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Duality of eta product identities: a new idea?

2012-03-25T00:07:41Z

Looking at the collection of Eta Function Product Identities by Michael Somos, it seems like generally those identities come in pairs: let's call two eta product identities $\sum\limits_{i=1}^r a_iP_i=0$ and $\sum\limits_{i=1}^r b_iQ_i=0$ dual if the eta products $Q_i$ can be re-labeled and the factors of each product re-arranged such that for all $i$, we can transform $P_i$ into $Q_i$ by replacing each argument $q^k$ or $k\tau$ of an eta function ($k\in\mathbb N$) by $q^{n/k}$ or $(n/k)\tau$. Note that the scalar coefficients $a_i$ and $b_i$ need not be the same, but they should be non-zero of course.

Example: The first two identities on the n=14 page, both of degree $12$, are

q14_12_44 = +1*u1^7*u2*u7^3*u14 +7*q*u1^3*u2*u7^7*u14 -1*u2^8*u7^4 -7*q^3*u1^4*u14^8 ; 
q14_12_64 = +1*u1^8*u14^4 +7*u2^4*u7^8 -56*q^2*u1*u2^3*u7*u14^7 -8*u1*u2^7*u7*u14^3 ;

re-arranged and written in a more humanly-readable form, putting $s_k=\eta(k\tau)$, we have

q14_12_44$\iff\qquad s_1^7s_2^{\ }s_7^3s_{14}^{\ } +7s_1^3s_2^{\ }s_7^7s_{14}^{\ } -s_2^8s_7^4 -7s_1^4s_{14}^8=0$

q14_12_64$\iff -56s_{14}^7s_7^{\ }s_2^3s_1^{\ }-8s_{14}^3s_7s_2^7s_1^{\ } +7s_7^8s_2^4+s_{14}^4s_1^8=0$

so both identities are duals of each other.

It should be easy to see that an eta product identity (edit: which is not dual to itself) cannot have two (linearly independant) duals. Is that really easy to see?

An eta product identity can also be self-dual, e.g. the known linear identities mentioned in my recent MO thread or the one equivalent to the well-known theta identity $\theta_3^4(q) = \theta_4^4(q) + \theta_2^4(q)$, which becomes, when written in terms of $\eta$'s, $$s_2^{24}=s_1^{16}s_4^8 +16s_4^{16}s_1^8 .$$

This concept of duality seems so logical that I wonder: Has there already been any research on that? Has each eta product identity a dual counterpart, unless it is self-dual ?

In the collection mentioned above, for $n=22$ there is only one identity listed, and that one is not self dual, it is even highly asymmetrical. Its dual should have the label x22_19_124. Maybe that one is missing just because the computer search had to stop at some point.

Given that self-dual identities have a higher degree of symmetry: can they always be expressed in terms of Ramanujan's psi, phi, and/or chi functions?

I would think the answer is yes, as there are even some not self-dual ones that can be expressed in that way.

Edit concerning the first question: Even this may be trickier than expected. For example, there are two self-dual identities, labeled q6_14_36a and q6_14_36b. They are:

$s_1^9s_3^{\ }s_6^4 +\ \ 6s_1^4s_2^{\ }s_6^9 +2s_2^9s_3^4s_6^{\ } -3s_1^{\ }s_2^4s_3^9=0$

$s_1^9s_3^{\ }s_6^4 +12s_1^4s_2^{\ }s_6^9 -4s_2^9s_3^4s_6^{\ } +3s_1^{\ }s_2^4s_3^9=0$.

They contain exactly the same $\eta$-products and so can also be considered as dual to each other (and to linear combinations of them). Moreover, by linear combinations we may eliminate any one of the four terms, resulting in four 3-term identities with coefficients

 0  1 -1  1
-1  0 -8  9
 1  8  0 -1
-1 -9  1  0.

Note that each of those has exactly one dual counterpart again, as they come in two dual pairs.
So the above conjecture of a unique dual doesn't necessarily hold if the given identity is already self-dual.

When can a family of polynomials get a weight function to be made orthogonal?

