What the figures prove
A dissection whose pieces overlap by two pixels looks exactly as convincing as one that works. A graph captioned as having an Eulerian circuit looks exactly like one that has an Eulerian circuit. The reader cannot tell, and for a long time neither could this site — a figure here carried a printed vertex degree that was simply wrong, through every gate, from the day it was written.
So the assertions are part of the drawing machinery, and the generators call them while drawing. A dissection must tile its target with no overlap, no gap and nothing outside it; congruent pieces must really be congruent; a quoted number must be computed from the drawing rather than typed beside it. A figure that does not prove its caption refuses to be drawn at all.
116 of 116 generator families reach an assertion. Between them they make 9189 distinct kinds of claim, tested 54691 times in the build that produced this page.
Every line below was collected by running each family at the parameters the essays actually use and recording what it asserted. A family that stops checking something loses its line here; a claim whose wording changes changes here with it. That is the only way this page is worth anything: a hand-written list of what code does drifts in exactly one direction, which is towards claiming more than exists.
What it does not say: that the claim is the right claim. An assertion compares a drawing against the arithmetic that produced it, and when a generator is handed an argument outside the range its picture means, both are wrong in the same way and they agree. A separate check covers that, and it still names families that would accept one.
Geometry
13 families
circle-angle
117 kinds of claim · 52 placements
- at x = 0.00 the signed product is d² − r² ×21
- at x = -9.00 the signed product is d² − r² ×20
- the distance between points 1 and 2 is the fraction 3/5 ×10
- point 1 of the nine is the same distance from the centre ×9
- the chord for inscribed angle 10° is 2R sin 10° ×9
- the chord at 0° splits into pieces whose product is r² − d² ×4
- the half-chord over the split 2 + 8 has square 16 ×4
- the apex at 120° sees AB at the same angle ×3
- the secant at 180° has near × far = 144 ×3
- a corner of a piece is α+60° ×2
- a corner of a piece is β+60° ×2
- a corner of a piece is γ+60° ×2
- each third at A is 26° ×2
- each third at B is 18° ×2
- each third at C is 16° ×2
- the assembled triangle's angle at A is 78° ×2
- the assembled triangle's angle at B is 54° ×2
- the assembled triangle's angle at C is 48° ×2
- a corner of a piece is 60°
- a corner of a piece is α
- a corner of a piece is β
- a corner of a piece is γ
- A lies on the circle
- a triangle whose incentre is on the Euler line has two equal sides
- a trisector and the far side are not parallel
- AB = cos α
- abc/4K is the radius of the circle through the corners
- AD = cos β
- and its second and third
- and its y coordinate
- and PA·PB = PC·PD
- and so do the other two
- and so is the nine-point centre with two of them
- and that distance is half the radius of the circle through the corners
- and the angle at C in the diameter's triangle is right
- and the centroid sits twice as far from the orthocentre as from the circumcentre
- and the incentre stays off the line on every scalene triangle in it
- and the same construction on the other trisectors is not
- and the same on the other side
- and the tangent meets the radius at a right angle
- and their radii differ
- and touching each escribed one
- B and D lie on the circle with diameter AC
- B lies on the circle
- BC = sin α
- BD = sin(α + β): a chord of a unit-diameter circle is the sine of the angle it subtends
- CD = sin β
- each side over the sine of the opposite angle is the diameter
- eighteen of the twenty-seven are equilateral
- equal products put D on the circle through A, B and C
- equilateral exactly when the three choices do not add to 2 mod 3
- every apex on the arc gives the same angle
- every side of every equilateral one makes (B − C)/3 with the base, give or take 60°
- every triangle in the sweep has its nine-point circle touching its inscribed one
- every triangle in the sweep puts its nine points on one circle
- every trisector triangle in the sweep is equilateral
- one angle, not several
- opposite angles add to a straight angle
- Ptolemy: the diagonals' product is the sum of the products of opposite sides
- so BC / sin A is the diameter
- so do the angles at C and B
- the angle at D stands on the same arc BC as the angle at A
- the angle between the chord and the tangent equals the inscribed angle
- the angle on a diameter is a right angle
- the angles add to 180°
- the angles at A and D stand on the same arc and are equal
- the apex is on the major arc
- the apex lies on the circle
- the centre's angle is twice the apex's
- the chord is a diameter
- the circumcentre, centroid and orthocentre are collinear
- the circumcentre, the centroid and the orthocentre lie on one line
- the collinearity holds across the sweep
- the exterior angle is twice the base angle
- the incentre is not on that line, which the determinant reports rather than being told
- the inner triangle's first two sides are equal
- the lines from A and B are not parallel
- the lines from B and C are not parallel
- the lines from C and A are not parallel
- the nine-point centre is the midpoint of that line
- the nine-point circle touches the escribed opposite A circle from outside
- the nine-point circle touches the escribed opposite B circle from outside
- the nine-point circle touches the escribed opposite C circle from outside
- the nine-point circle touches the inscribed circle from inside
- the point is inside the circle
- the quadrilateral's angles add to 360°
- the shape sweep is between 6 and 60 steps on a side
- the side is 8R sin α sin β sin γ
- the side opposite A makes (B − C)/3 with BC
- the splitters at A and B are not parallel
- the splitters at B and C are not parallel
- the splitters at C and A are not parallel
- the sweep covers a decent share of the shape space
- the sweep found enough non-degenerate triangles to mean something
- the sweep found enough triangles
- the sweep found enough triangles to mean something
- the sweep runs over between 20 and 2000 triangles
- the tangent length squared is the same number
- the tangent ray on the far side of the chord makes the inscribed angle with it
- the three angles are the angles of a triangle
- the three angles are those of a triangle
- the three angles are those of a triangle and none is a sliver
- the triangle has three corners and some area
- the triangles are similar, so PA/PD = PC/PB
- the two circles are not concentric
- the two halves are the whole angle
- the two sides agree at every angle
- the two sides of a piece are not parallel
- the view is one the family draws
- the weights for the centroid give its x coordinate
- the weights for the circumcentre give its x coordinate
- the weights for the incentre give its x coordinate
- the weights for the nine-point centre give its x coordinate
- the weights for the orthocentre give its x coordinate
- triangle OPA is isosceles
- triangle OPB is isosceles
- unequal products put D off it
conic
91 kinds of claim · 38 placements
- the two curves cross at a right angle at (0.543, 0.919) ×28
- the point found by straightedge along line 1 lies on the conic through the five ×18
- the ellipse with semi-axis 1.55 has the same foci ×9
- the hyperbola with semi-axis 0.42 has the same foci ×8
- line 1 is tangent to the ellipse ×6
- the solved conic passes through point 1 ×5
- every point of the 0.55 curve keeps the ratio ×3
- the foci are between 0.3 and 2.5 from the centre ×2
- a circle needs a cut that produces one
- a ellipse needs a cut that produces one
- a hyperbola needs a cut that produces one
- a parabola needs a cut that produces one
- all but a few of the lines give a finite construction
- and each point is as far from the focus as from the directrix
- and its sixth point lies on the conic
- and on its hyperbola
- and on the hyperbola
- and reaches it going forwards, not backwards
- and so they are for twenty-four other pairs
- and the image of Y on the one through Y
- and the other corner along the other hyperbola
- and the two distances differ by the same amount wherever it strikes
- and the two legs together are the same length wherever it bounces
- and travels towards it rather than away from it
- between 3 and 60 rays are traced
- between 6 and 60 lines are constructed
- between one and four curves
- each corner lies on its ellipse
- each curve is a name and six finite coefficients
- each ratio is between nothing and 2.4
- each vertex lies on the ellipse
- every ellipse's semi-axis is longer than the focal distance
- every hyperbola's semi-axis is shorter than the focal distance
- every pair of curves was checked at its crossing
- every parallel ray reflects to the focus
- five points in general position determine one conic
- five points, each within four units of the centre
- for any two points of the first ellipse, the distances across the stretch are equal
- moved off the ellipse, some corner decisively breaks the alignment
- neighbouring tangents meet at a finite corner
- no three of the five points are nearly collinear
- sides one-two and four-five meet at a finite point
- sides three-four and six-one meet at a finite point
- sides two-three and five-six meet at a finite point
- six distinct points of contact, given by their angles in degrees
- six distinct points, given by their angles in degrees
- the construction fits a drawable window
- the crossed distances across the stretch are equal
- the crossing lies on the ellipse
- the curve named a hyperbola really is one
- the curve named a parabola really is one
- the curve named an ellipse really is one
- the cutting plane is tipped by less than a quarter turn
- the discriminant and the count of crossings with a large circle agree that a hyperbola is a hyperbola
- the discriminant and the count of crossings with a large circle agree that a parabola is a parabola
- the discriminant and the count of crossings with a large circle agree that an ellipse is an ellipse
- the drawn construction is finite
