text:
# How can you use and convert Gauss-Krüger-Coordinates in Mathematica?

I am given geographic data in the form of Gauss-Krüger-Coordinates and would like to calculate distances with them and convert them to longitude/latitude coordinates in another system for plotting (e.g. WGS84).

Gauss-Krüger-Coordinates are essentially like UTM-coordinates based upon a transversal Mercator projection where positions are indicated by a Right-value (East-value in UTM; y-coordinate in the geodetic coordinate system) and a High-value (North-value in UTM; x-coordinate int the geodetic coordinate system)

Far from being an expert in geodesy the way I understand it Mathematica's geodetic functions allow entering coordinates as GeoPosition, as GeoPositionENU or as GeoGridPosition. GeoGridPosition essentially will reference to a position on a projection so probably should be the way the Gauss-Krüger-Coordinates are entered, but I am not so sure. Mathematica does not seem to know Gauss-Krüger-Projection (not listed among GeoProjectionData[]) and I do not know which parameters to use for Gauss-Krüger and how best to do this. Thus:

How can I enter the Gauss-Krüger-Coordinates into Mathematica so they can be used for distance-calculations and conversions using Mathematica's geodetic functions?

-
 For more information on coordinate reference systems and conversions one might turn to epsg.org where guidance on conversions and formulas is given under epsg.org/guides/G7-2.html. – gwr May 22 '12 at 10:39

With a little help from other sources and the links given as a comment to my question I came up with a solution. Essentially Gauss-Krüger is a variant of the Transversal Mercator projection so this projection can be used and made to fit the Gauss-Krüger specialties:

gaussKruegerPosition[{right_Integer,high_Integer},centralMeridian_Integer,
falseEasting_Integer]:=

GeoGridPosition[
{right-falseEasting-centralMeridian/3*10^6,high,0},
{"TransverseMercator","Centering"-> {0,centralMeridian},"ReferenceModel"->"Bessel1841"}
];

This short function will convert Gauss-Krüger-Positions into a valid Mathematica position which can then be converted using GeoPosition or referenced for calculation of distances using GeoDistance[pos1, pos2].

Here in Germany the central Meridians will be spaced 3° apart and one can recognize this in the first digit of the Right-value of the coordinates which has to be multiplied by 3 to find the central Meridian.

In order to avoid negative Right-values a False Easting is usually given; in Germany it is 500 000.

-
 As a side note: For some reason Mathematica seems not to be able to convert the gaussKruegerPosition into a WGS84-Position. GeoPosition[ gaussKruegerPosition[{3513495, 5405227}, 9, 500000], "WGS84"] returns the error "GeoPosition::invdtm:Unknown geodetic datum "WGS84". Use GeodesyData["Datum"] for a list of available named datums." This is strange as "WGS84" is of course explicitly listed under GeodesyData[] and accepted for GeoPositions. A bug? – gwr May 22 '12 at 14:30 Another observation is that it seems people at Wolfram Research have gotten the axes wrong. In geodesy as far as I have learned from web-research the x-axis should point north and the y-axis should point east. Have they missed that? – gwr May 23 '12 at 7:37


text:
Theoretical and empirical results for recovery from multiple measurements

E. van den Berg, M. P. Friedlander
IEEE Transactions on Information Theory, 56(5):2516–2527, 2010

Abstract

The joint-sparse recovery problem aims to recover, from sets of compressed measurements, unknown sparse matrices with nonzero entries restricted to a subset of rows. This is an extension of the single-measurement-vector (SMV) problem widely studied in compressed sensing. We analyze the recovery properties for two types of recovery algorithms. First, we show that recovery using sum-of-norm minimization cannot exceed the uniform recovery rate of sequential SMV using L1 minimization, and that there are problems that can be solved with one approach but not with the other. Second, we analyze the performance of the ReMBo algorithm [M. Mishali and Y. Eldar, IEEE Trans. Sig. Proc., 56 (2008)] in combination with L1 minimization, and show how recovery improves as more measurements are taken. From this analysis it follows that having more measurements than number of nonzero rows does not improve the potential theoretical recovery rate.

Reproducible research

To reproduce the experiments in this paper, download one of the following files:

To unpack the archive and run the experiments, execute the following commands from the Matlab commandline:

>> !unzip xxxx.zip
>> cd scripts
>> ScriptsSetup
>> GenerateFigure(...)

Note that you will need to have CVX installed. Our experiments are based on using Matlab 7.x and CVX v1.2m (and the included version of SDTP3).

BiBTeX

@Article{BergFriedlander:2010,
author =       {Ewout {van den} Berg and Michael P. Friedlander},
Title =        {Theoretical and empirical results for recovery from
multiple measurements},
journal =      "IEEE Trans. Inf. Theory",
year =         2010,
volume =       56,
number =       5,
pages =        {2516-2527},
DOI =          {10.1109/TIT.2010.2043876}
}

format: plain


text:
# Unit spacing with siunitx in an equation

I'm just starting to get to grips with the siunitx package, but I've noticed that there seems to be some issues with regards to unit notation spacing in an equation. With siunitx it is possible to put a space between the numerical value and the unit by leaving a space after the number of before the unit (e.g., \num{3.7d10} \si{\becquerel} or \num{3.7d10}\si{ \becquerel}).

However, when I do this in equation mode there doesn't seem to be a way to create a space between the number and the unit. Am I missing something, or is this an inherent fault/problem?

\documentclass[12pt,a4paper]{report}
% International language package - allows use of special characters
\usepackage[utf8]{inputenc}
% Line spacing package - allows user to set line spacing
\usepackage{amsmath}
% Graphics package - allows document to include various graphics files
\usepackage{chemformula}
% SI Unit package - allows for standard unit notation
\usepackage{siunitx}
\DeclareSIUnit\curie{Ci}

\begin{document}

% no spaces
\num{3.7d10}\si{\becquerel} \\

% space before unit
\num{3.7d10}\si{ \becquerel} \\

% space after number
\num{3.7d10} \si{\becquerel} \\

% no spaces
$$\num{1}\si{\becquerel} = \num{2.703d11}\si{\curie} \\$$

% spaces before units
$$\num{1}\si{ \becquerel} = \num{2.703d11}\si{ \curie} \\$$

% spaces after number
$$\num{1} \si{\becquerel} = \num{2.703d11} \si{\curie} \\$$

\end{document}

-

When you're typesetting a numerical quantity with units, you're supposed to use the \SI macro:

\SI{3.7d10}{\becquerel}

That effectively formats the first argument using \num, the second using \si, and inserts the proper spacing between them, even when typesetting an equation (i.e. when in math mode).

