TITLE: Transversality condition
QUESTION [4 upvotes]: From my textbook:
"If we examine linearizations of $f_\mu (x)$, (where $\dot{x}=f_\mu(x) \ x\in \mathbb{R}^n, \mu\in \mathbb{R}^2$) at the equilibria $f_\mu(x)=0$, then we can formulate a transversality condition which guarantees, for example, that no linearization of $f$, has a zero eigenvalue of multiplicity greater than two and that any equilibrium which does have a zero eigenvalue of multiplicity two has a Jordan normal form with the block
$$\left(
\begin{array}{cc}
  0 & 1 \\
  0 & 0
\end{array} \right)$$"
What is this transversality condition? I understand that this condition should include the Jacobian of $f$, even though I do not know how to proceed. I have already asked a question kept on transversality: Transversal intersection.
Thank you very much

REPLY [4 votes]: First, the authors introduce the set $\mathcal M=\{(x,\mu)\in\mathbb R^{n+2}: f_\mu(x)=0\}$. "Using transversality, we expect that $\mathcal M$ is a smooth two-dimensional surface" near $(\mu_0,x_0)$, they say. I think that the appropriate transversality condition for this part is $$\mathrm{rank}\, \frac{\partial f_\mu}{\partial (x,\mu)} (\mu_0,x_0)=n\tag{1}$$ In words, the total derivative of $f_\mu$ with respect to both $x$ and $\mu$, represented by an $n\times (n+2)$ matrix, has maximal rank $n$, and therefore the Implicit Function theorem applies.
Then we consider the set $$\mathcal J=\left\{ \frac{\partial f_\mu}{\partial x} : (x,\mu)\in\mathcal M\right\}\tag{2}$$
of all Jacobians (derivatives with respect to $x$ only) evaluated at the points of $\mathcal M$.
This is a parametric two-dimensional surface in the $n^2$-dimensional space of square matrices of size $n$.
Next, introduce a set $\mathcal B$ of bad square matrices. Specifically, $\mathcal B$ comprises of all matrices which have eigenvalue zero with multiplicity at least three, and also all matrices that have the block $\begin{pmatrix}0&0\\0&0\end{pmatrix}$ in their Jordan normal form. The authors leave it to the reader to verify that $\mathcal B$ has codimension greater than two. Informally, this is true because in order to qualify to be a member of $\mathcal B$, a matrix has to satisfy at least three different equations.
The idea of transversality now takes the following form: being two-dimensional, $\mathcal J$ will in general be disjoint from the set $\mathcal B$, because the latter set has codimension greater than two. I guess our new "transversality condition" is now $$\mathcal J\cap \mathcal B=\varnothing\tag{3}$$
As the authors say in the introduction, they are writing for mixed audience and are trying to avoid "mathematical pedantry". Thus, the lack of precise statements on these pages is supplanted by vigorous hand-waving and repeated mention of the words "transversality".


TITLE: Is the sequence of functions $g_n=ng_1(nx)$ a Cauchy sequence?
QUESTION [0 upvotes]: Given a function $g_1(x) \in \mathcal L^2(\Bbb R)$ that satisfies:
  $$\int_{-\infty}^{\infty}dx \space g_1(x)=1$$ one can define a
  sequence of functions  $g_n=ng_1(nx)$. Does $g_n(x)$ define a Cauchy
  sequence?

This is what I have tried:

Given that $\int_{-\infty}^{\infty}dx \space g_1(x)=1$ I "concluded" that $g_1(x)$ is $\delta$-function.
Under that assumption, the scaling property of the Dirac-delta function (distribution) yields $g_1(nx)=\frac{1}{n}g_1(x) \implies g_n=ng_1(nx)\iff g_n=\frac{1}{n}ng_1(x)\iff g_n=g_1(x)=\delta(x)$

I looked up the definition of a Cauchy sequence on wikipedia. As far as I understand I have to show that there exists an $N\in \Bbb N$ such that for every $m \ge n \ge N$, $g_m-g_n< \epsilon$
I basically have three questions now:

Is my reasoning correct so far?
How can I continue this proof?
What is the difference between a convergent sequence and a Cauchy sequence?

REPLY [2 votes]: No, the sequence $(g_n)$ need not be Cauchy in the $L^2$ topology.
Consider $g_1 = \begin{cases} 1, && x \in [0,1] \\ 0, && \text{otherwise}\end{cases}$. Then $g_n = \begin{cases} n, && x \in [0, \frac 1 n] \\ 0, && \text{otherwise}\end{cases}$.
One way of showing that $(g_n)$ is Cauchy is showing that $\| g_{n+p} - g_n \| \underset {p \to \infty} \longrightarrow 0 \ \forall n$. Let us show that this does not happen for our sequence.
Notice that $g_{n+p} - g_n = \begin{cases} p, && x \in [0, \frac 1 {n+p}] \\ -n, && x \in (\frac 1 {n+p}, \frac 1 n] \\ 0, && \text{otherwise} \end{cases}$. Then
$$\| g_{n+p} - g_n \| = \sqrt {\int \limits _{-\infty} ^\infty (g_{n+p} - g_n)^2} = \sqrt {\int \limits _{-\infty} ^\infty (g_{n+p} - g_n)^2} = \sqrt {\int \limits _0 ^{\frac 1 {n+p}} p^2 \ \textrm d x + \int \limits _{\frac 1 {n+p}} ^{\frac 1 n} (-n)^2 \ \textrm d x} = \\
\sqrt {\frac {p^2} {n+p} + \frac {n^2 p} {n(n+p)}} = \sqrt p$$
which clearly does not tend to $0$ for $p \to \infty$, therefore $(g_n)$ is not Cauchy in the topology of $L^2$.

