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# Dipole and electric field gradient

1. Jun 9, 2014

### Staff: Mentor

I'm developing a classical model of a dipolar ion in an external electric field. It consists of two charges $\delta_+$ and $\delta_-$, located at a fixed distance from each other. For the special case I'm considering, I end up with the potential energy
$$(\delta_+ + \delta_-) \Phi(\vec{R}) + \sum_{\xi = x,y,z} \left[ \delta_+ (\vec{r}_+')_{\xi} + \delta_- (\vec{r}_-')_{\xi} \right] \left. \frac{\partial}{\partial \xi} \Phi \right|_{\vec{R}} + \sum_{\xi = x,y,z} \left[ \delta_+ (\vec{r}_+')_{\xi}^2 + \delta_- (\vec{r}_-')_{\xi}^2 \right] \left. \frac{\partial^2}{\partial \xi^2} \Phi \right|_{\vec{R}}$$
where $\Phi$ is the electric potential, $\vec{R}$ is the position of the center of mass and $\vec{r}_\pm'$ the positions of the charges $\delta_\pm$ with respect to the center of mass. The notation $(\vec{r})_\xi$ means the $\xi \in \{x,y,z\}$ component of $\vec{r}$.

The first term I can identify as the total charge times the electric potential and the second I can rewrite as
$$\vec{\mu} \cdot \left. \vec{\nabla}\Phi \right|_{\vec{R}} = \vec{\mu} \cdot \vec{E}(\vec{R})$$
i.e., the interaction of the dipole with the electric field. But I'm struggling to find a physical interpretation to the third term. I've tried to rewrite it in other terms, but didn't come up with anything interesting. Is this simply because the center of mass is not exactly midway between the two charges?

Any input appreciated.

2. Jun 9, 2014

### dEdt

Well, the formula $$U=\mathbf{\mu}\cdot\nabla \Phi$$ is only appropriate for a point-like dipole, so it just seems like your third term is a correction term that arises when you avoid assuming that the dipole is point-like.

3. Jun 9, 2014


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## LaTeX forum ⇒ TeX Live and MacTeX ⇒ mactex paths texlive acmart

Information and discussion about TeX Live distribution for all platforms (Windows, Linux, Mac OS X) and the related MacTeX: installing, updating, configuring
David Epstein
Posts: 2
Joined: Fri Mar 03, 2017 11:11 am

### mactex paths texlive acmart

I have just joined Latex Community, and I am not familiar with it, so I hope I haven't missed the answer to my questions below.
I have searched but I have not been able to find answers.

I have just installed MacTeX. I'm using command line
1. latexmk -pdf filename.tex

to compile. The first line of my file is
1. \documentclass[format=acmsmall]{acmart}

which provokes the error message

That file is found on my system at
/usr/local/texlive/2016/texmf-dist/tex/latex/acmart/acmart.cls

How can I persuade latex to find this file? Or do I need to copy it to somewhere else?
Where do I find the paths along which this distribution of tex searches?
Should I set these path, or take some other action?
Are the paths set by MacTeX or by texlive?

Also what path will be used to look for my .bib files (called by \bibliography).

Thanks

Tags:

Stefan Kottwitz
Posts: 9586
Joined: Mon Mar 10, 2008 9:44 pm
Hi David,

welcome to the forum!

Try the kpsewhich tool at the command prompt, it shows the paths. Regarding bib files, take a look here: viewtopic.php?f=50&t=29206

Stefan

David Epstein
Posts: 2
Joined: Fri Mar 03, 2017 11:11 am
Thanks. The reply from Stefan Kottwitz is helpful in pointing to the reason for my problems, because I did in fact try kpsewhich but got null response.
However, my system's response to
1. which kpsewhich
is
1. /usr/local/texlive/2014/bin/universal-darwin/kpsewhich
.
Since the file I was searching for is in texlive/2016, kpsewhich didn't find it.

I had thought that installing MacTex would automatically change my path. I use tcsh. So now I'm assuming that the path needs to be changed manually. Here is the response to
1. echo \$path
:
1. /opt/local/bin /opt/local/sbin /Library/Frameworks/Python.framework/Versions/2.7/bin /opt/local/bin /opt/local/sbin /Users/dbae/bin/qepcad-B.1.65.MacOSX/qesource/bin /usr/local/texlive/2014/bin/universal-darwin /usr/local/texlive/2014/bin/x86_64-darwin /Users/dbae/bin /opt/local/bin /opt/local/sbin /usr/bin /bin /usr/sbin /sbin /usr/local/bin /opt/X11/bin /Library/TeX/texbin

What should be added to pick up commands defined in my recent installation of MacTex?

Stefan Kottwitz
Posts: 9586
Joined: Mon Mar 10, 2008 9:44 pm
On my Macbook it's simply:

% echo \$path
/usr/bin /bin /usr/sbin /sbin /usr/local/bin /opt/X11/bin /usr/texbin

Better remove the old 2014 entries from the path. Perhaps look into ~/.tcshrc or ~/.profile or what your shell uses to add path entries.

Check the folder /etc/paths.d. On my system, there's a TeX file containing just the line /usr/texbin. The contents of the files in that folder is added to the path. Perhaps you haven an older file there.

Stefan

format: plain


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# How to get greeks using Monte-Carlo for arbitrary option?

Let's assume I have an arbitrary option that I can price using Monte-Carlo simulation. What is the general approach (i.e. without relying on specific option type) to calculating the greeks in this case?

Edit: I woud like to add a few links on the topic that I found useful:

-
That vibrato Monte Carlo thesis is very interesting. –  Brian B Apr 27 '12 at 15:00

You need to compute your greeks as finite differences, but the full procedure may be pretty tricky. I will use vega $\aleph$ as the example here. Let's begin by designating your Monte Carlo estimator as a function $V(\sigma,s,M)$ where $\sigma$ is the volatility as usual, $s$ is the seed to your random number generator, and $M$ is the sample count.

To begin with, recall that the Monte Carlo estimate of any value converges with the square root of the sample count. In particular, if you choose, say, $M=100$, you can run your estimator $N=500$ times to get estimate $\{V_n\}_{i=1}^{500}$, obtaining the standard deviation $\Sigma_{100}$ of those estimates.

Having done this, we now know the standard error of the estimator for any $M$ to be

$$e_M \approx \Sigma_{100} \sqrt{\frac{100}{M}}$$

There are three possible cases:

1. You can control the random seed $s$, or the set of random samples, used by the Monte Carlo estimator
2. You cannot control $s$.
3. You cannot even control the sample count $M$.

In the first case, you can use the fact that $s$ has been controlled to get a reasonable estimate of vega with relatively little extra work.

Find an $M$ such that the error in option price $e_M$ is tolerable. Choose a seed $s_0$ and a small increment $\Delta\sigma$ in the volatility, and compute

$$\aleph^{(1)} = \frac{V(\sigma+\Delta\sigma,s_0,M)-V(\sigma-\Delta\sigma,s_0,M)}{2\Delta\sigma}$$ and $$\aleph^{(2)} = \frac{V(\sigma+\frac12\Delta\sigma,s_0,M) - V(\sigma-\frac12\Delta\sigma,s_0,M)}{\Delta\sigma}$$

If $\aleph^{(1)} \approx \aleph^{(2)}$ then you have a good estimate and you are done.

The reason this works so nicely is that, by controlling the seed, our difference computations

$$\delta=V(\sigma+\Delta\sigma,s_0,M)-V(\sigma-\Delta\sigma,s_0,M)$$ are direct Monte Carlo estimators of the vega, since the shared seed implies the samples $x_i$ match in the difference of sums. That is $$\delta = ( \frac1M \sum_{i=1}^M f(x_i, \sigma+\Delta\sigma) ) -( \frac1M \sum_{i=1}^M f(x_i, \sigma-\Delta\sigma) ) \\ =\frac1M \sum_{i=1}^M f(x_i, \sigma+\Delta\sigma)-f(x_i, \sigma-\Delta\sigma)$$

The second case where you cannot control the seed, on the other hand, is rather more difficult. Here, you will have a different error $e$ to the true value every time you run the function.

For brevity, let's let $$e_\pm = V(\sigma\pm \Delta\sigma,s_\pm,M).$$

Of course we do not know the value of $e_\pm$ or of $s_\pm$, but we do at least have our estimate of the size of $e_\pm$ as noted above. Therefore, the error in $\delta$ is approximately $e_M \sqrt{2}$. You need to choose $M$ so large that $$\delta \gg e_M \sqrt{2}.$$

Not knowing the value of $\delta$ a priori makes this difficult, but usually in a trading context one can specify an acceptable absolute error $\epsilon$ in vega. In that case, we can demand $$\epsilon < \frac{e_M \sqrt{2}}{\Delta\sigma}$$ which translates to $$M > \Sigma_{100}^2 {\frac{200}{\epsilon^2 \Delta\sigma^2}}.$$

The third case, where you can control neither the random seed $s$ nor the sample count $M$ should be treated as the second case above. You simply treat each run of the algorithm as a single sample.

-

The most general answer is to shift your input to approximate the first derivative. Given that you need Monte Carlo to price this, it may get expensive. But that's the way it goes as when you have no analytical solutions as there aint't no free lunch ...

-
Numerical derivatives are iffy business, but I agree that it seems to be your best choice. As you probably know; be aware of the how the precision decreases quickly(!) as higher orders are measures. –  AdAbsurdum Apr 24 '12 at 7:34

Download the code from http://fmsoption.codeplex.com to see how to do that for vanilla options. You are right, you need implementations for transcendental functions that are written for dual numbers. You will find them in the fmsdual project.

If you just want browse some source code, see http://fmsoption.codeplex.com/SourceControl/changeset/view/10924#145366. Note that eps is machine epsilon ~= 2e-16. (!)

-
For the curious, the "dual number" approach referenced here is an automatic differentiation package, where the bookkeeping mostly handled by C++ templates and extensions to the standard numeric types. Derivatives of transcendental functions are handled by automatic differentiation of the numerical analytic series approximations used to calculate function values. (Please correct any mistakes I have made in that) –  Brian B May 10 '12 at 14:12
I beleive AD refers to techniques for automatically generating functions for the derivatives. Dual numbers don't do that for you. –  Keith A. Lewis May 11 '12 at 14:05

if the pay-off is continuous, the standard approach is to use the path-wise method also known as IPA. This essentially means that you differentiate along each path. It is the limit as the bump size goes to zero of finite differencing.

The main downside of this method is that the differentiation can be fiddly and slow. The Smoking adjoints paper you mention made the observation that using adjoint/automatic differentiation makes it fast and indeed there are packages that will do the differentiating automatically for you. There is a survey article by Homescu on this topic.

