Coordinates of the regular heptagon

[quickfur]

So, yesterday and today I got bored, and decided to study the regular 2D heptagon. More specifically, to study it from the POV of deriving algebraic coordinates for the origin-centered regular heptagon with edge length 2. Thought I'd post it here because of potentially interesting points related to exact computations of polytope coordinates in the search for CRFs, CVP, and so on.

First step is to derive algebraic representations of sin(pi/7) and cos(pi/7), for obvious reasons. There are several approaches for this; the approach I adopted is to start with 0 = sin(pi) = sin(7*pi/7) = sin(8*pi/7 - pi/7). This then can be split using the angle addition trig identity into sines and cosines involving 8*pi/7 and pi/7, the former of which can be successively reduced to pi/7 using the double angle formulas. At first, I set about doing this by hand, but the expressions quickly got far too unwieldy to handle, so I decided on a different approach.

In my study of the Johnson solids, esp. the crown jewels, I've found the need to solve systems of polynomial equations, which is known to be a tough problem. Ultimately, this is impractical to do by hand for most cases, except for trivial cases like J84, so the go-to approach these days is to implement an algorithm to compute the Grobner basis for the system and then extract polynomial expressions from it that define the solution values. Since, in the course of working out algebraic coordinates of all the Johnson solids, I've developed the tools for doing this, I thought it would be a handy thing to do to solve our little problem.

So, I basically assigned each sin/cos expression as a variable, then encoded instances of the trig identities as polynomial relations. For example, sin(8*pi/7 - pi/7) = sin(8*pi/7)*cos(pi/7) - cos(8*pi/7)*sin(pi/7) = 0, so letting A = sin(8*pi/7), B = cos(8*pi/7), G = sin(pi/7), H = cos(pi/7), we get the polynomial equation AG - HB = 0. New variables are introduced for each different sin/cos argument (sin(4*pi/7), sin(2*pi/7), etc), and the relevant double-angle formulas are introduced as multiple polynomial relations between these variables. Then run the Grobner basis algorithm and extract the polynomials defining the target unknowns, G = sin(pi/7) and H = cos(pi/7). And here's what I got:

cos(pi/7) = H, where 8H³ - 4H² - 4H + 1 = 0, where 0.9 < H < 1.0;
sin(pi/7) = G, where 64G⁶ - 112G⁴ + 56G² - 7 = 0, where 0.1 < G < 0.2

The in-radius h of the heptagon therefore satisfies tan(pi/7) = 1/h (because we're assuming edge length 2), so h = 1/tan(pi/7) = cos(pi/7)/sin(pi/7). Now, in theory I could work out this value by hand, but since I already have a polynomial system solver at hand, it seemed easier to just make use of it. I.e., introduce h as a new variable that satisfies the equation h*G - H = 0, and then solve for h. The resulting polynomial is:

7h⁶ - 35h⁴ + 21h² - 1 = 0, 2.0 < h < 2.1

(Interestingly, the coefficients of this polynomial are multiples of 7 except for 1.)

Similarly, the out-radius r of the heptagon satisfies sin(pi/7) = 1/r, meaning r = 1/sin(pi/7). In this case there's no need to run the solver, since the reciprocal of a polynomial root is a root of the polynomial with the order of its coefficients reversed. So:

7r⁶ - 56r⁴ + 112r² - 64, 2.3 < r < 2.4

The height of the heptagon (from base edge to opposite vertex) is H=h+r.

Now, here's an interesting point. I tried to derive a single defining polynomial for H, but it turned out that the minimal such polynomial is a whopping degree-36 polynomial with gargantuan coefficients (about 40 digits or so per coefficient). Yeah, you read that right, the leading term is x³⁶. It only has even powers, though, so if you take it as a polynomial in x² it would be an 18th degree polynomial -- but that's still huge (and the coefficients remain gigantic 40+-digit numbers). And this nasty thing is irreducible. What's interesting about this is that h and r, by themselves, have very tame-looking polynomials. Yet their sum caused an explosion in polynomial degree and magnitude of coefficients. This makes me wonder if there any known algorithm or method for splitting an algebraic number of high degree into a sum of two algebraic numbers of significantly smaller degree, or at least smaller coefficients? Currently, the best polynomials I've found for, e.g., J88 are degree 16, with moderate-sized coefficients, and for J89 the polynomials are degree 12 with largish coefficients. It would be nice to find much smaller polynomials whose roots, when summed together, produce the same value. Is there any literature out there on this subject?