Normally spaces are ignored in math mode, so that you can write e.g. 3\times 10 or 3 \times 10 and the result will be the same.

-

Unless I'm missing something obvious, you can just use

\documentclass{article}

\usepackage{siunitx}

\begin{document}

$$\SI{1.234}{\m\per\square\s}$$

\end{document}

Another possibility (but not a recommended one) is

\documentclass{article}

\usepackage{siunitx}

\begin{document}

$$\num{1.234}\,\si{\m\per\square\s}$$

\end{document}

The second approach should only be used if you want to alignment points at both the numbers and the units, as shown here.

-


text:
#### IMAGES

1. Randomly Assign Numbers To Names

2. How to Assign Oxidation Numbers

3. Assign Oxidation Numbers To Each Element In The Following Compounds

4. Assign Oxidation Numbers To Each Element In This Compound No

5. 65 INFO PERIODIC TABLE ELEMENTS OXIDATION NUMBERS FREE DOWNLOAD PDF DOC

6. Assign Oxidation Numbers To Each Element In The Following Compounds

#### VIDEO

1. KennySTRUCT

2. Shopify

3. OXIDATION NUMBER AND STATE

4. How to import an SVG file into Xtool XCS Software

5. Oxidation numbers +2 and +4 are exhibited by

6. 2.Oxidation Number

1. How to find the Oxidation Number for N in N2O5 ...

To find the correct oxidation state of N in N2O5 (Dinitrogen pentoxide), and each element in the molecule, we use a few rules and some

2. What is the oxidation number of nitrogen in N_2O_5?

And there you have it, nitrogen has a +5 oxidation number in dinitrogen pentoxide. Answer link. Related topic. Oxidation Numbers Questions

3. How is the oxidation number of N2O5 determined?

There are some things you have to have in mind up front: N2O5 is a neutral molecule and oxygen almost always has a -2 oxidation state in molecules (only

4. What is the oxidation state of nitrogen in N2O5?

It is +5 .Even by the structure. enter image description here. Notice that it is a resonance structure of: enter image description here.

5. N2O5 Oxidation Number

The oxidation number of N in N2O5 is +5. The oxidation number of O in N2O5 is -2. FAQ Lewis Diagram. Element, Oxidation Number (Avg), Atoms, Count

6. What is the oxidation state of nitrogen in N2O5 ?

Let the oxidation state of N be x. Oxidation state of O is −2 as it is in oxide form. ⇒2x+5(−2)=0. ⇒x=210=+5. Solve any question of Some p-Block

7. Assign an oxidation number for N in N2O5(g).

Five oxygen atoms are present with a preferred oxidation state of -2 each. This means a total negative oxidation state of -10. The molecule has no net charge.

8. Find the oxidation number of N in N2O5

Find the oxidation number of N in N2O5 · Unless oxygen is combined with fluorine or isolated from other atoms the oxidation number of oxygen atoms is always

9. Assign oxidation numbers to all of the elements in the following

Assign oxidation numbers to all of the elements in the following compounds e) N2O5 f) GeCl2 g) HF h) Na2O2 i) SO42-. AI Recommended Answer: 1. Assign oxidation

10. What are the oxidation numbers of nitrogen in N2O5 and Ca(NO3)2?

And since it is neutral molecule charges zero. So the some of all the oxidation numbers of all the limits is to, so we write two plus two X, negative 12 is


text:
GeneralizedHexagonGraph - Maple Help

GraphTheory[SpecialGraphs]

 GeneralizedHexagonGraph
 construct generalized hexagon graph

 Calling Sequence GeneralizedHexagonGraph()

Description

 • The GeneralizedHexagonGraph() command returns the generalized hexagon graph.

Examples

 > $\mathrm{with}\left(\mathrm{GraphTheory}\right):$
 > $\mathrm{with}\left(\mathrm{SpecialGraphs}\right):$
 > $G≔\mathrm{GeneralizedHexagonGraph}\left(\right)$
 ${G}{≔}{\mathrm{Graph 1: an undirected unweighted graph with 126 vertices and 189 edge\left(s\right)}}$ (1)
 > $\mathrm{IsPlanar}\left(G\right)$
 ${\mathrm{false}}$ (2)
 > $\mathrm{DrawGraph}\left(G,\mathrm{style}=\mathrm{spring}\right)$


text:
# zbMATH — the first resource for mathematics

A matrix Poincaré formula for holomorphic automorphisms of quadrics of higher codimension. Real associative quadrics. (English) Zbl 0957.32016
The problem of the description of holomorphic automorphisms of quadrics dates back to the work of Poincaré. In this paper, the authors describe the holomorphic automorphisms of two infinite series of Hermitian quadrics. They give explicit formulas for the automorphisms in terms of rational maps. The quadrics in question are: quadric of real codimension two and a special class of quadric, the “real associated quadric”. The latest are quadrics that admit a ‘matrix Poincaré formula’ construction.

##### MSC:
 32V40 Real submanifolds in complex manifolds 32M05 Complex Lie groups, group actions on complex spaces


text:
# How can I circumvent this TikZ circuits bug when placing symbols at start or at end?

In the following MWE (essentially taken from page 603 of the pgf manual), when I try to place the symbols at start and at end, two unintended effects occur:

1. In each symbol there is an extra line extending from its center to its edge.
2. The the to path connects to the left side of the symbol at end. (I assume that TikZ "forgets" to transform the symbol coordinate system in order to make its x axis align with the path.)
\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{circuits}

\begin{document}

\begin{tikzpicture}[transform shape,circuit]
\draw (0,0) to [circuit handle symbol={draw,shape=rectangle,near start},
circuit handle symbol={draw,shape=circle,near end}] (0,3);
\begin{tikzpicture}[transform shape,circuit]
\draw (0,0) to [circuit handle symbol={draw,shape=rectangle,at start},
circuit handle symbol={draw,shape=circle,at end}] (0,3);
\end{tikzpicture}

\end{document}

There is a bug report about this issue. It is, however, from early 2017 which does not make me hopeful that this will be fixed in the near future.