On the other hand, if $g_1 \in L^1$ (not $L^2$, this is important!), then let us show that $g_n \to \delta$ in $\mathcal D '$ (the space of Schwartz distributions on $\Bbb R$). In particular, this will imply that $(g_n)$ is Cauchy in the topology of $\mathcal D '$.
Le $\varphi$ be a test function (i.e. a smooth function with compact support). Then
$$\lim \limits _n \int \limits _{-\infty} ^\infty g_n (x) \varphi (x) \ \textrm d x = \lim \limits _n \int \limits _{-\infty} ^\infty n g_1 (nx) \varphi (x) \ \textrm d x = \color{blue} {[\text{make } y = nx]} = \lim \limits _n \int \limits _{-\infty} ^\infty g_1 (y) \varphi \left( \frac y n \right) \ \textrm d x = \\
\int \limits _{-\infty} ^\infty \lim \limits _n \ g_1 (y) \varphi \left( \frac y n \right) \ \textrm d x = \int \limits _{-\infty} ^\infty g_1 (y) \varphi (0) \ \textrm d x = \varphi (0) \int \limits _{-\infty} ^\infty g_1 (y) \ \textrm d x = \varphi (0) \cdot 1 = \varphi (0) ,$$
which means exactly that $g_n \to \delta$ in $\mathcal D '$ (we have slipped the limit inside the integral with Lebesgue's dominated convergence theorem, and it is precisely here that you need $g_1 \in L^1$).


TITLE: Count of Disjoint subsets
QUESTION [1 upvotes]: Let $A$ be a set of $n$ elements. The number of ways, we can choose an ordered
pair $(B, C)$, where $B$, $C$ are disjoint subsets of $A$, equals
(A) $n^2$ (B) $n^3$ (C) $2^n$ (D) $3^n$
Doubt 1: What is a disjoint subset ?
My Attempt at the above problem: Assume 5 elements. Ordered pairs would be:
$$(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(3,5),(4,1),(4,2),(4,3),(4,4),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5)$$
25 elements. So, the answer would be: (A) $n^2$.
Is this the right approach ? Any better ways to solve this ?

REPLY [3 votes]: Take each of the $n$ elements of $A$ one by one and decide whether to put it in $B$ or in $C$.  No element can go in both $B$ and $C$ because they are supposed to be disjoint, so there are three options: in $B$, or in $C$, or neither.  The total number of pairs is $3^n$.


TITLE: Relationship between $su(4)$, $so(4)$ and $su(2)\oplus su(2)$
QUESTION [0 upvotes]: What is the relationship between the Lie algebras $su(4)$, $so(4)$ and $su(2)\oplus su(2)$ (if any)? I have read that $so(4)=su(2)\oplus su(2)$ but what is their relationship to $su(4)$?

REPLY [1 votes]: Yes, $\mathfrak{su}(2)\bigoplus\mathfrak{su}(2)\simeq\mathfrak{so}(4,\Bbb R)$. This comes from the fact that, if you see $SU(2)$ as the group of quaternions with norm $1$ and if $\Bbb H$ is the space of quaternions, then the image of the map$$\begin{array}{ccc}SU(2)\times SU(2)&\longrightarrow&\operatorname{Aut}(\Bbb H)\\(q,r)&\mapsto&\left(\begin{array}{ccc}\Bbb H&\longrightarrow&\Bbb H\\h&\mapsto&qhr^{-1}\end{array}\right)\end{array}$$is isomorphic to $SO(4,\Bbb R)$ (it is the set of the norm-preserving automorphims of $\Bbb H$), and its kernel is discrete (it is equal to $\{\pm(e,e)\}$).
And $SO(4,\Bbb R)$ is a strict subgroup of $SU(4)$.


TITLE: When is an improper Riemann integral equal to Lebesgue integral
QUESTION [1 upvotes]: My original problem is given $X_i\sim^{iid}U[0,1]$, find $$\lim_{n \rightarrow \infty} (X_1X_2 \cdots X_n)^{1/n} = \lim_{n \rightarrow \infty} (\prod_{i=1}^{n} X_i)^{1/n}$$
Well, $$\lim_{n \rightarrow \infty} (X_1X_2 \cdots X_n)^{1/n} = \lim_{n \rightarrow \infty} (\prod_{i=1}^{n} X_i)^{1/n} = \lim_{n \rightarrow \infty} \exp\{\frac{1}{n}\sum_{i=1}^{n} \log(X_i)\}$$ I'm thinking of using the weak law of large numbers.
So, I would like to compute $$E(\log(X_i))=\int^1_0 \log(x)\ dx$$
However, I'm supposed to be using the Lebesgue integral, and the only theorem that I know of that ensures me the Lebesgue integral will be equal to the improper Riemann integral is when the latter ($\int f \, dx$) exists and $f$ is non-negative, which is not my case $\because \log(x) < 0 \ \forall x \in (0,1)$
Therefore, what other theorems could ensure me that improper Riemann integral is the same as the Lebesgue integral?
Any help would be appreciated.