Adjoints and Automatic (Algorithmic) Differentiation in Computational Finance

Cristian Homescu

I have also written too many papers on this. See my SSRN page http://papers.ssrn.com/sol3/cf_dev/AbsByAuth.cfm?per_id=550354

I also devote a chapter of More Mathematical Finance to this.

When the pay-off is discontinuous or you want Hessians life is more complicated. Likelihood ratio is one approach but can lead to large variances. This is particularly the case for vegas.

See Glasserman (2003) Monte Carlo Methods in Financial Engineering for general discussion.

I have done various papers on the discontinuous case and on Hessians. My current favourites are

http://ssrn.com/abstract=2431580

http://ssrn.com/abstract=2011690

Essentially, these work by doing change of variables that remove the discontinuities and so then the pathwise method can be applied. They are designed to yield the most pathwise mix of likelihood ratio and pathwise that avoids the discontinuity.

-

Another way to do this is to use dual numbers. http://fmsdual.codeplex.com.

They let you calculate an arbitrary number of derivatives while running a single Monte Carlo. Here is an example of how to use it:

// Monte Carlo derivatives
void fms_test_monte(size_t N)
{
::srand(static_cast<unsigned int>(::time(0)));

double a = 0.5;
dual::number<double,3> A(a, 1);
dual::number<double,3> E(0.,1);

for (int i = 0; i < N; ++i) {
double x = 1.0*rand()/RAND_MAX;
E = E + (x - A)*(x - A);
}

E = E/(1.*N);

// X uniform [0,1]
// E(X - a)^2 = 1/3 - 2a 1/2 + a^2
ensure (fabs(E._(0) - (1./3 - a + a*a)) < sqrt(1./N));

// d/da E(X - a)^2 = -2 E(X - a) = 2a - 1
ensure (fabs(E._(1) - (2*a - 1)) < sqrt(1./N));

// d^2/da^2 E(X - a)^2 = 2
ensure (fabs(E._(2) - 2) < sqrt(1./N));
}

-
A novel method. It appears mathematically equivalent to finite differences. Also worth noting is that all samples need to remain in memory, at least if I read this stuff right. It would be informative to see code applied to a nontrivial estimator function. –  Brian B Apr 27 '12 at 15:36
You are reading it wrong. Not all samples need to remain in memory. It is not mathematically equivalent to finite differences, you seem to be completely missing the point of dual numbers. You have the source code, let me know if you need help applying it to a non-trivial estimator. Monte Carlo aside, dual numbers allow you to calculate derivatives down to machine precision. –  Keith A. Lewis May 7 '12 at 9:56
You're right, I now see it is not finite differences. I would find the simplicity of your example far more convincing if you demonstrated calculating, say, delta of an average strike option under the the Black-Scholes stochastic model. It seems to me the transcendental functions involved make this difficult even for vanilla options. Path dependencies will make the problem much worse. –  Brian B May 7 '12 at 16:24
I'll also note that, as @Dirk and I read the question, Alexey does not necessarily have the source code to the option pricer. –  Brian B May 7 '12 at 16:25


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# Chess checkmates - a simple neural network

## Introduction

Chess is a very old and very complex game. Humans have been playing chess and chess variants for thousands of years. I began playing chess somewhat seriously around 2015, and it has quickly become one of my favourite games. You can find a good primer for learning the rules of chess here.

Hobbyists, scientists, and professionals have long sought to look deeper into the nature of chess and explore it’s complexities. A natural course of study was to program computers to be able to play chess against humans and, eventually, beat them. This work reached a seminal moment in 1997 when IBMs Deep Blue supercomputer defeated the world champion of chess Garry Kasparov in a series of 6 games. From then on chess engines only became stronger and soon all top chess engines could easily defeat any human.

However, how typical chess engines play chess is very unlike how humans play. A chess engine such as Stockfish or Komodo will play chess through sheer calculation. That is, chess engines will utilize a computers ability to peform many ordered tasks quickly and efficiently to calculate the strength of thousands of series of moves in any given position, and then choose the best option. Such a strategey is far beyond any humans ability. Strong human chess players will choose their moves through a combination of experience, intuitition, and manual calculation.

In the past several years chess engines have been developed that do not rely purely on brute force calculation. The DeepMind company developed a program called AlphaZero that has mastered chess (along with go and shogi) through neural networks (NNs). A similar neural net based program, Leela Chess Zero, is open source and is an ongoing project. I will discuss neural networks in more detail later.

All of this recent work involving neural net based chess engines, as well as the (relatively) recent explosion of interest in artificial intelligence (AI), piqued my interest in further applications of AI in chess.

A particularly popular application of NNs is image classification, that is training a model to accurately predict a label across a given set of images. I have experience working with large sets of images for statistical applications before, and thought it would be fun to investigate a NN applied to large sets of chess images. I figured checkmates would be promising as a starting point. In chess, checkmate is the win condition of the game. It is achieved when a player’s king is attacked by an opposing piece, and every accessible ajoining square the king may move to is also attacked by an opposing piece.

I decided upon investigating checkmates because of their inherent simplicity. Any given chess position is either checkmate or not checkmate. This is a binary outcome, and binary classification is a well studied application of NN modelling. My current background is in undergraduate math and statistics, I have no formal training in AI and machine learning and have devoted little independent study to the field, so starting simple is prudent.

## Part One - Data sourcing and cleaning

After deciding what I wanted to investigate, I needed to figure out a way to obtain the necessary data. My initial idea was to just surf the internet for random repositories of checkmate positions in FEN files, which could then be used to generate images. However I quickly discovered two problems.

1. It is somewhat difficult to find repositories of FEN files that are of an actual checkmate (most are for checkmate in one move chess puzzles, more on this later).
2. After locating a repository there were far too few actual positions! Image classification needs many many images for training.

My next idea was to obtain a large dataset of chess games, identify which games ended in a checkmate position, and extract those positions to generate images. Enter lichess. Lichess is a completely free and open source chess website, and I highly reccomend it if you are interested in playing online chess. Lichess conveniently archives every single game played on their site by month, with games as far back as 2013. To reduce computing time, I chose the month with fewest games, January 2013, during which a mere 121,332 games were played.

An aside, to give you an idea of how popular lichess has become over the years I scraped the number of standard (non-variant) games played per month.

The huge spike at the end is due to the ongoing covid-19 global pandemic.

Moving on, the games are bundled into a single .bz2 file, and are all in PGN format. After I downloaded the January 2013 file I needed a way to parse through the games and extract desired features. I decided to use the python library python-chess as it was reccomended on the lichess game archive.

I have fairly minimal experience working with python, so I am sure my approach is by no means the best or most efficient.

#load library
import chess.pgn

pgn = open("lichess_db_standard_rated_2013-01.pgn", encoding="utf-8-sig")

We can look at all the categories in a game from the pgn file.

chess.pgn.read_headers(pgn)
## Headers(Event='Rated Classical game', Site='https://lichess.org/j1dkb5dw', White='BFG9k', Black='mamalak', Result='1-0', BlackElo='1403', BlackRatingDiff='-8', ECO='C00', Opening='French Defense: Normal Variation', Termination='Normal', TimeControl='600+8', UTCDate='2012.12.31', UTCTime='23:01:03', WhiteElo='1639', WhiteRatingDiff='+5')

The goal is to extract positions with checkmate on the board. Lichess tags games that end in resignation or checkmate as “Normal” termination in pgn files. The initial parse will identify the locations of all these games across the main .pgn file.

#Create empty array to store games that ended in checkmate
checkmate_games = []

#1st parsing: get game locations that end normally
while True:
offset = pgn.tell() #stores current file position

if headers is None: #if no value go to next file location
break
checkmate_games.append(offset)

Now that we have the file locations of games that ended normally, we need to store the actual games.

#2nd parsing: get games that ended normally
checkmate_games2 = []

for i in checkmate_games:
pgn.seek(i)
checkmate_games2.append(chess.pgn.read_game(pgn))

Now that we have stored all the actual games that ended normally, we need to find the subset of games that ended in checkmate.

#3rd parsing: get games that end in checkmate
checkmate_games3 = []

for i in range(len(checkmate_games2)):
if(checkmate_games2[i].end().board().is_checkmate()):
checkmate_games3.append(checkmate_games2[i])

len(checkmate_games3)
34474

Out of the initial 121,332 games in the file only 34,474 ended in checkmate.

The NN that will be trained still needs inputs, the next step is to get the checkmate positions in all 34,474 games and store these positions as images. Fortunately the python-chess library has a module that suits our needs, and will create an svg image of a specified chess position.

#Write an svg image of every checkmate position to disk
for i in range(len(checkmate_games3)):
outfilename = '%smate.svg' % (str(i)) #create unique file name, e.g. 0mate.svg, 1mate.svg, ...
outputFile = open(outfilename, mode = "w")
outputFile.write(chess.svg.board(board = checkmate_games3[i].end().board(), coordinates=False))
outputFile.close()

The .end() method will grab the position at the end of a game, .board() will generate a board representation for chess.svg.board() to render.

Now all the checkmate positions are saved as images as desired. However in order to train the NN, we will need more than just checkmate positions. To get chess positions that are not checkmate positions I looped through the stored games that ended normally, and extracted a random position from the game (that wasn’t checkmate), then saved that position again as a svg file.

import random as ran

for j in range(1,73552):

board1 = checkmate_games2[j].board() #store jth game as a board
moves = []

for move in checkmate_games2[j].mainline_moves(): #append move list
moves.append(move)

if(len(moves) - 2 <= 1): #filter games that end after 2 moves or less
continue

r1 = ran.randint(1,len(moves)-2) #random total number of moves that won't result in checkmate.

for i in range(r1): #play moves on current board
board1.push(moves[i])

outfilename = '%sNmate.svg' % (str(j))
outputFile = open(outfilename, mode = "w")
outputFile.write(chess.svg.board(board1, coordinates=False))
outputFile.close()

Now there are 73,551 chess positions saved as svg files that are not checkmate. I chose this number somewhat randomly, it is a little over twice the number of checkmate images. I wanted a balance between computation time and sufficient data, as for training and testing I wanted to have more checkmate than non-checkmate positions.

However I was still not done, as svg files are not exactly friendly to out of box NN modules. To convert my chess positions to an image file format easier to work with I used ImageMagick. This section is already somewhat lengthy so I will skip over the details, but there exist many informative tutorials on the web about using ImageMagick for batch processing.

Part 3 - NN fitting coming soon!


text:
Can ‘Shades’ of Carmichael Numbers be busted?