- the drawn line's direction is between nought and a half turn
- the ellipse has a longer axis and a shorter one
- the ellipse's semi-axes are between 0.3 and 4
- the hexagon's corners fall near enough to draw
- the hyperbola has both semi-axes positive
- the image of X lies on the confocal hyperbola through X
- the incoming ray has a length to normalise
- the mirror is an ellipse, a parabola or a hyperbola
- the normal to the curve has a length to normalise
- the order visits each of the six points once
- the outgoing ray has a length to normalise
- the point is on the upper half of the curve
- the radius to the first focus has a length to normalise
- the radius to the second has a length to normalise
- the ratio is between nothing and 2.4
- the reflected ray lies on the line through the near focus
- the reflected ray passes through the other focus
- the stretch carries a corner of the first ellipse along its hyperbola to the second
- the tangent direction has a length to normalise
- the tangent makes the same angle with both focal radii
- the three construction points lie on the Pascal line by construction
- the three diagonals pass through one point
- the three meeting points fall near enough to draw
- the three meeting points of opposite sides lie on one line
- the three values give three different kinds of curve
- the two diagonals of the curved quadrilateral are equal
- the view is one the family draws
- the window is between 1 and 8 wide
- tilted off the ellipse, some tangent decisively breaks the concurrency
- two clearly different points on the inner ellipse
- two ellipses, the second larger, both longer than the focal distance
- two hyperbolas, the second wider, both shorter than the focal distance
- with an outer ellipse of the same shape but different foci, the crossed distances differ
- with the second ellipse and hyperbola on different foci the diagonals differ
euclid
28 kinds of claim · 22 placements
- the smallest square has side gcd(34, 13) ×4
- the squares tile the 34 by 13 rectangle ×4
- a ≥ b ≥ 1 are whole numbers, a at most 90
- a ≥ b ≥ 1, a at most 12
- a and b are read by the tiling and the two game views
- and with the rule that the first run longer than one decides
- at is a short list of whole points
- at is read only by the sturmtable view
- consecutive Fibonacci positions alternate between win and loss
- every interval the count isolates holds exactly one real root found separately
- N is from 12 to 60
- N is read only by the gamegrid view
- poly is read only by the sturm, sturmtable, isolate views
- range is an interval [lo, hi]
- range is read only by the sturm view
- the count drops by exactly one across each real root
- the drawn verdict is the searched one
- the drop from −∞ to +∞ is the number of distinct real roots
- the halving isolates every real root
- the leftover strip is still there after every pass
- the polynomial is one of the named examples
- the search agrees with the golden-ratio rule
- the squares peeled at each pass are the continued fraction of the ratio
- the verdict follows the side of φ the ratio falls on
- the view is one the family draws
- upTo is from 5 to 11
- upTo is read only by the gamefib view
- V(lo) − V(hi) equals the number of distinct real roots found separately
figurate
29 kinds of claim · 19 placements
- the polynomial for power 1 gives the exact sum at n = 1 ×150
- 0 and 3: triangular triples match odd-square triples ×121
- for power 2 the remainder approaches m/12 ×4
- the first 6 odd numbers sum to 6² ×4
- a range from 60 to 400 ×2
- a number from 1 to 300
- a range from ten thousand to two million
- a triangular index from 1 to 7
- an even index from 10 to 40
- and it is the centre
- between 3 and 12 steps
- every number is a sum of three triangular numbers
- every odd Bernoulli number past the first is zero
- Faulhaber's polynomial gives the staircase's total
- n is two triangular numbers exactly when 4n + 1 is a sum of two squares
- only the centre cell is left over
- powers from 2 to 7
- powers up to between 3 and 8
- share × √(log x) changes little over two decades
- the kind is gnomon or one of triangular, galileo, staircase, faulhaber, correction, bernoulli, eight, trisum, twotri, twotridensity, threetri
- the L-shaped shells fill the square
- the leading coefficient is 1/(m+1), the integral's
- the next is 1/2, half the last step
- the power is between 1 and 6
- the share keeps drifting down
- the signs alternate, positive at 2, 6, 10, …
- the size matches 2(2k)!/(2π)^(2k)
- the staircase stands above the curve by about half its last step
- the three triangles add to the number
golden
46 kinds of claim · 15 placements
- generation 0: the counting rule agrees with the drawn subdivision on thin halves ×7
- … and one thick half
- … and two thick halves
- −2cos 144° = φ
- √2's takes it out twice, after the first step
- 1 to 7 generations
- 1 to 8 cuts
- 2cos 72° = 1/φ
- 2cos 72° is a root of w² + w − 1
- 4 to 14 subtraction steps
- 8/5's stops, because a common unit exists
- a thick half becomes one thin half …
- a thick half has a 108° apex
- a thin half becomes one thin half …
- a thin half has a 36° apex
- and as long as the piece of the side it cuts off
- and it climbs towards φ from below at every cut
- and on thick ones
- and so is 2cos 144°
- diagonal minus side is the smaller pentagon's diagonal
- diagonal over side is φ in every pentagon
- each inner pentagon is smaller by φ²
- each step shrinks the pair by φ
- each subtraction leaves something smaller than what it subtracted
- in the largest disc the ratio of thick to thin is within 0.05 of φ
- leg over base is φ
- one square per Fibonacci number
- one to five levels
- the bisector is as long as the base
- the chord to the second neighbour over the chord to the first is φ
- the convergents alternate about φ
- the diagonal's end piece is s/φ
- the last convergent is close to φ
- the long side is the next Fibonacci number
- the number of convergents plotted is between 2 and 20
- the number of whirling squares is between 1 and 12
- the pair at every step is in the ratio φ
- the pieces fill their parent exactly
- the ratio has converged to φ
- the rectangle is nearly golden
- the short side is the one before it
- the thick half built here has its 108° apex
- the view is one the family draws
- the whirling squares tile their rectangle
- φ's subtraction always takes the smaller out exactly once
inversion
75 kinds of claim · 30 placements
- the circles round vertex 0 subtend exactly a full turn at it ×7
- ring 1: and the circle it surrounds from outside ×2
- ring 1: and touches the containing circle from inside ×2
- ring 1: each circle touches the next ×2
- a circle missing the centre inverts to a circle
- a circle orthogonal to the mirror is carried to itself
- a circle through the centre inverts to a line
- a point inside goes outside and a point outside comes in
- a point on the mirror circle is left exactly where it is
- and a line missing the centre inverts to a circle
- and a point of the circle lands on it
- and conjugates the imaginary part
- and every edge of the triangulation is a tangency of two circles
- and every point of the straight line lands on its circle
- and its curvature is a whole number
- and one for each of the eight ways of choosing inside or outside
- and so do their images, because a circle's image is a circle
- and the two short gaps overshoot
- and touches each of them on the page
- between two and four points are inverted, none of them at the centre
- both images are circles here
- centre and all
- circles whose vertices are not joined do not overlap
- each drawn circle really touches each given one
- each image gap is the original gap divided by the two distances from the centre
- each new circle satisfies Descartes's relation with its three parents
- every curvature in the packing is a whole number
- every pair of the seed circles touches
- every point of the circle lands on its image
- every point of the circle through the centre lands on its image
- every walk round an interior vertex closes on the circles already placed
- four points on one circle have a real cross-ratio
- four points, in order round the circle and none of them nearly coincident
- in both parts
- neither boundary passes through the inversion centre
- no chain circle passes through the inversion centre
- no point being inverted sits at a mirror's centre
- no point being inverted sits at the centre
- off the circle the identity becomes a strict inequality
- off the circle the three images are not collinear
- on a circle the products satisfy Ptolemy's identity exactly
- one inversion leaves the real part of the cross-ratio alone
- so the two short image gaps add to the long one
- the angle between the two curves is the same after the map as before
- the chord's end sits on the mirror
- the circle's image is a circle
- the concentric ring closes exactly when the radius ratio is the sine
- the constructed point is the one inside the mirror
- the drawn ring is genuinely lopsided rather than a disguised concentric one
- the first mirror has a sensible radius
- the four points are not concyclic, so their cross-ratio has a real imaginary part to conjugate
- the fourth point is pushed off the circle by less than half a radius
- the gasket is drawn to between one and six generations
- the image of the centre is not the centre of the image
- the image of the crossing point is on both image circles
- the inner circle has a sensible radius
- the inversion centre is inside the inner circle and off its centre
- the line inverted does not pass through the centre
- the mirror circle has a sensible radius
- the ring has between three and twelve circles
- the ring round the centre closes up
- the second mirror is offset by a sensible amount
- the seed satisfies Descartes's relation
- the tangent construction lands on the same point the formula gives
- the three centres are not collinear
- the three given circles lie outside one another
- the three images fall on one straight line
- the two circles really cross
- the two distances from the centre multiply to the square of the radius
- the view is one the family draws
- three boundary radii, repeated round the outside