Since I need to typeset documents now: Is there a way for me to fix this locally while preserving the TikZ syntax? (I would like the drawings to be able to be typeset with TikZ once the bug is fixed.)

• How about \begin{tikzpicture}[transform shape,circuit] \draw (0,0) to [circuit handle symbol={draw,shape=rectangle,pos=0.05}, circuit handle symbol={draw,shape=circle,pos=0.95}] (0,3); \end{tikzpicture}? I am not sure I would call this a bug. After all this is a circuit and you are essentially moving stuff out of it. – user121799 May 8 '19 at 14:14
• @marmot: I absolutely disagree. Placing a symbol at the end of a path does not mean placing it outside a circuit. ground, for example, only ever makes sense at the end of a path. However, your comment led me to tinker around a bit and I found that pos=1.0 works (while pos=1 does not). I don't know why this is, but it is just the workaround I was looking for. – schtandard May 8 '19 at 16:55

Update: The Bug has been fixed in the latest release of pgf (version 3.1.3).

I do not know why, but

\tikzset{
at start/.style={pos=0.0},
at end/.style={pos=1.0},
}

seems to solve the problem. (The original definition is with pos=0 and pos=1.)

• How is that conceptually different from my comment? And I would never overwrite the standard TikZ definitions. – user121799 May 8 '19 at 17:01
• Well, in your comment you suggested to place the symbol close to (but not at) the start/end of the path. It turned out that one can actually correctly place a symbol at the end of a path by saying pos=1.0 instead of pos=1. I think those are conceptually different assertions. I still believe this to be a bug (after all 1.0 should not be treated differently here than 1; indeed the default definition expects 1 to work). – schtandard May 8 '19 at 17:08
• Regarding your concern about overwriting standard definitions: While I agree in principle, in this particular case (replacing 0 with 0.0 and 1 with 1.0) I am not very worried at all. (Though in general an append style would be safer, of course.) – schtandard May 8 '19 at 17:11
• OK, let's agree that we disagree with each other's judgement. I made a proposal that works, you say that you do not like the numerical values I used. (My proposal was also motivated by tex.stackexchange.com/a/431765/121799.) Not much I can do. You say there is a bug, I do not see that. You say that replacing the integers by floats should be fine, but at the same time you see that this can have important consequences. I guess each of us can defend their view, so it is best if we depart peacefully. ;-) – user121799 May 8 '19 at 17:23
• You do not have to agree with me, of course. But maybe this example will help you understand: \documentclass{article} \usepackage{tikz} \usetikzlibrary{circuits.ee.IEC} \begin{document} \begin{tikzpicture}[circuit ee IEC] \foreach \p [count=\i] in {0.9, 0.95, 0.99, 0.999999, 1, 1.0} { \draw (\i,0) node [above] {\p} to[resistor={near start}, ground={pos=\p}] (\i,-3); } \end{tikzpicture} \end{document} If you still see no bug and still propose using a numerical value close(ish) to 1, then I guess we disagree. – schtandard May 8 '19 at 18:53


text:
# Thread: Please help with this question!

1. ## Please help with this question!

Given ln 5 1.6

use the log laws to approximate ln 25

any help would be great!

2. ## Re: Please help with this question!

$\displaystyle \ln{(25)} = \ln{\left( 5^2 \right)} = 2\ln{(5)}$

Go from here.

3. ## Re: Please help with this question!

ok so ive got 2 ln 5 = 3.22?
do you think that would be the final answer or did i miss a step.

and the full working should show=

=ln (25)
=ln(5^2)
=2 ln 5
=3.22

thanks again for your help

4. ## Re: Please help with this question!

That is correct (although of course your final step should have an approximate sign, not an equals sign).

5. ## Re: Please help with this question!

thats great thanks very much for your support!!


text:
#### Archived

This topic is now archived and is closed to further replies.

# Fixed point divide/multiply macro's in gcc

This topic is 5653 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I''ve been having trouble dividing and multiplying 16.16 fixed point numbers. As far as i can see standard c++ does not provide methods for working with 64 bit numbers, so you have to write the routines in assembly. So far there is no problem. But obviously the methods to multiply and divide have to be inline functions or macro''s because calling a function every time you need a mul/div will really slow things down. Since i am using gcc the #pragma approach does not seem to work. What i have come up with so far are asm macro''s: [begin source code] #define fixmul(a, b) ({ \ long fixrv; \ asm volatile("imull %%edx\n" \ "shrd $16, %%edx, %%eax" \ : "=a" (fixrv) \ : "a" (a), "d" (b)\ ); \ fixrv; \ }) #define fixdiv(a, b) ({ \ long fixrv; \ asm volatile ("movl %%eax, %%edx\n" \ "shll$16, %%eax\n" \ "sarl \$16, %%edx\n" \ "idivl %%ecx" \ : "=a" (fixrv) \ : "a" (a), "c" (b)\ : "%edx"); \ fixrv; \ }) [end source code] Note that the asm in in at&t syntax not intel syntax, that is what gcc works with. These macro''s work fine when not optimizing, but when compiling with -O2 the results of calling these become unpredictable. Note that i did specify the clobbered register d for the div macro. When i do the same with and inline function problem still occurs. Does anyone know what goes wrong here, or does anyone know of macro''s with the same functionaly that do work with gcc? Thanks. Marijn

##### Share on other sites
I think you will be able to use ''long long'' for 64 bit ints in gcc. That''s what we used for 64 bit maths with the PS1 compiler (which is gcc).

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

##### Share on other sites
Ah, i didnt know that. Thanks.
Problem with the macro turned out to be that gcc assumes input registers do not get changed unless they are also specified as output registers, and in this case i shifted edx all over the place so when optimizing it wrongly assumed the old value still was in there and went on with a wrong value.

Marijn

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# Automatic Weights failing

I'm trying to rig a digitigrade model (which is already a challenge in itself), but I've hit an impasse where I try to generate Automatic Weights and it just fails, saying "Bone Heat Weighting: Failed to find a solution for one or more bones". How would I go about fixing this? I'll attach my Blender file.