Anyway, I decided to leave H as h+r and not try to poke at the beast too much more.

Now, armed with the values of sin(pi/7), cos(pi/7), h, and r, we are now ready to derive the coordinates of the regular heptagon of edge length 2. We start with the most obvious points: <0, r> and <±1, -h>, then derive the rest by multiplication with a suitable rotation matrix, the elements of which are ±sin(pi/7) and cos(pi/7). The resulting points have coordinates that are just linear combinations of sin(pi/7), cos(pi/7), h, and r; so they are easily encoded as equalities in the aforementioned system of polynomial equations and solved for. So here are the final results:

Regular origin-centered heptagon, edge length 2:

<0, r>
<±A, B>
<±C, D>
<±1, -h>

where A, B, C, D satisfy:

A³ + A² - 2A - 1 = 0; with -2 < A < -1
7B⁶ - 21B⁴ + 14B² - 1 = 0; with 1 < B < 2
C³ - 2C² - C + 1 = 0; with 2 < C < 3
D⁶ - 14D⁴ + 7D² - 1 = 0; with -0.6 < D < -0.5

These are all degree 3 and degree 6 polynomials, with the coefficients of A and C being particularly simple, interestingly enough. The coefficients of B and D have clear connections as multiples of 7, though their exact nature is somewhat obscure, being the result of the heavy machinery of the Grobner basis solver.

I'm not 100% certain, but this may be the first time anyone's posted algebraic coordinates for the regular heptagon.

Stepping back, the slightly larger picture here is that these coordinate values were derived by making use a polynomial system solver, which alleviated a lot of the complexity and tedium of manipulating the algebraic expressions directly. Although there are unsolved questions, such as how to automate the selection of specific polynomial roots in an actual computation, it seems that this could be a practical approach to the programmatic manipulation of the 3D crown jewels in 4D space, e.g., in a brute-force search for CRFs that contain 3D crown jewels, that uses exact coordinates instead of floating-point (which always run into a risk of unreliable results because of roundoff issues).

There's also the interesting possible relation to the idea of polytope CVP, the idea that polytopes whose coordinates require high-degree defining polynomials may have some kind of inherent complexity that makes it hard, if not impossible, to fit together into a CRF polytope. Particularly interesting is the case of H = h + r, where the sum of two seemingly-tame degree-3 algebraic numbers causes an explosion of complexity to 36th degree (or 18th degree in the squared value) and an explosion in coefficient size. It would be interesting to know if any of the 3D crown jewels with high-degree defining polynomials, like J88, J89, and J90, have some kind of alternative representation in which the high-degree polynomials are "split" into significantly simpler ones. An even more interesting question is, if such "splits" are possible, do they have any geometric meaning that might somehow pertain to assembling these polyhedra into CRF polytopes? Perhaps even such splits could serve as a kind of hint as to how to construct a CRF polytope containing a 3D crown jewel? Or conversely, the absence (impossibility) of such splits may possibly imply that such a construction might be impossible?

Lots of food for thought.

P.S. Although technically the 3rd degree polynomials above are solvable with the cubic polynomial formulas, I didn't use that route because basically all of these polynomials are casus irreducibilis; writing out the exact root value in terms of radicals would require taking cube roots of complex numbers, because they are not expressible in the reals alone. Furthermore, such expressions are large and unwieldy, and of limited use in computing the actual values as compared to a more general polynomial root-finding algorithm, so I decided it was not worth the effort.

[quickfur]

Just for kicks, here's the polynomial for H:

54116956037952111668959660849H³⁶ - 31284018300633296227243905167526H³⁴ + 7535265086355331616972415638496117H³² - 990225771789105195514482112130972936H³⁰ + 77900077648382757915178940605664835560H²⁸ - 3791724173407023183687086112687425756040H²⁶ + 114660790728461158267708112367743610538144H²⁴ - 2128399877984257076619148173489871099163800H²² + 23444275429232481667848617417681853364665970H²⁰ - 140249573996064768810280930767683415011883188H¹⁸ + 350467994071307649121060180978838222590011058H¹⁶ + 1380575710632541436428040426101819835533384H¹⁴ - 5893922159339073690311620369962773788768H¹² - 38168228917801058024215237706566210408H¹⁰ + 225911233517286258993061102752065000H⁸ + 412892382563408159325153819403608H⁶ + 1175109482659240348645973867221H⁴ - 59059363532192257095150518H² + 870527020973136269937489 = 0



Yeah, I don't know what's the deal with it either. Apparently, adding algebraic numbers is a very convoluted thing to do.

