The road that leaves the country
The journey
You are at a number u between 0 and 1, and you want the number x where a
Beta(2,2) distribution has accumulated exactly that much of itself. Statisticians
make this trip constantly. It is the quantile, the inverse of the CDF, the thing
under every “give me a random draw” and every error bar. For this particular
distribution the CDF is a cubic polynomial, 3x² − 2x³, so the trip is short.
You are inverting a cubic. People have been able to invert cubics since 1545.
Route one: the paved road
Every schoolbook gives you a formula. Move the square term out of the way, and
any cubic becomes t³ + pt + q = 0. Cardano’s rule then says: take
√(q²/4 + p³/27), add and subtract it from −q/2, take the cube root of each,
add the two. Square roots and cube roots of real numbers, four operations, done.
This is the road everyone assumes they are on.
For our trip it is p = −3/4, always, whatever u you started from. Only q
moves, and it moves as (u − 1/2)/2. So the quantity under the square root is
(u−1/2)²/16 − 1/64, and that is negative for every u strictly between 0 and 1.
The road ends at a wall about six steps in. Not for awkward values of u. For
all of them.
1545: the traveller who turned back
Cardano knew about the wall. The Ars Magna spends fifty-nine pages on the cubic
and splits it into thirteen cases, and the split exists because his rules simply
do not apply when that quantity goes negative. He does, in one chapter, allow
himself to write 5 + √−15 and 5 − √−15 as the two parts of ten whose product
is forty, and he checks that they multiply correctly, and then he backs away from
his own arithmetic and calls the thing sophistica. Everyone quotes him as
calling that arithmetic as subtle as it is useless, and I went looking for the
line and could not verify it, so here is what I can: “this is truly
sophisticated, since with it one cannot carry out the operations one can in the
case of a pure negative and other numbers.” His objection is not that the
quantity is unreal. It is that his tools do not run on it.
And then, in Chapter XII of the same book, he wrote an IOU. Where the discriminant goes the wrong way, he says, “the solution of this can be found by the aliza problem which is discussed in the book of the geometrical problems.” There was no such book. He was pointing his readers at a route he had not found, in a work that did not exist, in print, in 1545.
1570: the book nobody reads
Twenty-five years after the Ars Magna, Cardano published a whole further book
about the wall, and almost nobody has read it. The De Regula Aliza is an
attempt to get around the irreducible case without ever taking the square root of
a negative number. His tool is the substitution x = y + z, splitting the
unknown into two parts and then matching pieces of the expanded cube against the
coefficients. Sara Confalonieri catalogued every one of these splittings in a
2013 thesis that runs to seven hundred and sixty pages. Her verdict on the one
that comes closest is one sentence: unluckily, this leads back to the starting
point. The others land on equations of degree nine, or on cubics that can carry
the same bad sign he was fleeing.
So the IOU came due twenty-five years late and it was not payable. Every road curved back to where he set out, the volume that arrived was a miscellany that Confalonieri says was barely proofread, and the name of a problem he could not solve had been stretched over the whole of it.
And here is the part that stayed with me. The Ars Magna was reissued in 1570, bound together with the Aliza, and Cardano did not withdraw the promise in Chapter XII. He corrected the address. Where the 1545 text sends you to a book of geometrical problems that was never written, the 1570 text says “then consult the Aliza book appended to this work.” The pointer now resolves. It resolves to three hundred pages that do not contain the answer. He fixed the reference and left the claim standing, which is a thing I recognise, and he was two hundred and seventy-three years early to know it could not be fixed.
1572: the wild thought
Bombelli would not turn back. His L’Algebra takes the position that if these
quantities obey rules then you may as well use the rules, and he demonstrates it
on x³ = 15x + 4, an equation whose answer is 4 and which anybody can see is 4.
Cardano’s formula says the answer is ∛(2 + √−121) + ∛(2 − √−121). Bombelli
guesses that the cube root looks like a + √−b, multiplies the conjugate pair,
matches the real parts, and arrives at 4a³ − 15a − 2 = 0, which has the obvious
root a = 2. So the two cube roots are 2 + i and 2 − i. They add to 4. The
imaginary parts cancel on the way out of the country, the traveller arrives at a
real destination, and complex numbers have earned their first serious hour of
attention in Europe.
Now put x = 2a into that auxiliary equation of his.
You get x³ − 15x − 4, halved. It is the equation he started with. Bombelli’s
rescue works by assuming a root of the very cubic it is solving, which is why it
only ever runs on examples where the answer was visible from the doorstep. As a
proof it is circular. As an advertisement it worked for two hundred years.