2012-01-27T22:35:16Z

Let $\lbrace P_n(z)\rbrace_{n\in\mathbb N_0}$ be a family of polynomials defined by a generating function $g(t,z)=\sum\limits_{n=0}^\infty P_n(z)t^n$ or by a contour integral $P_n(z)=\frac1{2\pi i}\oint\frac{g(t,z)}{t^{n+1}}dt$. Are there known sufficient conditions on $g$ or on the $P_n$ themselves that guarantee the existence of a weight function $w:I\to \mathbb R^+_0$ (where $I\subset\mathbb R$ is an appropriate interval) such that the $P_n$ are orthogonal w.r.t. $w$?

Linear eta product identities - how many are there?

2012-03-19T22:47:22Z

For the Dedekind eta function, defined as usual by $\eta(q) = q^{\frac1{24}} \prod\limits_{n=1}^{\infty} (1-q^{n}) $, let for brevity $e_k:=\eta(q^k)$. With this notation, a blog entry of Michael Somos gives three beautiful identities for sums of $\eta$-products where all exponents are only $0$ or $1$:

$I_{60}:\qquad \ \ e_{1}e_{12}e_{15}e_{20} + e_{3}e_{4}e_{5}e_{60}=e_{2}e_{6}e_{10}e_{30} $

$I_{210}:\qquad e_{1}e_{30}e_{35}e_{42 }+ e_{3}e_{10}e_{14}e_{105} = e_{2}e_{15}e_{21}e_{70} +e_{5}e_{6}e_{7}e_{210}$

$I_{30}:\qquad\ \ e_{1}e_{3}e_{5}e_{15}+ 2e_{2}e_{6}e_{10}e_{30}=e_{1}e_{2}e_{15}e_{30}+ e_{3}e_{5}e_{6}e_{10} $

For $I_{60}$ and $I_{30}$ the structure is clear at one glance if we write the divisors of $60$ as vertices of a Cayley-like graph (here: 'union' of two cube graphs $Q_3$ with the common face $(2,6,30,10)$ ):

4                  20

    2          10

        1   5

        3  15

    6          30

12                 60

Alternatively, if we define

$a_0:=e_{1} e_{15} \qquad b_0:= e_{3} e_{5} $

$a_1:=e_{2} e_{30} \qquad b_1:= e_{6} e_{10} $

$a_2:=e_{4} e_{60} \qquad b_2:= e_{12} e_{20} $

then

$I_{60}\iff a_0b_2+b_0a_2=a_1b_1$

$I_{30}\iff a_0a_1+b_0b_1=a_0b_0+2a_1b_1$.

For $I_{210}$ the symmetry is a bit less obvious to see. We can identify the divisors of $210$ with the vertices of a tesseract graph $Q_4$ or write the factors of the four products as lines of a matrix and note $a_{i,j}a_{4-i,j}=210$ as well as the factor $3$ between the two pairs of lines:

  1  30  35  42   
  3  10 105  14   

 70  21   2  15   
210   7   6   5

I'd suggest to call identities of this type linear eta product identities. Their linearity seems to enforce a high degree of symmetry in the way these three identities $I_n$ feature the divisors of $n$, which makes them very special among the thousands of known eta product identities. It looks like there is something deeper behind. And: Why do all products have exactly $4$ factors?

So, more precisely, for naturals $a\ge b$ let's define a linear eta product identity of type $\mathbf{(a,b)}$ as an identity $L_1+\cdots+L_a=R_1+\cdots+R_b$, where each $L_i$ and each $R_i$ is a finite product of pairwise different terms of form $\eta(q^{\lambda})$ with $\lambda\in\mathbb N$. (The products $L_i$ and $R_i$ don't need to be all different, e.g. the above $I_{30}$ is of type $(3,2)$ with $L_2=L_3$. But of course we want $\{L_i\}\cap\{R_i\}=\emptyset$, and also that the gcd of all the $\lambda$'s is $1$.)
Somos conjectures that $I_{60}$ is the only linear identity of type $(2,1)$.

Is it possible that the three above identities are only the first ones of a whole (infinite?) set of linear eta product identities, and/or that for naturals $a\ge b$, there is at most one such identity of type $(a,b)$?

Any relationship between Viswanath's constant and the Khinchine-Lévy constant?

2012-03-11T14:07:45Z

It is well-known that if ${\{{F_n}\}}$ is a random Fibonacci sequence then we have almost certainly $\lim \limits_{n\to\infty}\sqrt[n]{|F_n|}=\tau$ where $\tau\approx 1.554682275$ is Viswanath's constant. There is also a simplified version of Viswanath's proof with some more motivation.