[wendy]

I normally use the double-form with the polygons, so

0 = a³+a²+2a-1 a lying near 1.801937736, b = a²-1, lies near 2.2469796037

The vertical axis is g^2 = -1+a+b, g = 0.867767478

The coordinates are then (2, 0), (a, ag), (1/b, bg), and (-b/a, g). The side is then 2g.

Because 7 is a prime or prime-power, you can't reach the centre of the heptagon by steps of the edge from a vertex.

The lengths of numbers of the form x+ya+zb, are an integer system, where primes of the form 14 +/- 1 have three indirect factors (ie the cube of any is not a multiple of the prime), and g as above (which g³ = 7ab, is a direct prime), and unfactorable primes.

When the cyclotomic order is taken to account, then you get this, modulo 7:

0 (direct prime of order 6)
1, (six indirect factors)
2, 4 (two indirect factors)
3, 5 (no further factorisation - direct and unitary)
6 (three indirect factors)

An indirect prime consists of p^k * v, where k is the inverse of the number of factors, eg 1/6, 2/6, ... v is a number that is never a product of units of the system.

The cyclotomic number system of the heptagon has five distinct units, of which two are currently known. I have not searched for the other three

[wendy]

For a laugh, years ago, i ran a proggie that spat out the algebraic equation for the shortchord every polygon up to class 162, that is up to the 162 power. For [162], the numbers end up in the range of 10,000 or so.

It;s a subject i've been offering here for a long while.

[mr_e_man]

First step is to derive algebraic representations of sin(pi/7) and cos(pi/7), for obvious reasons. There are several approaches for this; the approach I adopted is to start with 0 = sin(pi) = sin(7*pi/7) = sin(8*pi/7 - pi/7). This then can be split using the angle addition trig identity into sines and cosines involving 8*pi/7 and pi/7, the former of which can be successively reduced to pi/7 using the double angle formulas. At first, I set about doing this by hand, but the expressions quickly got far too unwieldy to handle, so I decided on a different approach.

There's an easier way to do this, using complex numbers. The 7th root of unity is

z = exp(2πi/7) = cos(2π/7) + i sin(2π/7)

which satisfies the equation z⁷ = 1, or

0 = z⁷ - 1 = (z - 1) (z⁶ + z⁵ + z⁴ + z³ + z² + z + 1).

Of course z =/= 1, so divide that out; and also divide by z³ to make the expression more symmetric and lower degree:

0 = z³ + z² + z + 1 + 1/z + 1/z² + 1/z³

0 = (cos³ + 3 cos² i sin - 3 cos sin² - i sin³) + (cos³ - 3 cos² i sin - 3 cos sin² + i sin³)
+ (cos² + 2 cos i sin - sin²) + (cos² - 2 cos i sin - sin²)
+ (cos + i sin) + (cos - i sin)
+ 1.

Notice that the imaginary terms all cancel out, so we're left with

0 = 2 cos³ - 6 cos sin²
+ 2 cos² - 2 sin²
+ 2 cos
+ 1

= 2 cos³ - 6 cos + 6 cos³
+ 2 cos² - 2 + 2 cos²
+ 2 cos
+ 1

= 8 cos³ + 4 cos² - 4 cos - 1.

The three roots of this equation correspond to 2π/7, 4π/7, 6π/7, giving the x-coordinates of the vertices of the unit-circumradius heptagon.

Also, cos 6π/7 = cos(π - π/7) = - cos π/7, so the above equation becomes

0 = - 8 cos³(π/7) + 4 cos²(π/7) + 4 cos(π/7) - 1.

[mr_e_man]

We can also do this for the enneagon:

z = exp(2πi/9) = cos(2π/9) + i sin(2π/9)

0 = z⁹ - 1 = (z - 1) (z⁸ + z⁷ + ... + z² + z + 1).

But this equation is also satisfied by w = exp(2πi/3) and exp(4πi/3), which instead should have the equation

0 = w³ - 1 = (w - 1) (w² + w + 1)

0 = w + 1 + 1/w.