For this instance, selecting all the vertices and doing Mesh->Vertices->Remove Doubles will solve it.


text:
## It's all about the Money, Money, Money

Visible to anyone in the world

Last year I talked about how the west was bankrupt and that part of this 'scamdemic' was to cover this. Here is an update on the world's finances, take 15 mins and watch it because there is a huge crash coming. The IMF and the Fed are printing money like confetti and there are hard times ahead. Please watch and share.

Share post

## Letters to the Paper Round 3

Visible to anyone in the world

Well, I waited in anticipation for their response in the hope that we could get this debate going but this was it under the banner - Hard to know who to believe

In response to last week’s letter – Respected doctors, scientist, epidemiologist and virologist with decades of study, research and practical experience behind them or someone who has spent six months watching Youtube videos.

It’s hard to know who to believe these days.

That was it, so this was my response -

The NCR still cannot put forward any argument to support masks, lockdowns or restrictions and can only demonstrate the same arrogance with their empty rhetoric and sarcastic comments. They talk about doctors, scientists and epidemiologists but cannot offer the name of one or cite their work.  So, here’s a few for them to look up – Dr Vernon Coleman, Dr Judy Mikovits Phd in Virology, Dr Rashid Bhuttar or any of the 640 German Doctors who recently spoke out against the WHO narrative, Covid 19 and their government as part of a global scam.

I would especially recommend watching the interview with Dr Mikovits on the London Real site and there is also a very interesting interview with Robert Kennedy Jr, son of the late Senator Bobby Kennedy and Chairman of the Children’s Health Defence, especially, if you have a child with autism.

I could name many more but I’m sure that the NCR won’t even bother to check them out  because even a few hours of ‘honest’ research would reveal glaring contradictions and discrepancies in the narrative being pumped out by the media and the government.

And if they are trying to work out who to believe, here are a few questions for the NCR to ask as they try to work out where the truth lies:

1 What is being said?

2 Who is saying it?

3 WHY are they saying it?

4 Where is the evidence

5 How do you know it is where or who they say it is?  Can it be verified by an independent body or person or someone you know is reliable?

6 And this is probably the most important one - Is there anything else going on in background that they may be trying to distract you from?

On the advice of the late great Carl Sagan, I try to ‘critically interrogate’ everything I am being told and when things don’t add up, I apply the scientific principle of ‘Occams Razor’ - when there are two competing narratives, the simplest explanation is probably the truest - and when it comes to politics (and health) the answer is usually money.

And coincidentally, back in March, in the world of financial trading, there was a lot happening. The west was/is bankrupt and was heading for another financial crash. The Saudi's and the Russians’ wanted to dump the petrodollar as the world’s trading currency. They fell out for a while about it which was why the price of oil went down but Russia was in the best position to take it over.  Saddam Hussein and Colonel Gadaffi had proposed this before and we all know what happened to them.

So, financial trading was going to change from petrodollars to whoever had a currency, and the wealth. to back it, the US, at that point, had a budget deficit of 1.3 trillion dollars, whereas Russia had no debt, lots of gold, natural resources and oil.  So, this is why, I think, we are where we are - the Globalist don't like that they have lost their hegemony and then along came Covid 19, just in time, to save the day.

So, we are being frogmarched towards the financial reset, the ‘new normal’ a cashless society where they can wipe out their debt with a bit of computer wizardry but not your debt, and we are all going to be paying for it, us, our children and our grandchildren.