Now, the Khinchine-Lévy constant $\gamma=e^{\pi^2/(12\ln2)}$ (sometimes also called Lévy constant) is obtained in a quite similar way as $\lim \limits_{n\to\infty}\sqrt[n]{q_n}$ for almost all reals $x$, where $\frac{p_n}{q_n}$ is the $n$th convergent of the continued fraction of $x$. (Note that we might as well take $\lim \limits_{n\to\infty}\sqrt[n]{p_n}$.)
So the question:

Is there a relationship between Viswanath's constant and the Khinchine-Lévy constant?

a limit by Gosper involving a product of arctan and $4^{1/\pi}$

2012-03-09T19:33:49Z

On the Wolfram page about pi formulas, there is this curious limit by R. W. Gosper (130) $$\lim\limits_{n\to\infty}\prod\limits_{k=n}^{2n}\dfrac{\pi}{2\arctan k}=4^{1/\pi}.$$

The only reference given is an entry from 1996 in some forum. Has anybody a proof or reference for this or similar formulas?

minimal polynomials of trig functions of ($k \pi/p$) and divisibility of coefficients by p

2012-01-24T21:27:47Z

Take an odd prime $p$ and put $x_0:=\sum\limits_{j=0}^{p-1}\left(a_{j}\sqrt{p}\cos\dfrac{j\pi}p+b_{j}\sin\dfrac{j\pi}p +c_{j}\tan\dfrac{j\pi}p\right)$, where the $a_{ij}$ are integers. If $f$ denotes the minimal polynomial of $x_0$, can we prove that $p$ divides all coefficients of $f$ except the leading one?

I have quite a bit of numerical evidence for this. Note that it obviously doesn't hold without the $\sqrt p$ factor, but more interestingly, it is also false if $\sqrt p$ goes with the other terms instead of the $\cos$ term. Moreover, it seems that in those cases, none of the non-zero coefficients is divisible by $p$.

(More generally, I think those coefficients are divisible by $p$ if we replace $\dfrac{j\pi}p$ by $\dfrac{j\pi}{p^r},\ r\in\mathbb N$ and do the sum over $j=0,...,p^r-1$.)

If all but one of the $a_j,b_j,c_j$ are $0$, the claim is quite easy to prove (and not new). For instance, for $x_0=\sin\dfrac{j\pi}p$ with any fixed $j$, we have explicitly $$f(x)=\sum\limits_{i=0}^k(-1)^k\dbinom p{2i+1}(1-x^2)^{k-i}x^{2i},$$ where $p=2k+1$. So the claim is obvious here.

Added: It should be clear from Galois theory that in general, the conjugates of $x$ are the sums obtained by replacing all the $j$'s by $kj$ for a fixed $k=2,...,p-1$.

Literature:

Beslin, S., de Angelis, V., 2004. The minimal polynomials of sin(2π/p) and cos(2π/p). Mathematical Magazine 77, 146–149.

Heierman, William E., Minimal polynomials for trig functions of angles rationally commensurate with π

Lang, Wolfdieter, Minimal Polynomials of sin (2π/n)

Surowski, David, and McCombs, Paul, Homogenous polynomials and the minimal polynomial of cos(2π/n)

W. Watkins and J. Zeitlin, The Minimal Polynomial of cos(2Pi/n), Am. Math. Monthly 100,5 (1993) 471-4.

Why are some q-analogues more canonical than others?

2012-03-01T23:25:27Z

It is striking that some q-analogs of functions, operators, identities and especially whole theorems seem quite "canonical", e.g.

the factorial and the q-Gamma function
the basic hypergeometric series (at least $_{r+1}\Phi_r$ and $_{r}\Psi_r$)
q-Pi and the q-Wallis formula

In a strict sense, q-analogs can of course not be canonical, as we might throw in almost everywhere powers of $q$ without changing the limit if $q\to1$.

I mean canonical in the sense that these are the forms that require the least extra powers of $q$, or, more importantly, that other q-identities/theorems using them also tend to avoid such extra powers at best, making the formulae shorter.

Does it really make sense to call these (and certain other) q-analogs "canonical"? And if so, is there an explanation why some are much more canonical than others?