If we want z⁹ = 1 but z³ =/= 1, we can get these simpler roots out of the way by writing

0 = (z³)³ - 1 = (z³ - 1) ((z³)² + (z³) + 1)

0 = z⁶ + z³ + 1

0 = z³ + 1 + 1/z³

= (cos³ + 3 cos² i sin - 3 cos sin² - i sin³) + (cos³ - 3 cos² i sin - 3 cos sin² + i sin³)
+ 1

= 2 cos³ - 6 cos sin² + 1

= 8 cos³ - 6 cos + 1.

The three roots of this equation correspond to 2π/9, 4π/9, 8π/9. And since cos(8π/9) = - cos(π/9), we also get

0 = - 8 cos³(π/9) + 6 cos(π/9) + 1.

[quickfur]

I have to confess that my grasp of complex arithmetic has rusted significantly since my college days, but this is certainly a pretty cool approach that works better than grappling with repeated applications of trig identities.

[wendy]

Most of my work here is done with chordal arithmetic.

It goes like this. You imagine rays from a point at steps of 1/n of the half-circle, or 2/n of the circle.

If you rotate a n-gon so that chords fall on the 0-ray, the edges are then k steps starting at k/2 or (k-1)/2. So c1 = r0, c2 = r1+r(-1), c3 = r2 + r0 + r'2 (r'm = r(-m).)

If you multiply c3 * c4, you get 12 chords, by replacing each of r3, r1, r'1, r'3, by r2, r0, r-2, ie r5, 2r3, 3r1, ~ (dashed forms(, or r6 + r4 + r2.). Fiddling around with this, gives first the chords, as 1, 10.1, 101.01, 1010.101, &c, and then you can construct powers of 10.1 in terms of the chords. It turns out that C(n-1)+C(n+1)=a.c(n), where a = the second or short-chord, as 10.1.

You fiddle around with this, and it gives a lot of interesting things.

[mr_e_man]

Now, here's an interesting point. I tried to derive a single defining polynomial for H, but it turned out that the minimal such polynomial is a whopping degree-36 polynomial with gargantuan coefficients (about 40 digits or so per coefficient). Yeah, you read that right, the leading term is x36. It only has even powers, though, so if you take it as a polynomial in x2 it would be an 18th degree polynomial -- but that's still huge (and the coefficients remain gigantic 40+-digit numbers). And this nasty thing is irreducible. What's interesting about this is that h and r, by themselves, have very tame-looking polynomials. Yet their sum caused an explosion in polynomial degree and magnitude of coefficients. This makes me wonder if there any known algorithm or method for splitting an algebraic number of high degree into a sum of two algebraic numbers of significantly smaller degree, or at least smaller coefficients? Currently, the best polynomials I've found for, e.g., J88 are degree 16, with moderate-sized coefficients, and for J89 the polynomials are degree 12 with largish coefficients. It would be nice to find much smaller polynomials whose roots, when summed together, produce the same value. Is there any literature out there on this subject?

This question appears on MSE, though the answers look unsatisfactory.

In general, if x and y are algebraic numbers of degree m and n respectively (so the fields ℚ(x) and ℚ(y) have dimensions m and n), then any polynomial combination of them such as x+y or x*y has degree at most m*n, because the field ℚ(x,y) has dimension m*n or smaller. (Anything in this field can be written as a linear combination, with rational coefficients, of powers of x multiplied by powers of y, where the exponents are less than m and n respectively.) To show that it can be smaller, just take x=y; or, less trivially, take x and y to be roots of two: ℚ(2^(1/6), 2^(1/10)) = ℚ(2^(1/30)), which has dimension 30 < 6*10.

Your heptagon numbers h and r have degree m=n=6, and their sum H has degree m*n=36, which is the maximum.

I found here a possible method for decomposing an algebraic number α, by finding sub-fields of ℚ(α). Take its minimal polynomial (which doesn't factor over ℚ), factor it into smaller polynomials with coefficients in ℚ(α), look at the sub-fields generated by the coefficients in each factor, and try to write α as a sum of numbers from two different sub-fields. I don't know if this is helpful.

[quickfur]

[...]I tried to derive a single defining polynomial for H, but it turned out that the minimal such polynomial is a whopping degree-36 polynomial with gargantuan coefficients (about 40 digits or so per coefficient). Yeah, you read that right, the leading term is x^36. It only has even powers, though, so if you take it as a polynomial in x2 it would be an 18th degree polynomial -- but that's still huge (and the coefficients remain gigantic 40+-digit numbers). And this nasty thing is irreducible. What's interesting about this is that h and r, by themselves, have very tame-looking polynomials. Yet their sum caused an explosion in polynomial degree and magnitude of coefficients. This makes me wonder if there any known algorithm or method for splitting an algebraic number of high degree into a sum of two algebraic numbers of significantly smaller degree, or at least smaller coefficients? [...]

