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# A story of block-ascending permutations

I recently had a combinatorics paper appear in the EJC. In this post I want to brag a bit by telling the “story” of this paper: what motivated it, how I found the conjecture that I originally did, and the process that eventually led me to the proof, and so on.

This work was part of the Duluth REU 2017, and I thank Joe Gallian for suggesting the problem.

## 1. Background

Let me begin by formulating the problem as it was given to me. First, here is the definition and notation for a “block-ascending” permutation.

Definition 1

For nonnegative integers ${a_1}$, …, ${a_n}$ an ${(a_1, \dots, a_n)}$-ascending permutation is a permutation on ${\{1, 2, \dots, a_1 + \dots + a_n\}}$ whose descent set is contained in ${\{a_1, a_1+a_2, \dots, a_1+\dots+a_{n-1}\}}$. In other words the permutation ascends in blocks of length ${a_1}$, ${a_2}$, …, ${a_n}$, and thus has the form

$\displaystyle \pi = \pi_{11} \dots \pi_{1a_1} | \pi_{21} \dots \pi_{2a_2} | \dots | \pi_{n1} \dots \pi_{na_n}$

for which ${\pi_{i1} < \pi_{i2} < \dots < \pi_{ia_i}}$ for all ${i}$.

It turns out that block-ascending permutations which also avoid an increasing subsequence of certain length have nice enumerative properties. To this end, we define the following notation.

Definition 2

Let ${\mathcal L_{k+2}(a_1, \dots, a_n)}$ denote the set of ${(a_1, \dots, a_n)}$-ascending permutations which avoid the pattern ${12 \dots (k+2)}$.

(The reason for using ${k+2}$ will be explained later.) In particular, ${\mathcal L_{k+2}(a_1 ,\dots, a_n) = \varnothing}$ if ${\max \{a_1, \dots, a_n\} \ge k+2}$.

Example 3

Here is a picture of a permutation in ${\mathcal L_7(3,2,4)}$ (but not in ${\mathcal L_6(3,2,4)}$, since one can see an increasing length ${6}$ subsequence shaded). We would denote it ${134|69|2578}$.

Now on to the results. A 2011 paper by Joel Brewster Lewis (JBL) proved (among other things) the following result:

Theorem 4 (Lewis 2011)

The sets ${\mathcal L_{k+2}(k,k,\dots,k)}$ and ${\mathcal L_{k+2}(k+1,k+1,\dots,k+1)}$ are in bijection with Young tableau of shape ${\left< (k+1)^n \right>}$.

Remark 5

When ${k=1}$, this implies ${\mathcal L_3(1,1,\dots,1)}$, which is the set of ${123}$-avoiding permutations of length ${n}$, is in bijection with the Catalan numbers; so is ${\mathcal L_3(2,\dots,2)}$ which is the set of ${123}$-avoiding zig-zag permutations.

Just before the Duluth REU in 2017, Mei and Wang proved that in fact, in Lewis’ result one may freely mix ${k}$ and ${k+1}$‘s. To simplify notation,

Definition 6

Let ${I \subseteq \left\{ 1,\dots,n \right\}}$. Then ${\mathcal L(n,k,I)}$ denotes ${\mathcal L_{k+2}(a_1,\dots,a_n)}$ where

$\displaystyle a_i = \begin{cases} k+1 & i \in I \\ k & i \notin I. \end{cases}$

Theorem 7 (Mei, Wang 2017)

The ${2^n}$ sets ${\mathcal L(n,k,I)}$ are also in bijection with Young tableau of shape ${\left< (k+1)^n \right>}$.

The proof uses the RSK correspondence, but the authors posed at the end of the paper the following open problem:

Problem

Find a direct bijection between the ${2^n}$ sets ${\mathcal L(n,k,I)}$ above, not involving the RSK correspondence.

This was the first problem that I was asked to work on. (I remember I received the problem on Sunday morning; this actually matters a bit for the narrative later.)

At this point I should pause to mention that this ${\mathcal L_{k+2}(\dots)}$ notation is my own invention, and did not exist when I originally started working on the problem. Indeed, all the results are restricted to the case where ${a_i \in \{k,k+1\}}$ for each ${i}$, and so it was unnecessary to think about other possibilities for ${a_i}$: Mei and Wang’s paper use the notation ${\mathcal L(n,k,I)}$. So while I’ll continue to use the ${\mathcal L_{k+2}(\dots)}$ notation in the blog post for readability, it will make some of the steps more obvious than they actually were.

## 2. Setting out

Mei and Wang’s paper originally suggested that rather than finding a bijection ${\mathcal L(n,k,I) \rightarrow \mathcal L(n,k,J)}$ for any ${I}$ and ${J}$, it would suffice to biject

$\displaystyle \mathcal L(n,k,I) \rightarrow \mathcal L(n,k,\varnothing)$

and then compose two such bijections. I didn’t see why this should be much easier, but it didn’t seem to hurt either.

As an example, they show how to do this bijection with ${I = \{1\}}$ and ${I = \{n\}}$. Indeed, suppose ${I = \{1\}}$. Then ${\pi_{11} < \pi_{12} < \dots < \pi_{1(k+1)}}$ is an increasing sequence of length ${k+1}$ right at the start of ${\pi}$. So ${\pi_{1(k+1)}}$ had better be the largest element in the permutation: otherwise later in ${\pi}$ the biggest element would complete an ascending permutation of length ${k+2}$ already! So removing ${\pi_{1(k+1)}}$ gives a bijection between ${\mathcal L(n,k,\{1\}) \rightarrow \mathcal L(n,k,\varnothing)}$.

But if you look carefully, this proof does essentially nothing with the later blocks. The exact same proof gives:

Proposition 8

Suppose ${1 \notin I}$. Then there is a bijection

$\displaystyle \mathcal L(n,k,I \cup \{1\}) \rightarrow \mathcal L(n,k,I)$

by deleting the ${(k+1)}$st element of the permutation (which must be largest one).

Once I found this proposition I rejected the initial suggestion of specializing ${\mathcal L(n,k,I) \rightarrow \mathcal L(n,k,\varnothing)}$. The “easy case” I had found told me that I could take a set ${I}$ and delete the single element ${1}$ from it. So empirically, my intuition from this toy example told me that it would be easier to find bijections ${\mathcal L(n,k,I) \rightarrow \mathcal L(n,k,I')}$ whee ${I'}$ and ${I}$ were only “a little different”, and hope that the resulting bijection only changed things a little bit (in the same way that in the toy example, all the bijection did was delete one element). So I shifted to trying to find small changes of this form.

## 3. The fork in the road

### 3.1. Wishful thinking

I had a lucky break of wishful thinking here. In the notation ${\mathcal L_{k+2}(a_1, \dots, a_n)}$ with ${a_i \in \{k,k+1\}}$, I had found that one could replace ${a_1}$ with either ${k}$ or ${k+1}$ freely. (But this proof relied heavily on the fact the block really being on the far left.) So what other changes might I be able to make?

There were two immediate possibilities that came to my mind.

• Deletion: We already showed ${a_1}$ could be changed from ${k+1}$ to ${k}$ for any ${i}$. If we can do a similar deletion with ${a_i}$ for any ${i}$, not just ${i=1}$, then we would be done.
• Swapping: If we can show that two adjacent ${a_i}$‘s could be swapped, that would be sufficient as well. (It’s also possible to swap non-adjacent ${a_i}$‘s, but that would cause more disruption for no extra benefit.)

Now, I had two paths that both seemed plausible to chase after. How was I supposed to know which one to pick? (Of course, it’s possible neither work, but you have to start somewhere.)

Well, maybe the correct thing to do would have to just try both. But it was Sunday afternoon by the time I got to this point. Granted, it was summer already, but I knew that come Monday I would have doctor appointments and other trivial errands to distract me, so I decided I should pick one of them and throw the rest of the day into it. But that meant I had to pick one.

(I confess that I actually already had a prior guess: the deletion approach seemed less likely to work than the swapping approach. In the deletion approach, if ${i}$ is somewhere in the middle of the permutation, it seemed like deleting an element could cause a lot of disruption. But the swapping approach preserved the total number of elements involved, and so seemed more likely that I could preserve structure. But really I was just grasping at straws.)

### 3.2. Enter C++

Yeah, I cheated. Sorry.

Those of you that know anything about my style of math know that I am an algebraist by nature — sort of. It’s more accurate to say that I depend on having concrete examples to function. True, I can’t do complexity theory for my life, but I also haven’t been able to get the hang of algebraic geometry, despite having tried to learn it three or four times by now. But enumerative combinatorics? OH LOOK EXAMPLES.

Here’s the plan: let ${k=3}$. Then using a C++ computer program:

• Enumerate all the permutations in ${S = \mathcal L_{k+2}(3,4,3,4)}$.
• Enumerate all the permutations in ${A = \mathcal L_{k+2}(3,3,3,4)}$.
• Enumerate all the permutations in ${B = \mathcal L_{k+2}(3,3,4,4)}$.

If the deletion approach is right, then I would hope ${S}$ and ${A}$ look pretty similar. On the flip side, if the swapping approach is right, then ${S}$ and ${B}$ should look close to each other instead.

It’s moments like this where my style of math really shines. I don’t have to make decisions like the above off gut-feeling: do the “data science” instead.

### 3.3. A twist of fate

Except this isn’t actually what I did, since there was one problem. Computing the longest increasing subsequence of a length ${N}$ permutation takes ${O(N \log N)}$ time, and there are ${N!}$ or so permutations. But when ${N = 3+4+3+4=14}$, we have ${N! \cdot N \log N \approx 3 \cdot 10^{12}}$, which is a pretty big number. Unfortunately, my computer is not really that fast, and I didn’t really have the patience to implement the “correct” algorithms to bring the runtime down.

The solution? Use ${N = 1+4+3+2 = 10}$ instead.

In a deep irony that I didn’t realize at the time, it was this moment when I introduced the ${\mathcal L_{k+2}(a_1, \dots, a_n)}$ notation, and for the first time allowed the ${a_i}$ to not be in ${\{k,k+1\}}$. My reasoning was that since I was only doing this for heuristic reasons, I could instead work with ${S = \mathcal L_{k+2}(2,4,3,2)}$ and probably not change much about the structure of the problem, while replacing ${N = 2 + 4 + 3 + 2 = 11}$, which would run ${1000}$ times faster. This was okay since all I wanted to do was see how much changing the “middle” would disrupt the structure.

And so the new plan was:

• Enumerate all the permutations in ${S = \mathcal L_{k+2}(1,4,3,2)}$.
• Enumerate all the permutations in ${A = \mathcal L_{k+2}(1,3,3,2)}$.
• Enumerate all the permutations in ${B = \mathcal L_{k+2}(1,3,4,2)}$.