Then again, maybe I am wrong, but coincidentally, last month, the Belarusian President said that the IMF and the World Bank offered him $940million to impose extreme lockdown on his people in the form of Covid Relief Aid. I wonder why? So, now, I think, the question has to be asked, have any of our political parties received money from the IMF or the World Bank to impose ‘lockdown’? Maybe the NCR could tell us or will I see what I can find on Youtube? But this letter wasn't printed, so the right of response has been denied by the editor of the paper, who I strongly suspect is the NCR. As well as editing the paper, he also has a column and the style and tone of writing is very similar. This week's paper was full of the fear-mongering again, with reports of a rise in cases so I don't think my letter would have fitted in with the narrative or the hysteria they were trying to generate. But I'll wait and see, maybe they were short of room and it might be in this week. However, right at the beginning of this, I sent a letter is which is on the blog, and it wasn't printed and very probably for the same reasons. Here is a link to an article on the Belarussian claim, this has been reported on other sites too. And just as a note to this, there are ongoing demonstrations now in Belarus against the President, and reports in August of an attempted coup, link to that story below as well. https://fort-russ.com/2020/09/belarusian-president-lukashenko-says-imf-offered-a-billion-usd-bribe-to-impose-covid-19-lockdown/ Share post ## Global warming Visible to anyone in the world Going back to the global - warming, climate change debate, I've been trying to figure out what is really going on because I couldn't understand why this false science has become so pervasive in the media and across the internet with real scientists being subjected to some serious abuse for trying to talk 'common sense' as David Bellamy put it when it is not proved or even close to being proved. Why is this big lie being promoted with such vigour across the entire world and what is the purpose? Just as an experiment, I tried to talk to a few 'Green' friends about the blog I had researched, going over the science and asking them to check it out. This propaganda is entrenched! They refused to even consider it, blinded by their own arrogance. Also, I think, the fact that I'm female and they are male, had a bearing. Here in NI the old attitudes prevail even amongst those who consider themselves open-minded and liberal. I've also been researching a lot into the Rothschilds, the Fed, the IMF/World Bank and how no matter who you vote for the Rothschilds are still in charge (a blog for another day). The powers that be are still pursuing their own agenda, creating wars as cover to gain control of the energy supplies, the oil and gas around Ukraine, Iraq, the rest of the world and of course, the big prize, Iran. I couldn't understand why they were still doing this while pushing the Co2 as a contributor to global-warming, across the media and the internet and call me paranoid, if you like, but I think I've got it! I think the whole Co2/climate-change agenda is a set-up which will be used to discredit the entire green movement and anyone connected to it. When it suits their purpose, of course, the tide of opinion will be turned through the media, which they mostly control and no one will listen to the environmentalists again. If you don't believe this is possible, I would ask you to look at the Adam Curtis documentary, Century of the Self, to see just how much we are, and have been, manipulated. This will be the big pull the rug out from underneath them and shut them up forever and this will happen when they have locked down the energy supplies of coal, gas and oil for themselves. Maybe I've been watching too many you tube documentaries but it's the only thing that makes sense to me. Anyway, all we can do is wait and see and remember you read it here first! Permalink 5 comments (latest comment by Aideen Devine, Saturday, 22 Jul 2017, 11:10) Share post ## Irony Visible to anyone in the world Edited by Aideen Devine, Sunday, 11 Oct 2020, 09:45 It's rather ironic that the year I was doing a 'writing course' is the one in which I was writing very little for the blog. I'm off work this week so I promised myself I would use this week to get back in the habit of more regular postings. There is so much going on in politics at the moment that it's a bit difficult to decide where to begin. Greece is the big news and good luck to them for defying the 'autocrats' and fear mongering of the troika and voting no. The biggest threat to the EU monolith is the chance that Greece would pull out and make a success of running their own country - I sincerely hope they do, then maybe Spain, Portugal and Ireland might consider doing the same. I support the idea of a common market for Europe but I don't support the idea of a European State controlled by corporations, the IMF and the World Bank. The creation of the single currency was a con that sucked in poorer nations who were never going to be able to compete economically with richer nations like Germany, and now it has come back to bite them on the ass and deservedly so - I've pasted this article by Chris Kanthan from nation of change.org as he can explain it all much better than me; Every single mainstream media has the following narrative for the economic crisis in Greece: the government spent too much money and went broke; the generous banks gave them money, but Greece still can't pay the bills because it mismanaged the money that was given. It sounds quite reasonable, right? Except that it is a big fat lie ... not only about Greece, but about other European countries such as Spain, Portugal, Italy and Ireland who are all experiencing various degrees of austerity. It was also the same big, fat lie that was used by banks and corporations to exploit many Latin American, Asian and African countries for many decades. Greece did not fail on its own. It was made to fail. In summary, the banks wrecked the Greek government and deliberately pushed it into unsustainable debt so that oligarchs and international corporations can profit from the ensuing chaos and misery. If you are a fan of mafia movies, you know how the mafia would take over a popular restaurant. First, they would do something to disrupt the business - stage a murder at the restaurant or start a fire. When the business starts to suffer, the Godfather would generously offer some money as a token of friendship. In return, Greasy Thumb takes over the restaurant's accounting, Big Joey is put in charge of procurement, and so on. Needless to say, it's a journey down a spiral of misery for the owner who will soon be broke and, if lucky, alive. Now, let's map the mafia story to international finance in four stages. Stage 1: The first and foremost reason that Greece got into trouble was the "Great Financial Crisis” of 2008 that was the brainchild of Wall Street and international bankers. If you remember, banks came up with an awesome idea of giving subprime mortgages to anyone who can fog a mirror. They then packaged up all these ticking financial bombs and sold them as "mortgage-backed securities” at a huge profit to various financial entities in countries around the world. [x] A big enabler of this criminal activity was another branch of the banking system, the group of rating agencies - S&P, Fitch and Moody's - who gave stellar ratings to these destined-to-fail financial products. Unscrupulous politicians such as Tony Blair got paid by Big Banks to peddle these dangerous securities to pension funds and municipalities and countries around Europe. Banks and Wall Street gurus made hundreds of billions of dollars in this scheme. But this was just Stage 1 of their enormous scam. There was much more profit to be made in the next three stages! Stage 2 is when the financial time bombs exploded. Commercial and investment banks around the world started collapsing in a matter of weeks. Governments at local and regional level saw their investments and assets evaporate. Chaos everywhere! Vultures like Goldman Sachs and other big banks profited enormously in three ways: one, they could buy other banks such as Lehman brothers and Washington Mutual for pennies on the dollar. Second, more heinously, Goldman Sachs and insiders such as John Paulson (who recently donated$400 million to Harvard) had made bets that these securities would blow up. Paulson made billions, and the media celebrated his acumen. (For an analogy, imagine the terrorists betting on 9/11 and profiting from it.) Third, to scrub salt in the wound, the big banks demanded a bailout from the very citizens whose lives the bankers had ruined! Bankers have chutzpah. In the U.S., they got hundreds of billions of dollars from the taxpayers and trillions from the Federal Reserve Bank which is nothing but a front group for the bankers.

In Greece, the domestic banks got more than \$30 billion of bailout from the Greek people. Let that sink in for a moment - the supposedly irresponsible Greek government had to bail out the hardcore capitalist bankers.

Stage 3 is when the banks force the government to accept massive debts. For a biology metaphor, consider a virus or a bacteria. All of them have unique strategies to weaken the immune system of the host. One of the proven techniques used by the parasitic international bankers is to downgrade the bonds of a country. And that's exactly what the bankers did, starting at the end of 2009. This immediately makes the interest rates ("yields”) on the bonds go up, making it more and more expensive for the country to borrow money or even just roll over the existing bonds.

From 2009 to mid-2010, the yields on 10-year Greek bonds almost tripled! This cruel financial assault brought the Greek government to its knees, and the banksters won their first debt deal of a whopping 110 billion Euros.

The banks also control the politics of nations. In 2011, when the Greek prime minister refused to accept a second massive bailout, the banks forced him out of the office and immediately replaced him with the Vice President of ECB (European Central Bank)! No elections needed. Screw democracy. And what would this new guy do? Sign on the dotted line of every paperwork that the bankers bring in.

(By the way, the very next day, the exact same thing happened in Italy where the Prime Minister resigned, only to be replaced by a banker/economist puppet. Ten days later, Spain had a premature election where a banker puppet won the election).

The puppet masters had the best month ever in November 2011.

Few months later, in 2012, the exact bond market manipulation was used when the banksters turned up the Greek bonds' yields to 50%!!! This financial terrorism immediately had the desired effect: The Greek parliament agreed to a second massive bailout, even larger than the first one.

[x]

Now, here is another fact that most people don't understand. The loans are not just simple loans like you would get from a credit card or a bank. These are loans come with very special strings attached that demand privatization of a country's assets. If you have seen Godfather III, you would remember Hyman Roth, the investor who was carving up Cuba among his friends. Replace Hyman Roth with Goldman Sachs or IMF (International Monetary Fund) or ECB, and you get the picture.

