This … is … “Maximal Jeopardy!”

What’s the largest possible amount of money you can win in a single game of TV’s “Jeopardy!”?

Read my note This … is … “Maximal Jeopardy!” to find out.

Posted 2 February, 2013 by Blue in Classroom, Misc. Math

A Hedronometric Theorem of Menger

In 1928, Karl Menger outlined necessary and sufficient conditions for a set of edge lengths to determine an actual, non-degenerate, tetrahedron. The conditions amount to dead-simple sanity checks that the consequent face areas and volume have to be positive real numbers.

In the short note “A Hedronometric Theorem of Menger”, I derive (as the title suggests) a hedronometric counterpart of Menger’s result, giving necessary and sufficient conditions for a set of face and pseudoface areas to determine a tetrahedron. It’s really about time I did.

The note came about when I observed a “Heronic duality” that I had never noticed (nor even suspected) before. Writing “\([\bullet]\)” for what I’ll call the Heronic product,

$$[x,y,z] := \left(x+y+z\right)\left(-x+y+z\right)\left(x-y+z\right)\left(x+y-z\right)$$

we know that the area, \(A\), of a triangle with edge lengths \(x, y, z\) is given by

$$16 A^2 = [x,y,z]$$

That’s Heron’s formula (hence, the term “Heronic product”), and it gives us areas from edge lengths.

As it turns out, Heron products give us tetrahedral edge lengths from face and pseudoface areas (and volume); for instance, if \(a\) is the edge between faces \(Y\) and \(Z\), and thus also a diagonal of pseudoface \(H\), then

$$9V^2 a^2 = \left[H,Y,Z\right]$$

Going back the other way, we can even express the areas of tetrahedral pseudofaces as Heronic products:

$$16 H^2 = \left[a,d,j\right] \qquad \text{where} \qquad j^2 = a^2-b^2+c^2+d^2-e^2+f^2$$

The Heronic product is quite the workhorse!

This note answers the an Open Question from my “Heron-like Results …” post, where I wrote that I had “other reasons to suspect Cayley-Menger-like ties” to the Pseudo-Heron volume formula. There, I wondered what matrix had that formula as its determinant; in the note, that matrix is \(\Delta^\star\). (There may be a better one, but that’s the best I have right now.)

Update. I’ve added a Corollary that gives conditions for just six faces —three standard, three pseudo— to determine a tetrahedron.

Posted 24 November, 2012 by Blue in Hedronometry, Open Question

What I know about hyperbolic tetrahedra

Inspired by Mednykh and Pashkevich’s “Elementary Formulas for a Hyperbolic Tetrahedron”, I have compiled most of my disparate notes about hyperbolic hedronometry into one document: “Hedronometric Formulas for a Hyperbolic Tetrahedron”. I consider this a “living document” that I will update as I learn more about the subject matter.

It’s primarily a formula look-up list for myself, so I’ve left out a lot of exposition and proof. There may be a few typos, math-os, or outright errors. (There are at least two places where my results disagree with Mednykh and Pashkevich. One is a sign discrepancy, and the other an errant square root.) So, caveat reador.

As for content, “Hedronometric Formulas” covers familiar territory, from the foundational Laws of Cosines to the roadblock that is volume. There are also a few never-before-noted results, primarily involving “pseudo-altitudes” (each of which is a segment perpendicular to a pair of opposite edges); perhaps the coolest of these is the “Law of Side-Angle-Side-Altitude Sines”, which asserts that the product

$$\sinh{(\text{edge})} \;\cdot\; \sinh{(\text{edge})} \;\cdot\; \sin{(\text{angle})} \;\cdot\; \sinh{(\text{altitude})}$$

is a metric invariant of a hyperbolic tetrahedron, having a constant value (related to the Gram determinant) for any choice of two distinct “edge”s and the “angle” and “altitude” they determine. (For adjacent edges, “angle” is the angle between them and “altitude” is the altitude to the face they bound; for opposite edges, “altitude” is the corresponding pseudo-altitude between them, and “angle” is the “twist” of those edges about the pseudo-altitude.) This is analogous to the Euclidean result in which “edge \(\cdot\) edge \(\cdot\) sin(angle) \(\cdot\) altitude” always gives “\(6 \cdot \text{volume}\)”, which is also valid across standard and pseudo elements.