I admit I never actually ran the enumeration with ${A}$, because the route with ${S}$ and ${B}$ turned out to be even more promising than I expected. When I compared the empirical data for the sets ${S}$ and ${B}$, I found that the number of permutations with any particular triple ${(\pi_1, \pi_9, \pi_{10})}$ were equal. In other words, the outer blocks were preserved: the bijection

$\displaystyle \mathcal L_{k+2}(1,4,3,2) \rightarrow \mathcal L_{k+2}(1,3,4,2)$

does not tamper with the outside blocks of length ${1}$ and ${2}$.

This meant I was ready to make the following conjecture. Suppose ${a_i = k}$, ${a_{i+1} = k+1}$. There is a bijection

$\displaystyle \mathcal L_{k+2}(a_1, \dots, a_i, a_{i+1}, \dots, a_n) \rightarrow \mathcal L_{k+2}(a_1, \dots, a_{i+1}, a_{i}, \dots, a_n)$

which only involves rearranging the elements of the ${i}$th and ${(i+1)}$st blocks.

## 4. Rooting out the bijection

At this point I was in a quite good position. I had pinned down the problem to a finding a particular bijection that I was confident had to exist, since it was showing up to the empirical detail.

Let’s call this mythical bijection ${\mathbf W}$. How could I figure out what it was?

### 4.1. Hunch: ${\mathbf W}$ preserves order-isomorphism

Let me quickly introduce a definition.

Definition 9

We say two words ${a_1 \dots a_m}$ and ${b_1 \dots b_m}$ are order-isomorphic if ${a_i < a_j}$ if and only ${b_i < b_j}$. Then order-isomorphism gives equivalence classes, and there is a canonical representative where the letters are ${\{1,2,\dots,m\}}$; this is called a reduced word.

Example 10

The words ${13957}$, ${12846}$ and ${12534}$ are order-isomorphic; the last is reduced.

Now I guessed one more property of ${\mathbf W}$: this ${\mathbf W}$ should order-isomorphism.

What do I mean by this? Suppose in one context ${139 | 57}$ changed to ${39 | 157}$; then we would expect that in another situation we should have ${124 | 68}$ changing to ${24 | 168}$. Indeed, we expect ${\mathbf W}$ (empirically) to not touch surrounding outside blocks, and so it would be very strange if ${\mathbf W}$ behaved differently due to far-away numbers it wasn’t even touching.

So actually I’ll just write

$\displaystyle \mathbf W(123|45) = 23|145$

for this example, reducing the words in question.

### 4.2. Keep cheating

With this hunch it’s possible to cheat with C++ again. Here’s how.

Let’s for concreteness suppose ${k=2}$ and the particular sets

$\displaystyle \mathcal L_{k+2}(1,3,2,1) \rightarrow \mathcal L_{k+2}(1,2,3,1).$

Well, it turns out if you look at the data:

• The only element of ${\mathcal L_{k+2}(1,3,2,1)}$ which starts with ${2}$ and ends with ${5}$ is ${2|147|36|5}$.
• The only element of ${\mathcal L_{k+2}(1,2,3,1)}$ which starts with ${2}$ and ends with ${5}$ is ${2|47|136|5}$.

So that means that ${147 | 36}$ is changed to ${47 | 136}$. Thus the empirical data shows that

$\displaystyle \mathbf W(135|24) = 35|124.$

In general, it might not be that clear cut. For example, if we look at the permutations starting with ${2}$ and ${4}$, there is more than one.

• ${2 | 1 5 7 | 3 6 | 4}$ and ${2 | 1 6 7 | 3 5 | 4}$ are both in ${\mathcal L_{k+2}(1,3,2,1)}$.
• ${2 | 5 7 | 1 3 6 | 4}$ and ${2 | 6 7 | 1 3 5 | 4}$ are both in in ${\mathcal L_{k+2}(1,2,3,1)}$.

Thus

$\displaystyle \mathbf W( \{135|24, 145|23\} ) = \{35|124, 45|123\}$

but we can’t tell which one goes to which (although you might be able to guess).

Fortunately, there is lots of data. This example narrowed ${135|24}$ down to two values, but if you look at other places you might have different data on ${135|24}$. Since we think ${\mathbf W}$ is behaving the same “globally”, we can piece together different pieces of data to get narrower sets. Even better, ${\mathbf W}$ is a bijection, so once we match either of ${135|24}$ or ${145|23}$, we’ve matched the other.

You know what this sounds like? Perfect matchings.

So here’s the experimental procedure.

• Enumerate all permutations in ${\mathcal L_{k+2}(2,3,4,2)}$ and ${\mathcal L_{k+2}(2,4,3,2)}$.
• Take each possible tuple ${(\pi_1, \pi_2, \pi_{10}, \pi_{11})}$, and look at the permutations that start and end with those particular four elements. Record the reductions of ${\pi_3\pi_4\pi_5|\pi_6\pi_7\pi_8\pi_9}$ and ${\pi_3\pi_4\pi_5\pi_6|\pi_7\pi_8\pi_9}$ for all these permutations. We call these input words and output words, respectively. Each output word is a “candidate” of ${\mathbf W}$ for a input word.
• For each input word ${a_1a_2a_3|b_1b_2b_3b_4}$ that appeared, take the intersection of all output words that appeared. This gives a bipartite graph ${G}$, with input words being matched to their candidates.
• Find perfect matchings of the graph.

And with any luck that would tell us what ${\mathbf W}$ is.

### 4.3. Results

Luckily, the bipartite graph is quite sparse, and there was only one perfect matching.

246|1357 => 2467|135
247|1356 => 2457|136
256|1347 => 2567|134
257|1346 => 2357|146
267|1345 => 2367|145
346|1257 => 3467|125
347|1256 => 3457|126
356|1247 => 3567|124
357|1246 => 1357|246
367|1245 => 1367|245
456|1237 => 4567|123
457|1236 => 1457|236
467|1235 => 1467|235
567|1234 => 1567|234

If you look at the data, well, there are some clear patterns. Exactly one number is “moving” over from the right half, each time. Also, if ${7}$ is on the right half, then it always moves over.

Anyways, if you stare at this for an hour, you can actually figure out the exact rule:

Claim 11

Given an input ${a_1a_2a_3|b_1b_2b_3b_4}$, move ${b_{i+1}}$ if ${i}$ is the largest index for which ${a_i < b_{i+1}}$, or ${b_1 = 1}$ if no such index exists.

And indeed, once I have this bijection, it takes maybe only another hour of thinking to verify that this bijection works as advertised, thus solving the original problem.

Rather than writing up what I had found, I celebrated that Sunday evening by playing Wesnoth for 2.5 hours.

## 5. Generalization

### 5.1. Surprise

On Monday morning I was mindlessly feeding inputs to the program I had worked on earlier and finally noticed that in fact ${\mathcal L_6(1,3,5,2)}$ and ${\mathcal L_6(1,5,3,2)}$ also had the same cardinality. Huh.

It seemed too good to be true, but I played around some more, and sure enough, the cardinality of ${\#\mathcal L_{k+2}(a_1, \dots, a_n)}$ seemed to only depend on the order of the ${a_i}$‘s. And so at last I stumbled upon the final form the conjecture, realizing that all along the assumption ${a_i \in \{k,k+1\}}$ that I had been working with was a red herring, and that the bijection was really true in much vaster generality. There is a bijection

$\displaystyle \mathcal L_{k+2}(a_1, \dots, a_i, a_{i+1}, \dots, a_n) \rightarrow \mathcal L_{k+2}(a_1, \dots, a_{i+1}, a_{i}, \dots, a_n)$

which only involves rearranging the elements of the ${i}$th and ${(i+1)}$st blocks.

It also meant I had more work to do, and so I was now glad that I hadn’t written up my work from yesterday night.

### 5.2. More data science

I re-ran the experiment I had done before, now with ${\mathcal L_7(2,3,5,2) \rightarrow \mathcal L_7(2,5,3,2)}$. (This was interesting, because the ${8}$ elements in question could now have either longest increasing subsequence of length ${5}$, or instead of length ${6}$.)

The data I obtained was:

246|13578 => 24678|135
247|13568 => 24578|136
248|13567 => 24568|137
256|13478 => 25678|134
257|13468 => 23578|146
258|13467 => 23568|147
267|13458 => 23678|145
268|13457 => 23468|157
278|13456 => 23478|156
346|12578 => 34678|125
347|12568 => 34578|126
348|12567 => 34568|127
356|12478 => 35678|124
357|12468 => 13578|246
358|12467 => 13568|247
367|12458 => 13678|245
368|12457 => 13468|257
378|12456 => 13478|256
456|12378 => 45678|123
457|12368 => 14578|236
458|12367 => 14568|237
467|12358 => 14678|235
468|12357 => 12468|357
478|12356 => 12478|356
567|12348 => 15678|234
568|12347 => 12568|347
578|12346 => 12578|346
678|12345 => 12678|345

Okay, so it looks like:

• exactly two numbers are moving each time, and
• the length of the longest run is preserved.

Eventually, I was able to work out the details, but they’re more involved than I want to reproduce here. But the idea is that you can move elements “one at a time”: something like

$\displaystyle \mathcal L_{k+2}(7,4) \rightarrow \mathcal L_{k+2}(6,5) \rightarrow \mathcal L_{k+2}(5,6) \rightarrow \mathcal L_{k+2}(4,7)$

while preserving the length of increasing subsequences at each step.

So, together with the easy observation from the beginning, this not only resolves the original problem, but also gives an elegant generalization. I had now proved:

Theorem 12

For any ${a_1}$, …, ${a_n}$, the cardinality

$\displaystyle \# \mathcal L_{k+2}(a_1, \dots, a_n)$

does not depend on the order of the ${a_i}$‘s.

## 6. Discovered vs invented

Whenever I look back on this, I can’t help thinking just how incredibly lucky I got on this project.

There’s this perpetual debate about whether mathematics is discovered or invented. I think it’s results like this which make the case for “discovered”. I did not really construct the bijection ${\mathbf W}$ myself: it was “already there” and I found it by examining the data. In another world where ${\mathbf W}$ did not exist, all the creativity in the world wouldn’t have changed anything.

So anyways, that’s the behind-the-scenes tour of my favorite combinatorics paper.


text:
tikz arrow tip thickness for double lines

I want to draw a double line with an arrow in both ends, but when I try to do it the thickness of the arrow tip becomes too big.