- three circles in general position admit exactly eight tangent circles
- three given circles, each with a sensible radius
- two inversions put the cross-ratio back exactly
- which passes through the centre of the mirror
iso
108 kinds of claim · 26 placements
- the 3-gon obeys Bonnesen's inequality ×10
- 2 pieces against a wall do exactly as well as a 4-gon does in the open ×7
- the 3-gon encloses less than a circle of the same perimeter ×7
- the 3-gon is drawn at the same perimeter as the rest ×7
- regular 3-gon does not beat the circle ×5
- the arc in the 60° corner is 300 long ×4
- 6 reflected copies of the 60° sector make a whole disc ×3
- the 2-piece fence is 300 long ×3
- the 2-piece fence's share of the half-disc is the 4-gon's quotient ×3
- 2:1 rectangle does not beat the circle
- a pentagon symmetrised has more than five corners
- a quotient from 0.9 to just below 1
- a triangle symmetrised about a side's perpendicular is a triangle
- a unit cube has six units of face
- a unit square cell needs two units of wall of its own
- after 60 it is nearly a disc
- an L does not beat the circle
- and every one holds less than the half-disc
- and holds twice the area
- and is still short of the circle's
- and it is the circle
- and its area is a quarter of the perimeter, squared
- and never lengthens the boundary
- and shortens the boundary
- and the area is strictly larger
- and the best angle between the pieces is a right angle
- and the circle's quotient is exactly one
- and the ring's square like the inverse fourth power
- and the sector holds the fence squared over twice the angle
- between 10 and 80 cells
- between 50 and 400 random shapes
- between one and four fences of 2 to 12 pieces
- each round of symmetrisation brings the shape nearer round
- each structure improves on the one before
- even the best rectangle falls short of the circle
- every extra side encloses strictly more
- every random convex shape obeys Bonnesen's inequality
- every shape with that quotient has the same allowance
- every step keeps the area
- exactly one shape in the list reaches one, and it is the circle
- fences of up to between 4 and 12 pieces
- five steps bring the triangle close to equilateral
- four corner angles from 30 to 180 degrees
- hexagons need the least wall, triangles the most
- more pieces hold more
- more sides always need less wall
- no cell beats the regular polygon with its number of sides
- reflecting the dent leaves the perimeter exactly as it was
- Reuleaux triangle does not beat the circle
- so the fence holds less than its length squared over 2π
- so the quotient rises
- so their average wall is at least the hexagon's
- symmetrisation keeps the area
- the 12-gon's circumradius is found
- the 12-gon's inradius is found
- the 12-gon's ring is no wider than its deficit allows
- the averaged pair is never longer
- the best split is an even one
- the bump is tuned to hold 0.99 of the circle's area
- the bump's ring is no wider than its deficit allows
- the cells average six sides
- the cells cover the square exactly once
- the circle does not beat the circle
- the circle of that perimeter encloses more than any of the polygons drawn
- the deficit falls like the inverse square of the number of sides
- the dent is between nothing and 160 units deep
- the doubled curve is twice the fence
- the doubled curve obeys the closed-curve inequality
- the ellipse is tuned to hold 0.99 of the circle's area
- the ellipse's circumradius is found
- the ellipse's inradius is found
- the ellipse's ring is no wider than its deficit allows
- the hexagon's circumradius is found
- the hexagon's inradius is found
- the hexagon's ring is no wider than its deficit allows
- the largest rectangle in the sweep is the square
- the polygons drawn have between 3 and 60 sides
- the reuleaux's circumradius is found
- the reuleaux's inradius is found
- the reuleaux's ring is no wider than its deficit allows
- the shared perimeter is between 60 and 900 units
- the sides of the cells add to exactly six per cell
- the square's circumradius is found
- the square's inradius is found
- the square's ring is no wider than its deficit allows
- the stadium is tuned to hold 0.99 of the circle's area
- the stadium's circumradius is found
- the stadium's inradius is a root of the quadratic
- the stadium's inradius is found
- the stadium's ring is no wider than its deficit allows
- the sweep takes between 20 and 2000 samples
- the triangle's circumradius is found
- the triangle's inradius is found
- the triangle's ring is no wider than its deficit allows
- the view is one the family draws
- turning directions approach a disc
- two fixed directions stop short, at a shape both leave alone
- two to four shapes
- which holds an eighth of the fence squared — half of a square
- whose two sides are equal
- πρ² − Lρ + A is not positive at the 12-gon's two radii
- πρ² − Lρ + A is not positive at the bump's two radii
- πρ² − Lρ + A is not positive at the ellipse's two radii
- πρ² − Lρ + A is not positive at the hexagon's two radii
- πρ² − Lρ + A is not positive at the reuleaux's two radii
- πρ² − Lρ + A is not positive at the square's two radii
- πρ² − Lρ + A is not positive at the stadium's two radii
- πρ² − Lρ + A is not positive at the triangle's two radii
ordinary
56 kinds of claim · 28 placements
- the even construction on 6 points has n/2 ×8
- Melchior's inequality holds for 8 random points of a 4 × 4 grid (1) ×3
- a line meets a cubic in at most three points
- a line with three points on it yields a strictly smaller distance
- a polygon of 3 to 10 sides
- a regular m-gon's chords point in exactly m directions
- and Csima and Sawyer's
- between four and twelve points on the curve
- Böröczky's set has exactly n/2 ordinary lines
- de Bruijn and Erdős: at least as many lines as points
- each pair of points lies on one connecting line only
- every construction respects Kelly and Moser's bound
- every family has at least as many lines as points
- every line through P either vanishes (if ordinary) or survives without P
- every point on a connecting line gives a dual line through its crossing
- every set with exactly as many lines as points is a near-pencil
- every triple is either collinear or clearly not — no decision rests on rounding
- every zero-sum triple of whole numbers is a row
- Melchior's inequality holds for a triangle, its midpoints and its centroid
- Melchior's inequality holds for five in a line and one off it
- Melchior's inequality holds for four points in general position
- Melchior's inequality holds for the nine-point grid
- Melchior's inequality holds for three points
- Melchior's inequality holds on every configuration
- no configuration beats one row per three pairs
- no configuration searched avoids ordinary lines altogether
- no set searched has fewer lines than points
- no two dual lines are parallel after the turn
- removing an end of an ordinary line removes that line entirely
- some connecting line holds exactly two of the points
- some point is off some connecting line
- some set of this size is not all in one line
- the census runs to between four and eight points
- the closest point-and-line pair uses a line with exactly two points
- the connecting lines between them account for every pair of points
- the cyclic group on the cubic meets Sylvester's count at every n
- the distance the argument produces is strictly the smaller one
- the drawn witch has the counted rows
- the five-by-five grid has 140 connecting lines
- the grid searched is three or four across
- the grid searched is three or four points across
- the highlighted triple adds to zero modulo n
- the highlighted triple is three point indices
- the ordinary line chosen exists
- the point set is one this family knows
- the points are drawn between 12 and 60 units apart
- the search runs to between four and seven points
- the set has an ordinary line to remove a point from
- the set is not all in one line
- the smaller set, not in one line, already has as many lines as points
- the three-point lines are the zero-sum triples of the cyclic group
- the view is one the family draws
- the window runs from −k to k, k between 2 and 5
- the witch never beats the known maximum
- three points of the witch are in line exactly when their indices add to zero modulo n
- three points of y = x³ + ax are in line exactly when their x add to zero
polyhedron
81 kinds of claim · 43 placements
- the order-4 axes carry 9 turns between them ×7
- a turn of order 3 sits on an axis of order 3 ×5
- and each such axis carries a turn of order 4 ×4
- 3 triangles at a vertex leave a gap ×3
- 5-cell: and the edge count agrees with that degree ×3
- 5-cell: every vertex is the same distance from the centre ×3
- 5-cell: every vertex meets the same number of edges ×3
- 3 hexagons at a vertex cannot leave a gap
- 3 pentagons at a vertex leave a gap
- 3 squares at a vertex leave a gap
- 4 pentagons at a vertex cannot leave a gap
- 4 squares at a vertex cannot leave a gap
- 6 triangles at a vertex cannot leave a gap
- a star polygon has between 5 and 13 points
- and 32 − 90 + 60 = 2, the Euler number of any sphere
- and at the depth found every edge is the same length
- and every one of them is the same triangle
- and sends every corner to a corner
- and steps by at least 2, less than half of p
- and steps by between 2 and 6
- and the angle at each point is π − 2πq/p
- as a surface it has thirty-two corners
- cuboctahedron: and lies on a circle, so it is regular
- cuboctahedron: and the walk round a corner visits every face at it once
- cuboctahedron: each face has equal sides
- cuboctahedron: Euler's formula still holds
- cuboctahedron: every edge is the same length
- cuboctahedron: every vertex is surrounded by the same faces in the same order
- each candidate is a turn rather than a reflection
- each polytope is one this family builds: 5-cell, tesseract, 16-cell, 24-cell
- each solid is one this family cuts: truncated-tetrahedron, truncated-cube, truncated-octahedron, cuboctahedron, icosidodecahedron, truncated-icosahedron
- every turn has an axis to be about
- exactly five vertex figures close up
- icosidodecahedron: and lies on a circle, so it is regular
