Generator

Completing the square, as a square

A generator in the algebra library, called 31 times across 6 essays. Below: what it draws with nothing chosen and at each mode an essay asks for, what it checks while drawing, and everywhere it is used.

algebra-tiles is one function. Everything below came out of it during this build, at parameters taken from the essays rather than invented for this page — so a figure here is the same figure a reader meets in an essay, and if the generator changes, this page changes with it.

With nothing chosen

Completing the square, as a square. An x by x square with the strip split in half and laid along two sides, leaving a square hole of side 1.5. Filling the hole costs 2.25 and buys a perfect square.

The same picture, solved

(a + b)², as four tiles

A quadratic in the exponent, completed

Every bell's area from one number

The bell's area squared, cut into rings

What it checks while it draws

Collected by running the family and recording what it asserted, not written here. The count is how many separate times the claim was put to the test while these drawings were made.

negative coefficients of A − 0.008I against eigenvalues below it ×530
negative coefficients of A − -7.000I against eigenvalues below it ×371
step 1 keeps the eigenvalue in its interval ×14
the diagonal point (0, 0) lands on the parabola ×13
the diagonal point (-3, -3) lands on the parabola ×12
decile 1's retest average is ρ times its first average ×10
and the line for root 0 never rises above it ×7
the line for 0 touches the cusped curve ×7
the measured slice spread at ρ = 0 is √(1 − ρ²) ×7
and the line for root -3 never rises above it ×6
order 012 has the eigenvalues' signs ×6
the line for -2.4 touches the cusped curve ×6
the line for root -3 touches the parabola at p = 6 ×6
the line for root 0.5 touches the parabola at p = -1 ×6
the area under e^(−1x² + 0x) is √(π/a)·e^(b²/4a) ×5
the area under e^(−x²)cos(0x) is √π·e^(−0²/4) ×3
the line for root 2 passes through the point ×3
the slice at x = 0 has spread √(1 − ρ²) ×3
the slice at x = 0 is centred at ρx ×3
the line for root -1.8100 passes through the point ×2
the sign of the discriminant says how many roots x² − 2x + 1 has ×2
the slice at x = -2 has spread √(1 − ρ²) ×2
the slice at x = -2 is centred at ρx ×2
x³ − 3x + 0.5: the discriminant's sign predicts the root count bisection finds ×2
a function on a torus has a pit, a peak and at least two passes ×1
a pair and its swap give the same p ×1
almost every sample completes without a zero pivot ×1
and its height is b²/4a ×1
and the reassembled rectangle has the same area ×1
and the rings out to 1.3 hold π(1 − e^(−1.69)) ×1
and the same q, so the map is two-to-one ×1
both binned slopes are near ρ, well under 1 ×1
completions disagreeing with the eigenvalues' signs ×1
each completion halves the interval ×1
each exponent opens downward ×1
each marked root is a root ×1
each x first direction carries the sign of its own square ×1
each y first direction carries the sign of its own square ×1
every critical point is non-degenerate ×1
every marked point is inside the plane drawn ×1
pits − passes + peaks on the torus ×1
the area under the parabola is 13/24 of the square ×1
the bell's area squared is π ×1
the census completes enough forms to mean something ×1
the completed form is the same function ×1
the drawn sample's share agrees with 13/24 ×1
the eigenvalues have one sign each ×1
the eigenvalues multiply to the determinant ×1
the exponent opens downward, or there is no bell ×1
the first pivot of the Hessian is not zero ×1
the form takes both signs and both orders can start ×1
the four tiles are the square ×1
the highlighted point is below the curve and has two roots ×1
the horizontal slice at y = 1.5 is centred at ρ·1.5 ×1
the larger length is larger ×1
the length really solves the equation ×1
the line for root −1 passes through the point ×1
the line for root 0 touches the parabola at p = 0 ×1
the marked pair has two different roots ×1
the missing corner is (b/2)² ×1
the negative root solves it too ×1
the order x, y, z meets no zero pivot ×1
the order x, z, y meets no zero pivot ×1
the order y, x, z meets no zero pivot ×1
the order y, z, x meets no zero pivot ×1
the order z, x, y meets no zero pivot ×1
the order z, y, x meets no zero pivot ×1
the picture is about positive lengths ×1
the pivots multiply to the determinant ×1
the positive root is the one a length can be ×1
the product is a constant times one bell ×1
the product is narrower than either factor ×1
the rings add to π ×1
the sampled general share agrees with (41 + 6 ln 2)/72 ×1
the sampled top of the exponent is at b/2a ×1
the second point is above the curve and has none ×1
the sign of the discriminant says how many roots x² − 2x − 3 has ×1
the square tile is x² ×1
the strips are bx however they are cut ×1
the three pieces and the corner make a square of side x + b/2 ×1
the two pieces are what is left of a² after b² is removed ×1
the x first coefficients multiply to the determinant ×1
the x first completion has one square of each sign ×1
the y first coefficients multiply to the determinant ×1
the y first completion has one square of each sign ×1
two to four members of the family ×1
which is completing the square, as arithmetic ×1
which is the identity ×1
x² + 3x + 1 lies below the curve and has that many real roots ×1
x² + x − 2 lies below the curve and has that many real roots ×1
x² − 2x + 1 lies on the curve and has that many real roots ×1
x² − 2x + 3 lies above the curve and has that many real roots ×1
x² − 2x − 3 lies below the curve and has that many real roots ×1

Where it is called

Every figure on this list is drawn by the same rule, so a change to the rule changes all of them at once. That is why the list is published.

Algebra

Completing the square, by completing a square

The step everybody is taught as an algebraic trick is a literal instruction about a literal square. There is a corner missing, its size is forced, and paying for it is the whole method.

Algebra

One number under every bell

The area under e^(−x²) has no formula in terms of the usual functions, and yet the area under e raised to any downward quadratic is known exactly. Completing the square in the exponent moves and squeezes every such curve into the same one, so a single number — √π — pays for all of them.

Algebra

Regression is a square completed halfway

A bell in two variables, sliced at a fixed x, is a bell in y — and completing the square in y alone says where it is centred: at ρx, on a line flatter than the ellipse's own axis. That gap is regression to the mean, and the same completion says it runs backwards in time just as well as forwards.

Algebra

The signs no completion can change

A quadratic in several variables can be completed square by square in many orders, and each order hands back different coefficients. What no order changes is how many come out positive and how many negative — and that count, read off a completion of A − tI, says how many eigenvalues lie below t without finding a single one.

Algebra

What the coefficients already know

Finding the roots of a polynomial is hard and often impossible in closed form. Reading off their sum, their product and how many of them are real is none of those things — those numbers are sitting in the coefficients, and no root-finding is required to get at them.

Algebra

Where two roots run into each other

Put every quadratic equation at a point of a plane, one coordinate per coefficient. Each possible root becomes a straight line there, every one of those lines touches the same parabola, and that parabola is the discriminant — the crease where the plane of roots is folded onto the plane of equations.

The whole library · What the figures prove