REPLY [2 votes]: $−\log{x}$ is non-negative, and the integral is linear.
In general, a function $f$ is Lebesgue-integrable on $X$ when it has an improper Riemann integral on $X$, and the (Lebesgue-)integrals of the positive and negative parts of $f$ are finite; this can be expressed as "$f$ is measurable and $\int_X \lvert f \rvert<\infty$", if I'm not forgetting some stupid special cases.


TITLE: Derangements of a multiset with "don't cares"?
QUESTION [0 upvotes]: Say I have a multiset with $n$ elements, with $n_1, n_2,..., n_k$ of each "type", and $n_0 = n-(n_1+n_2+...+n_k)$ of "type" "don't care".
I take a random permutation of the multiset.
What is the probability that none of the $n_1, n_2,..., n_k$ occupy a position taken by their respective "type" before the permutation?
As a simple example, say the multiset is $\{1,2,2,3,3,3,0,0,0,0,0,0,0\}$, where $0$ represents the "don't care" item.
Then $\{0,0,0,0,0,1,0,2,2,3,3,3,0\}$ is one such derangement - none of the  positive integers appears in a place previously occupied by one of itself, and I don't care about any zero that does.
There are some nice answers on the site for the case of multiset derangements, but none I've found deal with the "don't care" case.
Solution or reference(s) most appreciated.

REPLY [1 votes]: The answer here is basically the same as the standard derangement for multisets; the answer in Derangement formula for multisets provides the exact logic you need to solve this problem. The difference is that your "don't care" type is never used in the inclusion-exclusion; it simply needs to be added into the total number of permutations.
Following the major steps in that proof, specify $\beta_i$ instances of type $i$ to be fixed, for $1 \leq i \leq k$. Note we are specifically excluding type $0$, since we do not care about whether they are fixed points. The number of ways we can select these fixed points is
$${n_1 \choose \beta_1} {n_2 \choose \beta_2} \cdots {n_k \choose \beta_k}$$
We now permute the remaining elements arbitrarily. There may well be additional fixed points created, but the inclusion-exclusion will take care of that. The number of ways to permute the remaining elements is
$${{n-\sum_{i=1}^k \beta_k} \choose {n_0;n_1-\beta_1;\cdots;n_k-\beta_k}}$$
Finally, applying inclusion-exclusion gives us the answer
$$ \sum_{\beta_1 = 0}^{n_1} \cdots \sum_{\beta_k = 0}^{n_k} (-1)^{\beta_1 + \dots + \beta_k} \binom{n_1}{\beta_1} \cdots \binom{n_k}{\beta_k} {{n-\sum_{i=1}^k \beta_k} \choose {n_0;n_1-\beta_1;\cdots;n_k-\beta_k}}$$
Note that it is easy to adapt this argument to the case where you have multiple classes of "don't care" types.


TITLE: Blackbody radiation force (link), what are the fundamental forces behind it?
QUESTION [3 upvotes]: Here is a link to an article from phys.org describing what the scientists (M. Sonnleitner at the University of Innsbruck and Innsbruck Medical University in Austria, M. Ritsch-Marte at Innsbruck Medical University, and H. Ritsch at the University of Innsbruck) in the article call the blackbody force. http://phys.org/news/2013-07-blackbody-stronger-gravity.html
My question is, are they suggesting that this blackbody force is a fundamental force i.e. it belongs to group of gravity, weak/electromagnetic, and strong forces. And if not, what of the fundamental forces might or do make up this blackbody force?

REPLY [3 votes]: First, thanks for the link. It looks an interesting subject and in retrospect it's obvious that an atom in a field gradient will seek to minimise it's energy by moving to where the field is strongest, though I must admit it had never occurred to me.
The phenomenon is simply due to electromagnetism. Atomic energy levels will split in an electric field due to the Stark effect, so some levels will fall in energy while others will rise. If all the levels are occupied there is no net change, but if only some of the levels are occupied there will be a net decrease in energy. Since light is just an oscillating electric field, it causes a (oscillating) Stark splitting so it can lower the total energy of an atom. If the light intensity is completely even the atom won't move because the lowering of energy is the same everywhere. However if the light intensity is uneven (e.g. the usual 1/$r^2$ dependance on distance from the light source) then the energy of the atom is increasingly reduced as it moves towards the light source. Differentiate this to get the force, and you get the $1/r^3$ dependance mentioned in the article (for a first order Stark effect).
So it's not a new force, just a manifestation of electromagnetism, but it is an example of how physics can surprise me even after 52 years :-)


TITLE: If $\operatorname{Spec}(A)$ is a finite discrete set, then $A=\prod_i A_i$, where $A_i$ contains a unique maximal ideal.
QUESTION [3 upvotes]: Lemma 6.2 of Waterhouse's Introduction to Affine Group Schemes has a Lemma stating, for $k$ a field,

Let $A$ be a finite dimensional, commutative $k$ algebra. Then $A$ is a finite direct product of algebras $A_i$, each of which has a unique maximal ideal consisting of nilpotent elements.