- icosidodecahedron: and the walk round a corner visits every face at it once
- icosidodecahedron: each face has equal sides
- icosidodecahedron: Euler's formula still holds
- icosidodecahedron: every edge is the same length
- icosidodecahedron: every vertex is surrounded by the same faces in the same order
- no neighbour lies straight along the corner's own axis
- read as twelve pentagrams meeting five at a point, the number is −6
- tesseract: and the edge count agrees with that degree
- tesseract: every vertex is the same distance from the centre
- tesseract: every vertex meets the same number of edges
- the axes account for every turn, the one that does nothing included
- the corner polygon's edge starts shorter than the remaining edge and ends longer
- the faces at a corner close into a ring
- the number of distinct turns is twice the number of edges
- the path visits every point exactly once
- the path winds about the centre q times
- the projected solid has a measurable extent
- the solid is one of the five
- the spiky surface is sixty triangles
- the star has between 5 and 13 points
- the step and the number of points share no factor, so the path closes once
- the turn in four dimensions is at most half a circle
- the view is one the family draws
- truncated-cube: and lies on a circle, so it is regular
- truncated-cube: and the walk round a corner visits every face at it once
- truncated-cube: each face has equal sides
- truncated-cube: Euler's formula still holds
- truncated-cube: every edge is the same length
- truncated-cube: every vertex is surrounded by the same faces in the same order
- truncated-icosahedron: and lies on a circle, so it is regular
- truncated-icosahedron: and the walk round a corner visits every face at it once
- truncated-icosahedron: each face has equal sides
- truncated-icosahedron: Euler's formula still holds
- truncated-icosahedron: every edge is the same length
- truncated-icosahedron: every vertex is surrounded by the same faces in the same order
- truncated-octahedron: and lies on a circle, so it is regular
- truncated-octahedron: and the walk round a corner visits every face at it once
- truncated-octahedron: each face has equal sides
- truncated-octahedron: Euler's formula still holds
- truncated-octahedron: every edge is the same length
- truncated-octahedron: every vertex is surrounded by the same faces in the same order
- truncated-tetrahedron: and lies on a circle, so it is regular
- truncated-tetrahedron: and the walk round a corner visits every face at it once
- truncated-tetrahedron: each face has equal sides
- truncated-tetrahedron: Euler's formula still holds
- truncated-tetrahedron: every edge is the same length
- truncated-tetrahedron: every vertex is surrounded by the same faces in the same order
pythagoras
70 kinds of claim · 46 placements
- a larger exponent gives a ball that contains the smaller one
- a marked third side is inside the range drawn
- a triple whose squares add gives a right angle
- a triple whose squares do not add gives an angle that is not right
- and below one it is not, which is why p < 1 gives no distance at all
- and every one the search found is in the tree
- and half the rectangle — same base, apex on the parallel through the altitude
- and has not appeared before in the tree
- and is primitive
- at nine dimensions the inner sphere reaches the cube's faces
- at ten it is outside them
- consecutive knots are one unit apart
- each fixed ball drawn behind is between 1 and 12
- each leg at the corner is a positive length
- each leg is between nothing and a quarter turn
- each leg of the triangle is a positive length
- each side is a positive length
- each side of the box is a positive length
- every node of the tree is a right triangle in whole numbers
- every point drawn is at distance one in this norm
- four triangles and the tilted square fill the left square
- four triangles and the two upright squares fill the right square
- so the square and the rectangle are equal
- the altitude cuts the big square into exactly these two rectangles
- the angle at C is a right angle
- the angle at the corner is a right angle
- the ball meets the diagonal at 2^(-1/p) in each coordinate
- the box has three positive sides
- the constructed triangle's hypotenuse is the root of the sum of the squares
- the corner is square exactly when the squares add
- the dimensions drawn run from 1 to between 4 and 40
- the drawn third side is c
- the eight triangles are all the same
- the exponent is between 0.4 and 12 — below 1 it is drawn as a warning
- the exponent of the norm is between 0.4 and 12
- the first right face is half the product of its legs
- the fixed balls behind it are between 1 and 12
- the floor diagonal meets the vertical edge at a right angle
- the given triangle and the constructed right triangle are congruent
- the hyperbolic excess in c² is a²b²/3 at small size
- the inner circle touches each corner circle
- the largest triangle drawn has legs under a quarter turn
- the legs are positive lengths
- the long leg is b
- the proof is one the family draws
- the rope carries one knot per unit of its length
- the rope's three sides are whole numbers of knots, each under sixty
- the second right triangle has the floor diagonal and the height as its legs
- the sensitivity at the right angle is c/ab radians per unit
- the short leg is a
- the slanted face's area squared is the sum of the three right faces' areas squared
- the space diagonal squared is the sum of the three squares
- the sphere shortens the hypotenuse and the hyperbolic plane lengthens it
- the spherical hypotenuse is shorter than the flat one
- the spherical hypotenuse satisfies cos c = cos a cos b
- the spherical shortfall in c² is a²b²/3 at small size
- the square on the hypotenuse is c²
- the square on the leg has the leg's area squared
- the three faces at the corner meet at right angles
- the three lengths close into a triangle
- the tilted square equals the two upright squares
- the tilted square is c²
- the tree finds exactly as many primitive triples under 200 as the search does
- the tree is checked against a search up to a hypotenuse between 50 and 2000
- the tree is drawn to between 1 and 3 generations
- the tree of triples is drawn to between 1 and 3 generations
- the triangle is half the square — same base, apex on the opposite side
- the two floor edges meet at a right angle
- the two triangles are the same triangle turned
- with p at least one, the midpoint of any two points of the ball is inside it
reuleaux
51 kinds of claim · 25 placements
- the 3-sided curve is built at the common width ×4
- the 3-sided Reuleaux curve has one width at every angle ×4
- the integral of h round a 3-sided curve is π times the width ×3
- a Reuleaux polygon needs an odd number of sides
- all the shapes have one width
- and is convex
- and it holds for the solid of revolution, measured
- and it is the same in every direction
- and its radius of curvature stays positive, so it is convex
- and lands between the ball and the conjectured minimum
- and the circle's is a quarter of π times it
- and the drawn boundary really is that long
- and the drawn points confirm it, direction by direction
- and the largest is the edge times (√6 − 1)/√2, across two opposite curved edges
- and the Reuleaux triangle the least
- and the smooth family gets only part of the way to the triangle
- and the support function itself is not flat, so the flat sum says something
- and the sweep runs past the point where convexity fails
- and they do so in every direction, not only the one drawn
- at the width the function was given
- between one and four smooth amplitudes, each at most twelve
- between thirty and a hundred and sixty samples per angle
- between twenty and two hundred samples
- between two and five odd side counts, each at most eleven
- Blaschke's relation holds for the ball, which is the check that it is stated right
- convexity fails exactly where h + h″ first vanishes
- every arc bulges away from the centre
- every harmonic in the support function is odd
- every shape in the table has the same width
- ground and plank stay one width apart
- so the solid is not of constant width
- the area falls the whole way, so the minimum is on the constraint boundary
- the boundary is sampled between five hundred and twenty thousand times
- the built curve has constant width
- the circle encloses the most
- the circle is convex
- the four corners are equally spaced
- the generating curve has the stated width
- the harmonic is odd, between three and seven
- the least is about nine tenths of the most
- the plotted sum is flat at the width
- the smallest width is the edge itself
- the smooth curve has the common width
- the spun triangle encloses less than the ball
- the triangle's area is (π − √3)/2 times the square of the width
- the two support distances in opposite directions add to the width
- the view is one the family draws
- the width does not depend on how far it has rolled
- the width is positive
- the width is the arc radius
- though it misses by only a few per cent
unroll
20 kinds of claim · 17 placements
- and the circumscribed value is 6 tan(π/6) ×7
- and the inscribed value is 6 sin(π/6) ×7
- the 6-gons bracket pi ×7
- and along every ray the inner chord is exactly half the outer one while the areas are a quarter and three quarters
- and by ninety-six sides the bracket sits inside Archimedes' own two fractions
- and the whole ball is two of those
- at every height the dome's disc and the ring have the same area
- between three and seven doublings are tabulated
- each doubling tightens the bracket from both sides
- so the hemisphere's volume is the cylinder's less the cone's
- the areas are integrated at between four hundred and twenty thousand slices
- the drawn polygon has between three and twenty-four sides
- the rings account for the whole disc
- the shear leans by between a tenth and one and a fifth of the height
- the sheared stack has the same area
- the slice is taken strictly between the base and the top
- the triangle of base 2πr and height r has the disc's area
- the two regions together are the unit square
- the two solids are compared at between a hundred and four thousand heights
