text:
## Tristram-Levine signatures with Scilab

The signature of a Hermitian matrix ${A}$ can be defined either as the pair (number of positive eigenvalues, number of negative eigenvalues), or simply as the difference

$\displaystyle s(A)=\#\{\lambda_i: \lambda_i>0\}-\#\{\lambda_i: \lambda_i<0\}$

The function ${s(A)}$ hides some information when the matrix is degenerate, but this will not be of concern here. For a matrix of size ${n}$, the number ${s(A)}$ is between ${-n}$ and ${n}$ and has the same parity as ${n}$.

Given any square matrix ${V}$ with real entries and a complex number ${\omega}$, we can form ${V_\omega = (1-\omega)V+(1-\bar\omega)V^T}$, which is a Hermitian matrix. Then ${s(\omega):=s(V_\omega)}$ is an integer-valued function of ${\omega}$. Restricting attention to the unit circle ${|\omega|=1}$, we obtain a piecewise constant function with jumps at the points where ${V_\omega}$ is degenerate.

When ${A}$ is a Seifert matrix of a knot, the function ${s(\omega)}$ is the Tristram-Levine signature of the knot. To each knot ${K}$ there are infinitely many Seifert surfaces ${F}$, and to each Seifert surface there are infinitely many Seifert matrices ${A}$, depending on how we choose the generators of ${H_1(F)}$. Yet, ${s(\omega)}$ depends on ${K}$ alone.

Below I plot ${s(\omega)}$ for a few knots using Scilab for computation of signature and the knot data from

J. C. Cha and C. Livingston, KnotInfo: Table of Knot Invariants

Technical point: since Scilab enumerates colors using positive integers, the colors below correspond to the number of positive eigenvalues rather than the signature. Namely, black for 0, blue (1), green (2), and cyan (3).

As always, first comes the trefoil:

One of its Seifert matrices is

$\displaystyle \begin{pmatrix} -1 & 0 \\ -1 & -1 \end{pmatrix}$

and the Tristram-Levine signature is

Next, the knot ${8_5}$

with Seifert matrix

$\displaystyle \begin{pmatrix} -1& 0& 0& -1& -1& -1\\ 0& 1& 0& 0& 0& 0\\ -1& 0& -1& -1& -1& -1\\ 0& -1& 0& -1& -1& -1\\ 0& -1& 0& 0& -1& 0\\ 0& -1& 0& 0& -1& -1 \end{pmatrix}$

and the signature

And finally, the knot ${8_{19}}$:

with Seifert matrix

$\displaystyle \begin{pmatrix} -1& 0& 0& 0& 0& 0\\ -1& -1& 0& 0& 0& 0\\ -1& -1& -1& -1& 0& -1\\ -1& -1& 0& -1& 0& 0\\ 0& 0& -1& -1& -1& -1\\ -1& -1& 0& -1& 0& -1\end{pmatrix}$

and the signature

I experimented with a few more, trying to create more colors. However, guessing the complexity of the signature by looking at the Seifert matrix is one of many skills I do not have. So I conclude with the simple code used to plot the signatures.

 function sig(A) clf(); [n,n] = size(A); Npoints = 200; r = 2*%pi/Npoints; arcs = zeros(6,Npoints); colors = zeros(Npoints); for m = 1:Npoints x = cos(2*%pi*(m-1/2)/Npoints); y = sin(2*%pi*(m-1/2)/Npoints); omega = complex(x,y); B = (1-omega)*A+(1-conj(omega))*A'; signature = sum(sign(spec(B))); colors(m) = (signature+n)/2+1; arcs(:,m) = [x-r,y-r,2*r,2*r,0,360*64]'; end xfarcs(arcs,colors'); replot([-1,-1,1,1]); a=get("current_axes"); a.isoview="on"; endfunction

My university offers a Certificate of Advanced Study in Data Science:

As the world’s data grow exponentially, organizations need to understand, manage and use big data sets. These data spawned the term “big data,” which now monopolizes forward-thinking business dialogue.

Well, I have some medium-size data from last semester: the grades of business calculus students at the academic drop deadline. At that moment the available data consisted of 18 homework sets, 7 quizzes, and 1 midterm exam: 26 dimensions total. Based on these data, can we tell whether the student should drop the course?

Looks like we need some dimension reduction here. I achieve it by simply replacing 18 homework scores with their average, and likewise for quizzes. This has the effect of projecting 26-dimensional data to 3-dimensional space. I scaled each to the interval ${[0,1]}$, so that the data is represented by 139 points in the cube ${[0,1]^3}$.

Suppose that the minimal acceptable grade is C-; the grades of D or F are worse than dropping the class. Accordingly, the points are colored blue (C- or better) or red (D or F).

The goal is to find a plane ${Ax+By+Cz=D}$ that separates the red dots from the blue ones. Since there is no plane that separates the groups perfectly, the task calls for a soft-margin support vector machine. I decided to get my hands dirty with actual computations in a spreadsheet.

My objective is

$\displaystyle \mathcal{E}(A,B,C,D) = \sum_{i=1}^{139} ((0.1-\epsilon_i (Ax_i+By_i+Cz_i-D))^+)^2 \rightarrow \min$

subject to the normalization ${A+B+C=1}$. Here ${z^+=\max(z,0)}$, ${\epsilon_i=1}$ for blue dots and ${\epsilon_i=-1}$ for red dots. The logic is that I want ${Ax_i+By_i+Cz_i-D}$ to have the sign ${\epsilon_i}$, that is, to be on the correct side of the separating plane. Squaring the penalty terms helps with minimization. The term ${0.1}$ introduces a penalty for being too close to the separating plane; this ought to improve the quality of separation.

The gradient of ${\mathcal{E}}$ is easy to calculate: for example,

$\displaystyle \frac{ \partial \mathcal{E}}{\partial A}= -2 \sum_{i=1}^{139} \epsilon_i x_i (0.1-\epsilon_i (Ax_i+By_i+Cz_i-D))^+$

The spreadsheet recalculated both ${\mathcal{E}}$ and ${\nabla \mathcal{E}}$ every time I changed the values of ${A,B,D}$ (the value of ${C}$ was set to ${1-A-B}$). This made it easy to do gradient descent “by hand”, starting with ${A=B=C=1/3}$ and ${D=0.7}$.

At the final step the gradient of ${\mathcal {E}}$ is parallel to the gradient of the constraint function ${A+B+C}$: we are at a critical point. Since ${\mathcal{E}}$ is convex, the point is a minimum. The optimal plane has the equation

$\displaystyle 0.4703x+0.3285y+0.2012z=0.69035$

which tells us that Exam 1 score $x$ is the most effective predictor while homework average $z$ is the least effective. (Homework was done online, with unlimited attempts, and no control over who actually did the work.)

The plane gives the correct prediction in 123 out of 139 cases. Not bad, considering that the data is incomplete and somewhat biased; it does not include the students who made a forward-thinking business decision and dropped the class by the deadline.

## Gromov-Hausdorff convergence

The Hausdorff distance ${d_{ H}(A,B)}$ between two subsets ${A,B}$ of a metric space ${X}$ is defined by ${d_{ H}(A,B)=\inf\{r>0: A\subset B_r \text{ and } B\subset A_r \}}$, where ${A_r,B_r}$ are (open/closed does not matter) ${r}$-neighborhoods of the sets. Informally: ${d_{ H}(A,B) if no matter where you are in one set, you can jump into the other by traveling less than ${r}$. For example, the distance between letters S and U is about the length of the longer green arrow.

Generally, one assumes the sets to be closed (to avoid zero distance) and bounded (to avoid infinite distance). But I will not; in this post I’m not really interested in verifying all the axioms of a metric.

The Gromov-Hausdorff distance is defined between metric spaces ${X,Y}$ as follows: it is the infimum of ${d_{ H}(f(X),g(Y))}$ taken over all isometric embeddings ${f\colon X\rightarrow Z}$ and ${g\colon Y\rightarrow Z}$ into some metric space ${Z}$.

The infimum over all pairs of embeddings into all conceivable metric spaces does not sound like something you would want to compute in practice. Of course, the matter boils down to equipping the abstract union ${X\sqcup Y}$ with pseudometrics that are compatible with the original metrics on ${X}$ and ${Y}$.

A more directly computable notion of distance (not necessarily a metric) can be given as follows: ${\rho_{GH}(X,Y)}$ is the infimum of all ${\epsilon>0}$ for which there exist two maps ${f\colon X\rightarrow Y}$ and ${g\colon Y\rightarrow X}$ such that:

1. ${d_X(g\circ f(x),x)\le \epsilon}$ for all ${x\in X}$
2. ${d_Y(f\circ g(y),y) \le \epsilon}$ for all ${y\in Y}$
3. ${|d_Y(f(x_1), f(x_2)) - d_X(x_1,x_2)| \le \epsilon }$ for all ${x_1,x_2\in X}$
4. ${|d_X(g(y_1), g(y_2)) - d_Y(y_1,y_2)| \le \epsilon }$ for all ${y_1,y_2\in Y}$

This is not as elegant as “infimize over all metric space”, but is more practical. For example, it is easy to check that the sequence of one-sheeted hyperboloids ${H_n = \{x^2+y^2=z^2+1/n\}}$

converges to the cone ${C = \{x^2+y^2=z^2\}}$.

