R = 3438$, so arcs and sines are both in minutes. The correction is subtracted for $\kappa$ from $0^\circ$ to $180^\circ$ and added after.
Gross (sthūla) motion$\dfrac{\lambda(t + \Delta t) - \lambda(t)}{\Delta t}$
Bhāskara’s rule$\delta(R\sin\kappa) \approx \dfrac{R\cos\kappa}{R}\,\delta\kappa$
$\delta\mu \approx \tfrac{32}{360}\cos\kappa\;\delta\kappa$
So the instantaneous (tātkālika) motion is the mean motion less $\tfrac{32}{360}\cos\kappa$ times the daily motion of the anomaly: slowest at the apogee, fastest at the perigee, and exactly the mean motion where $\cos\kappa = 0$, which is where the equation is greatest.
History
In the Siddhānta Śiromaṇi (1150), Bhāskara II calls the difference between a planet’s longitudes at the same time on successive days its daily motion, and says this is only rough (sthūla); the fine motion is the instantaneous one. He finds it by replacing the change in the sine by the cosine times the change in the arc. Mañjula (932) had already used a rule of this kind for the Moon. Bhāskara also remarks that where the equation is greatest its instantaneous change is zero.
Numbers: the Moon’s epicycle of $32^\circ$ and its mean motions are those of the Sūrya Siddhānta, which also shrinks the epicycle to $31^\circ 40'$ at the quadrants (ignored here). Bhāskara’s own parameters differ by seconds of arc per day. Sines are computed exactly rather than read from a table; the drawing is a modern reconstruction, and the Moon’s latitude and second inequality are left out. Stars are placed at approximate sidereal longitudes.
The same number turns up three times: as the limit of compound interest (Bernoulli), as the point where the area under $y = \tfrac1x$ reaches $1$ (Saint-Vincent’s hyperbola), and as the base whose curve has gradient equal to its height (Euler).
Fourier built functions, even ones with jumps, out of sines and cosines. Add terms one at a time and compare the gold partial sum $S_N$ with the target in white; watch what happens next to the jumps.
A rectangle $n + 1$ wide and $1^k + 2^k + \dots + n^k$ tall, cut into a gold staircase and cyan columns. Counting its area two ways gives $(n+1)\sum i^k = \sum i^{k+1} + \sum_{p=1}^{n}\sum_{i=1}^{p} i^k$, which turns a known sum of powers into the next one. Ibn al-Haytham climbed from squares to fourth powers this way, then used them to find a volume by slicing.
Area counted two ways
From picture to formula
What he needed
History. Ibn al-Haytham (c. 965 to c. 1040), born in Basra and working in Cairo, wrote on measuring the paraboloid. Sums of squares and cubes were already known; the sum of fourth powers was new. He argued with $n = 4$ and then stated the rule in words; the argument works the same way for any $n$, which is why historians see it as an early form of induction. The figure drawn here follows modern accounts of his proof (Katz, 1995; Beery, Convergence).
Rotate $y = 1/x$, for $1 \le x \le a$, about the $x$-axis. Stretch the horn: its volume levels off at $\pi$ (as Torricelli showed), but its surface area grows without bound.
$\begin{aligned} V(a) &= \pi\displaystyle\int_1^a \frac{1}{x^2}\,dx \\ &= \pi\Big(1 - \frac1a\Big) \end{aligned}$
$\begin{aligned} S(a) &= 2\pi\displaystyle\int_1^a \frac1x\sqrt{1 + \frac{1}{x^4}}\,dx \\ &\ge 2\pi\displaystyle\int_1^a \frac{dx}{x} = 2\pi\ln a \end{aligned}$
Units are metres; $1\,\text{m}^3$ is $1000$ litres and a litre of paint covers $10\,\text{m}^2$. The area is computed accurately by Simpson’s rule on a log-spaced grid.