(Or is there a better definition of canonical?)

The canonicity of the q-binomial coefficients is obviously accounted for by their relationship with linear subspaces. (So this is not an analytical criteria using the limit $q\to1$.)

What about the q-Gamma function? We may consider it canonical because of the (?!) q-analogue of the Bohr-Mollerup theorem proved by R. Askey, which states that for $0\lt q\lt 1$, the only logarithmically convex function satisfying $f(1)=1$ and $f(x+1)=\frac{q^x –1}{q–1}f(x)$ is the q-gamma function $ \Gamma_q(z)=(1-q)^{1-x} \frac{(q\;;\; q)_{\infty}}{{(q^x;\; q) _\infty }}.$

Also note that this formula looks at least as elegant as Euler's definition of $\Gamma(z)= \dfrac{1}{z} \prod\limits_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}.$

On the other hand, one of the most basic identities, the recursion of binomial coefficients, has only an "asymmetric" q-analog and thus two of them: ${\genfrac[]00{n}{k}}_q=q^k{\genfrac[]00{n-1}{k}}_q+{\genfrac[]00{n-1}{k-1}}_q={\genfrac[]00{n-1}{k}}_q+q^{n-k}{\genfrac[]00{n-1}{k-1}}_q$.

For the classical orthogonal polynomials, it looks like there exist systematically "nice" q-analogs (see e.g. this survey), but it is not clear if there is a certain sense in which those can be considered canonical. Maybe for the Chebyshev polynomials, there is one "best" q-analog.

Is there a reasonable way of considering certain q-analogs of orthogonal polynomials "canonical", e.g. their uniqueness w.r.t. to an appropriate criteria, as for most classical orthogonal polynomials?
Has a q-analog of a polynomial identity (e.g. involving binomial coefficients) more chances of being canonical if it has a combinatorial interpretation?

For the q-derivative, there are at least two completely different approaches, both with their merits. So there is no use looking for canonicity there.
But nevertheless the next question:

Are q-analogs conceptually similar to an extension from $\mathbb R$ to $\mathbb C$?

By the latter I mean the following:
I wonder if generally speaking, the shift from an entity to its q-analog(s) can be likened, at least sometimes, to the shift of passing from $\mathbb R$ to $\mathbb C$, in the sense that some q-analogs provide a more complete picture than the entity itself (cp. for the $\mathbb R\to\mathbb C$ case the fundamental theorem of algebra or the meromorphic extension of the zeta function)?

Many features in $\mathbb C$, e.g. the residue theorem, cannot be reduced to $\mathbb R$. Likewise for example, identities of "infinite q-polynomials" (i.e. of generating functions), cannot be taken to the limit $q\to1$.
In situations where there are several useful q-analogs, e.g. for the q-exponential function or the q-cosine, we might consider those corresponding to different panes of a Riemann surface like the one of $\sqrt{z}$ or $\ln z$.
And we can take it further:
The next step after $\mathbb R$ to $\mathbb C$ are the quaternions.
The next step after q-analogs are p,q-analogs. It looks like they haven't been studied a lot yet.

Thank you for reading me so far.
Some of the questions may be rather subjective. As for the title, I had thought at first about "Why are most q-analogues canonical?" But then I figured that in the whole ocean of q-analogues, maybe there are just some "very" canonical islands, but for the vast majority the notion of canonicity is more or less fuzzy. Is that a feasible perception? Even though some of these thoughts are somewhat philosopical, anyway, here goes. Looking forward to your input!

Answer by spanferkel for Arrangement of integers 1..k^2 in k*k grid to minimize energy function

2012-02-24T11:21:10Z

This is not an answer but too long for a comment. It is a heuristical argument that $\frac{k+1}2$ should be best possible.

It is easy to see that for an optimal arrangement, $1$ and $k^2$ must both occupy corners. Supposing opposite corners, consider the (monotone) lattice paths between them. Each path covers $2(k-1)$ of the differences we are looking for, so the average difference of those is at least $\frac{k^2-1}{2(k-1)}=\frac{k+1}2$ for each such path (equal if the numbers on the path are strictly increasing, greater otherwise).

This is not a proof because summing over all those paths, the differences are not counted equally often. But maybe it can be made into a proof. Problem: if the corners of $1$ and $k^2$ are not opposite, such an argument won't work.