Route two: the road made of angles
There is another way out, and it was found by François Viète around 1593, printed
posthumously in 1615. It does not touch an imaginary quantity anywhere. Set
x = 2a·cos θ in a cubic of the form x³ − 3a²x = a²b, and the multiple-angle
identity 4cos³θ − 3cos θ = cos 3θ collapses the whole thing to
b = 2a·cos 3θ. Then you divide an angle by three and take a cosine.
That is the entire journey. You pay for it in trigonometry instead of arithmetic, which is a real price and not a trick: an arccosine is not something you build out of square roots and cube roots. But you never leave the real line, and the road runs the whole way.
I got to this road backwards, on the morning of the 27th of August 2026, without
knowing whose it was. I wanted a cheaper way to draw Beta(2,2) random numbers for
a Monte Carlo that was going to take five hours, and drawing them means inverting
that cubic, and I noticed that 3x² − 2x³ with x = 1/2 + t becomes
1/2 + (3t − 4t³)/2, and 3t − 4t³ is sin 3θ when t is sin θ. So:
x = 1/2 + sin( arcsin(2u − 1) / 3 )
I checked it on four million draws and it returned u to five parts in
10¹⁶. Then I went and read the history and found I had reinvented a
substitution from the reign of Henri IV.
1843: the survey
Wantzel settled what the first road actually is. If a cubic is irreducible over
a formally real field and has three real roots, then no expression in real
radicals gives you any of them. Not a hard road. Not a road nobody has found. The
road is not there, and the proof is almost embarrassing in its simplicity: any
tower of radicals that solves the thing has to contain a primitive cube root of
unity, that number satisfies x² + x + 1 = 0, and a field where every element is
comparable to zero has no room for it.
Six years earlier the same man proved that you cannot trisect a sixty-degree angle with straightedge and compass.
The milestone at one quarter
Ask for the lower quartile. Put u = 1/4 into the cubic and clear denominators
and you have 8t³ − 6t − 1 = 0, which has no rational root, which for a cubic
means it does not factor at all over the rationals. Under t = cos θ it says
2 cos 3θ = 1. So 3θ is sixty degrees, θ is twenty, and the three roots are
the cosines of 20°, 140° and 260°. The one that lands inside [0,1] gives
lower quartile of Beta(2,2) = 1/2 − sin 10° = 0.32635182233306964
and the upper is 1/2 + sin 10° by the symmetry of the distribution.
cos 20° is the number every textbook uses to prove the sixty-degree angle
cannot be trisected. It is sitting in the quartile of a distribution people reach
for when they want a mild opinion about a probability. The two theorems Wantzel
proved six years apart are not neighbours. They are the same obstruction, and
this small everyday number is on the seam.
What the road book is for
The discriminant of our cubic works out to (27/4)·u(1−u), which is the Beta(2,2)
density again, rescaled. That is not a cute coincidence and it explains the whole
geography. A discriminant vanishes when two roots collide; two roots collide when
the curve you are inverting goes flat; and this density goes flat exactly at 0 and
1, where it touches zero. So the irreducible case begins where the support begins
and ends where it ends. Any distribution with a cubic CDF and a density that dies
at both ends has the same map.
None of this is new. The closed form is the kind of thing that has been rediscovered by every generation of people who needed a fast sampler, and I claim nothing about it except that I derived it before I recognised it. What I did not expect, and what the morning actually bought me, is that a perfectly ordinary statistical convenience turns out to be standing on top of the oldest hole in algebra.
There is a version of this story where the moral is that you should look up the
literature first. That version is wrong. Had I looked it up I would have copied a
formula out of a reference and never once noticed that the thing under my hand was
cos 20°, or that the road I was on had a survey attached to it saying the other
road does not exist. The looking-up is what turned a shortcut into a place. Do
both, in that order.
Postscript: the next town over
Move one step up the parameter, to Beta(3,3), and the CDF becomes a quintic. There is a temptation to expect more of the same, a longer detour through a larger imaginary country. What happens instead is that the country stops.
The Beta(3,3) quantile satisfies 48t⁵ − 40t³ + 15t + 4 − 8u = 0, and this one
has only one real root, so nothing is forcing anybody through the complex plane
at all. Its Galois group over the rationals is the full symmetric group on five
letters, for every rational u I tested except one. That group is not solvable.
So the root cannot be written with square roots, cube roots, fifth roots or any
tower of them, over the reals or over anything else. Abel and Ruffini, not
Cardano, and there is no Viète waiting on the other side with a trigonometric
detour, because the identity that made the cubic surrender has no partner here.
The exception is u = 1/2. The density is symmetric, so the median is exactly a
half, t = 0 is a root, the quintic factors, and the answer is as plain as it
could be. One point in the interval is easy and every other point is beyond
radicals entirely.
Two distributions apart. On one side a road that goes through a country Cardano would not enter, and comes back. On the other, no road.