This question appears on MSE, though the answers look unsatisfactory.

In general, if x and y are algebraic numbers of degree m and n respectively (so the fields ℚ(x) and ℚ(y) have dimensions m and n), then any polynomial combination of them such as x+y or x*y has degree at most m*n, because the field ℚ(x,y) has dimension m*n or smaller.

Interesting. So any time you add two algebraic numbers together there's a chance the degree of the result will be the product of the starting degrees? Wow, that's a pretty raw deal.

Also, are there any bounds on the coefficients, or they can explode without bound when adding two arbitrary algebraic numbers?

[...]
I found here a possible method for decomposing an algebraic number α, by finding sub-fields of ℚ(α). Take its minimal polynomial (which doesn't factor over ℚ), factor it into smaller polynomials with coefficients in ℚ(α), look at the sub-fields generated by the coefficients in each factor, and try to write α as a sum of numbers from two different sub-fields. I don't know if this is helpful.

The theorem sounds promising, but factoring the minimal polynomial will not help, because since α is a root of its minimal polynomial, its minimal polynomial in ℚ(α) is nothing but x - α. What you want is to find some intermediate field L between ℚ and ℚ(α), which according to your link above, ought to have a degree that divides the degree of ℚ(α), that lies along the extension chain from ℚ to ℚ(α). If you can find such a field, then you could meaningfully talk about factoring the minimal polynomial of α over L.

However, finding L itself is more-or-less equivalent to solving the original problem in the first place, i.e., L = ℚ(β) for some β with the property that α = β + γ. So this doesn't really give us any further insight into how to find β in the first place.

Hmm, actually, on second thoughts, maybe there is a way: we know that the degree of L must divide the degree of ℚ(α). Let's say the degree is d = d1*d2. What if we assume root-taking operations of degree d1 and d2? I.e., consider the extension field M = ℚ(β1, β2, β3, ...) where the β's are all algebraic numbers of degree d1, and then try to factor the minimal polynomial of α over M. This should give us the product of two polynomials of degree d1 and d2, respectively, whose respective extension fields ought to be what we're looking for. I'm fuzzy about the details of how to actually do this, though.

[mr_e_man]

Also, are there any bounds on the coefficients, or they can explode without bound when adding two arbitrary algebraic numbers?

I don't know. But let's consider two general quadratic numbers for example:

x² = ax + b, y² = cy + d

z = x + y

z² = x² + 2xy + y²
= 2xy + ax + cy + b+d

z³ = (2xy + ax + cy + b+d) (x + y)
= 2x²y + ax² + cxy + (b+d)x + 2xy² + axy + cy² + (b+d)y
= 2axy + 2by + a²x + ab + cxy + (b+d)x + 2cxy + 2dx + axy + c²y + cd + (b+d)y
= 3(a+c)xy + (a²+b+3d)x + (c²+3b+d)y + ab+cd

z⁴ = (3(a+c)xy + (a²+b+3d)x + (c²+3b+d)y + ab+cd) (x + y)
= 3(a+c)x²y + (a²+b+3d)x² + (c²+3b+d)xy + (ab+cd)x + 3(a+c)xy² + (a²+b+3d)xy + (c²+3b+d)y² + (ab+cd)y
= 3(a+c)axy + 3(a+c)by + (a²+b+3d)ax + (a²+b+3d)b + (c²+3b+d)xy + (ab+cd)x + 3(a+c)cxy + 3(a+c)dx + (a²+b+3d)xy + (c²+3b+d)cy + (c²+3b+d)d + (ab+cd)y
= (4a²+6ac+4c²+4b+4d)xy + (a³+2ab+6ad+4cd)x + (c³+4ab+6bc+2cd)y + a²b+c²d+b²+6bd+d²

We should be able to invert these equations, using linear algebra, to write the basis xy,x,y,1 in terms of the new basis z³,z²,z,1, and thus to write z⁴ as a combination of z³,z²,z,1, with coefficients which are rational functions of a,b,c,d. Then we could write the original coefficients in terms of integers, like a=a₁/a₂, so the coefficients in the z⁴ equation are rational functions of these integers; and then just multiply by a common denominator to get a quartic equation in z with coefficients which are polynomial functions of a₁,a₂,b₁,b₂,c₁,c₂,d₁,d₂.