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{tikz}

\usetikzlibrary{arrows, positioning}

\begin{document}
\begin{tikzpicture}
\node[draw, rectangle, minimum size = 2cm] (a) {A};
\node[draw, rectangle, minimum size = 2cm, left = 3cm of a] (b) {B};
\node[draw, rectangle, minimum size = 2cm, left = 3cm of b] (c) {C};

\draw[angle 90-angle 90, line width=1pt, double, double distance=5pt] (a.west) -- (b.east);
\draw[angle 90-angle 90, line width=1pt] (b.west) -- (c.east);
\end{tikzpicture}
\end{document}

• to clarify, what I want is to decrease the arrow tip thickness for the double line – jagjordi Jan 28 at 15:58

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{tikz}

\usetikzlibrary{arrows.meta, positioning}

\begin{document}
\begin{tikzpicture}[>={Classical TikZ Rightarrow[]}]
\node[draw, rectangle, minimum size = 2cm] (a) {A};
\node[draw, rectangle, minimum size = 2cm, left = 3cm of a] (b) {B};
\node[draw, rectangle, minimum size = 2cm, left = 3cm of b] (c) {C};

\draw[<->, line width=1pt, double, double distance=5pt] (a.west) -- (b.east);
\draw[<->, line width=1pt] (b.west) -- (c.east);
\end{tikzpicture}
\end{document}

Little Improvement

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{tikz}

\usetikzlibrary{arrows.meta, positioning}

\begin{document}
\begin{tikzpicture}[>={Straight Barb[round]}]
\node[draw, rectangle, minimum size = 2cm] (a) {A};
\node[draw, rectangle, minimum size = 2cm, left = 3cm of a] (b) {B};
\node[draw, rectangle, minimum size = 2cm, left = 3cm of b] (c) {C};

\draw[{<._[sep=-13pt]}-{_[sep=-13pt].>}, line width=1pt, double, double distance=5pt] (a.west) -- (b.east);
\draw[<->,line width=1pt] (b.west) -- (c.east);
\end{tikzpicture}
\end{document}

• thankk you for the answer – jagjordi Jan 29 at 14:50

format: plain


text:
## Intermediate Algebra for College Students (7th Edition)

$-\frac{2}{y(y+1)}$
$\frac{6y+2}{y^2-1}*\frac{1-y}{3y^2+y}=\frac{2(3y+1)(1-y)}{(y+1)(y-1)(y)(3y+1)}=-\frac{2}{y(y+1)}$ NOTE: (1-y) and (y-1) cancel out to -1 because -1 can be factored out of either to make them equal.


text:
# RRB NTPC 27 April 2016 - Shift 1

Instructions

For the following questions answer them individually

Question 91

What is the local name given to the parliament of Pakistan?

Question 92

Which state of India has the largest number of cotton mills?

Question 93

Find the maximum area of a rectangular field which is surrounded by a rope of 100 meters?

Question 94

On the basis of Memory size & performance, which type of computer is known as "Big iron"?

Question 95

If 90% of A = 30% of B and B= x% of A, find x.

Question 96

Which of the following fractions is largest?

Question 97

The Value of $$sin 60$$ is:

Question 98

Which of the following hormone is not secreted by Pituitary gland?

Question 99

A, B and C are employed to complete a work in Rs. 784. If A and B together complete 23/28 of the work Calculate the wages (in Rs.) paid to C?

Question 100

Find the value of $$200 \times 20 \times 2 \times 0.2 \times 0.02 \times 0.002$$=?

OR


text:
## Deep Learning 15: Unsupervised learning in DL? Try Autoencoder!

There are unsupervised learning models in multiple-level learning methods, for example, RBMs and Autoencoder. In brief, Autoencoder is trying to find a way to reconstruct the original inputs — another way to represent itself. In addition, it is useful for dimensionality reduction. For example, say there is a 32 * 32 -sized image, it is possible to represent it by using a fewer number of parameters. This is called you are “encoding” an image. The goal is to learn the new representation, so it is also applied as pre-training; then a traditional machine learning models could be applied depending on the tasks — it is a typical “two-stage” way for solving problems.[3] by Hinton is a great work on this problem, which shows the ability of “compressing” in neural networks, solving the bottleneck of massive information.

### Typical Autoencoder Models

Suppose we have only input features, denoted by $x = \{x_1,x_2,....,x_n\}$. Think about PCA, we can easily find a way to reconstruct them into lower dimensional representations. Autoencoder could achieve the similar goal, but no restricted to a lower dimension.
The figure is from Andrew Ng’s lecture notes [1], a simple model of the autoencoder. It is a 3-layer fully-connected neural network with bias units. The model is to learn the function which let $h_{W,b}(x)\approx x$, instead of “=” in the middle — make no sense. Usually, we add some noise during training — will explain later in the post.

Usually, we could set a different number of units in the hidden layer: when smaller than that of the input layer, it is a compressed representation; also could be the same with or greater than that of the input layer, then it is a reconstruction, we will add a sparsity constraint.
Informally, the process L1->L2 is called ‘encoding’, L2->L3 is ‘decoding’. Think about you want to send a high-quality picture to your friend, you might want to compress it first to make it smaller for an email attachment (encoding); after receiving, your friend would decompress and view the picture (decoding). The below figure from [2] shows the model:

We focus on the hidden layer representation $z$. Here the input is denoted by $x$, output is $x'$. The error is $L = ||x-x'||^2$. During training, the error would be minimized.

### Autoencoder Variations

#### Denoising autoencoder

Firstly proposed by Vincent et al. in Extracting and composing robust features with denoising autoencoders. Denoising means we add noises: say original input is 64-dim, we randomly set few of them elements to be zero.

##### Model

This approach [4] can be used to train autoencoders, and these denoising autoencoders can be stacked to initialize deep architectures. The algorithm can be motivated from a manifold learning and information theoretic perspective or from a generative model perspective. Comparative experiments clearly show the surprising advantage of corrupting the input of autoencoders on a pattern classification benchmark suite.

“Our training procedure for the denoising autoencoder involves learning to recover a clean input from a corrupted version, a task known as denoising.” The Denoising Autoencoder is based on the idea of “unsupervised initialization by explicit fill-in-the-blanks training” on a deep learning model. Imagine that how our brain works, even though you cover a small part of an image, our brain could still distinguish the object. In other words, provided by clean inputs, we create some noise or blanks manually and attempt to learn the autoencoder with good robustness.

The model is shown in the above Figure form [4]. Given clean inputs $\textbf{x}$, try to corrupt through partially destroy. By means of mapping $\widetilde { \textbf{x} } \sim q_D(\widetilde { \textbf{x} } | \textbf{x})$ , we would have a “noisy” version $\tilde { \textbf{x} }$ of the original input. For example, some elements would be simply treated as zeros, pretending giving blank information. Then a hidden layer of representation is mapped from $\tilde { \textbf{x} }$ to  $\textbf{y}$ through  $f_\theta$. The next step is to reconstruct a clean input $\textbf{z}=g_\theta' (y)$, known as denoising. We expect the reconstructed output $\textbf{z}$is close to uncorrupted input  $\textbf{x}$. The parameters are optimizing the construction cross-entropy which is defined as:$L_{ H }(x,z)=H(B_{ x }||B_{ z })=-\sum _{ k=1 }^{ d }{ \left[ x_{ d }\log z_{ k }+(1-x_{ k })\log (1-z_k) \right] }$

Keep the trained $f_\theta$ parameters and use them directly on input $\textbf{x}$ yielding higher level representation. Then finish the first layer of denoising autoencoding. Treat the new representation layer  $\textbf{y}$ as the input of the next denoising autoencoding layer, we will get the second layer. Follow the same procedure, eventually we noticed they are stacked one by one. After a number of layers, the stacked denoising autoencoding is finished training in an unsupervised way, and below figure is from [4].

In the second procedure, the trained stacked dAE could be connected with a classifier, such as a linear Support Vector Machine (SVM). The choose of the classifier depends on the classification task. During training, it is possible to fine-tune the stacked dAE parameters for each layer $f_\theta^(i)$ by gradient descent.

##### Comments

Unsupervised: the function of the stacked dAE is to learn the sharing representations across all those domains. Since no labels are required in training, data from both domains could be helpful. The readers could also think about the Word2vec model, which trained from free-text and eventually give embeddings as representations. In addition, only a single round of training is required for the stacked dAE, which is an attractive point for the model reuse. Then the pre-trained stacked dAE can be fine-tuned and cooperate with other models for any specific tasks.
Scalability: it requires massive of data from domains to train the networks, so the scalability is always a consideration when measuring the task complexity. The stack dAE based deep learning model is proved to be scalable on an industrial-scale dataset of up to 22 domains on Amazon reviews. [4]

#### Sparse autoencoder

Another variant is sparse autoencoder. It is when we consider the sparsity of the hidden layer. Compared with an original autoencoder, we keep the loss but add one more penalty term( sometimes take K-L divergence) during training.  The penalty term would work by comparing the probability distribution of the hidden unit activations with some low desired value), or by manually zeroing all but the few strongest hidden unit activations ( check [1,5] for more).


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# Derivation of the Area of a Parallelogram

A parallelogram is a quadrilateral whose opposite sides are parallel.  In the figure below, $\overline{AB}$ is parallel to $\overline{CD}$ and $\overline{BC}$ is parallel to $\overline{AD}$ . Now, how do we find the area of a parallelogram?

Since the opposite sides of a $ABCD$ are parallel, if we slide $\overline{AB}$ to the right, it will coincide with $\overline{CD}$.

That means that if we cut along the height of the parallelogram, the cut portion will form a right triangle containing $\overline{AB}$. Again, if we slide the right triangle to the right such that $\overline{AB}$ coincides with $\overline{CD}$, it will form rectangle as shown below. Since the area of a rectangle is the product of its base and height, and we did not remove any portion of the parallelogram (remember, we just cut and slid the triangle), the area of the parallelogram is therefore equal to the area of the rectangle.  So, the area of a parallelogram is also equal to the area of the rectangle which is equal to the product of its base and its height.

Now, since we can form a rectangle from a parallelogram with any size and shape, the formula

area = base $\times$ height

applies to all parallelograms.

Points to Ponder

1. How do you know that the base of the quadrilateral that we assumed as a rectangle are congruent?
2. Why is it that the resulting figure after we slid the triangle is a rectangle? Justify your answer.


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# Math Help - Did I solve for h correctly?

1. ## Did I solve for h correctly?

The question asks: Solve for h

A=1/2h (b1+b2)

I think the solution is h= 2A/(b1+b2)

Is this correct?

2. Originally Posted by KarlosK
The question asks: Solve for h

A=1/2h (b1+b2)

I think the solution is h= 2A/(b1+b2)

Is this correct?
Indeed you are assuming the original equation is $A = \frac{1}{2}h(b_1+b_2)$

3. Yes that is what the original equation was. I am not very good at typing the problems out correctly in here!! So I did it correctly?