The torus case seems harder, it looks like optimal constuctions may be different for even and odd $k$.

Can elliptic integral singular values generate cubic polynomials with integer coefficients?

2012-02-04T22:20:34Z

For the elliptic integral of first kind, $K(m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m^2sin^2\theta}} $, it is well-known that $K(m)$ can be expressed in what Chowla and Selberg call "finite terms" (i.e. algebraic numbers and a finite product of Gamma functions of rational values) whenever $i\frac{K(\sqrt{1-m^2})}{K(m)}$ belongs to an imaginary quadratic field $\mathbb Q(\sqrt{d})$ (see Theorem 7 in S. Chowla and A. Selberg, On Epstein's zeta function, J. reine angew. Math. 227, 86-110, 196).

Examples for these so called elliptic integral singular values are given on this Wolfram page (with some small typos) and in the note of J.M. Borwein and I.J. Zucker, "Elliptic integral evaluation of the Gamma function at rational values of small denominator," IMA Journal on Numerical Analysis, 12 (1992), 519- 526.

See also what Tito Piezas has to say about this in his pleasant-to-read Collection of Algebraic Identities.

The following question arises:

For these singular values, is there (always, or, if not always: when?) a polynomial $P(t)$ of degree 3 with integer coefficients such that $K(m)=c\int\limits_{t_0}^\infty\dfrac{dt}{\sqrt{P(t)}} $ with $c\in\mathbb Q$?
(EDIT: After Noam Elkies' remark, introduced $t_0$, the biggest real zero of $P$, instead of $0$ as the lower limit. Only "complete" integrals make sense here.)

In particular for $d=-7$, we have by the Carlson symmetric form $$\frac12\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t+1)(t+\frac{8+3\sqrt 7}{16})}} =K(k_7)=\dfrac1{7^{1/4}4\pi}\Gamma\left(\dfrac17\right)\Gamma\left(\dfrac27\right)\Gamma\left(\dfrac47\right),$$ on the other hand I have seen somewhere (I can't remember the reference) $$\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+21t+112)}} =\dfrac1{4\pi\sqrt{7}}\Gamma\left(\dfrac17\right)\Gamma\left(\dfrac27\right)\Gamma\left(\dfrac47\right)=\frac{K(k_7)}{7^{1/4}}.$$

I would like to get it straight at least for this example:

Can the polynomial in the first integral be transformed into one with integer coefficients? And is there any sort of relationship between both above polynomials?
Note that the ratio of the discriminants of the two above polynomials is $-2^{24}\cdot7^3$, and both of them do not yield affirmative answers, as the second one would have to be divided by $\sqrt7$ to obtain $K(k_7)$ directly!
EDIT: Follow-up question: If $P(t)$ is an integer cubic polynomial such that $\int\limits_{t_0}^\infty\dfrac{dt}{\sqrt{P(t)}} $ (with $t_0$ its biggest real zero) can be written in "finite terms", is this value always an algebraic multiple of an elliptic integral singular value $K(m)$?

how many consecutive integers $x$ can make $ax^2+bx+c$ square ?

2012-02-11T22:17:47Z

The following problem was raised in a Mathlinks thread:

If $a,b,c\in\mathbb Z$ such that $a\ne0$ and $b^2-4ac\ne 0$, for how many consecutive integers $x$ can $ax^2+bx+c$ ba a perfect square ?

The polynomial $-15x^2+64$ is obviously a square for the five numbers $x=-2,...,2$, but the method used for finding this in the above thread cannot be extended further. Should $5$ really be the best possible answer?

Has this problem been treated somewhere else?

Is there any expression for the Feigenbaum constants ?

2012-02-06T19:06:08Z

It has puzzled me for a long time that the Feigenbaum constant $\delta$ and reduction parameter $\alpha$ do not seem to be related to other constants (that is, numerically), even not to each other. In fact I have never seen them expressed as an integral, any kind of series or product, a nested expression, ... The only thing I have found on the internet is this algorithmic approach which further links to this, but it seems rather like an "a posteriori" method, being interested more in the algorithm than in the nature of $\delta$. Anyway, I would not expect the prominent occurrence of the number $163$ in the algorithm to have a deeper meaning.