Using cylindrical coordinates, define ${f_n\colon H_n\rightarrow C}$ by ${f_n(r,\theta,z) = (\sqrt{r^2-1/n}, \theta,z) }$ and ${g\colon C\rightarrow H_n}$ by ${g_n(r,\theta,z) = (\sqrt{r^2+1/n}, \theta,z)}$, with an arbitrary choice of ${\theta}$ at the point ${g(0,0,0)}$. Now check the items one by one:

1. ${f_n\circ g_n}$ is the identity map on ${C}$
2. ${g_n\circ f_n}$ fixes all points of ${H_n}$ except for those with ${z=0}$. The latter are displaced by at most ${2/\sqrt{n}}$.
3. follows from ${|\sqrt{r^2-1/n} - r|\le 1/\sqrt{n}}$ on ${[1/n,\infty)}$
4. follows from ${|\sqrt{r^2+1/n} - r|\le 1/\sqrt{n}}$ on ${[0,\infty)}$

Note that the Gromov-Hausdorff convergence of manifolds is understood with respect to their intrinsic metrics. Although both ${H_n}$ and ${C}$ are naturally identified with subsets of ${\mathbb R^3}$, it would be a mistake to use the Hausdorff distance based on the Euclidean metric of ${\mathbb R^3}$. Even though this extrinsic metric is bi-Lipschitz equivalent to the intrinsic metric on both ${H_n}$ and ${C}$, bi-Lipschitz equivalence is too coarse to preserve the GH convergence in general. In general, the intrinsic metric on a manifold cannot be realized as the extrinsic metric in any Euclidean space.

In the example of hyperboloids we are lucky to have a Gromov-Hausdorff convergent sequence of unbounded spaces. Normally, bounded sets are required, and boundedness is imposed either by an exhaustion argument, or by changing the metric (moving to the projective space). For instance, the parabolas ${y=x^2/n}$ do not converge to the line ${y=0}$ as unbounded subsets of ${\mathbb R^2}$, but they do converge as subsets of ${\mathbb R\mathbb P^2}$. In the process, the degree jumps down from ${2}$ to ${1}$.

It seems that the Gromov-Hausdorff limit of algebraic varieties of degree at most ${d}$ is also an algebraic variety of degree at most ${d}$ (provided that we use the intrinsic metric; otherwise flat ellipses ${x^2+n^2y^2=1}$ would converge to a line segment). If this is true, I’m sure there’s a proof somewhere but I never saw one.

## Inner-variational equations

It’s been a while since the last time I posted a post-colloquium post. This one is based on a colloquium given by me, but don’t worry: none of my results are included here.

As a warm-up, consider the (trivial) problem of finding the shortest path between two points $a,b\in\mathbb R^n$. The naive approach is to minimize the length $L(f)=\int_0^1 |f\,'(t)|\,dt$ among all maps $f\colon [0,1]\to \mathbb R^n$ that are sufficiently smooth and satisfy the boundary conditions $f(0)=a$ and $f(1)=b$. This turns out to be a bad idea: $L(f)$ is neither strictly convex nor differentiable, and the set of minimizing maps is huge, containing some rather nonsmooth specimen.

The right approach is to minimize the energy $E(f)=\int_0^1 |f\,'(t)|^2\,dt$. While this functional is not immediately related to length for general maps, it is not hard to see that for minimizing maps $f$ we have $E(f)=L(f)^2$. Indeed, consider performing the inner variation $\widetilde{f}(t)=f(t+\epsilon \eta(t))$ where $\eta\colon [0,1]\to\mathbb R$ is smooth and vanishes at the endpoints. Expanding the inequality $E(\widetilde{f})\ge E(f)$, we arrive at $\int_0^1 |f\,'(t)|^2\eta'(t)\,dt=0$, which after integration by parts yields $\frac{d}{dt}|f\,'(t)|^2=0$. Thus, minimization of energy enforces constant-speed parametrization, and since for constant-speed maps we have $E(f)=L(f)^2$, the geometric nature of the variational problem has not been lost.

As a side remark, the inner-variational equation $\frac{d}{dt}|f\,'(t)|^2=0$ could be written as $f\,''\cdot f\,'=0$, which is a nonlinear second-order equation.

For comparison, try the first variation $\widetilde{f}(t)=f(t)+\epsilon \eta(t)$ where $\eta\colon [0,1]\to\mathbb R^n$ is smooth and vanishes at the endpoints. Expanding the inequality $E(\widetilde{f})\ge E(f)$, we arrive at $\int_0^1 f\,'(t)\cdot \eta'(t)\,dt=0$, which after integration by parts yields $f\,''\equiv 0$, a linear second-order equation. This Euler-Lagrange equation immediately tells us what the minimizing map $f$ is: there is only one linear map with the given boundary values. Obviously, $f\,''\equiv 0$ is a much stronger statement that $f\,''\cdot f\,'=0$. However, if $f+\epsilon \eta$ is not admissible for some geometric reason (e.g., the curve must avoid an obstacle), the Euler-Lagrange equation may not be available.

Moving one dimension up, consider the problem of parameterizing a given simply-connected domain $\Omega\subset \mathbb C$. Now we are to minimize the energy of diffeomorphisms $f\colon \mathbb D\to\Omega$ which is defined, as before, to be the sum of squares of derivatives. (Here $\mathbb D$ is the unit disk.) In complex notation, $E(f)=\iint_{\mathbb D}(|f_z|^2+|f_{\bar z}|^2)$. For definiteness assume $f$ is sense-preserving, that is $|f_z|\ge |f_{\bar z}|$. Minimizers ought to be conformal maps onto $\Omega$, but how can we see this from variational equations?

The Euler-Lagrange equation that we get from $E(f)\le E(f+\epsilon\eta)$ turns out to be the Laplace equation $\Delta f=0$. This is much weaker than the Cauchy-Riemann equation $f_{\bar z}=0$ that we expect. One problem is that $\eta$ must vanish on the boundary: otherwise $f+\epsilon\eta$ will violate the geometric constraint. We could try to move the values of $f$ in the direction tangent to $\partial\Omega$, but since does not necessarily make sense since the boundary of $\Omega$ could be something like the von Koch snowflake. And of course, it is not at all clear why the minimum of $E$ must be attained by a diffeomorphism. If the class of maps is expanded to include suitable limits of diffeomorphisms, then it’s no longer clear (actually, not true) that $f+\epsilon\eta$ belongs to the same class. All things considered, the approach via the first variation does not appear promising.

Let’s try the inner variation instead. For small $\epsilon$ the map $z\mapsto z+\epsilon\eta$ is a diffeomorphism of $\mathbb D$, hence its composition with $f$ is as good a candidate as $f$ itself. Furthermore, since the inner variation deals with the model domain $\mathbb D$ and not with the generic domain $\Omega$, it is easy to allow modification of boundary values: $\eta$ should be tangent to $\partial \mathbb D$, i.e., $\mathrm{Re}\,\bar z\eta(z)$ should vanish on the boundary. It takes a bit of computation to turn $E(f(z+\epsilon \eta(z)))$ into something manageable, but the final outcome is remarkably simple. The inner-variational equation says that the function $\varphi:=f_z\overline{f_{\bar z}}$ is holomorphic in $\mathbb D$ and $z^2 \varphi$ is real on the boundary $\partial \mathbb D$. (Technically speaking, $\varphi\, dz^2$ must be a real holomorphic quadratic differential.) What can we conclude from this? To begin with, the maximum principle implies that $z^2 \varphi$ is a constant function. And since it vanishes at $z=0$, the inevitable conclusion is $\varphi\equiv 0$. Recalling the sense-preserving constraint $|f_z|\ge |f_{\bar z}|$, we arrive at $f_{\bar z}\equiv 0$, the desired Cauchy-Riemann equation.

Executive summary

• Suppose we are to minimize some quantity (called “energy”) that depends on function (or a map) $f$
• We can consider applying the first variation $f+ \epsilon\eta$ of the inner variation $f\circ (\mathrm{id}+\epsilon \eta)$. Here $\eta$ is a small perturbation which is applied differently: in the first case it changes the values of $f$, in the second it shuffles them around.
• Inner variation applies even in the presence of geometric constraints that make first variation illegal. One example of such constraint is “$f$ must be injective”.
• Inner-variational equations are quite different from the Euler-Lagrange equations. Even for simple quadratic functionals they are nonlinear.
• Inner-variational equations are useful because they tell us something about the maps of minimal energy.