Proof: Using the fact that $A$ is finite dimensional, Waterhouse first shows that every prime ideal of $A$ is in fact maximal, and that there are only finitely many primes. Since each is maximal, it follows that in the Zariski topology $Z(P)=\{P\}$, so each point is closed. Hence $\operatorname{Spec}(A)$ is a finite discrete set.
I understand this, but I lose him when he then immediately states, that hence $A=\prod_i A_i$, (without mention of what $A_i$ is), and that the unique prime in $A_i$ is maximal, and its elements must be nilpotent.
I know that since $\operatorname{Spec}(A)$ is finite and discrete, each $Z(P_i)=\{P_i\}$ is clopen, hence $Z(P_i)=Z(e_i)$ for an idempotent $e_i\in A$. Does $\operatorname{Spec}(A)=Z(e_1)\sqcup\cdots\sqcup Z(e_m)$ imply somehow that $A\simeq Ae_1\times\cdots\times Ae_m$? I know that if $e$ is an idempotent, $\operatorname{Spec}(A)=Z(e)\sqcup Z(1-e)$ implies $A\simeq Ae\times A(1-e)$, but if there are more than two clopen sets, I'm not sure how it extends. For instance, is $\operatorname{Spec}(A)=Z(e_1)\sqcup Z(e_2)\sqcup Z(e_3)=Z(e_1)\sqcup Z(e_2e_3)$, then we have $1-e_1=e_2e_3$, so that
$$
A\simeq Ae_1\times A(1-e_1)\simeq Ae_1\times Ae_2e_3.
$$
But it doesn't seem like $Ae_2e_3\simeq Ae_2\times Ae_3$.
Edit Would it be fair to say
$$
\operatorname{Spec}(A)=Z(P_1)\sqcup\cdots\sqcup Z(P_m)\simeq\operatorname{Spec}(Ae_1)\sqcup\cdots\sqcup\operatorname{Spec}(Ae_m)\simeq\operatorname{Spec}(\prod_{i}Ae_i)
$$
so that $A\simeq \prod_i Ae_i$ since $\operatorname{Spec}$ reflects isomorphism as an equivalence of categories?

REPLY [1 votes]: I don't yet see how to get the direct approach to work, so here's an answer that uses the theory of Artinian rings. This is basically just Theorem 8.7 in Atiyah-MacDonald, but I'll try to add a little extra exposition.
First note that since $A$ is a finite-dimensional $k$-algebra, then it is Artinian: each ideal is a vector subspace, so the length of a descending chain of ideals is bounded by $\operatorname{dim}_k(A)$.  Now we apply the aforementioned theorem:
Theorem. Every Artinian ring $A$ can be written as a finite direct product of local Artinian rings.
Proof. One can show that every prime ideal is maximal in an Artinian ring and that there are only finitely many maximal ideals (c.f., Prop 8.1 and 8.3 in A-M; Waterhouse has already showed this in your particular case).  Denote these by $\newcommand{\m}{\mathfrak{m}} \m_1, \ldots, \m_n$.  Then
$$
\DeclareMathOperator{\Nil}{Nil}
\newcommand{\p}{\mathfrak{p}}
\Nil(A) = \bigcap_{\substack{\p \trianglelefteq A\\ \text{prime}}} \p = \bigcap_{\substack{\m \trianglelefteq A\\ \text{maximal}}} \m = \bigcap_{i=1}^n \m_i \, .
$$
By Prop. 8.4 the nilradical is nilpotent, so $\Nil(A)^r = 0$ for some $r$, hence
$$
\prod_{i=1}^n \m_i^r \subseteq \bigcap_{i=1}^n \m_i^r = \Nil(A)^r = 0 \, .
$$
Since the ideals $\m_i^r$ are pairwise comaximal, then $A \cong \prod_{i=1}^n A/\m_i^r$ by the Chinese Remainder Theorem.  Each $A/\m_i^r$ is a local Artinian ring with maximal ideal $\m_i/\m_i^r$.  That $\m_i/\m_i^r$ consists of nilpotent elements follows from the fact that prime and maximal ideals coincide, since this implies that the nilradical and Jacobson radical ($=\m_i/\m_i^r$) are equal.


TITLE: Possible fundamental dimensions in Pi Buckingham theorem
QUESTION [0 upvotes]: While the Pi Buckingham is often taught in a fluid dynamics context, in which the relevant fundamental dimensions are generally : Mass, Length, Time and Temperature, it can obviously be applied to much more complex physics.
Is there a source that gives an extensive list of the possible fundamental dimensions? I would presume something similar to Moles would also be a fundamental quantity?