 MAA source

Robert Carmichael was an American mathematician who worked and published in various areas of mathematics, but is best known for numbers that were named after him. It’s a rare treat to have your own numbers; our famous Manuel Blum has his:

$\displaystyle 21, 33, 57, 69, 77, 93, 129, 133, 141, 161, 177 \dots$

A Carmichael number is a product ${N}$ of two or more—actually three or more—distinct primes ${p}$, each of which is such that ${p-1}$ divides ${N-1}$—or equivalently, such that the least common multiple of all such ${p-1}$ divides ${N-1}$. Carmichael found the first example, ${N = 561 = 3 \cdot 11 \cdot 17}$. Note that 2, 10, and 16 all divide 560.

Today Ken and I want to talk about Carmichael numbers and their relationship to the factoring of integers.

The relationship is that Carmichael numbers ${N}$ masquerade as being prime by enjoying the property of Fermat’s Little Theorem: for all ${a}$ relatively prime to ${N}$,

$\displaystyle a^{N-1} \equiv 1 \bmod N.$

The significance is that ${N}$ fails the first part of the randomized compositeness test devised in the late 1970’s by Michael Rabin after earlier work by Gary Miller. Call a relatively prime number ${a}$ chosen at random a trial. Rabin proved that for all odd ${N}$ and ${3/4}$ of the trials, at least one of his two parts must succeed. Hence for Carmichael ${N}$ all the success probability is on the second part (and is higher). This is great because Rabin’s second part not only tells you ${N}$ is composite, it gives you a factor.

Of course factoring has been one of the world’s enduring mysteries for decades. Another is whether there will be a third “Ghostbusters” movie. Cryptographers needing to protect schemes whose security requires the hardness of factoring would like to “bust” heuristic attacks on factoring, by showing they have only negligible chance to succeed on numbers ${N}$ they care about, ones that are definitely not Carmichael numbers. But like ghosts in the two movies, attack ideas keep popping up. Here we talk about a variant of several old ideas in which we try to change an ${N}$ into an ${N^*}$ that is somehow a “shade” of a Carmichael number, such that getting a factor of ${N^*}$ might often give us one of ${N}$.

Carmichael himself wrote, with Edwin Smith, the book Mathematical Tables and Formulas published in 1931. It included five-place tables of many of the standard functions, such as logarithms. and other functions. Clearly, such books were once of great importance, used often by engineers and mathematicians.

What is the logarithm of 7 to base 2? We wonder if Volker Strassen used a table to look this up? The key was that it was less than 3, but knowing it is approximately 2.807354922… is quite important. This is of course the exponent on Strassen’s famous fast matrix product algorithm. By the way some web pages that discuss his algorithm quote the constant as approximately 2.808, which is a bit off.

You may find it hard to believe that we once had books on our shelves that contained tables, since now we can compute any of these functions instantly on our laptop or even on our phone—see here [urlsic] for computing logs. But once they were very much a necessary item for any well-equipped researcher.

The one problem with tables, however, is that at some point they run out. Computers can effectively give you the power to make bigger tables, but the problem may be merely deferred. An idea can look good on small numbers in tables, or even on fairly big numbers tested by computer, but utterly fail on large numbers that people would ultimately care about. So also take our small examples in this post with a grain of salt—a lot more work is needed to tell whether ideas like what follows are possible or can be “busted.”

## A Generic Attack

Let ${\cal M}$ be a subset of the natural numbers. They are required to have one property: there is a fast factoring algorithm ${\cal A}$ that works on elements of ${\cal M}$. We allow ${\cal A}$ to be probabilistic or even to not be an algorithm, but to be a circuit. By “fast” we would like polynomial time, yet even time just better than known methods would be acceptable. Since we can combine fast factoring algorithms, we can suppose ${\cal M}$ includes the Carmichael numbers.

The factoring method is quite simple:

1. Select a number ${r}$ and compute ${N^{*} = rN}$. Call ${r}$ a probe.
2. Try ${\cal A}$ on ${N^{*}}$.
3. If ${\cal A}$ returns a factor of ${N}$, then we are done.
4. If ${\cal A}$ returns a wrong answer or returns just a divisor of ${r}$, then go back to step (1) and repeat.

The probe ${r}$ might be a prime number, to minimize the chance of getting nothing in the last step, and might be chosen at random. What we wonder is whether probes ${r}$ can be chosen strategically to improve the prospects of ${N^{*}}$ hitting the success set ${\cal M}$ of ${\cal A}$. At least we know from work by Red Alford, Andrew Granville, and Carl Pomerance that there are infinitely many Carmichael numbers, indeed that the number ${C(X)}$ of Carmichaels below any given number ${X}$ satisfies

$\displaystyle C(X) > X^{2/7},$

for large enough ${X}$. In 1956 Paul Erdős conjectured that there were many more such numbers—note the time reversal. He reasoned there should be at least ${X^{1-o(1)}}$. Pomerance used heuristic arguments to guess that there might be many more. Steven Hayman and Andrew Shallue constructed a Carmichael number with over ten billion factors, drawing on a quantum algorithm by Greg Kuperberg. But all we actually need is a set ${\cal M}$ of numbers that are “Carmichael-like” enough for ${\cal A}$ to succeed, and either dense and divisor-rich enough for random probes to hit it with high probability, or distinctive in a way that successive probes can use earlier-gathered information to drive toward it.

One case where our probe idea is silly is when the success of ${\cal A}$ depends on ${N}$ having a factor ${p}$ with a special property. One property that has been used is when ${p-1}$ is smooth, meaning ${p-1}$ is a product of primes that are all small, or powersmooth, meaning also that none of those primes is raised to a high power. The point is that if this is true of ${N^{*} = rN}$ and ${p}$ is different from ${r}$, then it was already true of ${N}$, so the probe didn’t gain anything. Thus for instance our idea isn’t useful when ${\cal A}$ is an original form of the ${p-1}$ algorithm devised by John Pollard. But we think the variant of Pollard’s gcd-based idea used in the Rabin-Miller “part 2” might escape this problem.

## Part II

The Rabin-Miller idea is to write ${N - 1}$ uniquely as ${2^s v}$ with ${v}$ odd. Then given a trial ${a}$, we actually work with the number ${b = a^v \bmod N}$. The “Part II” is to compute

$\displaystyle \gcd(b - 1 \bmod N, N),\;\gcd(b^2 - 1 \bmod N, N), \;\dots,\;\gcd(b^{2^s} - 1 \bmod N, N).$

For good measure we can try ${b^{2^i} + 1}$ in the gcd’s too. If all of these return ${1}$ (or an immediate zero) we fail, but anything else gives us a factor of ${N}$.

We pause to note that Pollard’s idea was to try the gcd with ${N}$ and ${a^{u^e} - 1 \bmod N}$ where ${u}$ ranges over products of small primes and ${e}$ over small powers of them. Then a powersmooth ${p-1}$ would surely yield ${p}$ in the wash. But here we do not demand sureness, only a fighting chance of getting a factor.

One thing that intuitively helps is if we can probe ${N}$ to ${N^{*}}$ so that ${N^* - 1}$ is divisible by a large power of ${2}$. Then we get a large number of cracks at the gcd. The special role of powers of ${2}$ is similar to that in Miller’s original idea which we covered here.

Here is a small example to show that combining this with our probe idea at least isn’t silly. Consider ${N = 77}$ which is incidentally a Blum integer. We have ${N - 1 = 76 = 4*19}$ so we get three cracks at a gcd. Consider the trial ${a = 2}$. Alas this ${a}$ is a “miss” because:

1. ${b = 2^{19} \bmod 77 = 72}$ but ${71}$ is prime;
2. ${b^2 = 72^2 = 25 \bmod n}$ but ${24}$ is coprime to ${77}$;
3. ${b^4 = 25^2 = 9 \bmod 77}$ but ${8}$ is useless.

However, the probe ${r = 5}$ makes ${N^{*} - 1 = 384 = 128*3}$, which gives eight cracks. But for the same trial ${a = 2}$ we only need the first one: ${a^3 - 1 = 7}$ which divides ${77}$.

One can make tables of other small cases that can make our idea look good. But remember the caveat that we would need better reasons for optimism when ${N}$ gets large. The prospective “buster,” however, has the same problem. It’s not enough to argue that Carmichael numbers that are multiples of a given ${N}$ may not exist or may be too big or sparse for probing to hit them, because this part may need only some shadow of the Carmichael property and Rabin’s theorem to succeed.

Even we have ${N = pq}$ and a prime probe ${r}$ fails by returning just ${r}$, we can derive a kind of heuristic consolation: If we’d instead had ${N' = pr}$ and happened to hit on the probe ${q}$, then the same outcome ${r}$ would be a success. The same for ${N' = qr}$ with the probe ${p}$. When ${N^{*}}$ has ${m}$ prime factors and gives one of them, the density of ${N = pq}$ for which the given factor is ${p}$ or ${q}$ is still non-negligible—no worse than ${2/m}$. This is one reason we think our probe idea is worth considering.

## Open Problems

Can there be a Rabin-Miller “Part III”? Are there smart strategies for choosing successive probes that have a non-negligible chance of hitting the set ${\cal M}$ of numbers ${N^{*}}$ for which a non-negligible portion of trials succeed? How dense is such an ${\cal M}$ anyway? Or can this idea be “busted” along similar lines to the analysis of Pollard’s ${p-1}$ or other heuristics?

July 19, 2013 6:03 am

Do you think if factoring is easy, all one way functions can be broken?

I am unsure how Levin’s construction of Universal OWFs is related to factoring and discrete log problems. May be you can do a post sometime?

July 19, 2013 8:53 am

QuestionMan:

Well I think P=NP could be true, so yes. But even if not equal one-way is a much harder issue. Good idea on a possible topic.

Thamks

July 19, 2013 12:37 pm

You are not alone. Two of my professors think so and one of my professors said a Turing award winner also said so. It is not difficult to come up with a problem with parameters that when changed give problems in the following hierarchy:

P -> Integer Factor/Discrete Log -> NP complete

Firstly: My feeling is Levin’s Universal OWF should be sitting there after P and before NP complete. If it sits before Integer factor/Discrete Log and these problems are in P the no Universal OWF (someone is going to write a paper on this).
It possibly cannot sit after Integer Factor. Can it? After all OWFs are easier than NP complete. So they should philosophically be as close as possibly to P.

Secondly, for no such problem where one can come up with such a parametrization hierarchy (P -> Integer Factor/Discrete Log -> NP complete) there is a gap between the last two (if there is a gap we have proved P!=NP).

In fact, philosophically and probably technically some groups are not far from a possible outline solution for Integer factor/Discrete log. A modification of the problems I am hinting has parameters which has solution for the Integer factor/Discrete log in P but so far there is no map from those modified problems to Integer factor/Discrete log.

• July 19, 2013 12:58 pm

I believe P != NP, indeed in an exp(\Omega(n/log^k n)) circuit size lower bound for n-variable 3SAT (k = 1 or 2), but side with Dick that factoring is in BPP. I’m agnostic on getting k=0 (for any language in E) which would make BPP=P.