The painter’s paradox. The horn holds only $\pi$ cubic units of paint, yet its outside has infinite area, so it seems you could fill it but never paint it. The catch is that a mathematical surface has no thickness. A real coat of fixed thickness $t$ on the outside needs paint of cross-section $\pi\big((r+t)^2 - r^2\big) > \pi t^2$ at every $x$, so infinitely much over an infinite length. Pouring paint in does coat the inside, but there the coat is only as thick as the horn is wide, and that tends to $0$: infinite area times vanishing thickness can still give a finite volume. (With paint at $10\,\text{m}^2$ per litre, covering the outside only needs more paint than filling once $a > 10^{2171}$.)
Fit an exponential or a logistic model to real data, using only the points up to a cut-off. The solid points are used for the fit; the hollow points are held back, so you can see how well the model predicts the future. The strip underneath shows each residual, data minus model.
Retrace three Greek attacks on the circle. Antiphon fills it from inside with polygons, doubling the sides each time; Bryson traps it between inside and outside polygons; Archimedes turns the squeeze into numbers, from hexagons to 96 sides. The circle has radius $1$, so its area is $\pi$ and its circumference is $\pi$ times the diameter. The inset magnifies a small piece of the edge.
Results
With radius $1$ the polygon’s area is an estimate of $\pi$ from below. These numbers are a modern computation: Antiphon computed none.
History
Antiphon (Athens, c. 430 BC), as reported by Simplicius drawing on Eudemus, inscribed a square in a circle. On each side he set a triangle with its apex on the circle, making an octagon, and he kept doing this. He claimed that in the end the polygon would coincide with the circle; since any polygon can be turned into a square of equal area, the circle would be squared. Aristotle (Physics I.2) said a geometer need not even refute this, as it breaks the principles of geometry. Simplicius explains: a straight line never coincides with an arc, and magnitudes can be divided without end, so a sliver always remains.
Eudoxus (c. 370 BC) kept the useful part. Each new triangle is half of the rectangle on its side reaching up to the tangent, so it is more than half of the sliver it sits in: every doubling removes more than half of the gap left. Repeat often enough and the gap is smaller than any area you name. Euclid uses this in Elements XII.2 (circles are to one another as the squares on their diameters), a proof generally credited to Eudoxus. The Greek argument needs only more than half; the exact share (approaching $\tfrac34$) is a modern computation.
Modern aside
$A_n = \tfrac{n}{2}\sin\tfrac{2\pi}{n} < \pi$
Gap after doubling$\pi - A_{2n} < \tfrac12\,(\pi - A_n)$
Results
The circle’s area $\pi$ is trapped between the two polygon areas. Again the numbers are ours, not Bryson’s.
History
Bryson of Heraclea, of the same era or a little later, added polygons drawn outside the circle. The circle is larger than every inscribed polygon and smaller than every circumscribed one, so it is trapped from both sides. Aristotle called his squaring of the circle sophistical, because it rested on a general principle rather than a geometrical one (Posterior Analytics I.9, Sophistical Refutations 11). What Bryson actually argued is debated. Alexander, as reported by Philoponus, says he took a polygon between the inner and outer ones and declared it equal to the circle, since things greater and less than the same things are equal; Proclus, also reported by Philoponus, read the argument differently.
Modern aside
$\tfrac{n}{2}\sin\tfrac{2\pi}{n} < \pi < n\tan\tfrac{\pi}{n}$
Perimeter ÷ diameter
$\pi > \dfrac{6336}{2017\frac14} > 3\tfrac{10}{71}$
$\pi < \dfrac{14688}{4673\frac12} < 3\tfrac17$
So $3\tfrac{10}{71} < \pi < 3\tfrac17$, that is $3.14084\ldots < \pi < 3.14285\ldots$
History
Archimedes (Syracuse, c. 250 BC), Measurement of a Circle, Proposition 3: the circumference of any circle is less than $3\tfrac17$ and greater than $3\tfrac{10}{71}$ times its diameter. He worked with perimeters, not areas, starting from regular hexagons outside and inside the circle and doubling four times to 96 sides. Proposition 1 links the two: the circle’s area equals a right-angled triangle with the radius and the circumference as its legs, so the same bounds hold for the area.