REPLY [1 votes]: One authority on this is NIST; they have a webpage giving

length (in meter)
Mass (in kilogram)
time (in second)
electric current (in ampere)
thermodynamic temperature (in kelvin)
amount of substance  (in mole) and
luminous intensity (in candela)

as the SI base units, and also provide a long list of derived units expressed in terms of the base units.


TITLE: Simple (enough) equation for estimating travel time
QUESTION [1 upvotes]: I'm (hoping to) improve a GIS transportation model to determine fire station service areas.  Generally, I'm trying to model how far a fire engine (or other unit) can travel the road network in 4 minutes (a national standard).
I know the lengths of the road segments, and currently my model estimates that a unit can travel an average of 80% of the posted speed limit in an attempt to account for acceleration/deceleration (in layman's terms).  Since creating the model, I've determined that a fire engine can accelerate at approximately 2.6 ft/s^2 and decelerate at -3.8 ft/s^2.
I'm trying to create a formula that I can apply to the model that would account for:
1. A unit's acceleration to the posted speed limit (if the road segment is long enough).
2. A unit's travel at the posted speed limit (assuming it reaches the limit).
3. A unit's deceleration at the end of the segment.

I can account for network nodes that are passed thru without a change in speed, but I can't figure out a simple formula to account for the maximum speed and segment length.
This seems like there should be a simple, canon type solution, but google-jitsu hasn't offered one up.  Any guidance or support is appreciated.  And, in case it wasn't terribly obvious already, I'm an absolute layman when it comes to physics (and algebra), so I apologize that I've mixed up scalar and vector terms-feel free to correct/critique.
**EDIT:**Since the model is for a unit that starts at 0 ft/s that accelerates to some velocity (e.g. 51 ft/s), then decelerates to 0 ft/s, is it possible to calculate the ratio of accel/decel (approx 3:2) and use that as a constant for each segment, with a variable for the additional distance outside of the accel/decel area?

REPLY [1 votes]: The general case of what you want is to take the integral of velocity with respect to time, which gives you distance. Of course, that is using calculus which is above the level of mathematics you mention in your question, so we'll have to break it down further.  However, I think the reason you're having trouble finding the equations you want on Google is because the more general version of this problem is a classic calculus problem, so most people would approach it with calculus.  Let's see if we can break it down into algebra.
The key to this is that you've got what we call a "piecewise equation" for velocity, so naturally the distance traveled is going to be a piecewise equation as well.  Your velocity curve has 3 segments, so we can expect the distance to be 3 segments as well.  There won't be 1 equation, but 3 equations governing the 3 different regions.
We'll start off with the example you show, where you hit maximum speed, and then we'll handle the corner case where you don't reach maximum speed on that segment.
During the first period, accelerating, we can use the formula $x=\frac{1}{2}at^2 + v_0t + x_0$ to show how far we went, where $a$ is our acceleration, $t$ is time, $v_0$ is our velocity at the start of the segment, and $x_0$ is the position at the start of the segment.  Since we start off with no velocity and at our starting point, this simplifies to a much shorter equation, $x=\frac{1}{2}at^2$.
During the second phase, where we maintain a speed, the distance we have traveled is simply $x=v_{max}(t-t_1) + x_1$ where $v_{max}$ is our maximum velocity, $t_1$ is the time we reached that maximum velocity, and $x_1$ is the position we reached that velocity.  How do we find these?  Well, we know we got there by accelerating at $a$ from an initial velocity $v_0=0$ to a maximum velocity $v_1=v_{max}$.  Clearly this will take $\frac{v_{max}}{a}$  time (its simply the definition of a constant acceleration), so $t_1=\frac{v_{max}}{a}$.  Once we have this, we can plug that time into the first equation to determine $x_1$, which is the distance we traveled while accelerating for $t_1$ seconds.
The third phase is similar to the first, except we have a negative acceleration.  I'm going to assign this the letter $d$, for deceleration.  Now there can be some confusion as to whether we should have a positive or negative value for $d$, but you wisely chose to make it a negative number ($d=-3.8$).  Either way works, but your way is more consistent, so I'm more than happy to do it that way!  Our equation for position in the third phase is $x=\frac{1}{2}d(t-t_2)^2 + v_2(t-t_2) + x_2$, where $t_2$ $x_2$ and $v_2$ are the time, position, and velocity when we start decelerating.  If we can find these three variables, we've characterized the entire 3 part curve!  Of course, we save the best for last.  These are a little trickier than $t_1$ and $x_1$ were.
The first thing we will want to note is that we already know $v_2$, because $v_2=v_{max}$.  We know this because I mentioned at the start that we were only looking at the case where you hit maximum speed.  We'll handle the other case, where you never reach maximum speed, in a second part.  The next thing we know is that we have to decelerate to a stop.  If we call the stopping point $x_3$, which occurs at $t_3$, we can figure out how early we have to start applying the brakes.  Hopefully it should be easy to see that we will need to brake for $\frac{-d}{v_{max}}$ seconds to go from max velocity to stopped.  Using the variables we've been using all along, we can write that as $(t_3-t_2)=\frac{-d}{v_{max}}$.  I point this out because the above equation for where we are during the third phase has that $(t-t_2)$ term everywhere, which looks kinda scary, but we can see the same pattern here.
We're almost done!  The final step we need to finish characterizing the curve is to figure out $t_2$ and $x_2$.  Presumably we know $x_3$, because its the stopping point for that segment.  We can use the equation for the decelerating period to solve for $x_2$, given that we know $(t_3-t_2)$, $v_2$, and $x_3$.  Once we have $x_2$, we can plug that into either the steady state equation or the decelerating equation to find $t_2$ (the steady state one will be easier!).
Phew after all that math, we have several triplets of times, positions, and velocities. We have:

$t_0$, $x_0$, and $v_0$ which are the starting time position and velocity (and all have the value of 0, since we're just starting)
$t_1$, $x_1$ and $v_1$, which are the time, position and velocity where the truck finishes accelerating ($v_1=v_{max}$, and the other two are solved for)
$t_2$, $x_2$ and $v_2$, which are the time, position and velocity when the truck begins decelerating ($v_2=v_{max}$, and the other two are solved for)
$t_3$, $x_3$ and $v_3$, which are the time, position and velocity when the truck finishes stopping ($x_3$ is the length of your segment, and $v_3$ = 0.  $t_3$ is solved for)

Now for the final step, you don't care about how far you can travel in a given amount of time, you care about how long it takes to travel a given distance.  You want the inverse of this equation.  Fortunately, this is easy enough to do with a piece wise equation.  By comparing your desired distance, $x$ to $x_1$ and $x_2$, you can figure out which of the three regions you are in.  Once you know that, you can simply invert that one equation to solve for time given a distance!
Well, that was the second to last step.  All of that assumed you had enough time to accelerate to full speed.  What if the segment is short?  We need a different set of equations for that.  Fortunately, they're not all that different.  In fact, they're easier.  We only have two regions: $t_0$ to $t_1$ where the truck is accelerating, followed by $t_1$ to $t_2$ where it is decelerating.  All we have to do is find that midpoint where the truck lets off the gas and hits the brakes, and we're set!
Because you have a constant acceleration, we can easily relate the lengths of the two duration.  We know that $v_1=a(t_1-t_0)+v_0$ from the acceleration curve, and $v_2=d(t_2-t_1)+v_1$.  Dropping all of the terms which we know to be zero, we get $v_1=at_1$ and $0=d(t_2-t_1)+v_1$.  With a little bit of rearranging we get $at_1=-d(t_2-t_1)$ or $t_2=\frac{a+d}{d}t_1$.  This is a really nice equation to know: there's a clear relationship between $t_1$ and $t_2$ that we can exploit!
Okay, so how do we use this.  Well, we know the length of the segment, $x_2$ because that's a given.  Let's look at the equation for the second half: $x_2=\frac{1}{2}d(t_2-t_1)^2+v_1(t_2-t_1)+x_1$.  There's lots of unknowns here, but we can fill some of them in.  We can replace $t_2$ using the equation above, and we can get $v_1$ and $x_1$ from the acceleration curve.  Putting all of these together we get $x_2=\frac{1}{2}d(\frac{a+d}{d}t_1-t_1)^2+(at_1)(\frac{a+d}{d}t_1-t_1)+(\frac{1}{2}at_1^2)$.  That equation looks obnoxious, but if you look closely, there's only one unknown left: $t_1$.  You can solve for it!
After solving, we have the following triples:

$t_0$, $x_0$, and $v_0$ which are the starting time position and velocity (and all have the value of 0, since we're just starting)
$t_1$, $x_1$ and $v_1$, which are the time, position and velocity where the truck finishes accelerating and starts decelerating (all must be solved for)
$t_2$, $x_2$ and $v_2$, which are the time, position and velocity when the truck finishes stopping ($x_2$ is the length of your segment, and $v_2$ = 0.  $t_2$ is solved for)

From there it's all the same!
Now, after all of this, you should be able to see why it was hard to Google for!  None of these steps are particularly fancy, but they are all very specialized to your particular rules for acceleration and deceleration.  You also have two cases, one where you reach a max speed, and one where you do not.  There's just a lot of little steps, and it's not obvious that the steps are moving your forward!
Please check my math when you implement this.  I'd hate to misguide our first responders!


TITLE: Delta Epsilon Format limit Definition for a multi-variable function?
QUESTION [0 upvotes]: Show that $\lim\frac{x}{y}$ as $(x,y)\to(1,1)$ equals $1$ by the delta epsilon definition.
$\sqrt{(x-1)^2 + (y-1)^2} < \delta$
$|\frac{x}{y}-1| < \epsilon$
How do I proceed?

REPLY [0 votes]: Hint: $\sqrt{(x-1)^2 + (y-1)^2} < \delta$ implies $|x-1|<\delta$ and $|y-1| < \delta$.