Stage 4: Now, the rape and humiliation of a nation begin under the name of "austerity” or "structural reforms.” For the debt that was forced upon it, Greece had to sell many of its profitable assets to oligarchs and international corporations. And privatizations are ruthless, involving everything and anything that is profitable. In Greece, privatization included water, electricity, post offices, airport services, national banks, telecommunication, port authorities (which is huge in a country that is a world leader in shipping) etc. Of course, the ever-manipulative bankers always demand immediate privatization of all media which means that the country gets photogenic TV anchors who spew establishment propaganda every day and tell the people that crooked and greedy banksters are saviors; and slavery under austerity is so much better than the alternative.

In addition to that, the banker tyrants also get to dictate every single line item in the government's budget. Want to cut military spending? NO! Want to raise tax on the oligarchs or big corporations? NO! Such micro-management is non-existent in any other creditor-debtor relationship.

So what happens after privatization and despotism under bankers? Of course, the government's revenue goes down and the debt increases further. How do you "fix” that? Of course, cut spending! Lay off public workers, cut minimum wage, cut pensions (same as our social security), cut public services, and raise taxes on things that would affect the 99% but not the 1%. For example, pension has been cut in half and sales tax increase to more than 20%. All these measures have resulted in Greece going through a financial calamity that is worse than the Great Depression of the U.S. in the 1930s.

After all this, what is the solution proposed by the heartless bankers? Higher taxes! More cuts to the pension! It takes a special kind of a psychopath to put a country through austerity, an economic holocaust.

If every Greek person had known the truth about austerity, they wouldn't have fallen for this. Same goes for Spain, Italy, Portugal, Ireland and other countries going through austerity. The sad aspect of all this is that these are not unique strategies. Since World War II, these predatory practices have been used countless times by the IMF and the World Bank in Latin America, Asia, and Africa.

This is the essence of the New World Order -- a world owned by a handful of corporations and banks; a world that is full of obedient, powerless debt serfs.

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Question

# A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1: 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to20 10 50 60

Solution

## The correct option is C 50 Since the given volume ratio for five solid cubes is 1:1:8:27:27. The corresponding side ratio of the same cubes is 1:1:2:3:3 and the volume of the big cube is 64k^3, where k is any constant. Hence, its side is 4k.  Percentage by which the sum of surface areas of five cubes exceed the surface area of original cube =100×[6×24k2−6×(4k)2]{6×(4k)2}=[(24−16)16]×100=50

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1.8: Division of Real Numbers

Difficulty Level: Advanced Created by: CK-12
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Practice Division of Rational Numbers

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The meteorologist on the local radio station just announced that a cold front caused the temperature to drop \begin{align*}12^{\circ}C\end{align*} in just four hours. What was the mean temperature change per hour over these four hours?

Guidance

There are two rules for dividing integers:

1. When you divide two integers that have the same signs, the answer is always positive.
2. When you divide two integers that have opposite signs, the answer iss always negative.

We can demonstrate these rules using counters. The rules for dividing integers apply to the division of all real numbers.

The division of fractions involves one more step than the multiplication of fractions. This step is to begin the solution by inverting the fraction after the operation sign. If \begin{align*}\frac{5}{7}\end{align*} is inverted, the new fraction is \begin{align*}\frac{7}{5}\end{align*}. This new fraction \begin{align*}\frac{7}{5}\end{align*} is called the reciprocal of \begin{align*}\frac{5}{7}\end{align*}. The product of a number and its reciprocal is one. When you have completed this step, change the division sign to a multiplication sign and follow the rules for multiplying fractions.

To divide decimal numbers, use the following steps:

• Write the divisor and the dividend in standard long-division form.
• Move the decimal point of the divisor to the right so that the divisor is a whole number.
• Move the decimal point of the dividend to the right the same number of places that you moved the decimal point of the divisor. If necessary, add zeros in the dividend.
• Place the decimal point in the quotient directly above the new decimal point in the dividend.
• The decimal points can now be ignored. Divide the numbers the same as you would divide whole numbers.

Example A

Miguel was doing a science project on weather and he reported a total temperature change of \begin{align*}-15^{\circ}F\end{align*} and a mean hourly change of \begin{align*}-3^{\circ}C\end{align*}. How many hourly temperature changes did Miguel record?

You can use color counters to determine the number of changes that Miguel recorded. Start with zero and add a group of 3 yellow counters. Continue adding groups of three counters until you have 15 yellow counters.

How many times did you need to add -3 to get -15?

You added 5 groups of three yellow counters. When you divide integers with the same sign, you can think of repeated addition. The result of \begin{align*}(-15)\div(-3)\end{align*} is 5.

Example B

i) \begin{align*}\left(\frac{6}{11}\right) \div \left(\frac{5}{7}\right)\end{align*}

ii) \begin{align*}\left(4 \frac{1}{3}\right) \div \left(2 \frac{5}{7}\right)\end{align*}

Solution:

i) \begin{align*}& \left(\frac{6}{11}\right) \div \left(\frac{5}{7}\right)\\ & \frac{6}{11} \times \frac{7}{5}\\ & \frac{6 \times 7}{11 \times 5}\\ & =\frac{42}{55}\end{align*}

ii) \begin{align*}& \left(4\frac{1}{3}\right) \div \left(2\frac{5}{7}\right) \ \text{Write the mixed numbers as improper fractions.}\\ & \left(\frac{13}{3}\right) \div \left(\frac{19}{7}\right)\ \text{Multiply by the reciprocal of} \ \frac{19}{7}.\\ & \frac{13}{3} \times \frac{7}{19}\\ & = \frac{91}{57}= 1 \frac{34}{57} \ \text{Simplify the fraction.}\end{align*}

Example C

i) \begin{align*}(0.68)\div(1.7)\end{align*}

ii) \begin{align*}0.365 \div -18.25\end{align*}

Solution:

i) \begin{align*}(0.68) \div (1.7)\end{align*}

\begin{align*}& \overset{ \qquad \ 0.4}{\underset{\ \ \rightarrow}{1.7} \overline{ ) {\underset{\ \ \rightarrow}{0.6} \ 8 \;}}}\\ & \quad \ \ \underline{- 6 \; 8}\\ & \qquad \quad \ 0\end{align*}

The decimal point of the divisor was moved one place to the right. The decimal point of the dividend was moved one place to the right. The decimal point was placed in the quotient directly above the new decimal point of the dividend.

ii) \begin{align*}0.365 \div -18.25\end{align*}

You have learned that when you divide a positive number by a negative number, the answer will always be negative.

\begin{align*}& \overset{ \qquad \qquad -.02}{\underset{\quad \ \ \longrightarrow}{-18.25} \overline{ ) {\underset{\ \ \longrightarrow}{0.36} \ 5{\color{blue}0}}}}\\ & \qquad \quad \ \underline{- 36 \; \; 50}\\ & \qquad \qquad \qquad 0\end{align*}

The decimal point of the divisor was moved two places to the right. The decimal point of the dividend was moved two place to the right. The decimal point was placed in the quotient directly above the new decimal point of the dividend.