## Linear independence, matroids, and fun with Fano

Take a finite set of vectors $v_1,\dots,v_n$ in some vector space. Call a set $A\subset \lbrace 1,2,\dots,n\rbrace$ independent if the corresponding vectors are linearly independent. The collection of all independent sets is denoted $\mathcal{I}$; what properties must it have? Well, it must be a matroid:

A matroid with ground set $E$ is a nonempty collection of subsets $\mathcal{I}\subseteq 2^E$ which is both hereditary and augmentable.

1. Hereditary: if $A\in \mathcal{I}$, then any subset of $A$ is in $\mathcal{I}$
2. Augmentable: if $A,B\in \mathcal{I}$ and $A$ has fewer elements than $B$, then $A$ can take an element from $B$ and still remain in $\mathcal{I}$.

The augmentation property brings to mind wealth redistribution (at least if you grew up in the USSR). Anyway, it is fairly obvious that any collection $\mathcal{I}$ that arises from vectors is a matroid. Is the converse true? Not obvious at all.

Behold the Fano plane, the smallest projective plane in existence: 7 points, 7 lines, any two lines meet at one point, through any two points there is a unique line.

The Fano matroid has these 7 vertices as the ground set. A set of vertices is independent if it has at most three points and is not a line. The hereditary property is clear. The augmentation isn’t hard either: given a 2-point set $A$, let $L$ be the line through it; since $B$ has 3 points and is not a line, it must contain a point outside of $L$, and this is the point we add to $A$.

Can the Fano matroid be realized by 7 vectors in a Euclidean space? Guess what… no.

Suppose we do have such a set of 7 vectors. Then they span a 3-dimensional space, since there is no linearly independent subset with 4 vectors. The three vertices of the triangle are independent, and we may assume they are realized by the standard basis vectors $v_1,v_2,v_3$. Enumerate the other points/vectors $v_4,\dots,v_7$ so that $v_{j+3}$ is the midpoint opposite $v_j$ for $j=1,2,3$. Observe that $v_4$ is a linear combination of $v_2,v_3$, etc. Thus, $v_1,\dots, v_7$ are the columns of the matrix

$\displaystyle M= \begin{pmatrix} 1&0&0&0&*&*&* \\ 0&1&0&*&0&*&* \\ 0&0&1&*&*&0&* \end{pmatrix}$

But wait, there’s more! Denote $v_7=(x,y,z)^T$ and observe that the linear dependence of $v_1,v_4,v_7$ forces $v_4$ to be a scalar multiple of $(0,y,z)^T$. Similarly for $v_5,v_6$. So the matrix $M$ must be of the form

$\displaystyle M= \begin{pmatrix} 1&0&0&0&bx&cx&x \\ 0&1&0&ay&0&cy&y \\ 0&0&1&az&bz&0&z \end{pmatrix}$

But wait, there’s even more! The vectors $v_4,v_5,v_6$ are also dependent (that inscribed circle is actually a line). The determinant of the 4,5,6 columns is $2abcxyz$. Can any of $x,y,z$ be zero? No, because that would make $v_7$ a linear combination of two of $v_1,v_2,v_3$, which is not supposed to happen. Can any of $a,b,c$ be zero? No, because that would create a zero vector, and all of the 1- and 2-subsets are independent. So, the Fano matroid cannot be realized by vectors…

unless 2=0. Over $\mathbb F_2$ there is no problem:

$\displaystyle M= \begin{pmatrix} 1&0&0&0&1&1&1 \\ 0&1&0&1&0&1&1 \\ 0&0&1&1&1&0&1 \end{pmatrix}$

(P.S. All of this stuff is in Whitney’s 1935 paper that introduced matroids.)

## Three knot invariants

This has been a long Monday, toward the end of which I confused knot width with its bridge number. This post is meant to remind me which is which. Here is a trefoil knot:

The crossing number is the smallest number of self-intersections that the knot can have when thrown on a plane. For the trefoil cr=3.

Here is the same knot on which points of locally maximal height are marked with red, and points of locally minimal height are in blue. There are two extrema of each kind.

The bridge number of a knot is the smallest number of maxima in any of its diagrams. The trefoil has b=2. A knot is in the bridge position if its diagram achieves b, and in addition all the maxima are above all the minima. This allows one to neatly cut the knot by its bridge sphere (in orange), leaving b untangled arcs hanging from either side.

The third invariant also has to do with the extrema of the height function. Now we pick a regular value in each interval between each pair of consecutive critical values, and count the multiplicity of said value. The sum of all these multiplicities is the width of the knot. For the trefoil w=8.

The knot is in a thin position when it attains w, as on the diagram above. Despite their apparent similarity, two different positions of the knot may be required to attain b and w. Also, the invariants behave in different ways under connected sums. Here is a connected sum of two knots, with trefoil on the left:

It’s obvious that the crossing number is subadditive: $\mathrm{cr}\,(K\# K')\le \mathrm{cr}\,(K)+\mathrm{cr}\,(K')$. It remains an open problem whether it’s in fact additive, i.e., whether the equality always holds.

The bridge number is the best behaved of the three: it’s known that $b(K\# K')=b(K)+b(K')-1$ (one-sided inequality is easy: connected sum can kill the absolute maximum on the lower-placed knot). Horst Schubert proved this equality in 1954. Much more recently (in 2004) Jennifer Schultens used modern machinery to give a 6-page proof of Schubert’s theorem.

The width turns out to misbehave. By putting one of the summands way below the other, one sees at once that $w(K\# K')\le w(K)+w(K')-2$. In many examples this is in fact an equality. However, very recently (2011) Blair and Tomova proved that strict inequality may hold; moreover, the Scharlemann-Schultens lower bound $w(K\# K')\ge \min(w(K),w(K'))$ is attained for infinitely many pairs of knots.

## The magic of cyclic sieving and q-analogs

In my previous post the 14 possible triangulations of hexagons were arranged by Whitehead moves. But if asked, most people would probably group them in this way:

Well, some may decide to put the S- and Z-shapes together since they are mirror images of each other. But I won’t. It’s easy to rotate a piece of paper with a triangulation on it; it’s not nearly as easy to fetch a mirror and study triangulations through it. So let it be recorded that the cyclic group $\mathbb Z_6$ acts on triangulations by rotating the hexagon. The orbits have lengths 6,3,3,2.

In the previous post I also mentioned that the number of triangulations of an (n+2)-gon is the Catalan number $C_n=\frac{1}{n+1}\binom{2n}{n}$. With n=4 we get

$\displaystyle C_4=\frac{1}{5}\,\frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2}=\frac{8\cdot 7\cdot 6}{4\cdot 3\cdot 2} \qquad (*)$

Amazingly, the formula (*) does not just count triangulations: it also knows how they behave under rotations of the hexagons. But it will only tell us if we don’t rush to its evaluation. Instead, replace each number on the right of (*) by its q-analog

$\displaystyle [n]_q=\frac{1-q^n}{1-q} =1+q+q^2+\dots+q^{n-1}$

The result is

$\displaystyle C_4(q) = \frac{(1-q^8)(1-q^7)(1-q^6)}{(1-q^4)(1-q^3)(1-q^2)} =1+q^2+q^3+2q^4+q^5+2q^6+q^7+2q^8+q^9+q^{10}+q^{12}\qquad (**)$

The fact that we get a polynomial can be explained: since the Catalan number is an integer, every prime divisor of the denominator in (*) also appears in the numerator (to equal or higher power), and this can be translated into the order of vanishing of polynomials in (**) at the roots of unity. It’s not as easy to explain why the coefficients end up nonnegative.

Now, both the cyclic group $\mathbb Z_6$ and the roots of unity were mentioned. Let’s bring them together by introducing $\zeta=\exp(2\pi i/6)$ which generates $\mathbb Z_6$ by multiplication. The cyclic sieving phenomenon (in this special case) is the following: $C_4(\zeta^d)$ is the number of triangulations fixed by the action of $\zeta^d$. Indeed, evaluating $C_4(q)$ at $q=\zeta,\zeta^2,\dots,\zeta^6$ we get $0,2,6,2,0,14$. So, no triangulations are invariant under rotation by $\zeta$, two are invariant under rotation by $\zeta^2$, etc. It’s easy to read the entire orbit structure from this:

• there are no fixed points
• 2 elements of order 2 form a single orbit of size 2
• 6 elements of order 3 form two orbits of size 3
• there are no elements of order 4, since their number is $C_4(\zeta^4)-C_4(\zeta^2)=2-2=0$
• there are $C_4(\zeta^6)-C_4(\zeta^2)-C_4(\zeta^3)=6$ elements of order 6, which form a single orbit of size 6

And it’s not just triangulations. Consider noncrossing matchings of 2n points — these can be observed experimentally by sitting 2n people at a round table and asking them all to shake hands with someone with no arms crossing.