4. Originally Posted by KarlosK
Yes that is what the original equation was. I am not very good at typing the problems out correctly in here!! So I did it correctly?
You did


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# Homework Help: Thermodynamics Air Cannon

1. Mar 22, 2010

### RazerM

1. The problem statement, all variables and given/known data
1.75 J is required to fire a spherical pellet of radius 8.75 mm at a maximum velocity of 100 m/s

Find appropriate values for the inital pressure and volume and final pressure and volume

2. Relevant equations
$$P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}$$

$$\gamma=1.40\qquad P_{2}=10^5\text{ Pa}$$

$$W_{\text{total}}=\frac{P_{1}V_{1}-P_{2}V_{2}}{\gamma-1}$$

\begin{align*} W_{\text{total}} &= W_{\text{atm}}-W_{\text{useful}} \\ \frac{P_{1}V_{1}-P_{2}V_{2}}{\gamma-1}&=10^{5}(V_{2}-V_{1})-1.75 \end{align*}

3. The attempt at a solution
\begin{align*} V_{2}&=V_{1}\left( \frac{P_{1}}{P_{2}} \right)^{\frac{1}{\gamma}}\\ V_{2}&=V_{1}\left( \frac{P_{1}}{10^{5}} \right)^{\frac{1}{1.4}}\\ \end{align*}
It is then possible to rearrange work equation into
$$V_{1}\left( [P_{1}+4\cdot 10^{4}]-\left[ 1.4\cdot 10^{5}\left( \frac{P_{1}}{10^{5}} \right)^{\frac{1}{1.4}} \right] \right)=0.7$$

$$\text{Then using an arbitary guess for }V_{1}\text{ and an iterative process to find }P_{1}\text{ I found }$$
$$V_{1} = 10^{-5}\text{ m}^{3}$$
$$P_{1} = 420465\text{ Pa}$$
$$V_{2} = 2.79\cdot 10^{-5}\text{ m}^{3}$$
Giving a ridiculous barrel length of
\begin{align*} L &= \frac{V_{2}-V{1}}{\pi r^{2}} \\ L &= 0.0744\text{ m} \end{align*}

I'm not asking for a full solution; just a hint as to how to find out values that satify the criteria of work and have a reasonably small V_1 so that there can be many shots for say a 0.001m^3 tank

2. Mar 22, 2010

### Andrew Mason

The pellet travels down a cylindrical barrel of length L whose cross-sectional area is $A = \pi r^2$ where r = 8.75 mm. So the final volume is LA. The minimum final pressure is 1 atm. Choose a reasonable length for the barrel, say .75 m. Find an initial volume and pressure such that the work done in adiabatically expanding to volume LA is equal to 1.75 J.

AM

Last edited: Mar 22, 2010


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# How can I implement the Cartesian harmonics?

I'm studying the quantum chemistry calculation and now trying to implement the basic methods.

For example, given a water molecule, $$M = {(\ce{O}, R_\ce{O}), (\ce{H}, R_\ce{H}), (\ce{H}, R_\ce{H})}$$, where $$R = [X, Y, Z]$$ is the atomic position, the Cartesian Gaussian-type orbital $$\psi$$ on $$r$$, where $$r = [x, y, z]$$ is an arbitrary position, can be written as follows: $$\psi(r) = N (x-X)^l (y-Y)^m (z-Z)^n \exp(-\alpha ||r-R||^2),$$ where $$N$$ is the normalization term, $$\alpha$$ is the orbital exponent, and $$||r-R||$$ is the distance between $$r$$ and $$R$$. The term of $$(x-X)^l (y-Y)^m (z-Z)^n$$ is the Cartesian harmonics.

However, I believe that, because a water molecule can be described by any Cartesian coordinates if all the atomic distances (i.e., $$\ce{O-H}$$, $$\ce{O-H}$$, and $$\ce{H-H}$$) are correct, $$\psi(r)$$ can not be determined to a single value.

For example, the following two Cartesian coordinates represent the same water molecule.

O 0.00000 0.00000 0.11779
H 0.00000 0.75545 -0.47116
H 0.00000 -0.75545 -0.47116

O -0.034 0.978 0.0076
H 0.065 0.021 0.0015
H 0.87 1.30 0.00069

However, $$\psi(r)$$ is different because the above equation has the term of $$(x-X)^l (y-Y)^m (z-Z)^n$$. If the equation has only the distance term (i.e., $$\exp(-\alpha ||r-R||^2)$$), $$\psi(r)$$ has always the same value and I can implement this very easily only using the distance matrix.

How can I deal with this problem and implement the Cartesian harmonics?

Postscript: Of course, I know the spherical harmonics but I believe that even if the angles, θ and φ, are used, the problem is the same; how can I determine the θ and φ or the orientation of the 2px (2py or 2pz) orbital of each atom in a molecule? For example, in the following figure, specifically how can I determine the 2px of the H orbital in a water molecule? Of course, I know the 2px (2py or 2pz) is along with the H-O bond in this case; however, how can I determine the θ, φ, and orientation of 2px, 2py, and 2pz orbitals of each atom in arbitrary other more complex molecule and calculate the LCAO?

• The wave function has translational and rotational invariance and you can therefore define it uniquely. – Bertram Mar 5 at 20:37
• My question is not the definition of the wave function; my question is how to calculate the same wave function from the above two different input data of Cartesian coordinates. Quantum chemical simulation softwares (e.g., Gaussian) can output the same result from two different input data if the data represent the same molecule. How can the simulation softwares deal with such input data how can I implement it? – Lin Lin Mar 5 at 23:22
• It may be useful to remember that quantum chemistry suites use the basis functions mostly in integrals. Since the integration is done over all space (once or twice, depending on the integral), the orientation of the molecule does not matter (within the limits of numerical accuracy). – TAR86 Mar 6 at 14:15
• I see, but for example when we consider the LCAO, the molecular orbital ψ is calculated from the atomic basis functions φ as ψ(r) = c1φ1(r-R) + c2φ2(r-R) ... and then the electron density will be calculated as ρ(r) = sum |ψi(r)|^2. In these calculations, the integral does not appear. Additionally, if the atomic basis function φ has the terms of (x-X)^l(y-Y)^m(z-Z)^n that depend on the coordinates themselves (of course the distance does not depend on the coordinates), the density ρ(r) will be different. I may have some mistakes but I do not understand yet. – Lin Lin Mar 7 at 1:04
• I found the following description in Stack Exchange Chemistry. chemistry.stackexchange.com/questions/73177/… When specifying a Gaussian basis set in quantum chemistry, one talks only about the radial part, independent of the angular part. The angular part is really considered an implementation detail, because its form usually doesn't have any significant effect on the results.'' Is it really? – Lin Lin Mar 7 at 7:52

The question is not very well-posed. However, the question appears to be about whether rotational invariance is honored by LCAO calculations.

The old-time approach is to use Cartesian basis functions. Translating our origin to the nucleus, we can write the Cartesian basis functions in the form $$\psi({\bf r}) = N x^k y^l z^m \exp(-\alpha r^2)$$, where $$\lambda:=k+l+m$$ is the angular momentum of the function.

Basis functions are added one shell at a time: there is one cartesian S function with $$(k,l,m)=(0,0,0)$$, three P functions with $$(1,0,0)$$, $$(0,1,0)$$, and $$(0,0,1)$$, and six D functions: $$(2,0,0)$$, $$(0,2,0)$$, $$(0,0,2)$$, $$(1,1,0)$$, $$(1,0,1)$$, $$(0,1,1)$$.

While Cartesian functions were used in the early days of quantum chemistry, and form part of e.g. the infamous 6-31G* basis set, they have mostly fallen out of use, since they contain angular degeneracies, the number of which increases rapidly in the angular momentum.

What are these degeneracies? We know that spherical harmonics $$Y_l^m$$, or equivalently their real counterparts $$Y_{lm}$$ form a complete set of angular distributions. For S states we have $$Y_{00}$$, for P states we have $$Y_{1,-1}$$, $$Y_{1,0}$$ and $$Y_{1,1}$$, and for D states we have $$Y_{2,-2}$$, $$Y_{2,-1}$$, $$Y_{2,0}$$, $$Y_{2,1}$$ and $$Y_{2,2}$$. That is, for S or P functions it doesn't matter if you use cartesians or pure spherical harmonics; but for D it starts to matter: we have 6 cartesians but only 5 pure angular functions.

If you write out $$x$$, $$y$$, and $$z$$ in spherical harmonics (you can find the expressions in the literature), you can use the closure properties to derive what the angular form for $$x^k y^l z^m$$ is. However, it's pretty simple to see that if you combine three of the six D functions $$(2,0,0) + (0,2,0) + (0,0,2)$$ you get $$x^2 + y^2 + z^2 = r^2$$ which is an S function, $$Y_{00}$$. This means that the cartesian D shell is formed of the 5 pure D functions, and a S-type contaminant.

In quantum chemical calculations, the pure basis turns out to be of the form $$\psi({\bf r}) = N r^l Y_{lm} \exp(-\alpha r^2)$$. This is very convenient, since one can express the solid harmonics $$Y_{lm}$$ in terms of the cartesian functions. You can find a table on Wikipedia, but it is also quite simple to program up the bare equations to get the translation for arbitrary angular momentum. The transformation has also been discussed by Schlegel and Frisch in Int. J. Quantum Chem. 54, 83 (1995).

So, finally.. why are LCAO electronic structure calculations rotationally invariant?$$^*$$ No matter how you rotate the molecule, the spherical harmonics span the same angular degrees of freedom: e.g. when you rotate $$Y_{10}$$ into a new coordinate system, it is still expressible in terms of $$Y_{10}$$, $$Y_{1,-1}$$ and $$Y_{11}$$; the expansion coefficients are given by the so-called Wigner's $$D$$ matrix. This means you will always get the same energy regardless of how you rotate your system; the individual components of the wave function will just change.

$$^*$$ Rotational invariance doesn't hold fully for e.g. density functional calculations, since the quadrature grids employed to integrate the density functional are not rotationally invariant. The error can, however, be made insignificantly small by using a large enough quadrature grid. This error is also removed by default by transforming the molecule into a standard orientation; see Chem. Phys. Lett. 209, 506 (1993).

Addendum based on postscript: sounds like your problem is just not understanding how the molecular orbitals are defined. You don't need to try to align the orbitals; they arise automatically by solving the self-consistent field equations. The molecular orbitals are linear combinations of the atomic orbitals, so they'll automatically figure out what is the best combination. There are no atomic orbitals in a molecule, they're just basis functions used in the expansion, the coefficients of which are found by minimizing the total energy of the resulting wave function.