And polynomial functions are bounded by powers of bounds on the inputs; for example, if a,b,c,d are integers, and we write m=max(|a|,|b|,|c|,|d|), then

|a³+2ab+6ad+4cd| ≤ |a|³+2|a||b|+6|a||d|+4|c||d| ≤ m³+2m²+6m²+4m² = m³+12m² ≤ m³+12m³ = 13m³.

The theorem sounds promising, but factoring the minimal polynomial will not help, because since α is a root of its minimal polynomial, its minimal polynomial in ℚ(α) is nothing but x - α. What you want is to find some intermediate field L between ℚ and ℚ(α), which according to your link above, ought to have a degree that divides the degree of ℚ(α), that lies along the extension chain from ℚ to ℚ(α). If you can find such a field, then you could meaningfully talk about factoring the minimal polynomial of α over L.

No, I meant to take the minimal polynomial over ℚ, and factor that over ℚ(α). Say α is the 6th root of 2; its minimal polynomial is x⁶ - 2, which factors as

x⁶ - 2 = (x³ - √2) (x³ + √2) = (x³ - α³) (x³ + α³)

= (x² - ∛2) (x⁴ + ∛2 x² + ∛2²) = (x² - α²) (x⁴ + α²x² + α⁴)

= (x - α) (x + α) (x² + αx + α²) (x² - αx + α²).

Any one of these factors (not only those irreducible over ℚ(α)) has coefficients which generate a sub-field, and every sub-field appears in this way. The coefficients in (x³ - α³) generate ℚ(√2), the coefficients in (x² - α²) generate ℚ(∛2), and the coefficients in (x² + αx + α²) generate ℚ(α) itself.

By definition, the minimal polynomial over L does not factor over L.

[quickfur]

Also, are there any bounds on the coefficients, or they can explode without bound when adding two arbitrary algebraic numbers?

[...]
And polynomial functions are bounded by powers of bounds on the inputs; for example, if a,b,c,d are integers, and we write m=max(|a|,|b|,|c|,|d|), then

|a³+2ab+6ad+4cd| ≤ |a|³+2|a||b|+6|a||d|+4|c||d| ≤ m³+2m²+6m²+4m² = m³+12m² ≤ m³+12m³ = 13m³.

Aha, so coefficients ought to grow up to some polynomial bound on the starting coefficients? Makes sense in retrospect. Still, for high-degree algebraic numbers this could translate to a very fast growth. If deg(x+y) ≤ deg(x)*deg(y), then it would seem that the size of the coefficients would also grow by O(n^(deg(x)*deg(y)), IOW, exponential in the degrees of the starting numbers. That would explain the huge coefficients I got when I attempted to sum those two heptagon-related numbers.

The theorem sounds promising, but factoring the minimal polynomial will not help, because since α is a root of its minimal polynomial, its minimal polynomial in ℚ(α) is nothing but x - α. What you want is to find some intermediate field L between ℚ and ℚ(α), which according to your link above, ought to have a degree that divides the degree of ℚ(α), that lies along the extension chain from ℚ to ℚ(α). If you can find such a field, then you could meaningfully talk about factoring the minimal polynomial of α over L.

No, I meant to take the minimal polynomial over ℚ, and factor that over ℚ(α). Say α is the 6th root of 2; its minimal polynomial is x⁶ - 2, which factors as

x⁶ - 2 = (x³ - √2) (x³ + √2) = (x³ - α³) (x³ + α³)

[...]

Ah, I see what you mean.

But doesn't that require you knowing what the subfields are beforehand? For example, if α is a degree 25 algebraic number whose minimal polynomial factors into something whose coefficients are roots of unsolvable quintics, then how would you know which quintics those are? For degree 4 or below, you can just factor into nth roots, but when presented with some arbitrary algebraic number, how do you construct the intermediate subfields?

[mr_e_man]

We don't need to know the equations satisfied by the coefficients; we only need to write them as polynomials in α, and to be able to factor polynomials in x with such coefficients. But I confess that I don't know how to factor polynomials in general, whether the coefficients are in ℤ or ℚ or ℚ(α). Note, this does not mean finding all n roots (one of which is α); there may be some quadratic or higher irreducible factors.