- the view is one the family draws
voronoi
50 kinds of claim · 32 placements
- the centre of cell 0 is nearest to site 0 ×20
- a planar triangulation of n points has at most 3n − 6 edges
- a point is under the chord exactly when it is inside the interval
- a site has a cell exactly when its lifted point is a corner of the lower hull
- a site inside the circle is nearer to both ends than the ends are to each other
- a spanning tree has one edge fewer than it has sites
- and at least enough to be connected
- and so the refuted edge is not in the shortest tree after all
- at least one pair of circles actually crosses
- at least one site has no cell, which is what the picture is about
- between 2 and 60 sites, each a point in or near the unit square
- by the largest count drawn, the pairs outnumber the triangulation's edges several times over
- each graph in the chain has no more edges than the next
- each panel is on the side of the threshold the figure claims
- every Delaunay triangle spans a face of the lower hull
- every edge of the minimum spanning tree is an edge of the triangulation
- every Gabriel edge is a Delaunay edge
- every Lloyd step lowers the total squared distance to the nearest site
- every lower face projects to a Delaunay triangle
- every nearest neighbour edge is a shortest tree edge
- every relative neighbourhood edge is a Gabriel edge
- every shortest tree edge is a relative neighbourhood edge
- every site keeps a cell
- every upper face's circumcircle contains every other site
- exactly one of the two triangulations is the Delaunay one
- more than half of the whole fall happens in the first two rounds
- most interior cells settle at six sides, by more than double the next count
- no point beats the centroid
- no triangle's circumcircle holds another site
- the bisector is where the two distances tie
- the cells account for the whole box, with nothing double-covered
- the cells still account for the whole box
- the chain is not a chain of equalities
- the circle on a tree edge as diameter holds no other site
- the circumcircle verdict and the lifted-hull verdict agree
- the cost falls at every round
- the edge chosen to be refuted really does have a site inside its circle
- the farthest-point triangulation uses only sites on the convex hull
- the gap between chord and parabola is −(x − a)(x − b)
- the highlighted cell is one of the sites
- the iteration has actually moved, rather than starting at its own fixed point
- the lift finds as many triangles as the circumcircle test does
- the middle cell is absent at the small weight and present at the large one
- the middle of each cell has that cell's site nearest in the power distance
- the moment about the site is the moment about the centroid plus area times the gap squared
- the settled cells are far closer to equal in area than the scatter's
- the settled diagram holds a larger share of six-sided interior cells
- the tested point is inside the interval, which is what makes the pair illegal
- the triangulation is not empty
- where two circles cross, the two power distances agree
Analysis
9 families
circle-to-sine
29 kinds of claim · 21 placements
- the 1:1 figure crosses itself 2pq − p − q times ×11
- at phase 1.178 the 1:2 figure has the generic crossing count ×7
- a coprime pair up to five
- a generic Lissajous figure crosses itself 2pq − p − q times
- a step up to 1 and two to six sectors
- and they stay on the hyperbola
- between three and seven offsets
- cosh and sinh are the even and odd halves of e^t
- every slice between neighbouring points has the same area
- the catenary's parameter is between 0.2 and 2
- the circle's parameter is between 0 and 2π
- the circular sector's area is half its parameter
- the folding phases really fold
- the grid runs to between 2 and 5
- the hyperbola's parameter is between 0 and 2.2
- the hyperbolic sector's area is half its parameter
- the offset is generic — the curve does not fold back on itself
- the parabola and the catenary really differ
- the point is on the circle
- the point is on the hyperbola
- the point's height is the wave's height there
- the range is between 1 and 3.2
- the ratio is between one and three
- the row shows both a folded and an open figure
- the time spent in each cell is close to the product of two arcsine densities
- the two coordinates are a point of the unit circle
- the view is one of sector, squeeze, waves, catenary, lissajous, lissgrid, lissphase, lissdense
- two whole frequencies up to six with no common factor
- x + y is e to the parameter
converge
88 kinds of claim · 30 placements
- so sin 2π1x and sin 2π2x differ by at least 1 somewhere ×90
- the mean square of sin 2π1x − sin 2π2x over [0, 1] is 1 ×90
- the grid never beats the stated modulus of sin(2πkx) at k = 1 ×40
- the grid never beats the stated modulus of sin(2πx + k) at k = 1 ×40
- the grid never beats the stated modulus of xᵏ at k = 1 ×40
- x − x²/2 + x³/3 − …: 1 terms stay within the bound ×30
- x − x³/3 + x⁵/5 − …: 1 terms stay within the bound ×30
- members 7 and 13 share an arc, so they are within 0.765 everywhere ×28
- the n = 3 approximant is 1/1 at 0/1 ×28
- at 0/2 every member is 1 ×10
- the sum of 1 terms is within 0.5000 of the function everywhere on [0, 1], x = 1 included ×10
- a member later than the last one picked lies in the arc at stage 0 ×9
- the two members agree at node 0/8 ×9
- 1 − 2x + 3x² − 4x³ + … at x = 1/2 is the function it is said to sum to
- 1 − x + x² − x³ + … at x = 1/2 is the function it is said to sum to
- 1 − x + x² − x³ + …: the largest gap on [0, 1] never drops below 1/2
- 2 arctan t − t solves y′ = cos(t + y)
- 4 to 12 nodes
- a member fits inside the tube exactly when the convergence is uniform
- and between the nodes they are nearly 2 apart
- and comes within two hundredths of it
- and N is the first such member
- at an irrational point the approximants settle towards 0
- at x = 1 the partial sums stay away from the function's value there
- between 16 and 400 members
- between 4 and 10 halvings
- between the multiples of 1/m! the members fall to 0
- between two and six levels of cutting
- between two and six member numbers, each a whole number up to 400
- each column of the picture settles on the limit
- each point is passed by the bump and left behind
- every level covers each point exactly once
- every shifted wave has the same modulus at every k
- every step has slope between −1 and 1
- finer polygons sit closer to the solution
- from N on, the members are inside the band off the strip
- no point of the interval beats the stated largest gap
- one half holds at least half the members
- one to four members are drawn, from inside the table
- one to three approximants, each with n from 2 to 40
- one to three sequences this family draws
- one to three values of m, each from 1 to 4
- refining the grid finds a larger gap, because the largest is reached nowhere
- term counts up to between 8 and 60
- the bars run to between 20 and 80
- the circle of phases is cut into 4 to 16 arcs
- the fullest arc holds at least its share
- the gap to the previous pick is within the previous arc's bound
- the grid finds the largest gap, because it is reached at a point
- the intervals have total length below ε
- the largest gap for x / n falls with n
- the largest gap for xⁿ does not fall with n
- the largest gap for xⁿ never drops below where it started
- the largest gap shrinks exactly when the convergence is uniform
- the members are given in increasing order
- the members considered number 500 to 20000
- the members run to between 10 and 60
- the partial sums at x = 1 do not settle
- the points jumping by at least 1/5 are the fractions with q ≤ 5
- the power family's modulus rises with k
- the removed length is the geometric sum
- the search for a member inside the tube runs to between 20 and 2000
- the second frequency is one more than a multiple of the node count, so the two agree at every node
- the sequence is one this family draws
- the series is one this family draws
- the sine family's modulus reaches 2
- the stated gap between two members is the one on the grid
- the stretch δ is between 0.02 and 0.2
- the sup distance is plotted for between 6 and 80 values of n
- the table of gaps runs to between 4 and 12
- the threshold is 1/k for k from 2 to 12
- the time runs to between 1 and 6
- the tolerance is between 0 and a half
- the tube has a half-width between 0.02 and 0.6
- the upper sums of a function with finitely many spikes fall like 1/N
- the view is one the family draws
- the weight on partial sums equal to 1 is 1/(1 + x)
- the weights add to one, the unshown tail included
- two or three series
- two to five levels
- two to five step counts, increasing, up to 256
- two to six term counts, increasing, up to 60
- two values of x between 0.3 and 0.995, increasing
- which is the function the series sums to
- x − x²/2 + x³/3 − … at x = 1/2 is the function it is said to sum to
- x − x²/2 + x³/3 − …: the largest gap on [0, 1] falls away
- x − x³/3 + x⁵/5 − … at x = 1/2 is the function it is said to sum to
- x − x³/3 + x⁵/5 − …: the largest gap on [0, 1] falls away
convex
66 kinds of claim · 37 placements
- A has three to eight corners
- A's corners are all on its own hull, so the set is convex
- and a function that is not convex fails the same test, so the test has content
- and all but a handful of borderline cases are one or the other
- and every corner of the second strictly on the other
- and never more than three
- and positive against the target, which is the refutation
- and the other has more than one
- and the second is outside it
- and with every three assumed, the whole family has a common point
- B has three to eight corners
- B's corners are all on its own hull, so the set is convex
- between ninety and thirty-six hundred directions swept
- between one and six targets
- between six and twenty points in the scatter
- between twenty and four thousand random quadruples
- between two and eight values are averaged
- between two and seven points are averaged
- between two and six touch points, all inside the interval
- both ends of the chord are inside the interval drawn