• I edited the first question and added the figure. I know the spherical harmonics but I believe that even if the angles, θ and φ, are used, the problem is the same; how can I determine the θ and φ or the orientation of the 2px orbital of each atom in a molecule? For example, in the following figure, specifically how can I determine the 2px of the H orbital in a water molecule? Of course, I know the 2px is along with the H-O bond in this case; however, how can I determine the θ, φ, and orientation of 2px orbital in arbitrary other more complex molecules and calculate the LCAO? – Lin Lin Mar 16 at 12:19
• @LinLin I've extended the answer. Shortly: you don't. There is no 2px H orbital in a water molecule. You just build your molecular basis set; a triple-zeta basis typically includes S, P, and D functions on the hydrogens and S, P, D, and F functions on the oxygen, and then you find the molecular orbitals that minimize the Hartree-Fock or density functional energy. – Susi Lehtola Mar 16 at 16:21
• Thanks for answering this Susi! @LinLin if you want a longer answer, I would recommend that you ask your follow-up questions in separate posts. I think that Susi's 10 paragraph answer in this case is quite substantial and sufficiently answers the original question! – Nike Dattani Mar 16 at 18:34
• @Susi Lehtola, Thank you very much for the answer. It is very helpful and I keep studying this topic. – Lin Lin Mar 17 at 1:37


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# Stopping time for a martingale

Let $X_1,X_2,\ldots$ be iid random variables where $X_i\in\{-1,0,1,2,...\}$, $P(X_i=0)<1$ and $E(X_1)=\mu$. Let $S_n=1+X_1+\cdots+X_n$ and $T=\inf \{n:s_n=0\}$. Show that $E(T)=\infty$ if $\mu=0$ and $E(T)<\infty$ if $\mu<0$.

I try to use the Wald's equation but in this case we have $S_0=1$ instead of $0$.

• The assertion is not correct. Consider for example $X_j := - \frac{2}{3}$ for all $j$. They are clearly iid, $\mu<0$, but $T=\infty$, hence in particular $\mathbb{E}(T)=\infty$. Probably, you want to define $T$ as $T=\inf\{n; S_n \leq 0\}$ instead...? – saz Mar 12 '13 at 9:10
• Sorry I forgot to add $X_i\in\{-1,0,1,2,...\} P(X_i=0)<1$ in the problem. – BigMike Mar 12 '13 at 17:27
• Unfortunately, the assertion is still not correct. Consider $X_j := -2$. – saz Mar 12 '13 at 17:38
• Here $X_i$ is in the set $\{-1,0,1,2,...\}$, so it can not equal to -2. I think I proved the case where $\mu=0$ by making a new martingale $Y_n=S_n-n\mu-1$, but still confused about the case when $\mu<0$ – BigMike Mar 12 '13 at 21:32
• Well, since $X_i$ is a random variable, it won't be "in a set". However, so $\mathbb{P}[X_j = j]>0$ for all $j \in \mathbb{Z}$ is another assumption. – saz Mar 12 '13 at 21:44

If $X_1=-1$, $T=1$. Otherwise, $X_1=k$ with $k\geqslant0$ and $T=1+\sum\limits_{i=1}^{k+1}T_i$ where $(T_i)$ is i.i.d. and distributed like $T$ (this is where the hypothesis that the only negative jumps of the random walk are $-1$ is used). Thus, for every $s$ in $(0,1)$, $\mathbb E(s^T)=\sum\limits_{k\geqslant-1}\mathbb P(X=k)s\mathbb E(s^T)^{k+1}$. Introducing $g(t)=\mathbb E(t^X)$, this reads $g(\mathbb E(s^T))=1/s$. Deriving this, one gets $\mathbb E(Ts^{T-1})g'(\mathbb E(s^T))=-1/s^2$. Considering the limit when $s\to1$ yields $\mathbb E(T)g'(1)=-1$, that is, $\mathbb E(T)=-1/\mu$.
• Thanks for the answer, but I found a easier way using the wald's equation and let $Y_n=S_n-1$. I got the exactly the answer as your. – BigMike Mar 13 '13 at 6:15
Let $S_n' := S_n -1$. Then $$T=\inf\{n \in \mathbb{N}; S_n'=-1\}$$
• $\mu=0$: Assume $\mathbb{E}(T)<\infty$. Then we have by Wald's equation $$\mathbb{E}S_T' = \mathbb{E}(T) \cdot \mathbb{E}X_1$$ This is clearly a contradiction since $\mathbb{E}X_1=0$ whereas $\mathbb{E}S_T' = -1$.
• $\mu<0$: By the strong law of large numbers we have $S_n' \to - \infty$ almost surely, thus $\mathbb{P}(T<\infty)$ almost surely (since jumps to the left have magnitude $1$). This implies that $Y_n := S_{n \wedge T}'$ is well-defined. $(Y_n)_n$ is a supermartingale (by optional stopping). By applying Wald's equation to the (bounded) stopping time $T \wedge n$ we obtain $$\mathbb{E}X_1 \cdot \mathbb{E}(T \wedge n) = \mathbb{E}S_{n \wedge T}' \geq \mathbb{E}S_T' = -1$$ i.e. $\mathbb{E}(T \wedge n) \leq - \frac{1}{\mu}$ since $\mu<0$. Monotone convergence theorem yields $\mathbb{E}T \leq - \frac{1}{\mu}< \infty$. By applying Wald's equation (this time to the stopping time $T$), we obtain $\mathbb{E}T=- \frac{1}{\mu}$.


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# System for Estimating the Maximum Adherence Coefficient by Measuring Stresses in a Tire Tread

Imported: 10 Mar '17 | Published: 27 Nov '08

Frederic Spetler

USPTO - Utility Patents

## Abstract

A system for estimating the maximal grip coefficient in the contact area of a tire on a road. A tire tread is fitted with at least one tread sensor for measuring at least the tangential stresses which the tread experiences locally when the tire rolls on the ground. a signal processing unit and means for transmitting to the processing unit at least one signal delivered by the tread sensor are provided. The processing unit comprises means for finding in the signal the values corresponding to the sensor passing through the contact area of the tire on the ground, means for calculating an instantaneous value of a preselected criterion as a function of a series of values chosen from those relating to the sensor passing through the contact area, and means for extracting an estimate of the maximal grip coefficient by using the said instantaneous value.

## Description

The present invention relates to the evaluation of the grip of a vehicle on a road. It relates more particularly to the determination of maximal grip characteristics between the road and a vehicle wheel fitted with an elastic tire, such as an inflated tire or a non-pneumatic elastic tire, which rolls on the road.

The present invention also relates to the various electronic assistance devices used, for example, for antilock control of the brakes of a vehicle or antiskid control of the drive wheels, control of the direction of a vehicle or alternatively for other forms of control or monitoring, for instance the tire pressures. It is known that such devices reconstruct the grip coefficient () of the tires on the road by calculation, without having carried out any measurement either of the grip coefficient or of the forces developed in the contact of the tires with the ground. Even though these devices provide significant assistance and extra safety, their operation would benefit greatly from the use of a measured value, or a value estimated on the basis of real measurements carried out on the tire during operation, principally a value representative of the maximal grip coefficient available in real time.

Patent Application FR2835919 proposes to obtain an estimate of the maximal grip coefficient on the basis of measurements in the sidewall. This approach is therefore indirect, since the grip phenomenon takes place in the contact imprint on the ground. Patent Application WO 02/32733 A1 proposes to calculate micro-slips of the tire in the contact area on the basis of the speed. This approach is even more indirect since it no longer operates in the tire per se, and it is not possible to see what gain in precision could thus be obtained relative to the means in current use. U.S. Pat. No. 5,864,056 proposes to carry out measurements in an element of the tread by means of strain gauges, but does not explain how to process the signals recorded in this way. Patent Application EP 0937615 A2 proposes to carry out deformation measurements in elements of the tread, and illustrates the shape of the curves obtained in the case of dry ground and wet ground. According to this document, the slope reduction of the S-shaped signal measured in the element of the tread, particularly the appearance of a horizontal plateau, should be linked with a decrease of the grip. This, however, is visible only when the grip is already very small or even zero, and cannot therefore provide a sufficiently preventive alarm warning.

For this reason, it is an object of the present invention to provide a way of evaluating the maximal available grip of a vehicle which runs on a road, and more precisely of its wheels or tires or non-pneumatic elastic tires, these terms being regarded as equivalent in the context of the present invention.

The various electronic assistance devices mentioned above would therefore usefully benefit from real-time indications of the grip conditions liable to affect the performance of a vehicle, especially when it undergoes acceleration due to a driving force or a braking force, or due to a change of driving direction. The invention aims to provide a method of achieving this efficiently.

With a view to carrying out the measurements necessary for estimating the maximal grip coefficient, it has already been proposed that the tread of a tire, or certain specially adapted elements of the tread, be fitted with sensors intended to measure or estimate the forces generated locally, in particular under slip conditions. The reader may thus refer to U.S. Pat. No. 6,666,079. It is furthermore necessary to be able to extract the information relevant for estimating the maximal grip coefficient from the signals delivered by such sensors. This is the object of the present invention.

The invention proposes to estimate the maximal grip coefficient of a tire on the basis of the stresses measured in the tread pattern, for example in a tread block located substantially at the centre of the tread, or on the basis of another measurement reflecting the stresses i.e. one which can be correlated with the stresses, such as the deformation, which should be regarded as identical in the context of the invention.

The invention provides a system for estimating the maximal grip coefficient p in the contact area of a tire on a road, comprising a tire whose tread is fitted with at least one tread sensor for measuring at least the tangential stresses which the tread experiences locally when the tire rolls on the ground, comprising a signal processing unit and means for transmitting to the processing unit at least one signal delivered by the said tread sensor, wherein the processing unit comprises means for finding in the signal the values corresponding to the sensor passing through the contact area of the tire on the ground, means for calculating an instantaneous value of a preselected criterion as a function of a series of values chosen from those relating to the sensor passing through the contact area and means for extracting an estimate of the maximal grip coefficient by using the said instantaneous value.

In other words, the invention provides the means for recovering the maximal grip coefficient information on the basis of acquiring measurements taken during the passage of a fixed point of the tread through the contact area on the ground. To this end, this acquisition of measurements is used for calculating a value conventionally referred to as the instantaneous value since it does not correspond to a particular point or a given time but to a calculation reflecting the situation prevailing in all or part of the contact area. This calculation resorts to a mathematical function referred to as a Criterion. A plurality of Criteria may be employed. One or more of them are preselected once and for all as a function of a particular application envisaged (for example a particular tire on a particular vehicle). The said instantaneous value or values are then inserted into a mathematical function (which may itself reflect the preselected Criterion or Criteria, or may be an appropriate correlation) in order to obtain an estimate of the maximal grip coefficient .

FIG. 1 schematises a tread (of a tire, of a non-pneumatic elastic tire, of a caterpillar track, etc.) having a block 1 fitted with a sensor 10 installed at the base of the block 1, and the reinforcing plies 2 above. For further details concerning the sensor used in the system illustrating the present invention, the reader is invited to refer to the U.S. Pat. No. 6,666,079 cited above, the description of which is incorporated here by reference. The diagram of FIG. 2 shows the system according to the invention, of which the block 1 uses an aforementioned sensor in order to obtain signals containing the stress measurements x (shear in the longitudinal direction), the stress measurements y (shear in the transverse direction) and the stress measurements z (compression in the direction normal to the surface of the ground) . A second block 2 performs various signal processing operations which will be explained in detail below, on the basis of signals delivered by the measurement block and optionally on the basis of other signals S1, S2, S3 . . . Sn giving information about the load of the tire, its inflation pressure, the driving or braking torque transmitted by the tire, the drift angle or the lateral force Fy, and the speed of rotation, which resort to the Criteria and the details of which will be given below.