I also provide an appendix describing a tetrahedron in coordinatized hyperbolic space. “Coordination” happens to have provided the route to proving key results about pseudo-altitudes and twists, but generally it seems even messier than Euclidean analytic geometry can often be.

Some Open Questions raised (or repeated) here:

  • What is the Pythagorean Theorem for Right-Corner Simplices?
  • Is there a “monolithic” integral (in the spirit of the Derevnin-Mednykh formula) for the volume of a tetrahedron parameterized by its face and pseudo-face areas rather than its dihedral angles?
  • What is the geometric interpretation of a hyperbolic pseudo-face?

Regarding the last: A Euclidean pseudo-face is the quadrilateral shadow of a tetrahedron projected into a plane parallel to a pair of opposite edges. Parallelism and projection are tricky concepts in hyperbolic space, but interestingly there is a shadow-like construction —valid only in certain cases— of a quadrilateral with a given pseudo-face area (so that it’s reasonable to recognize the quadrilateral as the pseudo-face). A construction that works in general eludes me.

Posted 16 September, 2012 by Blue in Hedronometry, Open Question

Master of the Puniverse

I halve the power!

… for non-negative \(x\) and real \(n\), not both zero.

Posted 15 August, 2012 by Blue in Classroom, Misc. Math

The Descartes Rule of Sweeps

Something about Descartes’ Rule of Signs had bothered me ever since my exposure to it in high school.

As you know, the Rule of Signs runs something like this:

For a polynomial with non-zero real coefficients, the number of positive roots is, at most, the number of sign changes in the coefficient sequence (ordered by power); more specifically, the difference between the number of positive roots and the number of coefficient sign changes is even.

For instance, the polynomial

$$p(x) = x^5 + 2 x^4 – x – 2$$

has a coefficient sequence \(\{1,2,-1,-2\}\) with one sign change, so that \(p\) has a single positive root.

The Rule is a curious and far-from-obvious result, and the oft-cited proof at Cut the Knot is effective, but not terribly insightful. It’s not clear why the Rule works. That’s bothersome, of course, but not what bothered me.

We can also use the Rule to get information about the number of negative roots of a polynomial; we define an auxiliary polynomial that reverses the sign of \(x\) … for instance:

$$q(x) := p(-x) = -x^5+2x^4+x-2$$

The negative roots of \(p\) are the negatives of the positive roots of \(q\), and there are either two of those, or none, since \(q\) has two coefficient sign changes.

At this point, we have accounted-for three, or maybe just one, of the roots of \(p\). The Fundamental Theorem of Algebra tells us that there should be a total of five. Clearly, none of the roots is zero; but, if we’ve exhausted positive, negative, and zero candidates, then what’s left? Non-real complex numbers, of course (in conjugate pairs).

That’s what bothered me.

Not that I have anything against complex numbers; quite the contrary! The thing is that, while the Descartes Rule of Signs forces us to confront the complex numbers as a fact —and ultimately, a completion— of algebraic life, it won’t address non-reals in any specificity. We get an idea of how many roots lie in either direction along the real axis, but no information about roots along, say, the imaginary axis. That doesn’t seem fair.

On one level, the limitation is understandable. Consider the trick that allows us to count negative roots of \(p\); tweaking this to count roots of along the “positive” imaginary axis would require us to a new auxiliary polynomial thusly …

$$r(x) := p\left(\frac{x}{i}\right) = p(-i x) = -i x^5 + 2 i x^4 + i x – 2$$

… and then simply to count sign changes: from “\(-i\)” to “\(2i\)” is one; from “\(2i\)” to “\(i\)” doesn’t count; and then from “\(i\)” to “\(-2\)”— Hey, waydaminnit! The notion of “sign change” from “\(i\)” to “\(-2\)” doesn’t make any sense! … and so, we give up and go home.