The number of such matchings is also the Catalan number $C_n$ (hence 14 matchings for 8 people). But the orbit structure is completely different! How can the same polynomial $C_4(q)$ work here? It can because we now act by $\mathbb Z_8$ and therefore take $\zeta=\exp(2\pi i/8)$. Evaluation of $C_4(q)$ at the powers of $\zeta$ yields $0,2,0,6,0,2,0,14$. As above, we see the orbit structure in these numbers:

• 2 elements of order 2 form a single orbit
• 4 elements of order 4 form a single orbit
• 8 elements of order 8 form a single orbit

And it’s not just the Catalan numbers. Suppose that among those 8 guests at the round table are 2 men and 6 women. The number of seating arrangements is $\displaystyle \binom{8}{2}=\frac{8\cdot 7}{2}$. Don’t evaluate to 28 just yet. Replace each integer with its q-analog and you’ll get the polynomial $\displaystyle 1+q+2q^2+2q^3+3q^4+3q^5+4q^6+3q^7+3q^8+2q^9+2q^{10}+q^{11}+q^{12}$, which will tell you that under rotation, 4 arrangements form an orbit of size 4, while the other 24 form 3 orbits of size 8.

And there is much more. The cyclic sieving phenomenon was established recently (in 2004) by three mathematicians from Minnesota (Reiner, Stanton, and White), and there are still paths waiting to be explored.


text:
# Kolmogorov scale why $d\sim \left( \frac{\nu^3}{\epsilon} \right)^{1/4}$ and not $d\propto \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$?

I am looking at the Kolmogorov scale and in numerous sources (e.g. this one). I have seen the following: $$d\sim \left( \frac{\nu^3}{\epsilon} \right)^{1/4}$$ for the Kolmogorov scale. I can see why we must have: $$d\propto \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$$ on dimension grounds. But to replace $\propto$ with $\sim$ we would have to be sure that the constant of proportionality is of order unity. I have not been able to find justification for this assumption. Does anyone now such a justification?

• Uh...what exactly is the difference between $\sim$ and $\propto$ here? I've known people who use the two interchangably. – ACuriousMind Aug 13 '16 at 16:16
• @ACuriousMind I am assuming that $a \sim b$ means that $a$ and $b$ are of the same order of magnitude and $a \propto b$ means that they are proportional. – Quantum spaghettification Aug 13 '16 at 16:50
• What basis do you have for that assumption? Without further context, I would always read both as "proportional to" first. – ACuriousMind Aug 13 '16 at 17:11
• @ACuriousMind In one of the places I have seen this (which unfortunately isn't publicly available) it uses both $\propto$ and $\sim$ in the two following expressions $E(k)\propto k^{-5/3}\varepsilon^{2/3}$ and $d\sim \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$ the fact the author uses $\propto$ and $\sim$ close together in the same document, indicates she/he associates a different meaning to them. Which I am assuming is the ones I have given above. – Quantum spaghettification Aug 13 '16 at 17:29
• Either way, you want to know if there's rationale, and what is it, for why the constant of proportionality is of order one. That's a fair question. – Bob Bee Aug 14 '16 at 5:33

I think the most intuitive but maybe not the most mathematically correct interpretation is the implicit assumption that viscous forces start dominating over inertial forces at the Kolmogorov scale. This is easily seen when defining the kolmogorov length and velocity scales: $$l\sim\left(\frac{\nu^{3}}{\epsilon}\right)^{1/4}\quad v\sim\left(\nu\epsilon\right)^{1/4}$$ The Reynolds number at the Kolmogorov scale is then evaluated as: $$\mathrm{Re}=\frac{vl}{\nu}\sim 1$$ For a proportionality sign, the proportionality constant may be much larger than order unity. The Reynolds number would become much larger than order unity as well which would contradict the assumption of viscous forces being of the same order as inertial forces. Hence it suffices to make the constant of order unity and introduce the 'on the order of' sign.

More mathematically, according to the wiki:

The definition of the Kolmogorov time scale can be obtained from the inverse of the mean square strain rate tensor and the definition of the energy dissipation rate per unit mass.

Perhaps from these definitions it follows automatically that the proportionality constant is of order unity, however, i have not investigated this further.

• Hi thanks for your answer. I don't think your answer (correct me if I am wrong) excludes the possibility that constant of proportionality for $l$ be $\sim A$ and that for $v$ be $\sim A^{-1}$ where $A$ can be any (even large) arbitrary constant. Is their a way to show that this cannot be the case e.g. by showing that $l\sim A$ and $v\sim A$? – Quantum spaghettification Aug 14 '16 at 17:28
• @Quantumspaghettification you are right and unfortunately I cannot argue against your case except that in my experience dimensionless proportionality constants are by definition order of unity otherwise the dimensional analysis was not done correctly. Examples are the drag and lift coefficients. I understand that my experience is not a good argument ;), I'll have to think about this some more. – nluigi Aug 14 '16 at 18:00
• Just to let you know I have found a way of doing this (I think :) ). It would have been far to long to put in the comments, so I have put it in the form of an answer. – Quantum spaghettification Aug 15 '16 at 18:17

Throughout this answer $\sim$ means 'of the order of'.

The viscous force takes the form: $$F_V\sim \nu\frac{v}{l^2}$$ So the rate of energy desperation (per unit mass) is: $$\varepsilon \sim \nu\frac{l}{t} \frac{v}{l^2}$$ $$\sim \nu\frac{v^2}{l^2} \tag{1}$$ Now the inertial force is given by: $$F_I \sim \frac{v^2}{l}$$ On the assumption that $F_V \sim F_I$ we have: $$\nu \frac{v}{l^2} \sim \frac{v^2}{l}$$ $$\nu \frac{v^2}{l^2} \sim \frac{v^3}{l}$$ So: $$\varepsilon \sim \frac{v^3}{l} \tag{2}$$ From $(1)^3/(2)^2$ We have: $$\varepsilon \sim \frac{\nu^3}{l^4}$$ Giving us: $$l \sim \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$$

References

1) Fluid Mechanics: An Introduction to the Theory of Fluid Flows By F.Durst (page 544, link to Google Books).

2) nluigi's answer to this question.

• Are you satisfied with this answer? Does it answer for you why $\tilde$ is used rather than $\propto$? – nluigi Aug 16 '16 at 6:07
• @nluigi to me it does, although I am assuming the order of magnitude of $F_V$ and $F_I$. Whether it is valid or not is another matter... – Quantum spaghettification Aug 16 '16 at 6:32
• It is equivalent to my statement that Reynolds number is order of unity which is generally accepted to be true at the Kolmogorov scale. – nluigi Aug 16 '16 at 6:38

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text:
Method of Eigenfunction Expansion

The solution of a PDE can be represented by a Fourier cosine series $$u(x,t)=\sum_{n=1}^\infty A_n(t)\cos\frac{n\pi x}L.$$

Applying a given initial condition

$$u(x,0)=100,$$

lets us solve for $A_n(0)$ through the orthogonality of cosines:

\begin{align} u(x,0)&=\sum_{n=1}^\infty A_n(0)\cos\frac{n\pi x}L=100\\ &\Rightarrow A_n(0)=\frac{400}L\int_0^L\cos\frac{n\pi x}L\,dx=0. \end{align}

This does not seem correct, however. What am I doing wrong?

-
You are missing the $n=0$ term in your cosine expansion.
If you choose to expand your $f(x)$ in terms of cosine, then you need to have $$f(x) = \sum_{n=0}^{\infty} f_n \cos(n \pi x/L)$$
But $n=1,2,\dots$, or am I missing something? –  Josué Molina Oct 26 '12 at 19:25
Oh, the problem is given without an $A_0$ term. Could this be a typo? –  Josué Molina Oct 26 '12 at 19:27
@JosuéMolina If you choose to expand in cos then you need to have the $n=0$ term. However, if you choose to expand in $\sin$ you will not have the $n=0$ term. –  user17762 Oct 26 '12 at 19:29


text:
# Solutions of the congruence $x^2 \equiv 1 \pmod{m}$

For $m>2$, if a primitive root modulo $m$ exists, prove that the only solutions of the congruence $x^2 \equiv 1 \pmod m$ are $x \equiv 1 \pmod m$ and $x \equiv -1 \pmod m$.

Thanks.

#### Solutions Collecting From Web of "Solutions of the congruence $x^2 \equiv 1 \pmod{m}$"

Here is an indirect solution – a solution that doesn’t use any specific primitive root. I think a better solution exists, but I would still like to post it because it gives a different perspective.

The solutions $1$ and $m-1$ are obviously always solutions. So the difficulty is proving there aren’t any more.

By the primitive root theorem, the possibilities for $m$ are $m=2$, $m=4$, $m=p^k$, or $m=2p^k$ where $p$ is an odd prime and $k\ge 1$.

For $m=2$ we actually have $1=m-1$, and there is a unique square root of $1$.

For $m=4$ you may check directly that the claim holds, just by calculating.

For $m=p^k$, we prove by induction on $k$. For $k=1$ there are no zero divisors, so $(x-1)(x+1)=0$ implies $x-1=0$ or $x+1=0$. For $k>1$, assuming there are exactly two solutions modulo $p^{k-1}$, we use Hensel’s lemma to claim that these lift to exactly two solutions modulo $p^k$.