The block 2 is a signal processing unit which may be installed in the vehicle, for example, whereas the sensor 10 and optionally other sensors used for one or other of the signals S1 to Sn are installed in the tire. The system according to the invention therefore comprises means for communication between the tire and the vehicle, the various possible forms of which are known to the person skilled in the art. By way of illustration, it will merely be indicated that Patent Application EP 1350640 illustrates an important element of these communication means, namely an antenna to be installed in a tire.

Lastly, the block 3 makes it possible to determine an estimate of the maximal grip coefficient p on the basis of one or several instantaneous values. The system according to the invention comprises a stage (block 3) for making a correlation between the value or values of one or more criteria and the value of the maximal grip coefficient , by virtue of stored curves established experimentally or by virtue of mathematical models (for example: linear, polynomial model or neural network) whose coefficients have been fitted experimentally.

Returning now to the tangential stress measurements, and in particular the longitudinal i.e. circumferentially oriented stress measurements. FIG. 3 illustrates simulation results of the shear stress (x) in the longitudinal direction as a function of the angular position of the sensor in a conventional reference frame, according to which the sensor is at 0 when it lies directly opposite the contact area and crosses the vertical raised perpendicularly to the ground, which passes through the rotation axis of the tire. Simulation tools employing finite element calculations are used, a field which is well known to the person skilled in the art.

The results are those obtained for the same tire, identical to a Michelin Energy XH1 195-65 R 15 tire except for its tread which is adapted to receive a force sensor below a tread pattern element of square general shape (see FIG. 1), under the same rolling conditions (more precisely here at 60 km/h, subjected to a load of 430 daN in free rolling, i.e. without a driving or braking torque or camber or drift) and for different grip levels. It is seen that the sensor enters the contact area on the ground when it reaches approximately the azimuth 165, and leaves it approximately at the azimuth 195. The reference A indicates that the choice has been made to employ the measurements acquired between the azimuths 162 and 192, as will be explained below. On a central (along y) tread block of the contact area, FIG. 3 gives the shape of the X stresses corresponding to the measurement point, according to the whose values are indicated in FIG. 3, where is the azimuth in degrees (180 is the longitudinal centre of the contact area on the groundalso referred to as the imprint), x is the X stress in daN/mm2 and is the grip level.

FIG. 3A illustrates the measurement recordings of shear stress (x) which are established under the same conditions except for the grip coefficient which, for the real tracks used, varies from 0.40 to 1.20 as indicated in the box shown in the upper right-hand corner of the figure. The numbers on the abscissa axis can be translated directly into azimuth, to within a multiplicative coefficient. It is seen that the various curves, which correspond to different ground surfaces, have a shape very similar to those obtained by simulation.

Before going into details concerning the operations carried out in block 2 we will anticipate somewhat what follows by illustrating the influence of parameters which should preferably be taken into account. For a Criterion referred to as 1st order polynomial of (x), FIGS. 4A to 4F respectively illustrate the influence of the parameters: transmitted torque (significant influence, FIG. 4A), drift angle (or force Fy, which would be equivalent - significant influence, FIG. 4B), camber angle of the tire (negligible influence, FIG. 4C), load of the tire (FIG. 4D), pressure of the tire (FIG. 4E, pressure and load in fact are two expressions of the same physical phenomenon, the influence of which is significant), speed of the tire (FIG. 4F, little influence in the speed range tested here).

It should be noted that the measurements corresponding to these FIGS. 4A to 4F were carried out on a test machine referred to as a flat surface roller, on which the rolling surface consists of a steel belt tensioned by two rollers in order to advance horizontally while presenting a flat contact surface for the tire, the load generated by the tire on the steel belt being supported by a pressurised water carpet. This rolling mode generates a local phenomenon of flattening the tread of the casing, which is substantially different from that observed in simulation or on a real surface. This is why the shape of the correlation between our criterion referred to as 1st order polynomial of (x) and the grip in FIGS. 4A to 4F differs slightly from that represented in FIGS. 9A to 9D, even though the grip criterion is identical. This difference concerns only the high grip zone and yet does not prevent quantifying the influence of the rolling parameters (load, pressure, etc.) which, for its part, results from the overall mechanics of the tire and is influenced only little by different local flattening.

In order to be able to correlate the measurements to grip, it is useful to have at least certain pieces of information about the current rolling conditions of the tire, unless these conditions can be assumed constant. This would be the case with a measuring trailer such as those used by laboratories, for example, these trailers being fitted with systems for keeping the rolling conditions constant or referencing the rolling phases in which of these conditions are constant.

In applications to moving vehicles, however, it is thus preferable to take into account at least those of the parameters which have a fairly great influence on the correlation between the criteria and the grip. In this scope it is possible to use the same signals delivered by the sensor 10 to which an ad hoc treatment is applied, or alternatively it is possible to use other sensors installed in the tire (for example, see how to obtain Fx, Fy, Fz, P and the camber from stress measurements in the bottom zone by consulting Patent Application WO2003/014687-A1, or from extension measurements in the side wall by consulting Patent Application WO2003/014693-A1) or elsewhere (dynamometric measurements on the wheel, information coming from the drive engine management, ABS sensors, sensors installed in the suspension system).

As concerns the signals S1 to Sn, the information sought about the operating parameters of the tire can be obtained by ad hoc treatments carried out on the signals delivered by the sensor 10, knowing that further to the measurements of (x), the sensor proposed in the U.S. Pat. No. 6,666,079 cited above gives measurements of (y)direction parallel to the rotation axis of the tireand (z)direction perpendicular to the ground; note that:

• A good indicator of the force Fx applied to the centre of the wheel is given by

$c 1 = CA X x$

The notation CA signifies. Contact Area. Both for the indicator proposed above and for all the ones which follow, the integration may in fact be carried out in all or part of the contact area.

• A good indicator of the force Fy due to the drift, applied to the centre of the wheel, is given by

$c 2 = CA Y x$

• A good indicator of the load Fz applied to the centre of the wheel is given by

$c 3 = CA Z x$

• the inflation pressure can be obtained very easily by a specific sensor or obtained on the basis of the load (for which c3 is a good indicator) and the length of the contact area, which is detected easily when inspecting the min/max threshold signals x and y and the signal z, when it is greater than a threshold.
• the rolling speed is given by ABS sensors or on the basis of the sensor 10, which makes it possible to count the number of wheel revolutions, and knowledge of the perimetric development of the tire and the time taken to make a revolution.
• the camber could be provided by a specific sensor or by an ad hoc treatment, or it is also possible to use other sensors installed in the tire (see, for example, how to obtain (see the references to Patent Applications WO2003014687A1 and WO2003014693A1 above)).

Other pieces of information, such as the level of wear of the tire, may further contribute to improving the precision for the estimate of the maximal grip coefficient .

Various Criteria used to calculate the said instantaneous value will now be illustrated. All these criteria are illustrated with the aid of FIGS. 5A to 19D from which the correlations (or the lack of correlation for the last ones) can be read, and are applied only to the signal x. Each graph represents a different condition for obtaining the data:

• Tracks 30 km/h or Tracks 60 km/h signifies a real test on a test track by means of the tire indicated above which is run at the speed indicated, the load being that indicated above (see description of FIG. 3A),
• Simulation Z=240 daN or (Simulation Z=400 daN signifies that it is a result obtained by simulation, the load adopted being that indicated.

FIGS. 5A to 5D illustrate the correlation between the value of a first criterion called standard deviation of (x) and the value of the maximal grip coefficient . The standard deviations are calculated for the values acquired in the contact area, i.e. the zone contained between the two extrema (of the X signal) (minimum then maximum or maximum then minimum, depending on the reference frame):

$Criterion_ standard deviation = 1 L CA CA ( X - _ X ) 2 x$ $where$ $_ X = 1 L CA CA X x$

As already indicated in respect of the indicators, the notation CA signifies Contact Area. Both for the criterion proposed above and for all the ones which follow, the integration may in fact be carried out in all or part of the contact area. In the system according to the invention, for example, the means for finding in the signal the values corresponding to the sensor passing through the contact area of the tire on the ground comprise detection of the maximum and the minimum of (x). Advantageously, the values are taken between the said maximum and the said minimum.

FIGS. 6A to 6D illustrate the correlation between the value of a first criterion called energy of (x) and the value of the maximal grip coefficient . The energies in question are calculated in the contract area. An energy corresponds to the sum of the squares of the values, whence:

$Criterion_ energy = CA X 2 x$

FIGS. 7A to 7D illustrate the correlation between the value of a second criterion called max value of (x) and the value of the maximal grip coefficient . The following is calculated:

Criterion_max value=maxCA(x)

FIGS. 8A to 8D illustrate the correlation between the value of a third criterion called min value of (x) and the value of the maximal grip coefficient . The following is calculated:

Criterion_minvalue=minCA(x)

FIGS. 9A to 10D illustrate the correlation between the value of a fourth criterion called nth order polynomial of (x) and the value of the maximal grip coefficient . The criterion is the weight a of the highest order term, i.e. the value of the weight a of an nth order polynomial n (x)=axn+bxn1+cxn2+ . . . +u, fitted over at least a part of the contact area.

A first example of this latter criterion is illustrated in FIGS. 9A to 9D. A straight line is fitted over the second half of the contact area. The highest order term is then considered (Criterion_1st order polynomial=a), which corresponds to a kind of average slope of the signal. Instead of taking the 2nd half of the contact area, any continuous zone of the contact area could be taken. The following is calculated:

• (x)=ax+b, fitted over all or part of the contact area.

A second example of this latter criterion is illustrated in FIGS. 10A to 10D, which illustrate the correlation between the value of the criterion (fifth criterion) called 2nd order polynomial of (x) and the value of the maximal grip coefficient . A 2nd order polynomial is fitted over the second half of the contact area. The highest order term is then considered (Criterion_2nd order polynomial=a), which corresponds to a kind of average convexity of the signal. Instead of taking the 2nd half of the contact area, any continuous zone of the contact area could be taken. The following is calculated:

• (x)=ax2+bx+c, fitted over all or part of the contact area.

In the same spirit, FIGS. 11A to 11D illustrate the correlation between the value of a sixth criterion called 3rd order polynomial of (x) and the value of the maximal grip coefficient . A 3rd order polynomial is fitted over the contact area. The highest order term is then considered (Criterion_3rd order polynomial=a), which corresponds to a kind of average cubicity of the signal. The following is calculated:

• (x)=ax3+bx2+cx+d, fitted over all or part of the contact area.