Of course, that defeatist attitude runs contrary to the proper mathematical philosophy toward apparent nonsense; our instincts should be to make sense out of what we encounter. Besides, to turn away now is to ignore one of the fundamental conceptual insights of complex algebra: a sign change isn’t the result of a discrete flip across the origin, but rather the result of a continuous half-turn about it. Hmmm …

Adapting the notion of “sign change” to apply to the transition from “\(i\)” to “\(-2\)” in a coefficient sequence would seem to be very-nearly obvious: measure the “angular sweep” … which in this case is a quarter-turn (in the standard direction), or \(\pi/2\). Does this help?

I’ll cut to the chase: Yes. Yes, it does help.

We compute the angular sweep of polynomial \(r\) thusly: “\(-i\)” to “\(2i\)” is a sweep of \(\pi\); from “\(2i\)” to “\(i\)” doesn’t count; and then from “\(i\)” to “\(-2\)” is a sweep of \(\pi/2\). All together, that’s a total sweep of \(3\pi/2\).

In the original Rule of Signs, each sign change contributes one to the tally of possible positive roots; it also contributes \(\pi\) to this angular sweep value. Thus, according to Descartes, the tally of possible positive roots is exactly equal to the number of half-turns in the sweep.  (And the Rule admits that the tally may be off by an even number … An even number of half-turns is some number of full circles; we’re used to this kind ofmeasurement when measuring angles! Coincidence?) When the angular sweep lies between half-turns, we (very reasonably) round down; In the case of polynomial \(r\), the rounded-down count of half-turns in the sweep is one, so the tally of possible positive roots —equivalently, the tally of \(p\)’s roots along the positive imaginary axis— is also exactly one. And, in point of fact, \(p\) has \(i\) as its sole “positive imaginary” root.

To look for roots along the “negative imaginary” axis, we construct yet another auxiliary polynomial:

$$s(x) := p\left(\frac{x}{-i}\right) = p(ix) = i x^5 + 2 x^4 – i x – 2$$

The angular sweep from “\(i\)” to “\(2\)” is \(3\pi/2\); from “\(2\)” to “\(-i\)” is another \(3\pi/2\); from “\(-i\)” to “\(-2\)” is yet another \(3\pi/2\).  That’s a total sweep of \(9\pi/2\), which covers four half-turns. We happen to know that the polynomial \(p\) has a single root —namely, \(-i\)— along the negative imaginary axis, so the root count is certainly bounded above by the half-turn tally, although it doesn’t differ from that tally by an even number.

Interestingly, if we sweep from coefficient to coefficient in the clockwise direction, \(s\)’s angular sweep is \(\pi/2 + \pi/2 + \pi/2 = 3\pi/2\), which covers a single half-turn, and thus accurately predicts the sole negative-imaginary root of \(p\). It makes sense that we should consider oppositely-directed sweeps —and (again, very reasonably) adopt the smaller value as “the” angular sweep— because a polynomial knows nothing of our bias toward counter-clockwise rotations.

In any case … We now have a version of the Rule of Signs that does justice to the non-reals.

My note “The Descartes Rule of Sweeps and the Descartes Signature” explores these ideas in more depth; in particular, the “Signature” provides a holistic approach to seeking roots in a given direction of the complex plane. Unfortunately, my proof of the Rule of Sweeps is based directly on the Rule of Signs; consequently, we still don’t really know why this all works. (That’s my current source of bother in this area.)

The Open Question:

What is the independent proof of the Rule of Sweeps that establishes the conceptual intuition for believing it?

The sweep notion —with its rotational aspect and its quotients by \(\pi\)— seems awfully-closely tied to winding numbers and Cauchy’s Integral Formula. I get the sense that we’re one clever contour integral away from enlightenment … but apparently I’m insufficiently clever.

For interactive engagement with this stuff, see my “Descartes Signature Explorer” at the Wolfram Demonstrations Project.

Posted 20 July, 2012 by Blue in Misc. Math, Open Question