- every corner of the first set is strictly on one side
- every point is inside the interval drawn
- every point of the hull lies in the hull of three of the points
- every three of the family have a common point
- every two of the family have a common point
- every value is positive and inside the interval drawn
- no line in any of the 720 directions separates the two sets
- no vector is both a non-negative combination and refuted by a direction
- one positive weight per point
- so a refuting direction exists for the second
- some chord of this function dips below it
- the average of the values is at least the value at the average
- the average of the values is at most the value at the average
- the chord runs left to right
- the chord stays above the curve at every sampled point
- the chord stays below the curve at every sampled point
- the conjugate of x² is p²/4, found by search
- the convex function has exactly one local minimum on the interval
- the crescent's hull swallows its own mouth, so the crescent is not convex
- the curve stays above the tangent everywhere, not only nearby
- the direction is non-positive against every generator
- the first target is inside the cone
- the function drawn is a convex one
- the function is one this family draws
- the gap the line leaves is exactly the distance between the sets
- the geometric mean is at most the arithmetic mean
- the great majority of quadruples were non-degenerate
- the mean of the logarithms is at most the logarithm of the mean
- the mean of the logarithms names the geometric mean
- the scatter has a two-dimensional hull
- the search examined some pair of ends
- the search runs over between 20 and 200 candidate ends
- the small set lies outside the crescent
- the transform is drawn for a convex function
- the two generators are independent
- the two means agree exactly when every value is the same
- the two parts' hulls share the point the dependence produces
- the two sets are disjoint
- the view is one the family draws
- the weakened hypothesis is on or off
- the weights add to one
- transforming twice returns the original function
- two functions, side by side
- two generating vectors
- while the sweep does find a wall when one exists, so it is not simply blind
- with only the pairs assumed, the whole family can have nothing in common
exponential
33 kinds of claim · 26 placements
- so over 0.7702 exactly the factor is 2 ×4
- a base that is not e does not have slope equal to height
- and at base e the slope is the height
- and is what the closed form gives over that interval
- and the gap left is smaller than the last term added
- and the limit of the staircase is e itself
- and the place where that area is exactly 1 is e
- and the ratio falls towards 1 every step
- both factors are above 1 and their product fits
- both intervals lie inside the integrated range
- each solution starts at a positive height of at most 4
- every partial sum is below e
- more splits is closer to e
- no number of splits reaches e
- so by the end it is under one per cent
- so the two areas add to the area of the product
- the approximation is always an underestimate here
- the area is taken from 1 up to somewhere between 1 and 12
- the area under 1/x from 1 is the logarithm
- the base is a positive number of at most 20
- the curve integrated from the equation is the exponential
- the factor is the same wherever along the curve the interval starts
- the gap closes like one twelfth of 1/n, measured at the last row
- the growth rate is between a twentieth and three in size
- the interval over which the height changes by that factor is inside the window drawn
- the region from a to ab has the same area as the region from 1 to b
- the region runs from 1 to somewhere between 1 and 12
- the table runs to between 4 and 20
- the table shown gets e right to three decimals
- the tangent drawn is the curve's own slope
- the two factors are both above 1 and their product is at most 12
- the view is one the family draws
- the year is split between 2 and 6 ways, each a whole number up to a million
fourier
74 kinds of claim · 43 placements
- the 0-th line of the period-2 train sits on the envelope ×94
- harmonic 1 has decayed by exp(−1²t) at t = 0.005 ×18
- sine 1 against sine 2 integrates to nothing ×12
- the partial sum of Q₂₅₆ at 0 up to frequency 0 ×7
- the 1-term curve is sampled fast enough to be the function rather than an alias ×5
- the partial sum of Q₂ at 0 up to frequency 0 ×5
- sine 1 against sine 1 integrates to π ×4
- the lines of the period-2 train are 1/T apart ×4
- harmonic 1 of the heat profile has decayed by exp(−m²π²t) ×3
- the projection onto sine 1 is the harmonic this family draws ×3
- between 6 and 40 harmonics ×2
- the projection onto sine 2 cancels to nothing ×2
- a longer period puts the lines closer together
- a missing harmonic has a node at the pluck
- a ramp width up to 0.2
- a time from 0 to 2
- and every Lₙ past ten is within 0.02 of the logarithm
- and it is never negative
- and its finest oscillation is more than three pixels wide
- and so is its absolute area
- and the slopes grow faster than the amplitudes shrink, which is the whole condition
- and the terms it leaves out are below a twentieth of what the window shows
- between 5 and 121 harmonics carry the profile
- between two and eight times, each from 0 to 4
- between two and five times, each from 0 to 2
- between two and four harmonics, each between 1 and 12
- between two and four periods, each positive
- between two and six times, none of them negative
- by the last frame the profile is a single sine to within two percent
- each block's peak stands higher than the last
- each window is between a hundred-thousandth and two wide
- each window is drawn from between 200 and 4000 samples
- each window is sampled fast enough for the terms it draws
- every term count is a whole number between 1 and 200
- halfway through, the shape is the pluck inverted and mirrored
- its signed area is 1
- L₁ is 1/3 + 2√3/π
- Lₙ − (4/π²) ln N has settled to its limit
- Lₙ rises with N
- magnifying ten times does not flatten the curve ten times
- more terms is closer, away from the jump
- N from 2 to 40
- one to three kernel sizes from 1 to 60
- one to three pluck positions, each away from the ends
- the amplitudes shrink
- the area of |D₁₂| by quadrature matches Fejér's closed form
- the area of |D₄| by quadrature matches Fejér's closed form
- the averaged kernel is the mean of the first N + 1 Dirichlet kernels
- the closed-form b₁ is the projection of the tent onto sin(1πx)
- the closed-form b₂ is the projection of the tent onto sin(2πx)
- the closed-form b₅ is the projection of the tent onto sin(5πx)
- the coefficients of the sawtooth wave fall like 1/m^1
- the coefficients of the square wave fall like 1/m^1
- the coefficients of the triangle wave fall like 1/m^2
- the construction is the one with no tangents
- the curve drawn is the sum of the harmonics
- the deepest zoom still has something to look at
- the fine detail fades faster than the coarse shape
- the frequencies multiply by an odd whole number
- the function never leaves [−1, 1]
- the motion repeats after a time of 2
- the partial sum at 0 comes within a tenth of Lₙ and never exceeds it
- the pluck is at least a twentieth from either end
- the pulse fits inside every period drawn
- the range runs to between 100 and 100,000
- the signed area of Dₙ is 1
- the spectrum has something in it
- the steepest chord gets steeper as the spacing shrinks
- the string keeps a slope at least half the pluck's steepest at every time
- the travelling halves and 400 terms of the series agree to within the series' own tail
- the view is one the family draws
- the wave is one the family draws
- the Wilbraham–Gibbs integral
- two to four blocks — the fourth sits at 2⁶⁴
harmonic
201 kinds of claim · 95 placements
- 1/1 + … + 1/2 is not a whole number ×190
- one number in 1..2 carries the most factors of 2 ×190
- the density is positive at 0.00 ×56
- run 3 of the q = 0.5 series is under its own bound ×24
- 1 golden signs land on F(4) − 1 points ×20
- the histogram bar at -2.47 matches the density ×19
- the histogram bar at 0.07 matches the density ×19
- the vector for λ^1 is λ^1 ×19
- the denominator of H(1) carries 2 to the power ⌊log₂ 1⌋ ×16
- H(2) is not a whole number ×15
- term 2 is under the envelope ×14
- the first 1 terms add to the closed form ×14
- φ's powers close in on whole numbers by a factor 1/φ at 3 ×14
- piece 1 is 0.50 of what was left ×13
- and p³ does not, for p = 5 ×9
- p² divides the numerator of H(4) ×9
- stage 0 has two to the 0 intervals ×8
- and the upper bound on block 0 is the k-th power of 2^(1−p) ×7
- and their lengths add to (2/3) to the 0 ×7
- block 0 sits between its own two bounds ×7
- block 1 reaches a half ×4
- H(4) lies strictly between 2 and 3 ×4
- and that bound is 1/(k log 2)^0.5 ×3
- between 50 and 2000 terms ×2
- the partial sum differs from 1/(1 − 2) by exactly 2 to the power of its number of terms ×2
- the tail of 1/n^1.1 past 100000 is bounded by an integral ×2
- the transform at λ = 0.65 dies away along its own powers ×2
- 2 is first passed at n = 4
- 3 at n = 11
- a ratio with even numerator has odd denominator
- and between them they are all 32 strings of 5 bits
- and by the last drawn stage it is already tiny
- and by this stage they already cover most of the interval
- and ends at one
- and it closes on it
- and it is still coming down towards the limit
- and it is the largest power of 2 not above n
- and lists none of them twice
- and no address uses the digit one
- and points of different classes are different points
- and the denominator of the sum carries exactly that many
- and the numerator is odd
- and the remainder is the ratio to the power of the number of pieces
- and they add to what the series says
- and two of the series have ratios climbing to one, one of which converges and one of which does not
- and what has gone is the geometric series it looks like
- and what is left is exactly the tail
- at 1/3 the count is the Cantor set's log 2 / log 3
- at one half every bar has height a quarter
- at one half it is exactly one
- at one half the distribution function is a straight line