July 24, 2013 5:55 am

Well it is interesting that you would not take sides in the battle of P and BPP.

Razborov shows if we cant say much about OWF we cannot say much about P and NP either. I would be really interested in knowing how far Levin’s UOWF stands next to P and if BPP can contain it (like you are suspecting contains Integer Factoring). Given the equivalence to existence of PRNGs, may be this question can be attempted with known mathematical machinery.

July 24, 2013 6:11 am

Or in other words where exactly does Levin’s UOWF sits in the Ladner hierarchy (if it exists)? Whether in the bottom of bottom close to P or to the top of the top close to NP-complete (I suspect these two boundary conditions as the only reasonable ones where Levin’s UOWF can exist). I would suspect the second case may not be possible simply because again with the close connection with PRNGs where simple polynomial complexity suffices to create randomness. So if the second case is true we may be giving randomness huge power to solve all problems just below NP-complete. This then will wildly go against conventional wisdom that P=BPP.

Just my suspicion. Your opinion welcome.

2. July 19, 2013 7:16 am

Unfortunately Miller Part 2 will not work very often. The clearest way to see this is by considering the order of 2 mod p1 and mod p2. If N*-1=2^s*v and the order of 2 mod p1 is 2^i1*k1 and 2^i2*k2 mod p2 then I think that Miller Part 2 works iff i1 is not equal to i2 and at least one of k1 and k2 divides v. Since k1 and k2 will tend to be large the chance of success will essentially be the same the chance of success of just trying to guess a multiple of p1-1 or p2-1 and using the algorithm you discussed a while ago in A Lemma on Factoring.

3. July 19, 2013 7:36 am

This is with respect to a=2 to be concrete, but the case for general a’s is the same.

• July 19, 2013 1:18 pm

Thanks. As I read this, the import with further regard to our “probing” idea is that in going from N to N* = rN = 2^t w, say, we would be effectively trying to guess r so that w becomes a multiple of the unknown k1 or k2. The nub is whether it is possible to gather any partial information from earlier probes that would help in finding such an r. To answer “Tester” too, we haven’t made a concrete strategy and are most immediately interested in (and grateful for) statements of obstacles. Moreover I found the fascinating Hayman-Shallue–>Kuperberg connection only yesterday while doing a final pass on the post, but several “insane” chess-cheating developments this week and work on the FIDE-ACP commission I’ve been appointed to are taking precedence.


text:
# How to construct a controlled-Hadamard gate using single qubit gates and controlled phase-shift?

How can I construct a controlled-Hadamard gate using single qubit gates and controlled phase-shift?

I am stuck in this and any help would be appreciated.

One strategy is to find a single-qubit rotation $$R_{\hat n}$$ such that

$$R_{\hat n}(\theta) \, P(\phi) \, R_{\hat n}(-\theta) = H\tag1$$

where $$\hat n = (n_x, n_y, n_z)^T$$ is a real 3-vector of unit norm and $$P(\phi) = \mathrm{diag}(1, e^{i\phi})$$ is a phase gate. The motivation behind this approach is the observation that when the phase gate $$P(\phi)$$ is controlled by another qubit the sequence turns into the Hadamard gate when the control is in the $$|1\rangle$$ state and into $$R_1(\theta) R_1(-\theta) = I$$ when it is in the $$|0\rangle$$ state.

Now, we recognize $$(1)$$ as matrix diagonalization which means that $$P(\phi) = Z$$ and $$R_{\hat n}(\theta)$$ is the matrix whose columns are the eigenvectors of the Hadamard.

At this point, we could compute the eigenvectors. However, we notice that both the eigenvalues and the entries of the Hadamard matrix are real numbers. Therefore, the eigenvectors have real entries, too. We recall that among the rotations around $$X$$, $$Y$$ and $$Z$$ axes, one has real entries

$$R_y(\beta) = \begin{pmatrix} \cos\frac{\beta}{2} & -\sin\frac{\beta}{2} \\ \sin\frac{\beta}{2} & \cos\frac{\beta}{2} \end{pmatrix}$$

so we are led to guess that a $$Y$$ rotation might work. We try it out by substituting into $$(1)$$

$$\begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \cos\frac{\theta}{2} & \sin\frac{\theta}{2} \\ -\sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix} \\= \begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix} \begin{pmatrix} \cos\frac{\theta}{2} & \sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & -\cos\frac{\theta}{2} \end{pmatrix} \\= \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}$$

and we see that we obtain the Hadamard if $$\theta = \frac{\pi}{4}$$.

Therefore,

$$R_y\left(\frac{\pi}{4}\right) \circ CZ \circ R_y\left(-\frac{\pi}{4}\right) = CH\tag1$$

where $$CZ$$ and $$CH$$ denote the controlled-$$Z$$ and controlled-Hadamard gates.

• I think you meant to write the second rotation be $R_y(-\pi/4)$ instead of $R_{\hat{n}}(-\pi/4)$ on the last equation. :) – KAJ226 Jan 29 at 2:58
• Indeed! Thanks! Fixed. – Adam Zalcman Jan 29 at 2:59


text:
## introduction

Added later: there is an Edit near the very end; I removed some silly text.

In this post I want to “finish off” syllogisms. No, this is certainly not the last word on syllogisms, and is almost certainly not my last word on them… but when I pick them up next, it will probably be to look at them using Boolean algebra.

What I propose to do in this post is

• prove the four valid syllogisms in figure 1;
• reduce the number of essentially distinct valid syllogisms from 15 to 8;
• reduce that number from 8 to 6;
• prove the two additional valid syllogisms;
• suggest practical guidelines.

Let me begin by summarizing some of the information we will need to have at our fingertips.
Read the rest of this entry »

Posted in logic. Tags: , . 1 Comment »

## Introduction

In this post, I want to introduce us to the formalism of adding or removing quantifiers.

Let me say up front that what I’m about to do amounts to parking my car at a “vista point” along the road, and looking around. If I accomplish anything in this post, it may be to give us some background and context for taking a course in formal symbolic logic, or for working carefully through a text.

As we have seen before, in three or four of my logic posts, there are two quantifiers, $\exists$ and $\forall\$; they are read “there exists ” and “for all”. The first is called “the existential quantifier” and the second is called “the universal quantifier”. It seems that this material was first devised independently by Gentzen and Jaskowski in 1934, so it is surprisngly recent in logic.

What we want to do is to move comfortably between truth tables and tautologies and rules of inference, which do not contain quantifiers… and the kinds of statements which do contain quantifiers, which we find in mathematics and classical logic.

I will be using material I’ve posted about rules of inference and tautologies in one post, “all” and “some” in a second post, and syllogisms in a third prior post.

We have two quantifiers, and two operations to apply to each, so we should expect to find four rules.

Two of these laws remove quantifiers, and two of them add quantifiers. Their precise statements are covered with barbed wire; let’s begin with relatively vague statements. To put that another way, let’s start by being as vague as people usually are when they use these rules.

“The goal is not to make you a mathematical logician; the goal is to make you comfortable enough with quantifiers.” Exner, p. 68.


text:
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$5\left( 7a+11c+1 \right)$
The $GCF$ of the given expression, $35a+55c+5 ,$ is $5,$ since it is the highest number that can evenly divide (no remainder) all the terms. Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 5\left( \dfrac{35a}{5}+\dfrac{55c}{5}+\dfrac{5}{5} \right) \\\\= 5\left( 7a+11c+1 \right) .\end{array}


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How many prime factors does N have?

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How many prime factors does N have?

(1) The least prime factor of N is greater than √N.

(2) N has 2 positive factors.

Originally posted by piyush26 on 29 Oct 2018, 10:21.
Last edited by Bunuel on 29 Oct 2018, 20:47, edited 1 time in total.
Renamed the topic.
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29 Oct 2018, 14:19
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piyush26 wrote:
How many prime factors does the positive integer N have?

(1) The least prime factor of N is greater than √N.

(2) N has exactly 2 positive factors.

Very nice conceptual problem, piyush26 (kudos) !

$$?\,\, = \,\,\,\,\# \,\,{\text{prime}}\,{\text{factors}}\,\,{\text{of}}\,\,N\,\,\,\,\,\,\left( {N \geqslant 1\,\,\operatorname{int} } \right)$$

$$\left( 1 \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,N\,\,{\text{prime}}\,\,\,\,\, \Rightarrow \,\,\,\,? = 1\,\,\,\left( {N\,\,{\text{itself}}} \right)$$

$$\left( * \right)\,\,\left\{ \begin{gathered} N = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{no}}\,\,{\text{prime}}\,\,{\text{factors}} \hfill \\ N \geqslant 4\,\,{\text{not}}\,\,{\text{prime}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,2 \leqslant p\,\,{\text{prime}}\,\,{\text{factor}}\,\,{\text{of}}\,\,N\,\, \leqslant \sqrt N \,\,{\text{exists}} \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{contradicts}}\,\,\left( 1 \right)$$

$$\left( 2 \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,N\,\,{\text{prime}}\,\,\,\,\, \Rightarrow \,\,\,\,? = 1\,\,\,\left( {N\,\,{\text{itself}}} \right)$$

$$\left( {**} \right)\,\,\left\{ \begin{gathered} N = 1\,\,\,\,\, \Rightarrow \,\,\,{\text{only}}\,\,1\,\,{\text{positive}}\,\,{\text{factor}}\,\,\left( {N\,\,{\text{itself}} } \right) \hfill \\ N \geqslant 4\,\,{\text{not}}\,\,{\text{prime}}\,\,\,\,\, \Rightarrow \,\,\,{\text{more}}\,\,{\text{than}}\,\,{\text{the}}\,\,{\text{trivial}}\,\,{\text{two}}\,\,\left( {1\,\,{\text{and}}\,\,N} \right)\,\,{\text{positive}}\,\,{\text{factors}} \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{contradicts}}\,\,\left( 2 \right)\,\,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: How many prime factors does N have?  [#permalink]

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29 Oct 2018, 17:11
Hii
Statement 1: least prime factor of N is greater than √N, hence no there is no factor of N except 1 and the number N inself. It is a prime number. So no of prime factors =1. Sufficient.

Statement 2:
Since N has 2 prime factors, it is a prime number and hence the no of prime factors is 1. Sufficient.

Don't forget to tag me in case of any doubt.

piyush26 wrote:
How many prime factors does N have?

(1) The least prime factor of N is greater than √N.

(2) N has 2 positive factors.