 Then,
 $$-\frac{2\delta}{1+\delta}=\frac{1-\delta}{1+\delta}-1<\frac{x}{y}-1 < \frac{1+\delta}{1-\delta}-1 = \frac{2\delta}{1-\delta}.$$
 Choose $\delta$ small enough.


TITLE: Using induction to define Indexed family of HITs in agda
QUESTION [3 upvotes]: I am looking to define a family of HITs parametrized by $\mathbb{N}$. I want $(-)$-glob : $\mathbb{N} \to Type$, so that $n$-glob is the n-dimensional glob. I know how to construct the n-$glob$ by induction on $n$. The problem I am running into is that I do not know how to use induction on $\mathbb{N}$ in the middle of the definition of a HIT. For example, I have the code
data _-glob (n : ℕ) : Type where

and I want to case on $n$ so that I may specify what $n:\equiv 0$ is and what $n:\equiv suc (n)$ is, given what $n$ is. But, I cannot simply write
data _-glob (n : ℕ) : Type where
 H : ?

and then case on $n$. Does anyone know how to use induction in the middle of such a definition?
EDIT: Here is my (quite literal) "black board" inductive definition of _-glob. Its pretty messy and I'm sure there are more elgant ways to define it but this is what I'm trying to turn into agda code:
0-glob where

 G0 : 0-glob

1-glob where

 in : 0-glob -> 1-glob

 G0_2 : 1-glob

 G1 : in(G0) = G0_2

(n+2)-glob where

 in : (n+1)-glob -> (n+2)-glob

 G(n+1)_2 : in(in(Gn)) = in(Gn_2)

 G(n+2) : in(G(n+1)) = G(n+1)_2

Now, I'm pretty sure this give me what I want. 1-glob is equivalent to the interval. 2-glob contains a copy of 1-glob, via the constructor "in". But, 2-glob contains and additional path parallel to the preexisting one. And it contains a two path between these 1 paths. But, to actually define this as a HIT (in one go), i would need to case on n to specify the constructors.

REPLY [3 votes]: Use
data _-glob : ℕ -> Type where

Now you can just do something like H : blabla -> zero -glob. Warning: This is called an indexed inductive family, and cubical Agda still has some unimplemented functionalities about this feature yet. You might want to look at the standard library (not the cubical library) on the Fin type to see an example.
Another way to do it is as a recursive type. You first define zero-glob separately. Then you define suc-glob (n-glob : Type) : Type. Now take
_-glob : (n : ℕ) -> Type
zero -glob = zero-glob
(suc n) -glob = suc-glob (n -glob)


TITLE: closed substacks of cartesian powers of a stack
QUESTION [3 upvotes]: Let $\mathbb{Z}/2\mathbb{Z}$ act on $\mathbb{A}^1$ as $x \mapsto -x$, and let $\mathscr{X}$ be the quotient stack. It has coarse moduli space $\mathbb{A}^1$ and a residual gerbe $B\mathbb{Z}/2\mathbb{Z}$ at 0. There is then a surjective map from cartesian powers $\mathscr{X}^n$ to $\mathbb{A}^n$. What are the closed substacks of $\mathscr{X}^n$? Will there be anything other than pullbacks of closed subschemes of $\mathbb{A}^n$?

REPLY [4 votes]: A substack of $\mathcal{X}$ is given by a $Z/2Z$ invariant subscheme of $\mathbb{A}^1$ and hence given by a $Z/2Z$ invariant ideal $I\subset k[x]$. So for example, the ideals $(x^n)$ give substacks, but they only pullback back from subschemes of the coarse space for $n$ even: the coarse space is given by $Spec(k[x^2])$. The substack given by $(x)$ here is just a copy of $B(Z/2Z)$ embedded at the origin.


TITLE: Am I linearizing this non-linear system properly?
QUESTION [0 upvotes]: I have the following system:
$$x'=x(1-x)-xy$$
$$y'=y(1-\frac{y}{x})$$
I am trying to linearize it, and from my understanding, if the equilibrium points are hyperbolic, that is, its eigenvalues all have a non-zero real part, I can drop the higher order terms. If I compute $DF_{X_0}$, I get:
$$\begin{pmatrix} 1-2x-y & -x \\ \frac{y^2}{x^2} & 1-\frac{2y}{x} \end{pmatrix}$$
When I drop the higher order terms, I am left with:
$$\begin{pmatrix} 1-2x-y & -x \\ 0 & 1 \end{pmatrix}$$
Am I linearizing this correctly?

REPLY [2 votes]: No, you have misunderstood this, you should not drop higher order terms in the Jacobian like you are doing! What you should to is to substitute the equilibrium that you want to investigate into the Jacobian. That will give you a constant matrix containing the coefficients of the first-order terms in the Taylor expansion around that point, so the “dropping of terms” is automatically taken care of.