Concept Problem Revisited

The meteorologist on the local radio station just announced that a cold front caused the temperature to drop \begin{align*}12^{\circ}C\end{align*} in just four hours. You can use counters to show the mean temperature change per hour over these four hours.

Since it took four hours for the temperature to drop, put the counters into four equal groups.

There are three negative counters in each group. The mean temperature change per hour was \begin{align*}-3^{\circ} C\end{align*}. The result of \begin{align*}(-12)\div(+4)\end{align*} is -3. When you divide by a positive integer, you can think of sharing equally by the magnitude of the positive integer.

Vocabulary

Dividend
In a division problem, the dividend is the number that is being divided. The dividend is written under the division sign. \begin{align*} \overset{}{4 \overline{ ) {38}}}\end{align*}, 38 is the dividend.
Divisor
In a division problem, the divisor is the number that is being divided into the dividend. The divisor is written in front of the division sign. \begin{align*} \overset{}{4 \overline{ ) {38}}}\end{align*} 4 is the divisor.
Mixed Number
A mixed number is a number made up of a whole number and a fraction such as \begin{align*}4 \frac{3}{5}\end{align*}.
Reciprocal

The reciprocal of a number is the inverse of that number. If \begin{align*}\frac{a}{b}\end{align*} is a nonzero number, then \begin{align*}\frac{b}{a}\end{align*} is its reciprocal. The product of a number and its reciprocal is one.

Quotient

The quotient is the answer of a division problem.

Guided Practice

1. Use color counters to represent the division problem \begin{align*}(-24) \div(+6)\end{align*}.

2. Determine the answer to \begin{align*}\left(\frac{7}{10}\right) \div \left ( \frac{5}{6} \right )=?\end{align*} and \begin{align*}\left(6 \frac{2}{5}\right) \div \left(1 \frac{2}{3}\right)=?\end{align*} by using the rules for dividing fractions.

3. How many pieces of plywood 0.375 in. thick are in a stack of 30 in. high?

1. \begin{align*}(-24) \div (+6)\end{align*}

Begin with 24 yellow counters. Create 6 groups of yellow counters until you use all of the counters.

There are 6 groups of yellow counters. Each group contains four yellow counters.

\begin{align*}(-24) \div (+6)=-4\end{align*}

2. i) \begin{align*}\left(\frac{7}{10}\right) \div \left(\frac{5}{6}\right)\end{align*}

\begin{align*}\frac{7}{10} \times \frac{6}{5}\end{align*}

Multiply by the reciprocal.

\begin{align*}=\frac{7 \times 6}{10 \times 5}\end{align*}

Multiply numerators. Multiply denominators.

\begin{align*}=\frac{42}{50}=\frac{21}{25}\end{align*}

Simplify the fraction.

ii) \begin{align*}\left(6\frac{2}{5}\right) \div \left(1\frac{2}{3}\right)\end{align*}

\begin{align*}\left(\frac{32}{5}\right) \div \left(\frac{5}{3}\right)\end{align*}

Change the mixed numbers to improper fractions.

\begin{align*}\left(\frac{32}{5}\right) \times \left(\frac{3}{5}\right)\end{align*}

Multiply by the reciprocal.

\begin{align*}\frac{32 \times 3}{5 \times 5}\end{align*}

Multiply numerators. Multiply denominators.

\begin{align*}& =\frac{96}{25}\\ & =3\frac{21}{25}\end{align*}

Simplify the fraction.

3. To determine the number of pieces of plywood in the stack, divide the thickness of one piece into the height of the pile.

\begin{align*}& \overset{ \qquad \qquad \ 80}{\underset{\ \ \longrightarrow}{0.375} \overline{ ) {\underset{\ \ \longrightarrow}{30.000}}}}\\ & \qquad \underline{- 3000\;}\\ & \qquad \qquad \quad 0\\ & \qquad \ \underline{- \;\;\;\;\;\;\;0}\\ & \qquad \qquad \quad 0\end{align*}

There are 80 pieces of plywood in the pile.

Practice

Use color counters to find each quotient.

1. \begin{align*}(+14)\div (+2)\end{align*}
2. \begin{align*}(-14) \div (+2)\end{align*}
3. \begin{align*}(-9)\div (-3)\end{align*}
4. \begin{align*}(+16) \div (+4)\end{align*}
5. \begin{align*}(+25) \div (-5)\end{align*}

Calculate the quotient for each of the following:

1. \begin{align*}(-9)\times (7)\end{align*}
2. \begin{align*}(-8)\times (-8)\end{align*}
3. \begin{align*}(+4)\times (-7)\end{align*}
4. \begin{align*}(-10) \times (-3)\end{align*}
5. \begin{align*}(+5) \times (+2)\end{align*}
6. \begin{align*}\left(\frac{5}{16}\right) \div \left(\frac{3}{7}\right)\end{align*}
7. \begin{align*}(-8.8)\div (-3.2)\end{align*}
8. \begin{align*}(7.23)\div (0.6)\end{align*}
9. \begin{align*}\left(2\frac{3}{4}\right) \div \left(1\frac{1}{8}\right)\end{align*}
10. \begin{align*}(-30.24) \div (-0.42)\end{align*}

For each of the following questions, write a division statement and find the result.

1. A truck is delivering fruit baskets to the local food banks for the patrons. Each fruit basket weighs 3.68 lb. How many baskets are in a load weighing 5888 lb?
2. A wedding invitation must be printed on card stock measuring \begin{align*}4 \frac{1}{4} \ in\end{align*}. wide. If the area of the invitation is \begin{align*}23 \frac{3}{8} \ in^2\end{align*}, what is its length? (Hint: The area of a rectangle is found by multiplying the length times the width.)
3. A seamstress needs to divide \begin{align*}32 \frac{5}{8} \ ft\end{align*}. of piping into 3 equal pieces. Calculate the length of each piece.
4. The floor area of a room on a house plan measures 3.5 in. by 4.625 in. If the house plan is drawn to the scale 0.25 in. represents 1 ft, what is the actual size of the room?
5. How many hair bows of \begin{align*}3 \frac{1}{2} \ in\end{align*}. can be cut from \begin{align*}24 \frac{3}{4} \ in\end{align*}. of ribbon?