Posted from my mobile device
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29 Oct 2018, 18:48
Why statement 1 is sufficient? is because of 2?
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29 Oct 2018, 19:00
1
we need to check for factors only upto $$\sqrt{n}$$ because the factors repeat after this. See attached file
Attachment:

2018_10_30 08_26 Office Lens.jpg [ 98.8 KiB | Viewed 623 times ]

jorgetomas9 wrote:
Why statement 1 is sufficient? is because of 2?

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Re: How many prime factors does N have?  [#permalink]

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30 Oct 2018, 09:27
piyush26 wrote:
How many prime factors does N have?

(1) The least prime factor of N is greater than √N.

(2) N has 2 positive factors.

Stmt 2 is obviously sufficient

let's look at stmt 1:

In GMAT world we don't have sq rt of -ve numbers ... thus we will consider only whole numbers.

0 and 1 are out of commission because they don't have prime factors.

2 --> least prime factor pf 2 is 2 which is greater than sq rt of 2 i.e. 1.41
3 and other prime numbers --> same as 2
4 --> least prime factor = sq rt of 4 = 2 .... this doesn't satisfy our condition ... thus we can't consider 4

For all even numbers > 4, least prime factor = 2 will be lesser than sq rt of that number thus we can't consider those numbers

9 --> least prime factor = sq rt of 9 = 3 .... this doesn't satisfy our condition ... thus we can't consider 9

Next odd non prime number = 15 --> least prime number = 3 which is lesser than sq rt of 15 .... thus we can't consider odd non prime numbers ....

This leaves us with ONLY PRIME NUMBERS.

This is sufficient to answer our question.

I hope that helps!
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30 Oct 2018, 12:11
piyush26 wrote:
How many prime factors does N have?

(1) The least prime factor of N is greater than √N.

(2) N has 2 positive factors.

Either statement suggests that N is prime.

D
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01 Nov 2018, 21:25
How many prime factors does N have?

(1) The least prime factor of N is greater than √N.

(2) N has 2 positive factors.

1. Imagine, this is possible only when the greatest prime factor is the prime number itself. Hence, N is a prime number and has only 1 prime factor- N itself.

2. This is possible only for a prime number. Only a prime number has 2 factors. Any prime number has one prime factor (the number itself). Hence, the answer is 1. Sufficient.

D.
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Re: How many prime factors does N have?  [#permalink]

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02 Nov 2018, 03:12
Aamirso wrote:
piyush26 wrote:
How many prime factors does N have?

(1) The least prime factor of N is greater than √N.

(2) N has 2 positive factors.

Either statement suggests that N is prime.

D

Hi,

Please try to post meaningful posts. The option D in DS questions means that both the statements are individually sufficient to answer the question.

Best.

Posted from my mobile device
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Re: How many prime factors does N have? &nbs [#permalink] 02 Nov 2018, 03:12
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text:
40m QIL Cryo_Lab CTN SUS_Lab TCS_Lab OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop
 40m Log, Page 337 of 337 Not logged in
ID Date Author Type Category Subject
16950   Mon Jun 27 13:25:50 2022 CiciUpdateGeneralCharacterizing the Transfer Loop

[Deeksha, Cici]

We first took data of a simple low pass filter, and attempted to perform a fit to both the magnitude and phase in order to find the Z of the components. Once we felt confident in our ability to measure tranfer functions, we took data and plotted the transfer function of the existing control loop of the AUX laser. What we found generally followed the trend of, but was lower than, 10^4/f, which is what we hoped to match, and also had a strange unexplained notch ~1.3 kHz. The magnitude and phase data both got worse after around 40-50 kHz, which we believe is because the laser came out of lock near the end of the run.

Edit:

[Attachment 2 and 3] are the frequency response of the low pass filter, curves fitted using least squares in python.

[Attachment 1 and 4] is the same measurement of OLTF of the actual AUX circuit, and the control diagram pointing out the location of excitation and test point.

16951   Mon Jun 27 13:39:40 2022 DeekshaUpdateElectronicsSetting up the MokuLab

[Cici, Deeksha]

On Friday Cici and I set up the Mokulab to take readings of our loop. The aim is to characterise the PZT, in a similar manner as before, by exciting the circuit using our input noise (a swept sine) and recording the corresponding changes in the output. We used the MokuLab to observe the beat note created by the signals of the AUX and PSL, as well as the ASD of the output signal. The MokuLab simplifies the entire process.

Pictured : The beat note as observed by Cici

16952   Mon Jun 27 18:54:27 2022 yutaUpdateLSCModulation depths measurement using Yarm cavity scan

[Yehonathan, Yuta]
EDITED by YM on 22:11 June 27, 2022 to correct for a factor of two in the modulation index

Since we have measured optical gain in MICH to be an order of magnitude less compared with Yehonathan's FINESSE model (40m/16923), we measured the power at AS55 RF PD, and measured the modulation depths using Yarm cavity scan.
We found that 50/50 beam splitter which splits AS55 path into RF PD and RF QPD was not included in the FINESSE model. Measured modulation index were as follows:

TEM00 peak height: 0.6226 +/- 0.0237
RF11 peak height: 0.0067 +/- 0.0007
RF55 peak height: 0.0081 +/- 0.0014
RF11 modulation index: 0.208 +/- 0.012
RF55 modulation index: 0.229 +/- 0.020
RF11 modulation index: 0.104 +/- 0.006
RF55 modulation index: 0.114 +/- 0.010

Here, modulation depth m is defined in E=E_0*exp(i*(w*t+m*sin(w_m*t))), and m m/2 equals to square of the intensity ratio between sidebands and TEM00.

Power measurement at AS55 RF PD:
- ITMY and ITMX single bounce reflection was measured to be 50-60 uW at the front of AS55 RFPD.
- In the FINESSE model, it was expected to be ~110 uW with 0.8 W input to PRM (0.8 W * 5%(PRM) * 50%(BS) * 50%(BS) * 10%(SRM) * 10%(AS2) gives 100 uW)
- In AP table, AS55 beam was split into two paths with 50/50 beam splitter, one for AS55 RF PD and one for AS WFS and AS110. This will be included in the FINESSE model.

Modulation depth measurement using Yarm cavity scan:
- Aligned Yarm using ASS, and unlocked Yarm to get the 2sec scan data of C1:LSC-TRY_OUT_DQ, C1:LSC-POY11_I_ERR_DQ, C1:LSC-AS55_I_ERR_DQ.
- TRY data was used to get TEM00 peak heights
- POY11/AS55 data was used to find RF11/RF55 sideband peaks, and height was measured at TRY (see attached).
- If we define m to be E=E_0*exp(i*(w*t+m*sin(w_m*t))), the amplitude of TEM00 I_00 is proportional to J_0(m) and the amplitude of upper/lower sideband I_f1 is proportional to J_1(m), where J_n(m) is the bessel function of the first kind.
- m can be calculated using 2*sqrt(I_f1 / I_00).
- Results were shown above. Error is calculated from the standard deviation of multiple measurements with multiple peaks,
- The code for doing this lives in https://git.ligo.org/40m/measurements/-/blob/main/LSC/YARM/modulationIndex.ipynb

Discussion:
- Power at AS55 account for the factor of 2, In the FINESSE model, modulation index of 0.3 was used (could be m=0.3/2 or m=0.3; needs check). These combined can explain a factor of 3 at least (or 6).
- Gautam's measurement in Jan 2021 (40m/15769) gives almost double modulation index, but I'm not sure what is the definition Gautam used. It agrees with Gautam's measurement in Jan 2021.

16953   Tue Jun 28 09:03:58 2022 JCUpdateGeneralOrganizing and Cleaning

The plan for the tools in 40m

As of right now, there are 4 tool boxes. X-end, Y-end, Vertex, and the main tool box along the X-arm. The plan is the give each toolbox a set of their own tools. The tools of X-end, Y-end, and Vertex toolboxes will be very similar containing the basic tools such as pliers, screwdrivers, allen ball drivers. Along with this, each tool box will have a tape measure, caliper, level, and other measuring tools we find convinient.

As for the new toolbox, I have done research and found a few good selections. The only problem I have ran into with this is the width of the tool box corresponding with the prices. The tool cabinet we have now is 41" wide. The issue I have is not in finding another toolbox of the same width, but for a similar price we can find a 54" wide tool cabinet. Would anyone be objected to making a bit more space for this?

How the tools will stay organized.

I the original idea I had was to use a specified color of electrical tape for each tool box. Then to wrap the corresponding tools tools with the same color tape. But it was brought to my attention that the electrical tape would become sticky over time. So, I think the using the label maker would be the best idea. with the labels being 'X' for X-end, 'Y' for Y-end, 'V' for vertex, and 'M' for main toolboxes.

An idea for the optical tables:

Anchal brought it up to me that it is a hassle to go back and forth searching for the correct sizes of Hex Keys and Allen Wrenches. The idea of a pouch on the outside of each optical table was mentioned so I brought this up to Paco. Paco also gave me the idea of a 3D printed stand we could make for allen ball drives. Does anyone have a preference or an idea of what would be the best choice and why?

A few sidenotes:

Anchal mentioned to me a while back that there are many cables that are laying on the racks that are not being used. Is there a way we could identify which ones are being used?

I noticed that when we were vented that a few of the chamber doors were leaning up against the wall and not on a wooden stand like others. Although, the seats for the chamber doors are pretty spacious and do not give us much clearance. For the future ones, could we make something more sleek and put the wider seats at the end chambers?

The cabinets along the Y-Arm are labelled, but do not correspond with all the materials inside or are too full to take in more items. Could I organize these?

16954   Tue Jun 28 14:24:23 2022 yutaUpdateBHDBHD DC PD signals now also sent to c1lsc to circumvent IPC error

[JC, Yuta]

To circumvent IPC error sending BHD DC PD signals from c1sus2 to c1lsc, DB9 cable from BHD DC PD box sent to c1sus2 is now split and sent also to c1lsc.
They are now available in both

C1:X07-MADC1_EPICS_CH16 (DC PD A) and CH17 (DC PD B)

C1:X04-MADC1_EPICS_CH4 (DC PD A) and CH5 (DC PD B)

Next:
- Add battery powered SR560 to decouple c1sus2 and c1lsc to avoid the ground loop

16955   Tue Jun 28 16:26:58 2022 CiciSummaryGeneralVector fitting open loop transfer function/Audio cancellation of optical table enclosure

[Deeksha, Cici]

We attempted to use vectfit to fit our earlier transfer function data, and were generally unsuccessful (see vectfit_firstattempt.png), but are much closer to understanding vectfit than before. Couple of problems to address - finding the right set of initial poles to start with has been very hard, and also however vectfit is plotting the phase data is unwrapping it, which makes it generally unreadable. Still working on how to mess with the vectfit automatically-generated plots. In general, our data is very messy (this is old data of the transfer function from last week), so we took more data today to see if our coherence was the problem (see TFSR785_28-06-2022_161937.pdf). As is visible from the graph, our coherence is terrible, and above 1kHz is almost entirely below 0.5 (or 0.2) on both channels. Figuring out why this is and fixing it is our first priority.