For $m=2p^k$ we use the Chinese remainder theorem to combine the two solutions modulo $p^k$ with the unique solution modulo $2$ to get two solutions modulo $m$.

$x^2$$\equiv1 mod m means that m|(x^2-1)…..what can you infer from this? Here is a direct solution involving a primitive root (see also my other answer for an “indirect” solution): Let r be a primitive root modulo m and let \{r^0, r^1, \ldots, r^s\} be the multiplicative group mod m. Let x satisfy x^2 = 1. Then x must be a member of the multiplicative group, since it’s invertible mod m (it is its own inverse). So we can write x = r^k for some integer k, and we know that r^{2k} \equiv r^0 \pmod m. This gives us 2k \equiv 0 \pmod {\phi(m)} where \phi is Euler’s totient function. This congruence has at most two solutions (this I leave as exercise), giving at most two solutions for x. Suppose \,w\, is a primitive root modulo \,m\, , and \,x=w^k\, , then$$x^2=1\pmod m\iff w^{2k}=1\pmod m\iff (w^k-1)(w^k+1)=0\pmod m\iff (x-1)(x+1)=0\pmod m\ldots$$The fact that any unit$\,x\,$modulo$\,m\,$is a power of$\,w\,$follows from the fact of the latter being a primitive root modulo$\,m\,$…If$\,m\,$has no primitive roots then the claim isn’t true. Warning: In the last step you still have to give a very little explanation… hint: an equation of degree 2, also in modular arithmetic, has at most two solutions if a primitive root modulo$m\$ exists


text:
## Damped Forced Oscillations and Beats

Hello,

I know it is possible with the undamped case but is it possible to have beats with a Damped Forced Oscillation?

Graphing a particular instance of the equation:
$y(t)=C_1 e^{- \alpha t}cos(\omega _o t)+C_2 cos(\omega t)$
Resulted in:

Which kind of looks like beats that are fading away.

But I can't find a mathematical representation that describes it like the undamped case:
$y(t)=C sin(\frac{\omega_o + \omega}{2}t)sin(\frac{\omega_o - \omega}{2}t)$
That pesky exponential term from the first equation above keeps on getting in the way.

Any help appreciated,

Scott.


text:
# (UJ tS force attractive or repulsive? 8. (10 pts) A 50-cm long solenoid is to produce...

###### Question:

(UJ tS force attractive or repulsive? 8. (10 pts) A 50-cm long solenoid is to produce a field of 0.88 mT at its center. How much current should the solenoid carry if it has 2000 turns of wire?

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text:
# Introduction to $\mathrm{G}_2$ geometry

@article{Karigiannis2020IntroductionT,
title={Introduction to \$\mathrm\{G\}\_2\$ geometry},
author={Spiro Karigiannis},
journal={arXiv: Differential Geometry},
year={2020},
pages={3-50}
}
These notes give an informal and leisurely introduction to $$\mathrm {G}_2$$ geometry for beginners. A special emphasis is placed on understanding the special linear algebraic structure in 7 dimensions that is the pointwise model for $$\mathrm {G}_2$$ geometry, using the octonions. The basics of $$\mathrm {G}_2$$-structures are introduced, from a Riemannian geometric point of view, including a discussion of the torsion and its relation to curvature for a general $$\mathrm {G}_2$$-structure, as… Expand

#### References

SHOWING 1-10 OF 47 REFERENCES


text:
## Concatenation Operators in Unimperative

The Unimperative language I posted about nearly a year ago has been reborn as a "concatenative" language, ala Joy. ( personally I would prefer to call it a compositional language, but who am I to buck trends ).

The Unimperative language is also a tiny subset of C++, so I am trying to decide on the concatenation operators. Currently I support separate left-concatenation (f << g == f . g) and right-concatenation (g >> f == f . g). In the end this gets a bit clumsy, and I find hard to read. So I want to overload the comma operator.

My question is this: which concatenative form (left or right) should the comma overload to? In other words should I treat it as a left to right sequencing operator, or as a composition operator?

## Comment viewing options

### Obvious?

f, g reads to me as "first f, then g". If you would have asked what would be the best character to use as a sequencing operator, I would have said "," (or ";").

### Obvious.

You are right, it is obvious. Weird thing, is that when you stare at too many different ways of doing things, your mind plays tricks on you and the obvious has a way of hiding.

Forget all the silly ">>" and "<<" I think I will support just two forms:

def<> h = g, f;

and its partner:

def<> h = f(g);

### Choosing operators in C++

C++ operators have fixed precedences. "." has the highest precedence, but "," has the lowest precedence. I think it is more important that the appearance.

Another consideration is whether the overloaded operators confuse to the original meanings. Overloading "()" is not the best choose, since it can be confused with the usual function call.

However, if it is not required to be a "tiny-subset of C++", you can simply create two new operators, and choose the operation much more freely.


text:
# Ph.D. in Finance Programs: If I have a Ph.D. will they waive the GRE/GMAT?

#### FEP

Interested in applying but the deadlines are approaching soon and I will not have enough time to take the tests.

I have a Ph.D. in applied mathematics from a first tier school and 5 years of postdoc experience. Still seems quite daunting to gain admission even with my previous experience.

#### albo

It seems a little bit confusing to me. If you already have a PhD, why do you want to get a second one?

#### FEP

job prospects in academia are terrible for math/applied math in general. even some of the lower level schools like the cal states are extremely competitive and unreasonably so.

just weighing some options given this years tenure track search does not go well. I understand many MFE programs waive the requirement.

#### Andy Nguyen

Yes, if you have a PhD, most MFE programs will waive the GRE/GMAT requirement. Email them and ask for a formal waiver. I don't know if it's the practice among PhD admission but if you plan to go for another PhD, GRE is the least of your worries.

#### morty

NYU, for instance, waives the GRE requirement for applicants who already have a PhD.

#### purplyboy

Shouldn't the GRE or GMAT math be a piece a cake for you if you have a PhD?

#### Shantanu Kumar

Shouldn't the GRE or GMAT math be a piece a cake for you if you have a PhD?
Actually its true for any science/engineering graduate too.

#### DominiConnor

To introduce myself, I'm a headhunter who specialises in people who do maths to banks...

I'm with Andy here, from the info you're sharing you will be late 30s /early 40s before starting work in a bank which is a definite possibliity but a more than slightly risky bet with your career.

Is it your aim to change tracks in academia or work in a bank ? (either can be goals, you don't need to choose yet)

I am sadly aware of the terrible career situation for math/science PhDs even if academic recruitment is something I rarely do, so this advice is aimed at banking careers.

Firstly, doing a PhD purely for career advancement is rarely optimal and is often quite damaging, some managers I know would think less of you for having 2 PhDs. Also the bank job market 3-5 years from now could be almost anything from wonderful to terrible. I'm actually currently being paid to model the future pay levels of PhD quants and one mainstream scenario is that in nominal terms it stays constant for the next decade. There was a time when that wasn't scary but inflation looks like rising.
Another risk is that this may not be what you're good at, it does happen and I will share that you will end up doing more programming, you OK about that ?

I believe your optimum is in one of the following:
1: Spending $1 K on a bunch of books, (Wilmott, Shreve, etc) working through them in your spare time. Asking questions on various fora when you get stuck. A bit of work on your stats and programming is also necessary. Pros: Really cheap, doesn't risk your current track or earnings, some hiring managers see this as the gold standard for applicants. Cons: You don't get a certificate saying you can do stuff, you need to manage your study plan and be really disciplined. 2: A part time MFE or the CQF Pros: Relatively cheap, no loss of earnings Cons: You have to be self disciplined and part time MFEs take twice as long as full time but cost just as much. 3: Full time MFE Pros: Nice certificate to say you've done it, more support in studying. Cons: Fees plus loss of earnings, also it is harder to move directly back into academia. 4: Squatting: auditing finance courses at your current university but taking no exams Pros: Really cheap, no impact upon your current career. Cons: No certificate and a rather random set of skills gained, best tried in conjunction with option (1) #### Andy Nguyen NYU, for instance, waives the GRE requirement for applicants who already have a PhD. It's program specific, not university-wide policy so if you mean the NYU Courant MSMF program, yes, it's confirmed. #### FEP thank you all for the information. I'd like to stay in the academic track with the fall back being employment as a quant at an investment bank or hedge fund (depending on the economy... sigh). I'm not overly concerned about the GRE, it's just the timing and prep. Most app's for the PhD programs are due in about 2 weeks time. If I had a full month, I'd be able to prepare and take the test. As far as programming, I live for it. We do modeling and implementation so computing is at least 70-80% of the work. OK, that being said, I just got a few emails back from the local schools here in LA: anderson and ross. Both require the GRE for admission. Claremont as well. Claremont actually requires one to be in the MFE program first to even be considered a candidate for the PhD program. So, one more question: Would enrolling in an MFE program (anderson for example) help with admission into the PhD finance program? I'm pretty set on doing quant finance for the PhD as well so there will be some overlap. Lastly, I'm getting into my late 30's so the light at the end of the tunnel is closing fast. Wish I had looked into this 5 or so years back. Just ridiculous what's happening in applied math/math right now regarding academic job prospects. Thanks again for all of the info everyone! Cheers. #### FEP To introduce myself, I'm a headhunter who specialises in people who do maths to banks... I'm with Andy here, from the info you're sharing you will be late 30s /early 40s before starting work in a bank which is a definite possibliity but a more than slightly risky bet with your career. Is it your aim to change tracks in academia or work in a bank ? (either can be goals, you don't need to choose yet) I am sadly aware of the terrible career situation for math/science PhDs even if academic recruitment is something I rarely do, so this advice is aimed at banking careers. Firstly, doing a PhD purely for career advancement is rarely optimal and is often quite damaging, some managers I know would think less of you for having 2 PhDs. Also the bank job market 3-5 years from now could be almost anything from wonderful to terrible. I'm actually currently being paid to model the future pay levels of PhD quants and one mainstream scenario is that in nominal terms it stays constant for the next decade. There was a time when that wasn't scary but inflation looks like rising. Another risk is that this may not be what you're good at, it does happen and I will share that you will end up doing more programming, you OK about that ? I believe your optimum is in one of the following: 1: Spending$1 K on a bunch of books, (Wilmott, Shreve, etc) working through them in your spare time. Asking questions on various fora when you get stuck. A bit of work on your stats and programming is also necessary.
Pros: Really cheap, doesn't risk your current track or earnings, some hiring managers see this as the gold standard for applicants.
Cons: You don't get a certificate saying you can do stuff, you need to manage your study plan and be really disciplined.