On the other hand, I wouldn't be too surprised to see $\delta$ written as a continued fraction with "defineable" terms, given the fact that it can be defined by $\delta=\lim\limits_{n\to\infty}\dfrac{\mu_n-\mu_{n-1}}{\mu_{n+1}-\mu_n}$, where the $\mu_n$ are the bifurcation points of an iterated map. But I have never seen one either. Any leads?

Answer by spanferkel for Product of sine

2012-02-02T16:13:35Z

I couldn't resist... A different construction, which is "irreducible" also for odd $n$ unlike the construction of Ace of Base, is already implied by the very first example the OP gives. (Did you find it by computer and didn't notice?) Look at the differences of the numerators... It can be straight away generalized to $$\boxed{\sin\left(\dfrac{\pi}{2^{n+1}+2} \right)\prod\limits_{k=1}^{n-1}\sin\left(\dfrac{2^n+2^k+1}{2^{n+1}+2}\pi \right)=\dfrac1{2^n}}.$$ Sure enough, denoting $\sin\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ and $\cos\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ by $s_k$ and $c_k$ respectively, we have $$LHS=s_1\prod\limits_{k=1}^{n-1}c_{2^k}
=\frac1{2c_1}s_2\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{4c_1}s_4\prod\limits_{k=2}^{n-1}c_{2^k} =\cdots=\frac{s_{2^{n}}}{2^nc_1}=\frac1{2^n}.$$

Integrating gamma products and quotients over a vertical line

2012-01-29T21:21:44Z

The Wolfram functions collection contains a small number of integrals of products and quotients of terms $\Gamma(a_i\pm t)$ over a vertical line, all of which can be evaluated in terms of only gamma functions.

Does somebody know references concerning formulae of this type? If they are obtained by the residue theorem, how?

The second one seems to be false. By putting $2b=a-c+1$, we would get from it, after using the doubling formula, $$\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma(a+t)\Gamma(c-t)dt=2\pi i\cdot2^{a+c}{\Gamma(a+c)},$$ whereas for $a,c\in\mathbb Z$ such that $a+c\ge1$, we have $$\int\limits_{\gamma-i\infty}^{\gamma+i\infty}\Gamma(a+t)\Gamma(c-t)dt=\pi i\int\limits_{-\infty}^{\infty}\frac{\overbrace{(1-a-t)\cdots(c-1-t)}^{(a+c-1)\ {terms}}}{\cosh \pi t}dt=\pi i\int\limits_{-\infty}^{\infty}\frac{\Gamma(c-t)\; dt}{\Gamma(1-a-t)\cosh \pi t}$$ which may be written in umbral form as $(1-a-\frac E2)\cdots(c-1-\frac E2)$, i.e. as a linear combination of terms $\int\limits_{-\infty}^{\infty}\dfrac{t^{k}dt}{\cosh \pi t}=\dfrac{E_k}{2^k}$, where the $E_k$ are the (absolute) Euler numbers, thus it can definitely not be written in terms of $\Gamma(a+c)$ etc.

Some others of the Wolfram identities look like they could be generalized, e.g. the first and third one. Is it true e.g. that for $n\ge3$, $$\int_{\gamma-i\infty}^{\gamma+i\infty}\prod\limits_{j=1}^n\Bigl(\Gamma(a_j+t)\Gamma(b_j-t)\Bigr)dt=2\pi i\frac{\prod_j\prod_k\Gamma(a_j+b_k)}{\Gamma(\sum a_j+\sum b_j)^{n-1}}$$ where all products and sums run from $1$ to $n$? (Note that again, this does not hold for $n=1$.)

Cubic polynomials with "nice" roots, which can be expressed by trig functions of rational angles

2012-01-22T21:04:07Z

Consider the cubic polynomial $x^3-ax+b$ for $a,b\in\mathbb N$.

It has three real roots which, by Cardano's formula, can of course be written in closed form using thirds of angles or cube roots of complex numbers.

Now, in some cases, there is a much nicer way (at least in my opinion, how objective can this be??) of writing the roots: as a rational expression of trig functions of rational angles. Let's call such a polynomial nice in honour of Mathlinks, where this kind of roots are traditionally referred to as "nice roots", presumably starting with this one.