In the process of taking new data, we also found out that the optical table enclosure at the end of the X-arm does a decent job of sound isolation (see enclosure_open.mp4 and enclosure_closed.mp4). The clicking from the shutter is visible on a spectrogram at high frequencies when the enclosure is open, but not when it is closed. We also discovered that the script to toggle the shutter can run indefinitely, which can break the shutter, so we need to fix that problem!

16956   Tue Jun 28 16:59:35 2022 PacoSummaryALSALS beat allan deviation (XARM)

[Paco]

I took ~ 7 minutes of XALS beatnote data with the XAUX laser locked to the XARM cavity, and the XARM locked to PSL to develop an allan deviation estimator. The resulting timeseries for the channel C1:ALS-BEATX_FINE_PHASE_OUT_HZ_DQ (decimated timeseries in Attachment #1) was turned into an allan variance using the "overlapped variable tau estimator":

$\sigma_y^2(n\tau_0, N) = \frac{1}{2n^2\tau_0^2(N - 2n)} \sum_{i=0}^{N-2n-1} (x_{i+2n} - 2x_{i+n} + x_i)^2$

Where $x_k$ represents the k-th data point in the raw timeseries, and $n\tau_0$ are the variable integration intervals under which two point variances are computed (the allan variance is a special case of M-point variance, where M=2). Then, the allan deviation is just the square root of that. Attachment #2 shows the fractional deviation (normalized by the mean beat frequency ~ 3 MHz for this measurement) for 100 integration times spanning the full duration (~ 7 min = 420 s).

The code used for this lives in Git/40m/labutils/measuremens/ALS/

If this estimate is any good, wherever the fractional beatnote deviation reaches a minimum value can be used as a proxy for the longest averaging time that give a statistical increase in SNR. After this timescale, the frequency comparison is usually taken over by "environmental instabilities" which I don't think I can comment further on. In our particular estimate, the 100 second integration gives a fractional deviation of ~ 0.44 %, or absolute deviation of 12.925 kHz.

16957   Tue Jun 28 17:07:47 2022 AnchalUpdateCalibrationAdded Beatnote channels in demodulation of c1cal

I added today demodulation of C1:LSC-BEATX/Y_FINE_I/Q in the c1cal demodulation where different degrees of freedom can be dithered. For McCal (formerly soCal), we'll dither the arm cavity for which we can use any of the DOFs (like DARM) to send the dither to ETMX/ETMY. Then with green laser locked as well, we'll get the calibration signal from the beatnotes in the demodulaed channels. We can also read right after the mixing in c1cal model and try differnt poles for integration .

I've also added medm screens in the sensing matrix part of LSC screen. These let you see demodulation of beatnote frequency signals.

16958   Tue Jun 28 18:19:09 2022 TomislavUpdateElectronicsElectronics noise

I measured electronics noise of WFSs and QPD (of the WFS/QPD, whitening, ADC...) by closing PSL and measuring the error signal. It was needed to put the offset in C1:IOO-MC_TRANS_SUMFILT_OFFSET to 14000 cts (without offset the sum of quadrants would give zero, and 14000 cts is the value when the cavity is locked). For WFS that are RF, if there is intensity noise at low frequencies, it is not affecting the measurement.

In the attachment please find the power spectrum of the error signal when the PSL shutter is on and off.

16959   Tue Jun 28 18:53:16 2022 ranaSummaryALSALS beat allan deviation (XARM)

what's the reasoning behind using df/f_beat instead of df/f_laser ?

 Quote: [Paco] I took ~ 7 minutes of XALS beatnote data with the XAUX laser locked to the XARM cavity, and the XARM locked to PSL to develop an allan deviation estimator.

16960   Tue Jun 28 22:27:11 2022 YehonathanUpdateBHDMeasurement of input and output electronics noise

{Yuta, Yehonathan}

For MICH noise budgeting we measure the input electronics noise which includes the AS55 RFPD, preamp, demod board, the whitening, and the AA filters, and the ADC noises. To do so we simply close the laser shutter and take the spectrum of C1:LSC-AS55_I_ERR_DQ and C1:LSC-AS55_Q_ERR_DQ shown in attachment 1.

Next, we measured the output electronics noise which includes the DAC, dewhitening and AI filters, and coil driver noises. We disabled the BS watchdog and went to 1X4 rack. We measured the spectrum of one of the lemo outputs on the BS coil driver module using an SR785. Attachment 2 shows the spectrum together with the SR785 dark noise.

16961   Tue Jun 28 23:10:34 2022 YehonathanUpdateBHDMeasurement of AS55 demod board conversion factor

{Yuta, Yehonathan}

We measured the AS55 demod board conversion from the amplitude of a 55MHz signal to a demodulated signal. We hooked the unused REFL55 LO into the PD input port on the AS55 demod board.

The REFL55 LO was measured to be 1.84 Vpp. The IQ outputs were: I = 0.86 Vpp, Q = 2.03 Vpp giving an amplitude of 2.205 Vpp. The overall conversion factor is sqrt(0.86**2+2.03**2)/(1.82/2)=2.422.

We also set to measure the loss in the RF cable from AS55 PD to the demod board on 1Y2. REFL55 was connected with a long BNC cable to the input of the cable under test. REFL55 at the input was measured to be 1.466 Vpp and 1.28 Vpp at the output signifying a transmission of 87.6%.

16962   Wed Jun 29 14:28:06 2022 PacoSummaryALSALS beat allan deviation (XARM)

I guess it didn't make sense since f_beat can be arbitrarily moved, but the beat is taken around the PSL freq ~ 281.73 THz. Attachment #1 shows the overlapping tau allan deviation for the exact same dataset but using the python package allantools, where this time I used the PSL freq as the base frequency. This time, I can see the minimum fractional deviation of 1.33e-13 happening at ~ 20 seconds.

 Quote: what's the reasoning behind using df/f_beat instead of df/f_laser ?

## Another, more familiar interpretation

The allan variance is related to the beatnote spectral density as a mean-square integral (the deviation is then like the rms) with a sinc window.

$\sigma^2_\nu = 2 \int_0^{\infty} S_\nu(f) \lvert \frac{\sin({\pi f \tau})}{\pi f \tau} \lvert ^2 df$

16963   Wed Jun 29 18:53:38 2022 ranaUpdateElectronicsElectronics noise

this is just the CDS error signal, but is not the electronics noise. You have to go into the lab and measure the noise at several points. It can't be done from the control room. You must measure before and afte the whitening.

 Quote: I measured electronics noise of WFSs and QPD (of the WFS/QPD, whitening, ADC...) by closing PSL and measuring the error signal. It was needed to put the offset in C1:IOO-MC_TRANS_SUMFILT_OFFSET to 14000 cts (without offset the sum of quadrants would give zero, and 14000 cts is the value when the cavity is locked). For WFS that are RF, if there is intensity noise at low frequencies, it is not affecting the measurement. In the attachment please find the power spectrum of the error signal when the PSL shutter is on and off.

16964   Thu Jun 30 17:19:55 2022 DeekshaSummaryElectronicsMeasured Transfer Functions of the Control Loop, Servo (OLTF); got Vectfit working

[Cici, Deeksha]

We were able to greatly improve the quality of our readings by changing the parameters in the config file (particularly increasing the integration and settle cycles, as well as gradually increasing our excitation signals' amplitude). Attached are the readings taken from the same (the files directly printed by ssh'ing the SR785 (apologies)) - Attachment 1 depicts the graph w/ 30 data points and attachment 2 depicts the graph with 300 data points.

Cici successfully vectfit to the data, as included in Attachment 3. (This is the vectfit of the entire control loop's OLTF). There are two main concerns that need to be looked into, firstly, the manner in which to get the poles and zeros to input into the vectfit program. Similarly, the program works best when the option to enforce stable poles is disabled, once again it may be worth looking into how the program works on a deeper level in order to understand how to proceed.

Just as the servo's individual transfer function was taken, we also came up with a  plan to measure the PZT's individual transfer function (using the MokuLab). The connections for the same have been made and the Moku is at the Xend (disconnected). We may also have to build a highpass filter (similar to the one whose signal enters the PZT) to facilitate taking readings at high frequencies using the Moku.

16965   Thu Jun 30 18:06:22 2022 PacoUpdateALSOptimum ALS recovery - part I

[Paco]

In the morning I took some time to align the AUX beams in the XEND table. Later in the afternoon, I did the same on the YEND table. I then locked the AUX beams to the arm cavities while they were stabilized using POX/POY and turned off the PSL hepa off temporarily (this should be turned on after today's work).

After checking the the temperature slider sign on the spectrum analyzer of the control room I took some out-of-loop measurements of both ALS beatnotes (Attachment #1) by running diaggui /users/Templates/ALS/ALS_outOfLoop_Ref_DQ.xml and by comparing them against their old references (red vs magenta and blue vs cyan); it seems that YAUX is not doing too bad, but XAUX has increased residual noise around and above 100 Hz; perhaps as a result of the ongoing ALS SURF loop investigations? It does look like the OLTF UGF has dropped by half from ~ 11 kHz to ~ 5.5 kHz.

Anyways let this be a reference measurement for current locking tasks, as well as for ongoing SURF projects.

16966   Thu Jun 30 19:04:55 2022 ranaSummaryPSLPSL HEPA: How what when why

For the PSL HEPA, we wanted it to remain at full speed during the vent, when anyone is working on the PSL, or when there is a lot of dust due to outside conditions or cleaning in the lab.

For NORMAL conditions, the policy is to turn it to 30% for some flow, but low noise.

I think we ought to lock one of the arms on IR PDH and change the HEPA flow settings and plot the arm error signal, and transmitted power for each flow speed to see what's important. Record the times of each setting so that we can make a specgram later

16967   Thu Jun 30 19:24:24 2022 ranaSummaryPEMeffect of nearby CES construction

For the proposed construction in the NW corner of the CES building (near the 40m BS chamber), they did a simulated construction activity on Wednesday from 12-1.

In the attached image, you can see the effect as seen in our seismometers:

this image is calculated by the 40m summary pages codes that Tega has been shepherding back to life, luckily just in time for this test.

Since our local time PDT = UTC - 7 hours, 1900 UTC = noon local. So most of the disturbance happens from 1130-1200, presumably while they are setting up the heavy equipment. If you look in the summary pages for that day, you can also see the IM lost lock. Unclear if this was due to their work or if it was coincidence. Thoughts?