2: A part time MFE or the CQF
Pros: Relatively cheap, no loss of earnings
Cons: You have to be self disciplined and part time MFEs take twice as long as full time but cost just as much.

3: Full time MFE
Pros: Nice certificate to say you've done it, more support in studying.
Cons: Fees plus loss of earnings, also it is harder to move directly back into academia.

4: Squatting: auditing finance courses at your current university but taking no exams
Pros: Really cheap, no impact upon your current career.
Cons: No certificate and a rather random set of skills gained, best tried in conjunction with option (1)

Hi DominiConnor,

thank you for the detailed advice. I completely agree on option #1 and 4. I can easily learn all the MFE requirements in 3-4 months I feel given my background. An MFE program would likely help the most in placement. Being in the latter part of my 30's, the job prospects are looking tighter and the options being more limited. I would like to stay in academia hence the desire to pursue a finance PhD. Unlike math for example, it is far more difficult to get into a finance program but the job prospects after are exponentially higher.

Cheers

They will not waive the GRE / GMAT requirement if you want to apply for Finance PhD. They will treat you just like any other applicant.
It is quite common for Math / Physics PhDs to do a second PhD in Finance. They normally get the Math PhD around age 26-27, so there is plenty of time for a second PhD.
You have very good chance at admission in the quant finance oriented schools like Chicago, Wharton, Berkeley. Even in the other schools there will be a group of Finance faculty (at least one or two) who specializes in quant finance research.
There are some schools like Stanford where they have shifted focus heavily toward Behavioral Finance research.
In most of the schools, the largest group of faculty will be working in Empirical Finance research, but there will always be one or two quant people.
Your starting salary in 2011 terms will be around $250 K. Since you already have a PhD, you will probably get admission in a good school and you already know what it takes to get a PhD. Getting tenure. That will be a big problem. From the first day of your PhD program you should start doing research as if you are on a 11-year tenure-track position (5 years of PhD + 6 years tenure-track). #### morty It's program specific, not university-wide policy so if you mean the NYU Courant MSMF program, yes, it's confirmed. Apologies, I should have been more specific. That is exactly what I intended to say. #### MiloRambaldi Okay, here are the answers to your queries. From the first day of your PhD program you should start doing research as if you are on a 11-year tenure-track position (5 years of PhD + 6 years tenure-track). Just curious: I guess there's no such thing as a part-time PhD in Finance? #### FEP Okay, here are the answers to your queries. They will not waive the GRE / GMAT requirement if you want to apply for Finance PhD. They will treat you just like any other applicant. It is quite common for Math / Physics PhDs to do a second PhD in Finance. They normally get the Math PhD around age 26-27, so there is plenty of time for a second PhD. You have very good chance at admission in the quant finance oriented schools like Chicago, Wharton, Berkeley. Even in the other schools there will be a group of Finance faculty (at least one or two) who specializes in quant finance research. There are some schools like Stanford where they have shifted focus heavily toward Behavioral Finance research. In most of the schools, the largest group of faculty will be working in Empirical Finance research, but there will always be one or two quant people. Your starting salary in 2011 terms will be around$250 K.
Since you already have a PhD, you will probably get admission in a good school and you already know what it takes to get a PhD.
Getting tenure. That will be a big problem. From the first day of your PhD program you should start doing research as if you are on a 11-year tenure-track position (5 years of PhD + 6 years tenure-track).

thanks trader joe, you hit the nail on the head. at least 11 years from start date of phd to tenure. can get snagged along the way but the backup is quant work in finance which is why there are a lot of options.

looked at the gre again and the quant part is quite straightforward. just the hassle of going in and taking the test and submitting scores. maybe a 1 week process at most, not a big issue.

I will say a few things, the research in computer science and many parts of applied math is quite a bit more interesting than the work as a researcher in finance, but the financial applications are more applicable in many ways.

guess I have some soul searching to do in the next few weeks. tough economy, even graduating from a top 5 applied program and doing 3 postdocs at big name schools can't even land a cal state tenure track job for example despite superlative teaching and excellent research. pretty random process this tenure track thing.

thanks again. i will at least apply to a few mfe programs as a backup and time permitting, some phd programs in finance although at my age in my mid 30's I'm hesitant to go through the process all over again.

Cheers.

ps. to answer the last question, the phd program is very demanding, it is unlikely there are any part time opp's there. maybe with some TA support, otherwise a full time affair.

Just curious:
I guess there's no such thing as a part-time PhD in Finance?
In think that EDHEC is the only recongnized part-time PhD. I don't think that you will learn to do research. The good thing is that all your classmates will be MDs, Country Heads, or CEO level people.

#### DominiConnor

"recognized" is a tricky term and about as subjective as it gets.
I "recognize" part time finance PhDs from Birkbeck but not Imperial even though on most rankings Imperial is rather higher.

Also there is a grey area where some people switch from full to part time PhDs, so I "recognize" a couple of people who are doing part time maths PhDs at Imperial.
EDHEC has the networking qualities that TJ cites and in I am all for that as part of your career management, but never lose sight of the fact that there has to be meat in the burger else you just expose your failings to a wider and more powerful audience.

But my perspective is that of a banking headhunter, I know that academia has a very different evaluation process (though they don't recognise part time Imperial finance PhDs either).

I'm vaguely concerned at the idea of finance academia as a safe option that pays adequately well.
Certainly it has grown and some have managed to acquire positions that enjoy a sweet spot of earning bank-like money but academic lifestyles but that is a result of growth in banking driving the demand for financial education driving the demand for academics.
That's quite a heavy gearing, hence the growth and therefore it could go down as hard as it grew.

I guess that part of my general view of labour markets is that of mean reversion and the mean in academia has been mean to many.

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×

# $$abc=1$$

$$\textbf{(IMO 1995)}$$

Let $$a, b, c$$ be positive real numbers such that $$abc=1$$. Prove that

$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}.$

Note by Victor Loh
2 years, 11 months ago

Sort by:

Let $$x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$$. Then by the given condition we obtain $$xyz=1$$. Note that

$\sum \frac{1}{a^3(b+c)}=\sum \frac{1}{\frac{1}{x^3}(\frac{1}{y}+\frac{1}{z})}=\sum \frac{x^2}{y+z}.$

Now by Cauchy-Schwarz inequality

$\sum \frac{x^2}{y+z} \geq \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2} \geq \frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2},$

where the last inequality follows from AM-GM. · 2 years, 11 months ago

You may use Cauchy-Schwarz in Engel Form as well.

By Cauchy-Schwarz in Engel Form, we have

$\sum_{cyc} \dfrac{1}{a^{3}(b+c)} \geq \dfrac{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^{2}}{2(ab+bc+ca)}$

And you can continue from here. · 2 years, 11 months ago

Yeah :D · 2 years, 11 months ago

Did you see this from a book? :D (if I won't reply, I'm in school :) ) · 2 years, 11 months ago

$\sum_{\text{cyc}} \frac{1}{a^3(b+c)}=\sum_{\text{cyc}} \frac{b^2c^2}{a(b+c)}$

$\stackrel{\textbf{Cauchy-Schwarz}}\ge \frac{(ab+bc+ca)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}\stackrel{\textbf{AM-GM}}\ge \frac{3}{2}$ · 2 years, 11 months ago


text:
Mathematics
OpenStudy (anonymous):

how do you prove 2sin^2(t)=2-3sin(2t)-4cos^2(t)

OpenStudy (anonymous):

so far i have 2-2cos^2t = 2-6sintcost-4cos^2t

OpenStudy (anonymous):

({1-\cos(2t)} =

OpenStudy (anonymous):

$({1-\cos(2t))} =$ ... Need to fix your identity on the left.