- at p = 0.4 it is large and growing with n
- at p = 1 the variance left after the first tenth is tiny
- at x = 0 every sign is plus and the value is H(digits)
- at x = 1/3 the digits alternate and so do the signs
- below one half most of the interval is empty (λ = 1/3)
- Bertrand: a prime lies between 20/2 and 20
- between 12 and 40 rationals are covered
- between 2,000 and 40,000 samples
- between 3 and 20 random runs
- between 3 and 7 ternary digits are listed
- between 3 and 9 stages are drawn
- between 8 and 16 binary digits
- between eight and twenty-two terms are drawn
- between four and twelve blocks are drawn
- between ten and forty terms are drawn
- between three and nine translates are drawn
- between three and six exponents between 0.3 and 3 are compared
- between three and six exponents between nought and three are compared
- between three and ten stages are built
- between two and five series the family knows are compared
- between two and four values of λ
- between two and six classes are drawn
- by 4 it is more than ten times thinner than the normal curve
- consecutive partial sums differ by the next term
- each address names the left end of its own interval
- each block leans further out than the one below it
- each class is sampled at between 6 and 30 points
- each rational is inside its own interval
- every matrix drawn has its eigenvalues inside the unit circle
- every partial sum drawn is under the head plus the geometric tail's total
- every partial sum falls short of the limit
- every partial sum sequence reaches (I − A)⁻¹
- every random run has settled down over its second half
- every such sum has an even denominator
- every triangle is isosceles with its two longest sides equal
- every whole number from 2 up is crossed
- exactly one number up to 12 carries the highest power of 2
- from one half up the whole interval is used (λ = 1/√2)
- from one half up the whole interval is used (λ = 1/2)
- from one half up the whole interval is used (λ = 1/φ)
- inside the circle the partial sums settle on the inverse
- it starts at nought
- just below 1 every sign is minus
- n runs to between 10 and 30
- no candidate length puts the union between one and three
- no point is in two classes at once
- no term is used twice
- no two addresses read the same in binary
- no two seeds is a small rational away from another, which would be a typo
- on the circle they stay bounded without settling
- one to four Pisot numbers this family knows
- one to three of the roots 2, 3 and 4
- only one number up to 20 is a multiple of 19
- outside it they grow without bound
- so 19 divides the denominator of H(20) exactly once
- so its total is finite
- so the translates of the selection do not overlap
- the 2-adic digits times the denominator give the numerator, to 24 places
- the alternating run is heading for log 2
- the answer is caught between an even and an odd partial sum
- the bars carry all the probability
- the bound on each block is exactly a half
- the bounding series is geometric exactly when p is above one
- the comparison holds a series from each side of the threshold
- the comparison holds a series that converges and one that does not
- the covered rationals leave no wide gap, so the covering meets every part of the line
- the covering budget is between a hundredth and a half
- the crossings of 2 to between 4 and 11 are drawn
- the denominator of H(12) carries exactly 2^3
- the density at 0 misses 1/4 by a few millionths
- the density at 2 is 1/8 to double precision
- the density falls away from the centre
- the diagonal matrix's powers only shrink
- the distribution function climbs to one
- the drawn lean is the sum of the drawn steps
- the drawn running total is the partial sum
- the entropy per sign has settled for the golden ratio
- the entropy per sign has settled for the pentanacci
- the entropy per sign has settled for the tetranacci
- the entropy per sign has settled for the tribonacci
- the enumerated bars match the closed form at λ = 1/√2
- the enumerated bars match the closed form at λ = 1/∛2
- the envelope dominates the series from some term onwards
- the error is smaller than the last term used
- the exponent is between a half and three
- the first middle removed is between a third and an eighth
- the function is constant across each removed interval
- the function never decreases
- the gap to ln n has settled on γ
- the geometric series' ratio is the same at every term
- the golden dimension is Alexander and Zagier's 0.99571
- the golden ratio's distribution has dimension below one
- the golden transform holds level at 2πφⁿ
- the halves share bars at λ = 0.6
- the halves share no bar at λ = 0.4
- the last term is small
- the length left is the product of what each stage keeps
- the lengths add to less than the budget
- the list holds the rationals it was asked for
- the longest surviving interval shrinks at every stage
- the middle-thirds set of the same depth has far less left
- the n-th root of the size of Aⁿ never falls below the spectral radius
- the neglected tail is under an eighth of a bar
- the neglected tail is under half of a bar
- the other two grow a great deal before they shrink
- the p = 1 row is the harmonic sum, which is a logarithm plus Euler's constant
- the patterns are different
- the patterns disagree only in a leading block
- the pentanacci's distribution has dimension below one
- the pieces and what is left fill the square exactly
- the Pisot values sit just below the line
- the plotted partial sums are the series
- the prime is 2, 3 or 5
- the ratio is strictly between nothing and one
- the rearranged total settles near the number it was aimed at
- the removed intervals so far have the length the construction gives
- the run is 1 to between 4 and 30
- the sample variance is near Σ 1/n² = π²/6
- the series is one the family knows
- the signs are the ones the series has
- the smallest interval is still wide enough to be represented
- the special number is found by powers of two or by a prime
- the staircase is flat over most of its interval
- the sums run to between a thousand and a million terms
- the sums run to between ten thousand and ten million terms
- the surviving length is bounded away from nothing
- the table runs to between 6 and 20
- the tail is drawn to between 3 and 4.5
- the target is visibly away from the sum the same terms give in order
- the tetranacci's distribution has dimension below one
- the total is not small
- the triangle has side between 8 and 24
- the tribonacci's distribution has dimension below one
- the two patterns land on the same point exactly
- the union is no longer than the sum of the pieces
- the view is one the family draws
- there is one address for each surviving interval
- this view is for a series that does converge
- three to ten primes between 3 and 61
- two different translates never carry the same point of a class twice
- two or three values of λ
- two sign patterns of equal length
- two to four exponents in (0, 1.5]
- two to four values of λ
- what is left and what has gone add to the whole interval
- λ = 1/θ satisfies its relation for the golden ratio
- λ = 1/θ satisfies its relation for the pentanacci
- λ = 1/θ satisfies its relation for the tetranacci
- λ = 1/θ satisfies its relation for the tribonacci
- λ is between 0.2 and 0.9
riemann
107 kinds of claim · 62 placements
- the area under x^0e^(−x) is 0! ×8
- at 1 steps the length is still 2 ×7
- the area under 1/x^0.9 from 1 to 10 matches its closed form ×6
- the integral of sin to the power 2 is (2 − 1)/2 times the one for power 0 ×5
- the area under 1/x^1.25 beyond 1 is 1/(p − 1) ×4
- the horn to 10 holds π(1 − 1/T) ×4
- and it equals the far-end area of the reflected exponent 1.75 ×3
- the area under 1/√x from 1 to 10 matches its closed form ×3
- the area under 1/x from 1 to 10 matches its closed form ×3
- the area under 1/x² from 1 to 10 matches its closed form ×3
- the staircase of 2 steps is exactly 2 long ×3
- between 256 and 8192 samples ×2
- the area under 1/x^0.25 between 0 and 1 is 1/(1 − p) ×2
- a fine enough polygon reaches the curve's length
- a known increasing curve over a real stretch
- adding corners never shortens the inscribed polygon
- after 20,000 lines the estimate is within its own error band of the true length
- and are within 1/N of it after N factors
- and by this stage it is already small
- and every one of them falls short of the curve's own length
- and is nearly nothing by the end
- and its length is nowhere near the curve's
- and the fat one keeps a length in the limit
- and the finest is strictly longer than the coarsest, so the family is climbing
- between 20 and 20,000 lines
- between three and five corner counts, each up to 64
- both parts only ever increase
- each area is n times the one before
- each curve has the same length
- each curve's mean crossing count is 2L/(πd), whatever its shape
- each panel is at least as close as the one before it
- each staircase runs closer to the curve than the last
- every parametrisation traces the same point set
- lines is read only by the crofton view
- middle quarter, halving: no cell lies wholly inside the set, so the lower sum is nought
- middle quarter, halving: refining the partition never raises the upper sum
- middle quarter, halving: the upper sum never falls below the set's own length
- middle thirds: no cell lies wholly inside the set, so the lower sum is nought
- middle thirds: refining the partition never raises the upper sum
- middle thirds: the upper sum never falls below the set's own length
- more bars means less error
- no width, no area
- out and back covers the image a whole number of times
- out, back, out again covers the image a whole number of times
- refining a partition never lowers the variation
- seed is read only by the crofton view
- so in the limit the middle-thirds set has no length at all
- so the first upper sum is on its way down to nothing
- stage is read only by the cantorlen view
- straight through, accelerating covers the image a whole number of times
- straight through, once covers the image a whole number of times
- the accumulation only ever climbs, because f is positive
- the area between the two shrinks every time the steps are halved