ELOG V3.1.3-


text:
### Carla Froehlich

University of Basel, Switzerland

### Core Collapse Supernovae: New Aspects of Nucleosynthesis and their Imprint on Galactic Evolution

In 1D, a self-consistent treatment of core collapse supernovae with presently known input physics and computer simulations does not yet lead to successful explosions, while 2D models show promise. On the other hand there is a need to provide correct nucleosynthesis abundances for the quickly evolving field of galactic evolution and observations of low metallicity stars. The innermost ejecta are directly affected by the explosion mechanism, most strongly the yields of Fe-group nuclei for which an induced piston or thermal bomb treatment will not provide the correct yields because the neutrino interactions are not included. I will show how parametrized variations in the neutrino reaction cross sections can mimic effects of a multi-D convection treatment and cause explosions with an electron fraction Y_e>0.5 in the innermost zones. Nucleosynthesis predictions from these models show improved Fe group abundances and enhanced abundances of 45Sc, 49Ti, and 64Zn. In addition, I will present a new nucleosynthesis process, the $\nu p$-process. This process occurs in proton-rich ejecta which are subject to large neutrino fluxes (e.g. in core collapse supernovae and possibly in gamma-ray bursts). This process allows for the nucleosynthesis of nuclei with mass number $A>64$, making it a possible candidate for the origin of light p-nuclei abundances and to explain the large Sr abundance observed in hyper-metal-poor stars. Finally, I will focus on the metallicity dependence of the ejected element abundances of core collapse supernovae. The observational features in low metallicity stars will be summarized, the complementarity of type Ia and type II supernovae addressed, and a special emphasis will be put on the explanation of specific abundance features at very low metallicities.

Back to the theory seminar page.


text:
# What value of the constant $\alpha$ makes $Y$ and $Z$ uncorrelated?

The random variables $$X_1$$ and $$X_2$$ have means 1 and 2 respectively, with same variance 4 and covariance 0.8. Let $$Y = X_1 − 3X_2$$ and $$Z = \alpha X_1 + X_2$$. What value of the constant α makes Y and Z uncorrelated?

It is given that covariance of $$X_1$$ and $$X_2$$ is $$0.8$$ Hence $$EX_1X_2-(EX_1) (EX_2)=0.8$$. So $$EX_1X_2 =0.8+(1)(2)=2.8$$. Now can you proceed?

• Now, how can I relate $E[X_1X_2]$ with $E[YZ]$? – Mark Jacon Jan 24 at 13:36
• $EYZ=E(X_1-3X_2)(\alpha X_1+X_2)=\alpha EX_1^{2}-3\alpha EX_1X_2+EX_1X_2-3EX_2^{2}$ – Kavi Rama Murthy Jan 24 at 23:13
• Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :) – Mark Jacon Jan 25 at 8:17
• $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$. – Kavi Rama Murthy Jan 25 at 8:19

Solve:

$$E[X_1]=1$$, $$E[X_2]=2$$, $$\operatorname{Var}(X_1)=4$$, $$\operatorname{Var}(X_2)=4$$ and $$\operatorname{Cov}(X_1,X_2)=0.8$$

$$Y, Z$$ are uncorrelated if $$\operatorname{Cov}(Y,Z)=0$$ where $$\operatorname{Cov}(Y,Z) = E[YZ] − E[Y]E[Z]$$.

$$E[Y]=E[X_1]-3E[X_2]=1-3*2=-5$$

$$E[Z]=\alpha E[X_1]+E[X_2]=\alpha+2$$

$$Cov(X_1,X_2)=E[X_1X_2]-E[X_1]*E[X_2]=0.8$$ so $$E[X_1X_2]=0.8+2*1=2.8$$

$$E[X_1^2]=Var(X_1)+(E[X_1])^2=4+1^2=5$$ and $$E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$$

$$E[YZ]=E[(X_1 − 3X_2)(\alpha X_1 + X_2)]=\alpha E[X_1^2]+(1-3 \alpha)E[X_1X_2]-3E[X_2^2]=5\alpha+(1-3\alpha)*2.8-3*8$$

$$Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$$, $$5\alpha+(1-3\alpha)*2.8-3*8-(-5)*(2+\alpha)=0$$, $$\alpha=7$$


text:
Search by Topic

Resources tagged with Area similar to Griddy Region:

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There are 83 results

Broad Topics > Measures and Mensuration > Area

Same Height

Stage: 4 Challenge Level:

A trapezium is divided into four triangles by its diagonals. Suppose the two triangles containing the parallel sides have areas a and b, what is the area of the trapezium?

Dividing the Field

Stage: 4 Challenge Level:

A farmer has a field which is the shape of a trapezium as illustrated below. To increase his profits he wishes to grow two different crops. To do this he would like to divide the field into two. . . .

Bound to Be

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Four quadrants are drawn centred at the vertices of a square . Find the area of the central region bounded by the four arcs.

Percentage Unchanged

Stage: 4 Challenge Level:

If the base of a rectangle is increased by 10% and the area is unchanged, by what percentage (exactly) is the width decreased by ?

Diagonals for Area

Stage: 4 Challenge Level:

Can you prove this formula for finding the area of a quadrilateral from its diagonals?

Six Discs

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Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?

Gutter

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Manufacturers need to minimise the amount of material used to make their product. What is the best cross-section for a gutter?

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Investigate the properties of quadrilaterals which can be drawn with a circle just touching each side and another circle just touching each vertex.

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Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .

Two Shapes & Printer Ink

Stage: 4 Challenge Level:

If I print this page which shape will require the more yellow ink?

Uniform Units

Stage: 4 Challenge Level:

Can you choose your units so that a cube has the same numerical value for it volume, surface area and total edge length?

From All Corners

Stage: 4 Challenge Level:

Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.

Semi-detached

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A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.

Of All the Areas

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Can you find a general rule for finding the areas of equilateral triangles drawn on an isometric grid?

Max Box

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Three rods of different lengths form three sides of an enclosure with right angles between them. What arrangement maximises the area

Squ-areas

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Three squares are drawn on the sides of a triangle ABC. Their areas are respectively 18 000, 20 000 and 26 000 square centimetres. If the outer vertices of the squares are joined, three more. . . .

Equilateral Areas

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ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.

Inscribed in a Circle

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The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?

Two Circles

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Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap?

Crescents and Triangles

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Triangle ABC is right angled at A and semi circles are drawn on all three sides producing two 'crescents'. Show that the sum of the areas of the two crescents equals the area of triangle ABC.

Stage: 4 Challenge Level:

In this problem we are faced with an apparently easy area problem, but it has gone horribly wrong! What happened?

Square Pizza

Stage: 4 Challenge Level:

Can you show that you can share a square pizza equally between two people by cutting it four times using vertical, horizontal and diagonal cuts through any point inside the square?

Semi-square

Stage: 4 Challenge Level:

What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?

Perimeter Possibilities

Stage: 3 Challenge Level:

I'm thinking of a rectangle with an area of 24. What could its perimeter be?

Warmsnug Double Glazing

Stage: 3 Challenge Level:

How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?

Can They Be Equal?

Stage: 3 Challenge Level:

Can you find rectangles where the value of the area is the same as the value of the perimeter?

Salinon

Stage: 4 Challenge Level:

This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?

Towering Trapeziums

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Can you find the areas of the trapezia in this sequence?

Changing Areas, Changing Perimeters

Stage: 3 Challenge Level:

How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?

Compare Areas

Stage: 4 Challenge Level:

Which has the greatest area, a circle or a square inscribed in an isosceles, right angle triangle?

Trapezium Four

Stage: 4 Challenge Level:

The diagonals of a trapezium divide it into four parts. Can you create a trapezium where three of those parts are equal in area?

Pick's Theorem

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Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.

Curvy Areas

Stage: 4 Challenge Level:

Have a go at creating these images based on circles. What do you notice about the areas of the different sections?

Growing Rectangles

Stage: 3 Challenge Level:

What happens to the area and volume of 2D and 3D shapes when you enlarge them?

Areas of Parallelograms

Stage: 4 Challenge Level:

Can you find the area of a parallelogram defined by two vectors?

Rati-o

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Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?

Lying and Cheating

Stage: 3 Challenge Level:

Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!

Rhombus in Rectangle

Stage: 4 Challenge Level:

Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.

Isosceles Triangles

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Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?

Making Rectangles

Stage: 2 and 3 Challenge Level:

A task which depends on members of the group noticing the needs of others and responding.

Maths Filler

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Imagine different shaped vessels being filled. Can you work out what the graphs of the water level should look like?

Tiling Into Slanted Rectangles

Stage: 2 and 3 Challenge Level:

A follow-up activity to Tiles in the Garden.

Extending Great Squares

Stage: 2 and 3 Challenge Level:

Explore one of these five pictures.

Biology Measurement Challenge

Stage: 4 Challenge Level:

Analyse these beautiful biological images and attempt to rank them in size order.

Stage: 3 Challenge Level:

Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?

Maths Filler 2

Stage: 4 Challenge Level:

Can you draw the height-time chart as this complicated vessel fills with water?

Hallway Borders

Stage: 3 Challenge Level:

A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.

Tilted Squares

Stage: 3 Challenge Level:

It's easy to work out the areas of most squares that we meet, but what if they were tilted?

Cylinder Cutting

Stage: 2 and 3 Challenge Level:

An activity for high-attaining learners which involves making a new cylinder from a cardboard tube.

Fence It

Stage: 3 Challenge Level:

If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?


text:
List of Local to Global principles

What are some of the local to global principles in different areas of mathematics?

-
community wiki? –  lhf Apr 20 '11 at 11:55
am I supposed to do that? –  quanta Apr 20 '11 at 11:56
when posting a big-list question, please flag it for moderator attention with a request to make it community wiki. @lhf, J.M: thanks for the ping. –  Willie Wong Apr 20 '11 at 12:56

Differential Geometry

The existence of partitions of unity allows one to transfer local results to global ones.

The Gauss-Bonnet Theorem relates the Gaussian curvature (a local quantity) to the Euler characteristic (a global one).

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Diophantine Equations

Hasse Condition: If a Diophantine equation is solvable modulo every prime power (locally) as well as in the reals then it is solvable in the integers.

Hasse Principle is that the Hasse Condition holds for all quadratic Diophantine equations.

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See also the Grunwald–Wang theorem and this question –  lhf Apr 20 '11 at 12:05

Graph Theory

A graph has an Eulerian circuit (global) iff every node has even degree (local).

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Don't you also need that the graph is connected? –  JSchlather May 14 '11 at 20:33

Number Theory

The "original" (in terms of giving rise to the name) local-global principle for quadratic forms over number fields, due to Hasse, has already been mentioned. Here are two further local-global principles in which Hasse was involved.