For example, we can write

the roots of $x^3-3x+1=0$ as $2\cos\dfrac{k\pi}9,\ k=2,4,8$,
the roots of $x^3-21x+7=0$ as $\dfrac3{2\cos\frac{k\pi}7}+2,\ k=2,4,8$,
the roots of $x^3-19x+19=0$ as $\dfrac7{2(\cos k\omega+\cos7k\omega+\cos49k\omega)}-2,$ where $\omega=\dfrac{\pi}{19}$ and $k=1,3,9$, or equivalently,

$ \dfrac{\sqrt{19}}{2(\sin k\omega+\sin 7k\omega+\sin 49k\omega)} $ or $ \dfrac{2\sqrt{19}}{\csc k\omega+\csc 7k\omega+\csc 49k\omega}. $

The last example (see also here) shows that it makes sense to also admit square roots. (Probably that doesn't change anything, by virtue of identities of type $ \tan\frac{{3\pi }}{{11}}+4\sin\frac{{2\pi }}{{11}}=\sqrt{11} $, but it can obviously allow to write roots in a more elegant way.)

Question:

Is it possible to characterize the pairs $(a,b)$ that yield nice polynomials?

Some ideas

For any angle $\frac{k\pi} n$ involved in a nice root, $\mathbb Z_3$ must be a subgroup of the multiplicative group $\mathbb Z_n^*$.
For a nice polynomial $x^3-ax+b$, its discriminant $\Delta=4a^3-27b^2$ is always a square. (See theorem 2 here).
It appears that if the involved fractions of $\pi$ have a common prime denominator, this prime divides both $a$ and $b$. (Why?)
The article Kurt Girstmair, On root polynomials of cyclic cubic equations, ARCHIV DER MATHEMATIK, Vol. 36, N° 1, 313-326, DOI: 10.1007/BF01223707 may possibly be helpful, but I can't access it.
Maybe nice polynomials have something in common with the ones mentioned in this MO thread, e.g. in the sense that in the definition of a nice cubic polynomial, if we replace "trig functions of rational angles" by "roots of unity", nothing essential changes?

convex polytopes with many faces and edges but few cells and vertices

2012-01-20T18:32:40Z

For a convex polytope $P$ in $\mathbb R^4$, denote by $N_0,N_1,N_2,N_3$ respectively the number of vertices, edges, faces, cells. By Euler's formula, we know $N_0+N_2=N_1+N_3$, which means there is a sort of equilibrum among the $N_i$. But I wonder if there exist upper and/or lower bounds for $f(P):=\frac{N_1+N_2}{N_0+N_3}$ where $P\subset\mathbb R^4$ is any convex polytope.

E.g. for the regular one called 24-cell, we have $f(P)=4$.

How many vertices/edges/faces at most for a convex polyhedron that tiles space?

2012-01-18T22:38:06Z

I wonder if this problem has already been examined before:

Consider a convex polyhedron that tiles $\mathbb R^3$. What is the maximum of vertices/edges/faces that such a polyhedron can have?

Intuitively, it seems that the truncated octahedron is best possible for edges (36) as well as for vertices (24) Its packing is also known as "bitruncated cubic honeycomb".

For faces, we can do better than 14, as there is a polyhedron with 16 faces that can be obtained as follows: Take a truncated tetrahedron and add on each triangular face a small pyramid that is a quarter of a tetrahedron. The tessellation of it is known as the quarter cubic honeycomb, with each small tetrahedron "distributed" among its four neighbors.

Questions: Are these best possible? What about the corresponding problem in higher dimensions?

In $\mathbb R^4$, it looks like the polytope yielded by the equivalent of the "Quarter cubic honeycomb" tiles it. This one, based on the truncated 5-cell has 25 cells, 60 faces, 60 edges, and 25 vertices, and so for cells and vertices, it does again (slightly) better than the 24-cell with its 96 faces, 96 edges, and 24 vertices.

Euler's divergent series sum n!*(-1)^n: what is known about the resulting constant?

2012-01-14T18:13:34Z

Much of the theory of continued fractions has been developped by Euler in the 18th century. The little survey "Euler: continued fractions and divergent series (and Nicholas Bernoulli)", mentions towards the end the continued fraction $$f(x)=\dfrac{1}{1+\dfrac{x}{1+\dfrac{x}{1+\dfrac{2x}{1+\dfrac{2x}{1+\dfrac{3x}{1+\dots}}}}}}$$ which Euler has "derived" from the divergent power series $1-x+2x^2-6x^3+24x^4-+...$ .