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square root equation calculator

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text:
## Top quark pair production cross-section results at LHC in final states with tau or no leptons    [PDF]

Gia Khoriauli, for the ATLAS, CMS Collaborations
ATLAS and CMS measured the top quark pair production inclusive cross section using proton-proton collision data collected at 7 TeV c.m. energy at the LHC during the year 2011 in several channels. This article presents the measurements of the cross section using selections of top quark pair production events with one hadronically decaying {\tau} lepton or no leptons at all.
View original: http://arxiv.org/abs/1305.6952


text:
# I Pressure and Lift around a Wing

Gold Member
You are sort of running in circles here. The best way to address that question again goes back to the consideration of energy. A given fluid parcel has a finite total energy (represented by total pressure in Bernoulli's equation). If it is accelerated to some constant velocity further downstream (e.g. after a pipe constriction) and experiences no non-conservative forces, then it still has the same total energy but it now has higher kinetic energy. The energy stored as static pressure must therefore decrease.

#### FactChecker

Gold Member
2018 Award

Given a fluid parcel moving along a straight streamline in the positive x-direction, I see that a pressure difference between two points (one before and one after point $P$) would cause a force $F$ on the fluid parcel which would accelerate. But what if the parcel is moving at constant speed? The force would zero and the pressure gradient would be zero. In that case, being pressure isotropic, the parcel experiences the same force from the back, from the front, from the top and from the bottom. This isotropic pressure is smaller compared to the static pressure at the same point $P$ when the fluid is not moving at all. I am not sure how to justify the fact that the speed makes the pressure smaller without using Bernoulli's equation.
After the velocity has increased and the pressure has decreased, it is hard to figure things out if you ignore what happened to cause the velocity to increase. You can't intuitively understand the difference between point A and point B while ignoring everything in between.

Suppose you wanted to compare the velocity of a car at the top of a hill with the same car after it has rolled down the hill without considering the obvious speed up as it rolled down. You want a reason why the velocity at the bottom is faster. In addition, you are trying to rule out consideration of the potential vs kinetic total energy trade-offs. That is asking for too much.

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#### fog37

Ok, thank you. I look forward to boneh3ad insight article :)

Gold Member
I have a rough draft made up but it needs some figures, which may take a while to actually put together, especially with the end of the semester rapidly approaching. It might have to wait until mid-May.

#### Joseph M. Zias

For your information: I have found a very good web book on flying that includes a very readable section of wings and lift. Look up "See How it Flies" by
John S. Denker, a physicist and flight instructor.

#### sophiecentaur

Gold Member
For your information: I have found a very good web book on flying that includes a very readable section of wings and lift. Look up "See How it Flies" by
John S. Denker, a physicist and flight instructor.
That guy knows a lot about flying and planes and the book is certainly entertaining as well as informative. I didn't read from cover to cover, of course, but one bit leaped out of the screen at me:

"Downwash behind the wing is relatively easy to understand; the whole purpose of the wing is to impart some downward motion to the air."

He is obviously aware of Newton's Third Law so he is, by definition 'A Good Lad" and gets the general principle.

#### DarioC

Gold Member
I don't understand fog37's initial statement that the air flowing over the top of the wing causes a downward pressure on the top of the wing.

None the less:There was (is) an excellent video online of wind tunnel airflow over an airfoil with the flow shown by pulsed smoke. It really changed my view of what happens. What I saw was that the air velocity near the wing speeds up compared to the general tunnel flow as it goes up to the high point on the airfoil and then slows back down as it curves down to the trailing edge of the wing. This is displayed by the vertical lines of the smoke pulses.

The second part that conflicted with what I thought I knew about wings was that the "smoke lines" indicated the flow slowed down as it passed under the wing. Interestingly enough, if that is correct, it would still produce a downward vortex, and therefore lift, from the trailing edge of the wing.

I fail to see any conflict between the downward flow of air from a helicopter, the rearward flow from a propeller, or the lift from a wing. I can personally vouch for the vortex from a wing as I well remember having my light airplane flipped almost 90 degrees by a vortex from a larger corporate aircraft just as I was turning final following it

Don't have a link for that video now as it was lost on my older computer when it crashed. I'll look around and post if I find it.

#### fog37

Hello DarioC,

Let me see if I can be more clear if what I was trying to describe: if the air above the wing far from the surface is at regular atmospheric pressure and the air right above the wing surface is at pressure lower than atmospheric pressure, I would envision a pressure gradient directed toward the wing surface.

#### Joseph M. Zias

That guy knows a lot about flying and planes and the book is certainly entertaining as well as informative. I didn't read from cover to cover, of course, but one bit leaped out of the screen at me:

"Downwash behind the wing is relatively easy to understand; the whole purpose of the wing is to impart some downward motion to the air."

He is obviously aware of Newton's Third Law so he is, by definition 'A Good Lad" and gets the general principle.

Dr. Denker also has a LOT of comments on theorems in physics and reviews (very critical at times) on text books. Take a look at his website: http://www.av8n.com/physics/

Gold Member
That guy knows a lot about flying and planes and the book is certainly entertaining as well as informative. I didn't read from cover to cover, of course, but one bit leaped out of the screen at me:

"Downwash behind the wing is relatively easy to understand; the whole purpose of the wing is to impart some downward motion to the air."

He is obviously aware of Newton's Third Law so he is, by definition 'A Good Lad" and gets the general principle.
I am amused by his view of the purpose of a wing. I'd argue that the purpose of the wing is to hold up the plane, and imparting some downward motion to the air is the means by which it achieves this purpose, not the other way around. Otherwise I'd agree.

#### sophiecentaur

Gold Member
it achieves this purpose, not the other way around.
How would you feel about saying that the purpose of a planes propellor is to push air backwards or that an aircraft's engine is there to turn the propellor or that the fuel is put in to make the engine go? Or even that the pilot is there to make the passengers arrive at JFK.
Causes and effects.

Gold Member
How would you feel about saying that the purpose of a planes propellor is to push air backwards or that an aircraft's engine is there to turn the propellor or that the fuel is put in to make the engine go? Or even that the pilot is there to make the passengers arrive at JFK.
Causes and effects.
Well, I might argue that the pilot is there so he can earn his paycheck and eat.

#### sophiecentaur

Gold Member
Well, I might argue that the pilot is there so he can earn his paycheck and eat.
The toe bone's connected to the foot bone, the foot bone's connected to the ankle bone . . . . . now hear de word of de lawd.

#### Joseph M. Zias

I am amused by his view of the purpose of a wing. I'd argue that the purpose of the wing is to hold up the plane, and imparting some downward motion to the air is the means by which it achieves this purpose, not the other way around. Otherwise I'd agree.
From the frame of reference of the pilot - yep, keeping that plane up is Primary. Take a look at his physics site:

http://www.av8n.com/physics/

#### Joseph M. Zias

Well, I might argue that the pilot is there so he can earn his paycheck and eat.
Well as a private pilot I spend my money to fly - unfortunately the money tree has lost most of its leaves.

#### sophiecentaur

Gold Member
Well as a private pilot I spend my money to fly - unfortunately the money tree has lost most of its leaves.
I sit in my garden and watch guys like you, flying overhead every day. I can hear the cool person of hard earned cash disappearing down a black hole. As a former boat owner, I can sympathise (by a factor of ten, at least).

#### DarioC

Gold Member
Fog37, In reply I think you might want to look at what happens at the leading edge of the wing. I am getting close to the limits of my understanding here.
The air running into the leading edge curve is flung/forced upwards which compacts the air above the wing, at the same time making a lower pressure near the surface of the wing. The pressure layers higher up cannot exert downward force because they are being constantly compacted and separated from the wing by the up flow from the curved leading edge.
Thanks
DC
Added: Are you familiar with the triangular strips of metal that are put on the leading edge of the wings, near the fuselage to make that part of the wing stall before the outboard part where the ailerons are?
Rather interesting little detail on real-world wings.

Last edited:

#### FactChecker

Gold Member
2018 Award
This should "get you over the top" then... .

.
I think this is an excellent video. As an interested amateur, I wish that I had seen this decades ago. It really gives one a clue that any simple explanation necessarily omits too much and is misleading when predicting lift and drag. That is why CFD programs are used in spite of the huge amount of computer power required and the limitations of the answers.

Last edited:
OCR

#### DarioC

Gold Member
I think this is the video I mentioned. Pulsed smoke flow in wind tunnel. Check at about 47 seconds.

