Generator

A rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like n

One function in the borderrank library, called 9 times across 2 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 18 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

At its defaults it draws a rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like n. Aₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00169 at n = 1024; the rising one is the norm of the larger of its two rank-one terms, 1024. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

border-approach is one function in lib/figures/borderrank.js — border rank — a nearest point that is not there, and a rank that is a sign. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

A rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like nAₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00169 at n = 1024; the rising one is the norm of the larger of its two rank-one terms, 1024. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.110¹10²10³10⁻³10⁻²10⁻¹110¹10²10³n‖Aₙ − A‖ and the largest term's norm‖Aₙ − A‖the larger of its two termsan infimum that is not attainedn1024‖Aₙ − A‖0.0017largest term1024their product1.7√31.7the distance goes to zeroand nothing reaches it

Aₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00169 at n = 1024; the rising one is the norm of the larger of its two rank-one terms, 1024. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

upTo: 10

The arguments are the ones A nearest point that is not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

A rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like nAₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00169 at n = 1024; the rising one is the norm of the larger of its two rank-one terms, 1024. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.110¹10²10³10⁻³10⁻²10⁻¹110¹10²10³n‖Aₙ − A‖ and the largest term's norm‖Aₙ − A‖the larger of its two termsan infimum that is not attainedn1024‖Aₙ − A‖0.0017largest term1024their product1.7√31.7the distance goes to zeroand nothing reaches it

Aₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00169 at n = 1024; the rising one is the norm of the larger of its two rank-one terms, 1024. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

upTo: 4

The arguments are the ones A nearest point that is not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

A rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like nAₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.108 at n = 16; the rising one is the norm of the larger of its two rank-one terms, 16.09. Their product runs 2.519, 1.917, 1.777 … 1.74334, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.110¹10⁻¹110¹n‖Aₙ − A‖ and the largest term's norm‖Aₙ − A‖the larger of its two termsan infimum that is not attainedn16‖Aₙ − A‖0.11largest term16their product1.7√31.7the distance goes to zeroand nothing reaches it

Aₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.108 at n = 16; the rising one is the norm of the larger of its two rank-one terms, 16.09. Their product runs 2.519, 1.917, 1.777 … 1.74334, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

upTo: 6

The arguments are the ones A nearest point that is not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

A rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like nAₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.0271 at n = 64; the rising one is the norm of the larger of its two rank-one terms, 64.02. Their product runs 2.519, 1.917, 1.777 … 1.73276, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.110¹10⁻¹110¹10²n‖Aₙ − A‖ and the largest term's norm‖Aₙ − A‖the larger of its two termsan infimum that is not attainedn64‖Aₙ − A‖0.027largest term64their product1.7√31.7the distance goes to zeroand nothing reaches it

Aₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.0271 at n = 64; the rising one is the norm of the larger of its two rank-one terms, 64.02. Their product runs 2.519, 1.917, 1.777 … 1.73276, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

upTo: 12

The arguments are the ones A nearest point that is not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

A rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like nAₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 4.23·10⁻⁴ at n = 4096; the rising one is the norm of the larger of its two rank-one terms, 4096. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.110¹10²10³10⁻³10⁻²10⁻¹110¹10²10³10⁴n‖Aₙ − A‖ and the largest term's norm‖Aₙ − A‖the larger of its two termsan infimum that is not attainedn4096‖Aₙ − A‖4.2·10⁻⁴largest term4096their product1.7√31.7the distance goes to zeroand nothing reaches it

Aₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 4.23·10⁻⁴ at n = 4096; the rising one is the norm of the larger of its two rank-one terms, 4096. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

upTo: 16

The arguments are the ones A nearest point that is not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

A rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like nAₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 2.64·10⁻⁵ at n = 65536; the rising one is the norm of the larger of its two rank-one terms, 65540. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.110¹10²10³10⁴10⁻⁴10⁻³10⁻²10⁻¹110¹10²10³10⁴10⁵n‖Aₙ − A‖ and the largest term's norm‖Aₙ − A‖the larger of its two termsan infimum that is not attainedn6.6·10⁴‖Aₙ − A‖2.6·10⁻⁵largest term6.6·10⁴their product1.7√31.7the distance goes to zeroand nothing reaches it

Aₙ = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖Aₙ − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 2.64·10⁻⁵ at n = 65536; the rising one is the norm of the larger of its two rank-one terms, 65540. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

What it checked while drawing

Every figure above checked its own claims on the way to being drawn, and a claim that failed would have stopped the picture rather than shipped a wrong one. Those checks used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

18 distinct claims across 6 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

the residual at n = 2 is the one the expansion gives — checked 16 times

a range of n the sequence is drawn over

and the product of the two descends onto √3

Against the rule

The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.

Across the library: the rule bites on 217 of 397 generators — 199 print a residual and 18 are exempt with a published reason; 180 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

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