For $x=1$, the continued fraction converges to a limit $f(1)\approx 0.5963475922$. I was wondering: what is known about the values $f(x)$, in particular $f(1)$? Are they known to be transcendental for $x\in\mathbb N$? Can they be expressed by other known constants?

Answer by spanferkel for Rearrangement-style inequality with lots of terms and little evidence

2012-01-13T16:00:23Z

Edit: the following is false. (see my comment below) The Sum-maximum conjecture is easily proved by Fedja's observation. Put $S_a:=\frac1n\sum a_k $ and for big $N$, define $a_k^':=1+\frac{a_k-S_a}N$, similarly for $S_b$ and $b_k^'$. According to Fedja, the Product-sum conjecture holds for $a_k^'$ and $b_k^'$, and this implies the Sum-maximum conjecture for $a_k^'$ and $b_k^'$. Now we can replace $a_k^'$ and $b_k^'$ by $a_k$ and $b_k$ because the linear transformations don't change anything for the difference of the two sums.

Too bad we can't say the same for the Product-sum conjecture as well.

Comment by

2012-03-25T21:31:39Z

I suppose that's about the question of unique duals? It's not that simple. Even once we have established that a dual always exists, a simple "infinite descend" argument may not necessarily work. See my edit above.

Comment by

2012-03-13T21:27:13Z

Thank you for sharing these rather deep insights!

Comment by

2012-03-09T20:43:25Z

yes, this was intended.

Comment by

2012-03-02T08:41:28Z

@Johann: and you don't think it is worth giving a special consideration to those "doubly nice ones" where both properties coincide?

Comment by

2012-02-25T23:34:27Z

For the partial products, it even seems like the polynomials $$\frac{(-1)^N}{(q-1)^{3N-2}}\frac{d^2}{dq^2}\prod_{n=1}^N (1-q^{2n-1})^2(1-q^{2n})$$ have positive and unimodal coefficients. See e.g. tinyurl.com/WolframForNEq8 :)

Comment by

2012-02-24T17:42:56Z

For the torus, the filling in by rows also yields the same average $\frac{k^2-1}k$ as, for $k=2^n$, the Morton layout!

Comment by

2012-02-23T18:54:38Z

Note that for $k=2^n$, the "filling in row by row" yields exactly the same average difference as for the Morton layout, viz. $(k+1)/2$. (assuming you mean horizontal and vertical neighbors only). So after all, I'd think there is no better arrangement than "filling in row by row" for the non-torus case.

Comment by

2012-02-19T21:16:11Z

This is really cool. Thanks again!

Comment by

2012-02-05T13:16:11Z

Thank you very much. I have edited the lower limit, as it is exactly examples like yours I'm interested in. I suppose the curve is uniquely determined by $d$, up to translation? So again the question: What about $n=-7$?

Comment by

2012-02-05T00:03:33Z

I didn't necessarily think of that restriction. So if we restrict them to that, which polynomial(s) would that be for the $d=-7$ case? And how to find them? Unique?

Comment by

2012-01-31T20:37:41Z

No, that would be the complement of $3K_{2n}$.

Comment by

2012-01-31T19:15:35Z

@Ira: But the trivalent graphs may have multiple edges (but no loops). And even counting those wouldn't help to solve the problem I think. If I get it right, the problem asks for the number of 1-factors of the complement of $2nK_3$.

Comment by

2012-01-30T22:54:36Z

Thank you very much.

Comment by

2012-01-28T10:24:36Z

Thank you. That's about what I expected: that one of the necessary conditions is essentially sufficient. Like Noam D. Elkies, I didn't really expect there to be any hope of deriving anything directly from a generating function. Let alone the even more hopeless question: If only $g(t,z)$ is given, is there even a way to know, without doing a Taylor expansion, that the coefficients will actually be polynomials in $z$?

Comment by

2012-01-25T18:34:21Z

I have just edited the expression for $x$ to make it (slightly) more general in a quite obvious way, and I think what you suggest should still apply. Only I don't really understand what you presume, particularly what do you mean by "early terms of the expansion"?