Also, the quartic formula only works when everything has square and cube roots, which is not true in ℚ(2^(1/6)); there is no √(√2)=∜2 in this field.

[wendy]

It always amuses me that mathematicians will bring to the task, the most inappropriate tools for the problem at hand. And then choose to measure the wrong things.

There is a power-like function that runs (2.cos a)^^n = 2 cos (an), or (2 cosh a)^^n = 2 cosh (an). The general iterative for it is a.t(n) = t(n+1)+t(n-1), or in a progression, t(n+1) = a.t(n)-t(n-1). It generalises to t(n) = isoqd(t_0, t_1, a, n) for the nth term. It has other uses, eg the process of finding large primes involves running isoqd(2, 4, 4, 2^p, 2^p-1), where the fifth field is a modulo used in the main loop multiply. This is how the Messerine primes are found.

a is the shortchord. This is the third side of a triangle, formed by two consecutive edges. The necessary values to remember are a and a². Where a polygon appears at a vertex, then the vertex-figure will contain a chord 'a'. The chords of a polygon are isoqd(0, 1, a, n), and the shortchords of p/n run isoqd(2, a, a, n). The circumdiameter of a polygon is sqrt(4/(4-a)), or where b, c, d, stand for the following chords, sqrt(2*2/(3-b)).

Having devised that, the isoqd() routine runs at 3 measures of time, where the power() runs at two. That is, it's quite known to run things like isoqd(2, 3, 3, 4147200000), in logrithmic time.

A general polygon 'p' satisfies some relation that a(p)^^n = -2. In terms of cyclotomic numbers, where p^n=-1, we have a(p)=10.1, a(p/2)=100.01 etc, and eventually a(p/p) = 10^p + 10^(-p) = 2. The algebraic factors of this can be exactly found with a calculator, since these follow the factors of the general b^n-a^n polynomial. The decimal example is to put b=10, a=1, and note that 9, 99, 999, 9999, ... are divisible by others, if the number of nines go into each other, so 99 goes into 9999 but not 999. You get a unique factor for each number, the 9's is the product of the f(n), where n divides the number. The factors are 9, 11, 111, 101, 11111, 91, 1111111, ... You can look at the decimal factorisations at the cunningham project for a hint here.

Having produced such a table, we can then convert these into equations, by adding a stream of 5's to it, trimming any leading 5. The list becomes 64, 66, 666, 656, 66666, 646, 6666666 etc. When 5 is subtracted from each digit, you end up with the polynomial, eg 656 = 101 = x^2+1.

The first two are irregular, and we treat the rest as foldable. That is, x = reverse(x) in rexx. They are then folded, and we use eg A30 = 109889011. => 665444566. Folding it pairs x^n with x^-n, so we get 66544, ie b^^4+b^^3-b^^1-1. = 0 for the polygon of 15 sides. In essence, A30 is the large factor in 10^15-1 = 0, and it is the factor involved as 0.

Using the isoseries on 2, a

a^^0 = 2 (halve this) 4-5 -1
a^^1 = a 4-5 -a
a^^2 = a²-2 5-5 0
a^^3 = a³ - 3a 6-5 a³-3a
a^^4 = a^4 - 4a²+2 6-5 a^4-4a²+2
total a^4+a³-4a²-4a+1

The polygon of 15 sides has a short-chord a, that solves this relation.

This set of equations have been carried out to all polynomials with a first-term not bigger than a^^81, There are values in the 5-added form that produce 3 and 7 (eg 210).

An equation like a^3 = a^2+2a-1 defines an integer system, in the form of (z2, z2, z0), where powers greater than 3 are reduced by formula. This is the system of numbers designated Z7, that is, the 'span of chords of the heptagon'. c(2) = (1,0), c(3) = (1,0,-1), c(4)=(1,0,-2,0) = (1,0,-1) = c3. Indeed, subtracting c4 from c3 gives (1,-1,-2,1), or (1,0,0,0)=(1,2,-1).

The numbers described by the span of the heptagon, provide integers, with the following factorisations. For 7 itself, there is a direct factor -1+a+b = (1,1,-2), for which the cube is 7ab. For primes equal to ±1, nod 7, there is a factorisation into three non-direct numbers. 13, for example, has three factirs, the cube of none of them is a multiple of 13. For the remaining primes, there is no further factorisation.