FIGS. 12A to 12D illustrate the correlation between the value of a seventh criterion called first harmonic of (x) and the value of the maximal grip coefficient . A Fourier series expansion is performed on the signal (x) representing the change in the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area, the operation of Fourier series expansion being well known to the person skilled in the art, and the amplitude of the first harmonic (coefficient a1) is taken as the criterion,

$( x ) = a 0 + i = 1 a i sin ( 2 f i + i ) with f i = i * 1 A$

(where A is the length of the contact area zone chosen for the calculation).

FIGS. 13A to 13D illustrate the correlation between the value of an eighth criterion called second harmonic of (x) and the value of the maximal grip coefficient . A Fourier series expansion is performed on the signal (x) representing the change in the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area, and the amplitude of the second harmonic (coefficient a2) is taken as the criterion,

$( x ) = a 0 + i = 1 a i sin ( 2 f i + i ) with f i = i * 1 A$

(where A is the length of the contact area zone chosen for the calculation).

FIGS. 14A to 14D illustrate the correlation between the value of a ninth criterion called mean value of the derivative of (x) and the value of the maximal grip coefficient . The average value of the derivative of the X stresses over the contact area is calculated:

• Criterion_average value of the derivative

$= 1 L CA CA x x x$

FIGS. 15A to 15D illustrate the correlation between the value of a tenth criterion called standard deviation of the derivative of (x) and the value of the maximal grip coefficient . The standard deviation is calculated over the contact area:

• Criterion_standard deviation of the derivative

$= 1 L CA CA ( x x - x _ x ) 2 x$ $where$ $x _ x = 1 L CA CA x x x$

FIGS. 16A to 16D illustrate the correlation between the value of an eleventh criterion called energy of the derivative of (x) and the value of the maximal grip coefficient . The energy is calculated over the contact area:

• Criterion_energy of the derivative

$= CA x 2 x x$

FIGS. 17A to 17D illustrate the correlation between the value of a twelfth criterion called max value of the derivative of (x) and the value of the maximal grip coefficient . Over the zone defined above, the maximal value is calculated (if the convexity is turned upwards) or else the minimum value is calculated (if the convexity is turned downwards):

• Criterion_max value of the derivative

$= max CA ( x x ) or min CA ( x x )$

Another possible preselected criterion is the value of the weight a of an nth order polynomial n d(x)/dx=axn+bxn1+cxn2+ . . . +u, fitted to the values of the derivative d(x)/dx of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area. For example, FIGS. 18A to 18D illustrate the correlation between the value of a thirteenth criterion called 2nd order polynomial of the derivative of (x) and the value of the maximal grip coefficient . A 2nd order polynomial is fitted over the zone defined above. The highest order term (Criterion_2nd order polynomial of the derivative=a) is then considered, which corresponds to a kind of average convexity of the signal. The following is calculated:

$x x ( x ) = a x 2 + b x + c$

fitted over all or part of the contact area.

FIGS. 19A and 19D illustrate an unsatisfactory criterion. This is the average value of the signal (x), calculated over all the contact area. There are several values of for the same value of the criterion, so that discrimination is impossible.

As concerns the block 3, it carries out the step of estimating the maximal grip coefficient p on the basis of one or several instantaneous values. The means for extracting an estimate of the maximal grip coefficient in the system according to the invention advantageously comprise the selection of a predetermined characteristic curve of the change in the maximal grip coefficient as a function of the values of the said preselected criterion and the identification on this characteristic curve of an estimate of the maximal grip coefficient on the basis of the said instantaneous value. In a particular embodiment, this is respectively the curve illustrating each of the criteria presented above. As a variant, the means for extracting an estimate of the maximal grip coefficient i comprise means for calculating an estimate of the maximal grip coefficient on the basis of the said instantaneous value and mathematical equations whose parameters have been fitted.

FIG. 20 presents an exemplary embodiment of the block 3. Note that C1, Cn represent one or more criteria containing the grip information, which are chosen for example among those presented above. It has been seen that it is useful (in various degrees) to provide indicators of certain of the operating parameters of the tire such as the load, the pressure of the tire, the camber angle of the tire, the speed of the tire, the transmitted torque and the drift angle or the lateral thrust Fy. IFx is an indicator for the X force applied to the wheel centre; IFy is an indicator for the Y force applied to the wheel centre; IFz is an indicator for the Z force applied to the wheel centre; IP is an indicator of the inflation pressure of the tire; IY is an indicator of camber applied to the wheel centre; IV is an indicator of the rolling speed; IU is an indicator of the wear of the tire. Lastly, a unit F comprises a function taking into account all the Indicator parameters or certain of them, as well as one or more Criteria in order to derive the maximal grip coefficient therefrom.

This function comprises means for selecting as a function of the chosen indicators a predetermined characteristic curve of change in the maximal grip coefficient p as a function of the values of the said preselected grip criterion, and this function comprises means for deriving an estimate of the maximal grip coefficient p from the said instantaneous value (see the explanation of each criterion above).

As a variant, this function comprises for example mathematical models (for example: linear, polynomial model or neural network) whose coefficients have been fitted experimentally or theoretically.

## Claims

1. A system for estimating the maximal grip coefficient t in the contact area of a tire on a road, comprising:
at least one tread sensor adapted to be fitted in a tire tread for measuring at least the tangential stresses which the tread experiences locally when the tire rolls on the ground;
a signal processing unit; and
means for transmitting to the signal processing unit at least one signal delivered by said tread sensor,
wherein the processing unit comprises
means for finding in the signal the values corresponding to the sensor passing through the contact area of the tire on the ground,
means for calculating an instantaneous value of a preselected criterion as a function of a series of values chosen from those relating to the sensor passing through the contact area, and
means for extracting an estimate of the maximal grip coefficient by using the instantaneous value.
at least one tread sensor adapted to be fitted in a tire tread for measuring at least the tangential stresses which the tread experiences locally when the tire rolls on the ground;
a signal processing unit; and
means for transmitting to the signal processing unit at least one signal delivered by said tread sensor,
wherein the processing unit comprises
means for finding in the signal the values corresponding to the sensor passing through the contact area of the tire on the ground,
means for calculating an instantaneous value of a preselected criterion as a function of a series of values chosen from those relating to the sensor passing through the contact area, and
means for extracting an estimate of the maximal grip coefficient by using the instantaneous value.
means for finding in the signal the values corresponding to the sensor passing through the contact area of the tire on the ground,
means for calculating an instantaneous value of a preselected criterion as a function of a series of values chosen from those relating to the sensor passing through the contact area, and
means for extracting an estimate of the maximal grip coefficient by using the instantaneous value.
2. The system according to claim 1, wherein said means for extracting an estimate of the maximal grip coefficient comprise means for selecting a predetermined characteristic curve of the change in the maximal grip coefficient as a function of the values of the preselected criterion and identification on this characteristic curve of an estimate of the maximal grip coefficient on the basis of the instantaneous value.
3. The system according to claim 1, wherein said means for extracting an estimate of the maximal grip coefficient comprise means for calculating an estimate of the maximal grip coefficient on the basis of the said instantaneous value and mathematical equations whose parameters have been fitted.
4. The system according to claim 1, wherein the preselected criterion is a standard deviation of the values (x) of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
5. The system according to claim 1, wherein the preselected criterion is a sum of the squares of the values (x) of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
6. The system according to claim 1, wherein the preselected criterion is a maximal value of the values (x) of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
7. The system according to claim 1, wherein the preselected criterion is a minimum value of the values (x) of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
8. The system according to claim 1, wherein the preselected criterion is a value of the weight a of an nth order polynomial (x)=axn+bxn1+cxn2+ . . . +u, fitted over at least a part of the contact area.
9. The system according to claim 1, wherein the preselected criterion is an amplitude a1 of the first harmonic of a Fourier series expansion of the signal (x) representing the change in the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
10. The system according to claim 1, wherein the preselected criterion is an amplitude a2 of the second harmonic of a Fourier series expansion of the signal (x) representing the change in the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
11. The system according to claim 1, wherein the preselected criterion is a mean value of the values of the derivative d(x)/dx of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
12. The system according to claim 1, wherein the preselected criterion is a standard deviation of the values of the derivative d(x)/dx of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
13. The system according to claim 1, wherein the preselected criterion is a sum of the squares of the values of the derivative d(x)/dx of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
14. The system according to claim 1, wherein the preselected criterion is a maximal value of the values of the derivative d(x)/dx of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
15. The system according to claim 1, wherein the preselected criterion is a value of the weight a of an n'h order polynomial d(x)/dx=axn+bxn1+cxn2+ . . . +u, fitted to the values of the derivative d(x)/dx of the tangential stresses measured in the longitudinal direction, the values being taken in at least a part of the contact area.
16. The system according to claim 2, wherein the means for finding in the signal the values corresponding to the sensor passing through the contact area of the tire on the ground comprise detection of the maximum and the minimum of (x).
17. The system according to claim 16, wherein the values are taken between the maximum and the minimum.
18. The system according to claim 1, comprising indicators for certain of the operating parameters of the tire, chosen from the group consisting of the load and pressure of the tire and the camber angle of the tire and the speed of the tire and the transmitted torque and the drift angle and the lateral thrust Y, and comprising means for selecting as a function of the chosen indicators a predetermined characteristic curve of change in the maximal grip coefficient p as a function of the values of the said preselected grip criterion, and comprising means for deriving an estimate of the maximal grip coefficient from the said instantaneous value.
19. The system according to claim 1, comprising indicators for certain of the operating parameters of the tire, chosen from the group consisting of the load and pressure of the tire and the camber angle of the tire and the speed of the tire and the transmitted torque and the drift angle and the lateral thrust Y, and comprising means for calculating an estimate of the maximal grip coefficient as a function of the chosen indicators and the values of the said preselected grip criterion, with the aid of mathematical equations whose parameters have been fitted.


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# dense subspace of $c_0( \mathbb N)$

Prove that

$$Y= \left\{ x=(x_n)_{n \in\mathbb{N}} \in c_{0}(\mathbb N )~ \Bigg | ~\sum_{n=1}^{\infty} x_n = 0 \right\}$$

is a dense linear subspace of $c_0( \mathbb N)$.

where $\displaystyle{c_0( \mathbb N) = \left\{ x=(x_n)_{n \in\mathbb{N}} \in \mathbb R ^{\mathbb N} : \lim_{n \to \infty} x_n =0 \right\}}$

I cannot prove that it is dense.

Any help?