1. Two (finite-dimensional) central simple algebras over a number field $K$ are isomorphic if and only if their base extensions to central simple algebras over $K_v$ are isomorphic for every completion $K_v$ of $K$. This is essentially the Albert-Brauer-Hasse-Noether theorem.

2. Class field theory can be formulated both for number fields and for their completions, called respectively global class field theory and local class field theory. (It can formulated also for function fields over finite fields, but let's not worry about that here.) Historically global class field theory came first and the proofs of local class field theory originally depended on global class field theory. Eventually Hasse was able to develop local class field theory in a self-contained way and then use it to prove global class field theory.

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Complex Analysis

Analytic continuation might be viewed as a local-to-global principle.

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Real analysis

The extreme value theorem can be read as: if a function on a compact set is locally bounded then it's globally bounded. (This is only part of the theorem.) The key word here of course is compact.

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I just wanted to add that compactness is a very important property precisely because it is often used (in analysis) to extend local properties to global properties. The extreme value theorem is one example among many. –  svenkatr Apr 20 '11 at 18:25
Another example of this type comes from Differential Geometry. For a smooth vector field on a smooth manifold we can define flows locally. On a compact manifold, we can define a flow globally. –  sjm.majewski Feb 2 at 19:23

Euler Characteristic

If you count the faces vertices and edges (local information) of a polygonal shape then you can compute the Euler Characteristic which tell you how many holes the shape has (global information).

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Number Theory

If an integer $n \not \equiv 0 \pmod m$ (for any $m$) then $n \not = 0$.

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Complex Analysis

If a $\mathbb C \rightarrow \mathbb C$ function can be differentiated (local) then it can be integrated! (Inspired by Qiaochus answer here).

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Number Theory

Zeta L-functions and Birch-Swinnerton Dyer conjecture: These are quantifies which are defined in terms of local things (like multiplying together a function on primes) and global information (like class numbers) is extracted from them.

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Set Theory

The axiom of choice is local, while Global choice is global.

(Global choice, or Axiom E in the NGB set theory, is equivalent to saying that all proper classes are of the same "cardinality", or there is a well-ordering of the universe)

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$C^*$-Algebras

Let $\mathfrak{A}$ be a unital $C^*$-algebra, $\mathfrak{C}$ a $C^*$-subalgebra of the center of $\mathfrak{A}$ which contains the unit of $\mathfrak{A}$ and for any maximal ideal $x$ of $\mathfrak{C}$ let $I_x$ be the smallest closed two-sided ideal of $\mathfrak{A}$ containing $x$.

Now the local principle by Allan and Douglas says, that $a\in\mathfrak{A}$ is invertible in $\mathfrak{A}$ if and only if $a+I_x$ is invertible in the quotient algebra $\mathfrak{A}/I_x$ for every maximal ideal $x$ of $\mathfrak{C}$.

One can use this result to tackle questions of invertibility and Fredholmness in operator algebras, e.g. one can characterize Fredholmness properties of Toeplitz operators with piecewise continuous symbols by means of easily checked properties of their respective symbols.

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Analysis

$$\prod_{p}|x|_p = |x|^{-1}$$ gives a way to piece together all local norms to find the value of a global norm.

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I don't agree with this. The standard absolute value |x| is also a local norm (it's a norm on a completion of Q), so the standard way to express this equation is to move the right side to the left side and say that the product of all normalized absolute values of any nonzero rational number is 1. –  KCd Sep 10 '11 at 19:25

And i was studying completely about Hasse's local-Global principle,which has many applications like ,there is a separate fantastic group called the Tate-Shafarevich Group that measures the extent of Failure of Hasse's local-global Principle.it is given by $Ш(E/K)=\mathrm{Ker}(H^1(K,E)\mapsto \prod_{v}H^1(K_v,E))$ which means literally ,"the non-trivial elements of the Tate-Shafarevich group can be thought of as the homogeneous spaces of $A$(where $A$ is an Abelian Variety defined over $K$) that have $K_v$-rational points for every place $v$ of $K$, but no $K$-rational point." and the TS group has many important applications in BSD conjecture,Iwasawa theory....

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If $a$ is a square (or $n$th power) modulo every prime power then, since its $p$-adic valuation is even (or a multiple of $n$) for every $p$, $a$ is a square (or $n$th power).
Not really true for $n>2$. See the Grunwald–Wang theorem mentioned in another comment of mine. –  lhf May 6 '11 at 16:59
16 is not an 8-th power but it is mod $p$ for all primes $p$. See math.stackexchange.com/questions/6758/… –  lhf May 6 '11 at 20:22


text:
• # question_answer     A uniform rod of length$l$is free to rotate in a vertical plane about a fixed horizontal axis through point B. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle$\theta$its angular velocity co is given as A)  $\sqrt{\frac{6g}{l}}\sin \theta$                              B)  $\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}$ C)  $\sqrt{\frac{6g}{l}}\cos \frac{\theta }{2}$                          D)  $\sqrt{\frac{6g}{l}}\cos \theta$

The fall of centre of mass $h=\frac{l}{2}(1-\cos \theta )$ $\therefore$Decrease in potential energy$=mgh$ $=mg\frac{l}{2}(1-cos\theta )$ From law of conservation of energy, KE$o$ rotation = decrease in PE $\frac{1}{2}I{{\omega }^{2}}=\frac{mgl}{2}(1-\cos \theta )$ $\frac{1}{2}\frac{m{{l}^{2}}}{3}{{\omega }^{2}}=\frac{mgl}{2}(1-\cos \theta )$ ${{\omega }^{2}}=\frac{6g}{2l}(1-\cos \theta )$ $=\frac{6g}{2l}2{{\sin }^{2}}\frac{\theta }{2}$ $\omega =\sqrt{\frac{6g}{l}{{\sin }^{2}}\frac{\theta }{2}}$ $=\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}$


text:
## How can we sum up sin and cos series when the angles are in arithmetic progression?

How can we sum up sin and cos series when the angles are in A.P (arithmetic progression) ?For example here is the sum of cos series:

k=0n1cos(a+kd)=sin(n×d2)sin(d2)×cos(2a+(n1)d2)
There is a slight difference in case of sin ,which is:
k=0n1sin(a+kd)=sin(n×d2)sin(d2)×sin(2a+(n1)d2)
How do we prove the above two identities?

Let
S=sin(a)+sin(a+d)++sin(a+nd)
Now multiply both sides by sind2. Then you have
S×sin(d2)=sin(a)sin(d2)+sin(a+d)sin(d2)++sin(a+nd)sin(d2)
Now, note that
and
sin(a+d)sin(d2)=12[cos(a+dd2)cos(a+d+d2)]
Then by doing the same thing you will have some terms cancelled out. You can easily see which terms are going to get Cancelled. Proceed and you should be able to get the formula.
I tried this by seeing this post. This has been worked for the case when d=1. Just take a look here:
• http://mathforum.org/library/drmath/view/72777.html


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# Tutor profile: Jack G.

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Jack G.
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## Questions

### Subject:Applied Mathematics

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Question:

State the Navier-Stokes equations for incompressible viscous fluid flow, defining the variables and constants used.

Inactive
Jack G.

$$\rho \frac{D\vec u}{Dt}= \rho \vec F - \vec \nabla p + \mu \nabla^2 \vec u$$ $$\vec \nabla \cdot \vec u=0$$ Here, $$\rho$$ is the density of the fluid, $$\vec u$$ is the velocity field of the flow, $$t$$ is time, $$\vec F$$ is the resultant force field on the fluid, $$p$$ is the scalar pressure field of the fluid, and $$\mu$$ is the fluid's viscosity.

### Subject:Environmental Science

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Question:

What is Ocean Acidification? Describe the cause, effects, and likely projected damage in the 21st century.

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Jack G.

The Ocean absorbs 90% of excess atmospheric heat and around 30% of the carbon dioxide emitted into the atmosphere. When carbon dioxide is dissolved in the ocean, it reacts with water to create carbonic acid, as follows:  $$CO_2 + H_2O ⇌ H_2CO_3$$ This lowers the pH of our oceans, making them more acidic, with potentially catastrophic effects on marine ecosystems. Through the absorption of both excess heat and carbon dioxide, increasing the temperature and decreasing the pH, coral reefs around the tropics are bleaching at an alarming rate. The Great Barrier Reef, for example, experienced back-to-back extreme bleaching events for the first time in history in 2016 and 2017. This loss of essential high-biodiversity ecosystems is felt in the health of our oceans worldwide. Although ocean acidification is still green, if we follow RCP8.5, we could see consecutive extreme bleaching events every year by 2100.

### Subject:Physics

TutorMe
Question:

A projectile is launched from ground-level, at an angle $$\theta$$ to the horizontal. For what value of $$\theta$$ is the maximum horizontal range of the projectile reached? Note, $$0<\theta<\frac{\pi}{2}$$ (in radians).

Inactive
Jack G.

Using $$suvat$$ we split these quantities into vertical and horizontal components. Vertical: $$s=0, u=usin\theta, v=-usin\theta, a=-g, t=t$$ (1) Horizontal: $$s=s, u=ucos\theta, v=ucos\theta, a=0, t=t$$ (2) Using: $$v=u+at$$ (1): $$2usin\theta=-gt$$ Using: $$s=ut+\frac{1}{2}at^2$$ (2): $$s=ucos\theta t$$ We can now sub the value of t from (1) into (2): $$s=ucos\theta \times \frac{-2usin\theta}{g}$$ Hence $$s=\frac{u^2sin(2\theta)}{g}$$ (3) Now we have found the horizontal displacement, we must find the maximum by differentiating (3) w.r.t $$\theta$$, setting equal to zero, and solving for $$\theta$$: $$\frac{ds}{d\theta}=\frac{2u^2cos(2\theta)}{g}=0$$ Hence: $$cos(2\theta)=0$$ Thus $$2\theta=\frac{\pi}{2}+2\pi k$$, for $$k = 1,2,3.....$$ Which gives $$\theta=\frac{\pi}{4}$$, given the constraints on $$\theta$$ provided. Hence the angle of projection that achieves the maximum horizontal distance is $$\frac{\pi}{4}$$ radians, or 45 degrees to the horizontal.

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## “Bulletin Board”

School of Particles and Accelerator - December 19, 2007

### Determination of the Strong Coupling Constant $\alpha_s(M_z)$ from DIS Experimental Data

#### Dr. Ali N. Khorramian from IPM, Semnan University

 Abstract: Various QCD analysis of the world unpolarized and polarized deep-inelastic data on charged lepton?nucleon and neutrino?nucleus scattering were performed in order to measure the QCD scale ΛQCD and αs(Mz), from the nucleon structure func-tions. In this talk I will give a brief overview on the status of these analysis. Also I will present our results for the values of ΛQCD and αs(Mz) inbased on the polynomials expansion method. Time: December 19, 2007