TITLE: Show the transformation matrix in relation to the canonical basis of the respective spaces.
QUESTION [3 upvotes]: $W = \{A \in M_{2\times2} (\mathbb{R}): A_{11} =  A_{12}\text{ and }A_{22} =  A_{21}\}$ is isomorphic to $P_1 (\mathbb{R})$.
Show the transformation matrix in relation to the canonical basis of the respective spaces.
I'm in doubt about the "canonical basis" of $W$ and $P_1$.
Well, I know that the caninocal basis of $P_1$ is $\{1, x\}$.
Intuitively, I think the canonical basis of $W$ is the set of the matrices
\begin{bmatrix}
   1 & 1\\
   0 & 0
\end{bmatrix}
and
\begin{bmatrix}
   0 & 0\\
   1 & 1
\end{bmatrix}
I can understand the transformation and write it with $1 \times 2$ matrices (and the transformation as a $2 \times 2$ matrix), for example. but I don't know how to show the transformation matrix in the canonical basis of each set.

REPLY [1 votes]: A canonical bases is something like a "natural/obvious choice" of a basis. So you cannot prove, that something is a canonical bases, you can just state, that most of the mathematicians would use take a given basis as a canonical one.
Your choice for the canonical bases are right. Note, that if you take a transformation $f$ which maps
$\begin{align}\left(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right) & \mapsto 1 \\
\left(\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right) & \mapsto x \end{align}$
then the matrix of $f$ is $\left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)$.


TITLE: How to interpret this formula from dsp?
QUESTION [1 upvotes]: Can you describe me what this formula does and what result in the end?
$$\delta(x)=\begin{cases}
0,  & \text{if $x \neq 0$} \\
1, & \text{if $x = 0$}  \\
\end{cases}$$
$$x(t)=\sum_{k=-\infty}^{\infty}x(t_k)\delta(t-t_k)$$

REPLY [1 votes]: Just to keep things sorted out, let us denote the expression $x(t)$ occurring on the left by $\hat x(t)$.  Presumably $t$ is one of the $t_k$, say $t_l$; so then the formula reads
$\hat x(t_l) = \sum_{k = - \infty}^{k = \infty} x(t_k) \delta(t_l - t_k), \tag{1}$
and if the given definition of $\delta$ is applied in this equation, we see that $\delta(t_l - t_k) = 0$ unless $l = k$; then it is $1$.  So evidently
$ \sum_{k = - \infty}^{k = \infty} x(t_k) \delta(t_l - t_k) = x(t_l), \tag{2}$
whence
$\hat x(t_l) = x(t_l); \tag{3}$
apparently the formula yields the identity map, so we may as well have written $x(t_l)$ on the left in any event; the notation $\hat x$ now looks somewhat superfluous, though I think it was a useful distinction while things were being sorted out.
Well, I hope this helps.  Cheers,
and as always,
Fiat Lux!!!


TITLE: How does the dot product "remove" unit vectors?
QUESTION [2 upvotes]: From my understanding from university physics from Pearson (Page ~20):
Dot product multiplies "like" unit vector terms and cross product multiplies "unlike"  unit vector terms. So, why does the cross product retain its unit vector (ñ), but dot product does not?
Example:
$A\cdot B = AB\cos(\theta) =$ magnitude of $A$ * The component of $B$ that is parallel to $A$.
So shouldn't the unit vector be going in the direction of $A$? Do mathematicians drop unit vectors when stuff only goes in one direction and then call it a scalar?

REPLY [0 votes]: Asker: If the dot product was a vector, then it should be in the direction that was multiplied by the like components. Thus, for Newton's second law:
F = mr = xi + yj + zk
(F)•j  = (xi + yj + zk )•j
(F_y)*j = (yj)*j
Cancel out j from both sides, which is our "residue" unit vector:
F_y = yj

Chris


TITLE: A Banach space is reflexive if a closed subspace and its quotient space are both reflexive
QUESTION [10 upvotes]: Let $X$ be a Banach space. Let $Y$ be a closed subspace. Suppose that the normed spaces (in fact Banach spaces) $Y$ and $X/Y$ are both reflexive. I need to show that $X$ is reflexive.
I cannot show this but I feel that I could use the fact that a Banach space is reflexive if and only if its closed unit ball is weakly compact. So in this case we know $B_Y$ and $B_{X/Y}$ are weakly compact. Let $\mathcal{U}$ be a weakly open cover for $B_X$. I feel that a finite subcover can be obtained by considering the sets of the form $a+Y\cap B_X\subset X$, which are certainly compact, as $a+Y\cap B_X\subset a+nB_Y\cap B_X$, which is w-closed in the weakly compact set $a+nB_Y$ for some sufficiently large $n$, and the corresponding $a+Y\in X/Y$.
Could anybody suggest anything? Thanks.

REPLY [8 votes]: A property of a Banach space is a three-space property if whenever $E$ is a Banach space, $F\subseteq E$ is a closed linear subspace and two of the spaces $E$, $F$ and $E/F$ have the property, then all three of the spaces $E$, $F$ and $E/F$ necessarily have the property.
Reflexivity is a three-space property. For the proof see Theorem 1.11.19 in An Introduction to Banach Space Theory Graduate Texts in Mathematics. Robert E. Megginson.