Gold Member
Fog37, In reply I think you might want to look at what happens at the leading edge of the wing. I am getting close to the limits of my understanding here.
The air running into the leading edge curve is flung/forced upwards which compacts the air above the wing, at the same time making a lower pressure near the surface of the wing. The pressure layers higher up cannot exert downward force because they are being constantly compacted and separated from the wing by the up flow from the curved leading edge.
Air running into the leading edge is directed both over and under the wing, not simply over it. You are correct that the pressure away from the surface does not impact it directly (only the surface pressure acts on the surface), but it is related.

Consider the streamlines as they move around the airfoil over the top. Near the leading edge they are deflected upward and have a curvature whose center points away from the airfoil. Curved motion requires a centripetal force, which here must be provided by a pressure gradient. Since the pressure far from the airfoil must be atmospheric, this means that the pressure in that region must be locally higher than atmosphere. This shouldn't be surprising because this is the result of being near the stagnation point.

After that point, the curvature flips direction as the streamlines bent to follow the curve of the airfoil over the top. Now the center of curvative is in or below the wing and the pressure must therefore decrease as you approach the surface. There is therefore a low pressure near the surface at that point.

So the pressure far from the surface does not directly impact the lift force, but it definitely plays an indirect role here.

Added: Are you familiar with the triangular strips of metal that are put on the leading edge of the wings, near the fuselage to make that part of the wing stall before the outboard part where the ailerons are?
Rather interesting little detail on real-world wings.
That's not what those triangles do. Those are vortex generators and they actually delay stall. The turbulence they induce energizes the boundary layer near the wall and makes it more resistant to separation.

I think this is an excellent video. As an interested amateur, I wish that I had seen this decades ago. It really gives one a clue that any simple explanation necessarily omits a too much and is misleading when predicting lift and drag. That is why CFD programs are used in spite of the huge amount of computer power required and the limitations of the answers.
Well, CFD is quite effective for lift. It isn't terribly accurate for drag. Computers do a terrible job of predicting laminar-turbulent transition, for example, and as a result do a terrible job predicting separation.

#### DarioC

Gold Member
On the triangles, I think somehow you have misunderstood which "triangles" I meant. The ones I am talking about are very small long strips that are located on the leading edge of the inboard part of the wing. Their purpose is to disturb the airflow at high angles of attack and cause the inboard part of the wing to "stall" before the sections outboard that have the ailerons. The desired result is the you dump a lot of the lift of the wing before you loose roll control when the aileron area stalls. It allows you to get a good comfortable sink rate during the final flare. A Beachcraft Bonanza has them as I remember and some others I am sure. I suspect they are called something like leading edge spoilers.

They are completely different in appearance and function from the various vortex generator "tabs."

Of course air goes under the wing as well as over it, but I was talking about the air going over the wing specifically, I wasn't discussing the air going under the wing, that is why I didn't write anything about that part of the airflow.

Anyone who has compared the landing characteristics of a Piper cub versus a Taylorcraft will understand the significance of the shape of the wing leading edge. Interestingly enough those two planes were designed by the same person.

#### Attachments

• 22.3 KB Views: 125
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#### DarioC

Gold Member
Yes that is one. Thanks for the photo.

#### fog37

Thank you DarioC , Boneh3ad and everyone else.

I am back to thinking about the fundamentals of pressure.

In a fluid at rest, the pressure at a hypothetical point $P$ could be measured using a Bourdon vacuum pressure gauge (the higher the spring compression the higher the pressure). The measured pressure at the infinitesimal point $P$ is the same no matter the orientation of the gauge (pressure is isotropic both in hydrostatics and fluid dynamics).

But when the fluid is moving, the pressure at a particular point along a specific streamline will be dependent on the fluid speed and on the way we measure it: if we placed Bourdon gauge facing the fluid upstream, the fluid would come to rest against the measuring surface of the gauge and the measured pressure would depend on the fluid speed (stagnation point): the higher the flow speed the higher the stagnation pressure. At that same stagnation point, with the first gauge remaining in place, the pressure measured by an extra Bourdon gauge oriented perpendicularly to the first gauge would be the same since the pressure is isotropic.

What if we don't create a stagnation point but use a Bourdon gauge oriented perpendicularly to the streamline? I guess it is called a tap if we measure that pressure at the surface of the pipe (pressure does not change across the streamlines). What pressure would the gauge measure in that case? This pressure would be smaller than the pressure at the stagnation point and the faster the fluid is moving along the streamline the smaller that pressure measured by the Bourdon gauge would be. In both case (fluid at rest or moving), the pressure $p$ in Bernoulli's theorem is the same. However, it is interesting that its value changes depending on the way we measure it (facing the fluid and creating a stagnation point or perpendicularly to a streamline).

In a pipe that has a narrowing, since the fluid must accelerate to maintain the same flow rate ($m^3 / s$), the pressure in the narrowing must be smaller to have a pressure gradient.

Thanks!


text:
abbracciopj

2022-06-20

Consider the following function
$f\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}{x}^{k}.$
Let us say I know ${a}_{k}$ belongs to a strict subset $F\subset \mathbb{R}$.
I want to characterize the set of sequences $\left\{{a}_{k}{\right\}}_{k=0}^{\mathrm{\infty }}$ for which $f\left(x\right)$ is a rational function in $x$.
Can someone point me to relevant results in literature?

jarakapak7

As
$f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},$
past a certain degree the coefficients of $f\left(x\right)q\left(x\right)$ vanish.
Hence except for the first few, the ${a}_{k}$ are constrained by a linear recurrence on $\text{deg}\left(q\right)+1$ terms.

Do you have a similar question?


text:
# SMARTS: distinguish between alcohol and carboxylic acid

I'm new to SMARTS and have to identify substructures with the Python package rdkit, so far so good. Let's say I have a SMILES with an carboxylic acid and a alcohol such as

OC(C1=CC(CO)=CC=C1)=O

and I want to get only the alcohol, then I would try something like this:

    mol = Chem.MolFromSmiles("OC(C1=CC(CO)=CC=C1)=O")
toSearch = Chem.MolFromSmarts("[#6][OX2H]")
bool = mol.HasSubstructMatch(toSearch)
subStruct = mol.GetSubstructMatches(toSearch)
print(bool)
print(subStruct)

which would return:

True
((1, 0), (5, 6))

However, the substructure search also identifies the OH of the acid which I don't want. How can I exclude that? My idea was to use some kind of recursive smarts for the alcohol to exclude all those "alcohols" where the carbon is bonded to another oxygen like "=O".

New contributor
dideldideldum is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Hello! You may also be interested in the Matter Modeling 'sister site'. (though please don't post questions twice) Aug 5 at 16:03

The simplest change to make for your SMARTS pattern would be specifying the total degree of the attached carbon is 4 with "[#6X4][OX2H]". This is enough to discriminate against the carboxylic acid.
The SMARTS pattern is "[OX2H][CX4;!$(C([OX2H])[O,S,#7,#15])]", which stipulates that the carbon attached to the oxygen not also be connected to O, S, N, or P. I made a little interactive widget that takes a molecule name or SMILES in the first field and a SMARTS pattern in the second. • Awesome!!!! Thanks for the answer and the two great links! I have one more question. This smart doesn't hit if the alcohol is on an aromatic carbon. In this Case this smarts "[OX2H][CX4,c;!$(C([OX2H])[O,S,#7,#15])]" should work, shouldn't it? thx! Aug 6 at 13:37


text:
Deals with image value superior to 0,1 range
2
0
Entering edit mode
@valerianmeline-12890
Last seen 4.3 years ago

Hello,

I use EBImage to analyse chlorophyll fluorescence images. The pixels values are contains between 0 and 1500. I can read these images but I can't write them while keeping the fluorescence values.

I can't export them with the intensity contrast  (0,1 range) because I need to keep these chlorophylll fluorescence range values.

Do you know how can i write these images without rescaling between 0 and 1 ?

Cheers,
Valérian

value ebimage writeImage error • 592 views
2
Entering edit mode
Andrzej Oleś ▴ 750
@andrzej-oles-5540
Last seen 8 months ago
Heidelberg, Germany

Dear Valérian,

you can save your images as 16 bit TIFFs and transform between the integer and double pixel representation on read/write. For this when saving your images you would divide the pixel values by the maximum value which can be stored as unsigned integer (UINT16), which is 2^16-1.  To transform the data back to integer range after loading the image multiply it by  2^16-1. The resulting image will contain integer values but stored internally as doubles, so in order to obtain output identical to your original data you want to change the storage mode back to "integer".  For convenient use it's probably a good idea to define some wrapper functions, say writeImageInt and readImageInt.

library(EBImage)

max16b = 2^16-1

# test image containing integer values
x = Image(1:max16b, c(2^8, 2^8))

f = "tmp.tiff"

# scale to double 0:1 range on write
writeImageInt = function(x, ...) {
writeImage(x/max16b, ...)
}

# rescale to integers on read
storage.mode(y) = "integer"
y
}

## [1] TRUE
0
Entering edit mode
@valerianmeline-12890
Last seen 4.3 years ago

Hello Andrzej,

Perfect it works for my data, Thank you very much for you fast answer

Valérian