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For denseness, you can show each unit vector is in the closure of $Y$. For example, to show $(1,0,\ldots)$ is in the closure of $Y$, consider vectors of the form $(1,\underbrace{-1/n,\ldots ,-1/n}_{n-\text{terms}},0,\ldots)$. –  David Mitra Oct 22 '12 at 13:15
O.K I can show that each $e_n$ is in the closure of $Y$ but I can't see how I can get density. –  passenger Oct 22 '12 at 13:26
@DavidMitra: Yes you are right! I think the most simple solution! Thank you for your time! –  passenger Oct 22 '12 at 14:44
Sorry, there was a "typo" in my previous comment (now deleted). I meant to say just use the fact that the linear span of the set of unit vectors is dense in $c_0$. So the closure of $Y$ contains the closure of the linear span of the set of unit vectors, and hence is all of $c_0$. –  David Mitra Oct 22 '12 at 15:53
Yes i understand that! Thank's again! –  passenger Oct 23 '12 at 12:28

The elegant proof is the following. Consider linear functional $$f:c_{00}(\mathbb{N})\to\mathbb{R}:x\mapsto\sum\limits_{n=1}^\infty x_n$$ Then

1. Show that $f$ is unbounded and $\mathrm{Ker}(f)\subset Y$.
2. Show that that kernel of each unbounded functional is dense in the domain space.
3. Recall that $c_{00}(\mathbb{N})$ is dense in $c_0(\mathbb{N})$.
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I'm confused. Consider $x = (1, 0 , 0 \dots) \in c_0$. If $Y$ is dense in $c_0$ then there is a sequence $y_k = (y_n)_k \in Y$ such that $\|y_k - x\| = \max_n |y_{kn} - x_n| \to 0$ as $k \to \infty$. What would be such a sequence? –  Rudy the Reindeer Oct 22 '12 at 13:00
The problem is that $f$ is not well defined on the whole $c_0$ (take $x_n=n^{-1}$). –  Davide Giraudo Oct 22 '12 at 13:01
@MattN. Consider $y_k=(1-k^{-1},k^{-1}2^{-1},k^{-1}2^{-2},k^{-1}2^{-3},\ldots)$ –  Norbert Oct 22 '12 at 13:03
@DavideGiraudo You are right. I'll try to salvage my proof see edits. –  Norbert Oct 22 '12 at 13:04
@Norbert: Can you prove step 2 ? –  passenger Oct 22 '12 at 13:35

The fact that $Y$ is a subspace is quite clear. To see density, we can use a corollary of Hahn-Banach theorem: we just need to show that each linear continuous functional on $c_0(\Bbb N)$ which vanished on $Y$ vanishes on the whole space.

Let $f$ such a functional. We have $f(e_n-e_m)=0$ if $m\neq n$, where $e_n$ is the sequence whose $n$-th term is $1$, the others $0$. So $f(e_k)=:K$. As $\left\lVert\sum_{k=0}^ne_k\right\rVert_{\infty}=1$, we show have $nK\leq \lVert f\rVert$ and $K=0$.

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Can you explain a little more the last line of your solution? –  passenger Oct 22 '12 at 13:09
$f(e_k)$ as the same value $K$. Then I use $l(\sum_{1\leq k\leq n} e_k)=nK, and it's supposed to be$\leq$the norm of$f$. – Davide Giraudo Oct 22 '12 at 13:20 Taking on Norbert's answer: since the linear functional$\,f\,$is obviously not the zero functional, we know$\,\ker f\,$is a maximal subspace of$\,c_0(\Bbb N)\,$, from which it follows that $$c_0(\Bbb N)=\operatorname{Span}\{\ker f\,,\,v\,\}\;\;,\;\forall\;v\notin\ker f$$ Perhaps this now will make it simpler to find the solution ( hint: a subset$\,A\,$of a topological space$\,X\,$is dense in it iff$\,A\cap Y\neq\emptyset\,$for every non-empty open subset$\,Y\subset X\,\$ )

-


text:
Please use this identifier to cite or link to this item: https://hdl.handle.net/20.500.11851/3659
DC FieldValueLanguage
dc.contributor.authorThe ATLAS Collaboration-
dc.contributor.authorSultansoy, Saleh-
dc.date.accessioned2020-08-24T14:14:59Z
dc.date.available2020-08-24T14:14:59Z
dc.date.issued2018-02
dc.identifier.citationATLAS collaboration. (2018). Measurement of longitudinal flow decorrelations in Pb+ Pb collisions at $$\sqrt {s_ {\text {NN}}}= 2.76$$ and 5.02 TeV with the ATLAS detector. The European Physical Journal C, 78(2), 142.en_US
dc.identifier.issn14346044
dc.identifier.urihttps://hdl.handle.net/20.500.11851/3659-
dc.description.abstractMeasurements of longitudinal flow correlations are presented for charged particles in the pseudorapidity range | η| < 2.4 using 7 and 470 μ b - 1 of Pb+Pb collisions at sNN=2.76 and 5.02 TeV, respectively, recorded by the ATLAS detector at the LHC. It is found that the correlation between the harmonic flow coefficients vn measured in two separated η intervals does not factorise into the product of single-particle coefficients, and this breaking of factorisation, or flow decorrelation, increases linearly with the η separation between the intervals. The flow decorrelation is stronger at 2.76 TeV than at 5.02 TeV. Higher-order moments of the correlations are also measured, and the corresponding linear coefficients for the kth -moment of the vn are found to be proportional to k for v3, but not for v2. The decorrelation effect is separated into contributions from the magnitude of vn and the event-plane orientation, each as a function of η. These two contributions are found to be comparable. The longitudinal flow correlations are also measured between vn of different order in n. The decorrelations of v2 and v3 are found to be independent of each other, while the decorrelations of v4 and v5 are found to be driven by the nonlinear contribution from v22 and v2v3, respectively. © 2018, CERN for the benefit of the ATLAS collaboration.en_US
dc.language.isoenen_US
dc.publisherSpringer New York LLCen_US
dc.relation.ispartofEuropean Physical Journal Cen_US
dc.rightsinfo:eu-repo/semantics/openAccessen_US
dc.subjectCollisionsen_US
dc.subjectionic collisionsen_US
dc.subjectflow harmonicsen_US
dc.titleMeasurement of longitudinal flow decorrelations in Pb+Pb collisions at ?sNN=2.76 and 5.02 TeV with the ATLAS detectoren_US
dc.typeArticleen_US
dc.departmentFaculties, Faculty of Engineering, Department of Material Science and Nanotechnology Engineeringen_US
dc.departmentFakülteler, Mühendislik Fakültesi, Malzeme Bilimi ve Nanoteknoloji Mühendisliği Bölümütr_TR
dc.identifier.volume78
dc.identifier.issue2
dc.authorid0000-0003-2340-748X-
dc.identifier.wosWOS:000425395800005en_US
dc.identifier.scopus2-s2.0-85042405179en_US
dc.institutionauthorSultansoy, Saleh F.-
dc.identifier.pmid31265002en_US
dc.identifier.doi10.1140/epjc/s10052-018-5605-7-
dc.relation.publicationcategoryMakale - Uluslararası Hakemli Dergi - Kurum Öğretim Elemanıen_US
dc.identifier.scopusqualityQ1-
item.grantfulltextopen-
item.openairecristypehttp://purl.org/coar/resource_type/c_18cf-
item.languageiso639-1en-
item.cerifentitytypePublications-
item.fulltextWith Fulltext-
item.openairetypeArticle-
crisitem.author.dept02.6. Department of Material Science and Nanotechnology Engineering-
Appears in Collections:Malzeme Bilimi ve Nanoteknoloji Mühendisliği Bölümü / Department of Material Science & Nanotechnology Engineering
PubMed İndeksli Yayınlar Koleksiyonu / PubMed Indexed Publications Collection
Scopus İndeksli Yayınlar Koleksiyonu / Scopus Indexed Publications Collection
WoS İndeksli Yayınlar Koleksiyonu / WoS Indexed Publications Collection
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text:
# Vectors

For vector w is true: w = 2u-5v. Determine coordinates of vector w if u=(3, -1), v=(12, -10)

Result

wx =  -54
wy =  48

#### Solution:

$w = 2u-5v = (2\cdot 3+(-5)\cdot 12, 2\cdot (-1)+(-5)\cdot (-10) ) = (-54, 48)$
$w_{ y }=(2) \cdot \ (-1)+(-5) \cdot \ (-10) = 48$

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#### Following knowledge from mathematics are needed to solve this word math problem:

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text:
NLO QCD+EW predictions for V + jets including off-shell vector-boson decays and multijet merging

Kallweit, S; Lindert, J M; Maierhöfer, P; Pozzorini, S; Schönherr, M (2016). NLO QCD+EW predictions for V + jets including off-shell vector-boson decays and multijet merging. Journal of High Energy Physics, 2016(4):1-42.

Abstract

We present next-to-leading order (NLO) predictions including QCD and electroweak (EW) corrections for the production and decay of off-shell electroweak vector bosons in association with up to two jets at the 13 TeV LHC. All possible dilepton final states with zero, one or two charged leptons that can arise from off-shell W and Z bosons or photons are considered. All predictions are obtained using the automated implementation of NLO QCD+EW corrections in the OpenLoops matrix-element generator combined with the Munich and Sherpa Monte Carlo frameworks. Electroweak corrections play an especially important role in the context of BSM searches, due to the presence of large EW Sudakov logarithms at the TeV scale. In this kinematic regime, important observables such as the jet transverse momentum or the total transverse energy are strongly sensitive to multijet emissions. As a result, fixed-order NLO QCD+EW predictions are plagued by huge QCD corrections and poor theoretical precision. To remedy this problem we present an approximate method that allows for a simple and reliable implementation of NLO EW corrections in the MePs@Nlo multijet merging framework. Using this general approach we present an inclusive simulation of vector-boson production in association with jets that guarantees NLO QCD+EW accuracy in all phase-space regions involving up to two resolved jets.

Abstract

We present next-to-leading order (NLO) predictions including QCD and electroweak (EW) corrections for the production and decay of off-shell electroweak vector bosons in association with up to two jets at the 13 TeV LHC. All possible dilepton final states with zero, one or two charged leptons that can arise from off-shell W and Z bosons or photons are considered. All predictions are obtained using the automated implementation of NLO QCD+EW corrections in the OpenLoops matrix-element generator combined with the Munich and Sherpa Monte Carlo frameworks. Electroweak corrections play an especially important role in the context of BSM searches, due to the presence of large EW Sudakov logarithms at the TeV scale. In this kinematic regime, important observables such as the jet transverse momentum or the total transverse energy are strongly sensitive to multijet emissions. As a result, fixed-order NLO QCD+EW predictions are plagued by huge QCD corrections and poor theoretical precision. To remedy this problem we present an approximate method that allows for a simple and reliable implementation of NLO EW corrections in the MePs@Nlo multijet merging framework. Using this general approach we present an inclusive simulation of vector-boson production in association with jets that guarantees NLO QCD+EW accuracy in all phase-space regions involving up to two resolved jets.

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Item Type: Journal Article, refereed, original work 07 Faculty of Science > Physics Institute 530 Physics Physical Sciences > Nuclear and High Energy Physics English 2016 30 Dec 2016 07:30 29 Apr 2020 22:41 Springer 1029-8479 Gold Publisher DOI. An embargo period may apply. https://doi.org/10.1007/JHEP04(2016)021