- the area left of the curve up to b
- the area under 1/√x between 0 and 1 is 1/(1 − p)
- the area under the curve up to a
- the Cantor, noodle and Koch views draw their own curves
- the counts increase
- the curve is one of quarter, parabola, wiggle
- the curve lies inside the disc the lines are thrown at
- the drawn partition has between 24 and 240 cells
- the fat construction's first middle is between a third and an eighth
- the finest one is within a twentieth of the curve everywhere
- the finest partition is a power of two between 64 and 16384
- the function is one of root, square, smooth
- the function is one the figure knows
- the function moves, so the decomposition is not trivial
- the functions are among root, square, smooth
- the inscribed polygon's length, measured, matches the closed form
- the interval runs from nought to at most 2
- the maps differ in length or in speed, so the panels are comparing something
- the maps run at different speeds
- the middle-thirds approximation's length is two-thirds to the power of the stage
- the panels really are in increasing order of rectangles
- the parametrisations are among once, back, thrice, slow
- the partial products rise towards π/2 from below
- the printed total is the area of the bars drawn
- the quoted exact area is the area under this curve
- the set is approximated at between 5 and 14 stages
- the share of the straight line, w, is between 0 and 1
- the signed area is within 1/T of π/2
- the slope of the accumulated area is the height of the curve
- the speed integral and the polygon agree for out and back
- the speed integral and the polygon agree for out, back, out again
- the speed integral and the polygon agree for straight through, accelerating
- the speed integral and the polygon agree for straight through, once
- the stage drawn is 1 to 7
- the stage-10 polygon is within 0.02 of the limit
- the stage-k Koch curve has length 1.8 × (4/3)^k
- the staircase is built against quarter or diagonal
- the staircase of 1 step is exactly 2 long
- the step counts are between 2 and 5 whole numbers up to 512
- the sweep runs over between three and eight partition sizes
- the sweep runs up to between 4 and 16384 steps, or an interval end for the variation view
- the two areas together are at least ab
- the two regions fill the big rectangle less the small one
- the unsigned area exceeds (2/π) ln T
- the variation of t sin(1/t) climbs by as much at the finest refinement as at the coarsest
- the variation of t² climbs less and less, so it settles
- the variation of t² sin(1/t) climbs less and less, so it settles
- the view is one the family draws
- the volume is below π and the surface above 2π ln T
- their difference is the function
- two to four exponents between 0 and 3
- w is read only by the cantorlen view
- while the second has already settled on the set's own length
- with equality when b = f(a)
secant
37 kinds of claim · 23 placements
- and the difference quotient closes on u'v + uv'
- and the gap as a fraction of the window falls with it
- and the rough curve's do not settle on anything
- between 1 and 4 mirrored pairs are marked
- between 1 and 4 pairs are marked
- between 2 and 4 windows are drawn
- between 4 and 14 spacings are measured
- both sides grow, so the picture is four pieces and not a signed sum
- both sides of the rectangle are positive lengths
- by the last halving the corner is under a twentieth of the increment
- each magnification is between 2 and 20
- each secant is closer than the last
- each step magnifies by between 2 and 20
- every secant from the left has slope −1
- every secant from the right has slope 1
- the bisection really inverts the function
- the corner's share of the increment falls at every halving
- the curve is one the figure knows
- the first window is at most 4 wide
- the four pieces fill the grown rectangle
- the function increases across the window it is drawn in
- the function is increasing where it is marked, so it has an inverse there
- the function is one the family knows
- the function is one the figure knows
- the gap to the tangent falls by the square of the magnification
- the increment is the two strips and the corner
- the last secant is near the tangent
- the marked point is inside the window
- the pair is one the figure knows
- the rectangle is drawn inside the domain of both functions
- the secants are drawn at a list of 2 to 24 positive offsets
- the smooth curve's chords close on its tangent
- the step is a positive length under 1.6
- the step the rectangle grows by is a positive length under 1.6
- the two sides disagree, so there is no limit
- the two slopes at a matched pair multiply to one
- the view is one the family draws
taylor
100 kinds of claim · 37 placements
- the [2/2] approximant reproduces coefficient 0 of the series exactly ×144
- with 2 terms the sum is within the first omitted term at x = 0.013 ×120
- with 1 terms at x = 0.05 the error is below the first term left out ×90
- the polynomial passes through node 1 ×21
- the size of coefficient 1 of 1/(1 − 2x) is what the exact value says ×20
- the size of coefficient 1 of 1/√(1 − 4x) is what the exact value says ×20
- the size of coefficient 1 of Euler's series is what the exact value says ×20
- Laguerre node 1 of 4 is a root ×12
- pole 1 of [3/4] is minus one over a Gauss–Laguerre node ×12
- pole 1 of [4/4] is minus one over a Gauss node ×12
- the [2/2] approximant exists with a denominator starting at one ×12
- the degree-1 sum matches to order 1 at the expansion point ×7
- the degree-2 sum about 1.2 matches the function to order 2 there ×6
- as x shrinks, the point settles at 1/(n+2) for degree 1 ×4
- the integral Euler's series belongs to agrees two ways at x = 0.05 ×4
- the [2/2] denominator does not vanish on the drawn interval ×3
- the best error at x = 0.05 has the size √(2π/x)·e^(−1/x) ×3
- the best number of terms at x = 0.05 is within two of where the terms are smallest ×3
- the degree-2 sum about -1.4 matches the function to order 2 there ×3
- well past the best at x = 0.05, every term added makes the error larger ×3
- at x = 0.5, 4 terms lose to 2 ×2
- between 3 and 41 nodes ×2
- between one and three orders, each between 1 and 12 ×2
- close to nought, 4 terms beat 2 ×2
- a degree between 1 and 20
- a function that is already a quotient of polynomials is reproduced exactly
- a higher-order approximant is at least as close at the probe point
- an amplification factor is never below one, since the nodes reproduce themselves
- and by the largest count drawn it is orders of magnitude worse
- and by the last coefficient they are within ten per cent of it
- and by the last they are past k/e, so no radius but nought is possible
- and every approximant is closer than the last
- and it dwarfs anything in the middle of the interval
- and the terms are smallest near 1/x
- between 6 and 40 terms
- between one and five degrees, each between 1 and 8
- between one and five series
- between one and four degrees
- between one and four singularities, each a point of the plane
- between one and four term counts, each between 1 and 20
- between one and six centres
- between one and three values of x, each between 0.04 and 1
- between three and fourteen node counts, each between 3 and 25
- Chebyshev nodes keep the error small across the whole interval
- each drawn pole is a root of the denominator
- each new centre lies inside the previous disc, so its coefficients are known there
- each node is the projection of its angle
- equally spaced nodes never amplify less than Chebyshev ones
- every pole lies on the cut, at or beyond −1
- every pole lies on the negative real axis
- every term count is a whole number between 1 and 40
- inside the new interval more terms help
- inside the radius, more terms help
- ln(1+x) is centred where it is defined
- ln(1+x) is expanded inside its own radius
- no centre sits on a singularity
- no division by nought
- no rational has a denominator of nought
- no singularity lies strictly inside a disc
- outside it they hurt
- outside it, more terms hurt
- past the radius every Taylor error is larger than the last
- past the radius, the Taylor sum of higher degree is the worse one
- the bound is a bound
- the centre is inside the drawn domain
- the Chebyshev end gaps are several times smaller than the middle one
- the drawn domain is an interval
- the equally spaced constant grows with every step
- the error never exceeds the bound Lagrange's form puts on it
- the evenly spaced nodes have one gap throughout
- the function has a closed form to compare against
- the function has a closed form to draw
- the function is one the figure knows
- the function is one whose derivatives the figure knows
- the interval drawn is inside the function's radius
- the interval drawn is not empty
- the interval is between 0.4 and 8 wide
- the largest order is between 3 and 12
- the last coefficient is between 8 and 200
- the mode draws the functions whose derivative is monotone
- the mode of the taylor family is one of sums, radius, remainder, xi, centre, discs, interp, lebesgue, nodes, pade, poles, paderate, root, asymptotic, asymsums
- the nodes are equally spaced or Chebyshev
- the numerator's degree is a whole number close to the denominator's
- the point is one where the function is defined
- the point moves steadily in one direction as x grows
- the pole picture is drawn for ln(1 + x), 1/(1 + x²), eˣ or Euler's series
- the polynomial has the degree it claims
- the probe point is inside the domain
- the radius is the distance to the nearest singularity
- the roots for 1/(1 − 2x) close on one over the distance to the singularity
- the roots for 1/√(1 − 4x) close on one over the distance to the singularity
- the roots for Euler's series grow at every step
- the series is one of log, runge, exp, geom2, central, lacunary, euler
- the smallest gap is at one of the ends
- the span is between 0.2 and 1.5
- the tail of the series is the difference between the function and its partial sum
- the Taylor degree is between 1 and 40
- the unknown point lies strictly between nought and x
- the window reaches past −1 and past 0.5
- with equally spaced nodes the worst error is out near the ends