In the cyclotomic numbers CZ7. there is a number whose sixth power is 7 times a product of units, for primes 1 mod 7, there are 6 indirect factors, for 6 mod 7, there are three, for 2,4 mod 7, there are two indirect factors, and for 3,5 one.

In essence, for the Gaussian numbers a+bi. we see that 1+i is a direct divisor, since its power includes a numeric multiple of another number, but it itself isn't. (1+i)² = 2i. The numbers like (2+i)(2-i) are indirect. Their product is 5, but the power of (2+i) is never 5 times a gaussian. In every polygonal system, there are not more than a finite number of direct divisors, although in larger systems, there are indirect primitives for every prime.

Of course, there is no need to descend to the full rationals for this exercise. The equation-spans provide a number system, and even if halves and thirds turn up, it does not mean that the full rational numbers are invoked, It's more 'bad coordinates' producing a body-centered array, such as Z{1,r5] also includes Z½[1,r5]. The general cover is called the sense. Ie there is some s that sS > T > S where s is a number, S the set in question, and T the coordinate system, > means 'is part of'.

[mr_e_man]

Does that have anything to do with simplifying the sphenomegacorona J88, which quickfur was asking about?

[mr_e_man]

Wendy, it seems that we argue or disagree on many things. But I think we can continue to communicate and learn from each other in spite of this.

[mr_e_man]

Huh?

The heptagon's height should be

H = cos(π/7)/sin(π/7) + 1/sin(π/7)

= (1 + cos(2π/14)) / sin(2π/14)

= (2cos²(π/14)) / (2sin(π/14)cos(π/14))

= cot(π/14)

= tan(π/2 - π/14)

= tan(3π/7) ≈ 4.381286

whose minimal polynomial, according to WolframAlpha, is simply

H⁶ - 21H⁴ + 35H² - 7 = 0.

(This is the same as that of tan(π/7) = 1/h which you already found.)

Where did that 36th-degree thing come from?

[quickfur]

[...]
Where did that 36th-degree thing come from?

Hmm, that's weird. I must've made a mistake somewhere, maybe a typo in the coordinates or something, I tried to feed that x^36 monster to Maxima to compute the numerical roots, but none of the roots match the value of H+h. How did I even get that monster in the first place??

Edit: OK, I think I figured it out. Here's the input I used:

7A^6 - 35A^4 + 21A^2 - 1 = 0
7B^6 - 56B^4 - 112B^2 + 64 = 0
H = A + B

Singular came back with the x^36 monster. Note that the coordinates I computed for the regular heptagon are for edge length 2. Note that the "usual" coordinates are for a heptagon of radius 1, but the relationship between the radius and the edge length is a high-degree polynomial (note: not the same thing as the relationship between the radius and the circumference, because the edge length is a chord of the circle). That's probably where the x^36 thing comes from.

[mr_e_man]

There's your typo! You flipped some signs; it should be 7B^6 - 56B^4 + 112B^2 - 64 = 0.

The relationship between the radius and edge length is just r/2 = B/2 = 1/(2sin(π/7)).

[quickfur]

OK, I finally figured out where that x^36 monster came from. I accidentally used the wrong syntax for the input to Singular: it was expecting 7*A^6, etc., but I wrote 7A^6... I've no idea how it got parsed, but it was definitely something very, very wrong.

After correcting the input syntax, the output looks a lot saner:

Code: Select all

7*C^6 - 147*C^4 + 917*C^2 - 1681
C^6 - 21*C^4 + 35*C^2 - 7
7*C^6 - 35*C^4 + 21*C^2 - 1
7*C^6 - 91*C^4 + 245*C^2 - 169


The second of which matches what you wrote.

[mr_e_man]

So is your corrected input

7*A^6 - 35*A^4 + 21*A^2 - 1 = 0
7*B^6 - 56*B^4 - 112*B^2 + 64 = 0,

or

7*A^6 - 35*A^4 + 21*A^2 - 1 = 0
7*B^6 - 56*B^4 + 112*B^2 - 64 = 0?

[quickfur]

7*A^6 - 35*A^4 + 21*A^2 - 1
7*B^6 - 56*B^4 + 112*B^2 - 64

[mr_e_man]

Well, does the first pair of equations still make the 36th-degree monster? It's not clear to me whether it came from the sign flip or the missing "*".

[quickfur]

It came from the missing '*'. With the corrected signs, I still get a x^36 monster. A different one, but equally hairy and ugly.