Bounds for $\pi$
| $n$ | lower | upper | |
|---|---|---|---|
| 6 | polygon | 3.00000 | 3.46410 |
| his | 3.00000 | 3.46415 | |
| 12 | polygon | 3.10583 | 3.21539 |
| his | 3.10577 | 3.21541 | |
| 24 | polygon | 3.13263 | 3.15966 |
| his | 3.13245 | 3.15973 | |
| 48 | polygon | 3.13935 | 3.14609 |
| his | 3.13922 | 3.14619 | |
| 96 | polygon | 3.14103 | 3.14271 |
| his | 3.14091 | 3.14283 |
“Polygon”: the true perimeter ÷ diameter, $n\sin\tfrac{\pi}{n}$ and $n\tan\tfrac{\pi}{n}$. “His”: what Archimedes’ rounded fractions give, always slightly further from $\pi$. He converted to perimeters only at 96 sides; the earlier rows are what his ratios imply.
His fractions
| $n$ | $a_n >$ | $b_n \le$ |
|---|---|---|
| 6 | $265 : 153$ | $1560 : 780$ |
| 12 | $571 : 153$ | $3013\tfrac34 : 780$ |
| 24 | $1162\tfrac18 : 153$ | $1838\tfrac{9}{11} : 240$ |
| 48 | $2334\tfrac14 : 153$ | $1009\tfrac16 : 66$ |
| 96 | $4673\tfrac12 : 153$ | $2017\tfrac14 : 66$ |
$a_n$ is the ratio radius : half-side for the outside polygon, so $\pi < n \div a_n$; $b_n$ is diameter : side for the inside polygon, so $\pi > n \div b_n$. He started from $\tfrac{265}{153} < \sqrt3 < \tfrac{1351}{780}$, without saying how he found these.
His doubling rule
$a_{2n} = a_n + \sqrt{1 + a_n^2}$
Inside$d_{2n} = d_n + b_n,\quad b_{2n} = \sqrt{1 + d_{2n}^2}$
Here $d_n$ is the other short side ÷ side in the right-angled triangle on a diameter. Archimedes states each step as a ratio of lines in his figure, from the angle bisector theorem and Pythagoras. Each square root he replaces by a fraction: slightly too small outside, slightly too large inside, so every bound stays safe. Twice he scales the inside numbers down, by $\tfrac{4}{13}$ and $\tfrac{11}{40}$, to keep them manageable.
Modern aside
$n\sin\tfrac{\pi}{n} < \pi < n\tan\tfrac{\pi}{n}$
Cut every solid with the same horizontal plane at height $h$ above the centre and compare the slices. Liu Hui (263) saw that the sphere’s slice is always the circle inside the square slice of the “double box lid”; Zu Gengzhi (about 500) found the lid’s volume by matching one eighth of it, slice by slice, against a pyramid. Drag the picture to turn it.
Slices at height $h$, with $r = 1$
The argument
History
Liu Hui (263). The Nine Chapters gives a rule equivalent to $V = \tfrac{9}{16}D^3$ for a sphere of diameter $D$. With $\pi = 3$, this treats the sphere as $\tfrac34$ of the cylinder around it, as if circle to square held between sphere and cylinder. Liu Hui showed that this is wrong: circle to square holds between the sphere and the double box lid (牟合方蓋, móuhé fānggài), which is smaller than the cylinder. He could not find the lid’s volume and, in MacTutor’s translation, wrote: “Let us leave the problem to whoever can tell the truth.”
Zu Gengzhi (about 500). His solution survives only because Li Chunfeng quoted it in his commentary on the Nine Chapters (656); the book he wrote with his father Zu Chongzhi, the Zhui shu, is lost. His principle reads 冪勢既同，則積不容異: if the areas at every height are the same, the volumes cannot differ. The pyramid is a yangma (陽馬), with a square base and its apex above one corner; three of them make a cube, so its volume is $\tfrac13 r^3$, a result already in the Nine Chapters and proved by Liu Hui. Zu worked with $\pi = 3$: his rule doubles the volume and takes the cube root to give the diameter, that is $V = \tfrac12 D^3$.
What is reconstruction. The text is in words only; no figures survive. The pictures here, like all modern ones, are reconstructions, and translations of the principle differ in detail. Archimedes had also stated the lid’s volume, $\tfrac23$ of its cube, in the preface to The Method, but that work was lost until 1906 and was unknown in China.