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# Middle man

Geometry Level 2

Diagram above shows two pieces of overlapping paper. The non-overlapping area of the smaller square corresponds to 52% of the area of this square and non-overlapping area of the larger square is 73% of the area of this square. What is the ratio between the sides of the smaller square and the largest square?

×


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# How to calculate RNA copies in qPCR using R?

I have these Ct data from qPCR for a series of samples A-H, I want to quantify the RNA copies of my samples using R:

Sample_ID   Replicate1_Ct Replicate2_Ct
A           22.92         24.21
B           24.29         22.16
C           24.29         22.16
D           27.91         23.76
E           21.44         22.81
F           16.50         18.31
G           17.79         16.56
H           17.34         19.09

In addition, I have measured a standard (STD) curve of copy number of the RNA in question, that I want to use as a reference:

STD_Ct      STD_copies      Log10(STD_copies)
33          4.00E+03        3
28          4.00E+04        4
24          4.00E+05        5
20          4.00E+06        6
15          4.00E+07        7
12          4.00E+08        8

It seems

This the original data form:

• This is pretty unclear- at a minimum we need some explanation of the data table. e.g. why do rows G and H not have standards when all other rows do? Is each row a different standard sample? Do you have a loading control reaction for each sample? Nov 28 '20 at 20:44
• The first two columns are the the mean of two replicates, I wanted to make it simple, the standard is only 6 rows, the standard has nothing to do with the samples, it was there just to build to the standard curve to use it for the absolute method @MaximilianPress Nov 28 '20 at 20:54
• I will edit the question to makes this clearer. Nov 28 '20 at 21:13
• I updated the question with the table from excel @MaximilianPress , it should reflect a better idea. Nov 28 '20 at 21:30
• Discussion of latest update moved to a chat. I think there is a numerical error in the way the predictions are being generated, but you seem ok with the results so I'm calling this done. Nov 29 '20 at 23:11

Now that we've straightened out the data, it seems like the simplest solution is to just linearly interpolate the copy number from your standard curve.

When I read your standard curve (std) and your sample data (ct) into R, I see that the log10(Ct) fits the copy number fairly well, whereas the linear is a bit harder:

So we can fit a model to predict copy number based on Ct on the STD data:

> fit = lm(std$$Log10.STD_copies. ~ std$$STD_Ct)
> fit

Call:
lm(formula = std$$Log10.STD_copies. ~ std$$STD_Ct)

Coefficients:
(Intercept)   std$STD_Ct 10.6847 -0.2357 # we plug those numbers into the linear equation, # taking the mean of the two replicates for a best estimate. > rowMeans(ct[,2:3]) * -.2357 + 10.6847 [1] 5.130429 5.210567 5.210567 4.595390 5.469837 6.582341 [7] 6.636553 6.391424  So those are log10(copy number) estimates, of a sort. You can get much more elaborate, of course. • I accepted your answer and upvoted for the efforts, however; I'm not getting the correct results, both intercept (45.29) and slope (-4.2435) are not accurate, I suppose the correct formula (to calculate RNA copies) would be Copy number = 10^((Ct - Intercept)/(Slope))  Nov 29 '20 at 1:27 • @user432797 I think that those parameters would come if you fit the model lm(std$STD_Ct ~std$Log10.STD_copies.), which is then intercept 45.26 and slope -4.23. I believe that you simply want to switch your independent and dependent variables to do your prediction. Nov 29 '20 at 3:42 • I checked another software and it shows intercept (45.29) and slope (-4.2435) @MaximilianPress so I will proceed with these numbers. Nov 29 '20 at 4:47 • @user432797 I honestly wouldn't recommend that- I would recommend instead fitting the model as lm(std$Log10.STD_copies. ~ std\$STD_Ct ), because you are trying to predict copy number from Ct. If you use those parameters that you mention, those are predicting Ct from copy number, which is the opposite of what you want. Obviously, you are free to do as you like, but that does not seem appropriate to me. Nov 29 '20 at 5:03
• @user432797 it sounds like there might still be some numerical error. Are the values that I've given in my answer implausible? I can't see your results. Nov 29 '20 at 15:40


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Home > Webshop > General interest > Maths Odyssey

# A Maths Odyssey

### Tracing a Line from Euclid to the Computer

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ISBN-13: 978-1-86317-131-1
Year Levels: 7 - 10
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Page Count: 48
Cover type: Soft cover
Publication date: 2010
Product number: PHM001
Keywords: Measurement and geometry, Classroom material


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+0

# Trigonometry Question

0
42
1

If the point P(0.5,k) lies on a circle with a radius of 1, then the exact value of k can be expressed as +/- square root of b. The value of b, to the nearest hundredth, is ___.

May 18, 2022

#1
+1768
+1

The equation of the circle is $$x^2 + k^2 = 1$$

We know that $$x = {1 \over 2}$$. Subsituting this in gives us: $${1 \over 2} ^2 + k^2 = 1$$

This means $$k^2 = {3 \over 4}$$.

Now, we have to take the square root, and round the result to the nearest hundredth.

May 18, 2022


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# Point generation in polygon

I know about the Halton sequence. But so far I can’t find the formulas by which points are generated. Also worried is the question Halton sequence generates points only in the rectangle? Or can I generate points in any polygon using the Halton sequence? If the Halton sequence generates points only in the rectangle, what other way is there to generate points in the polygon? I need an optimal method for generating points in a polygon. What do you advise? It is advisable to describe the method and give specific formulas.

• What is your criterion for "optimality"? – Gerry Myerson Nov 28 '19 at 11:28
• @Gerry Myerson as in jts testbuilder. RandomePointInPolygon for 6 milliseconds, 10,000 points generated – Ivan Triumphov Nov 28 '19 at 11:46
• I'm sure you could generate a billion points inside a polygon in six femtoseconds, but what makes one set of a billion points better than another? – Gerry Myerson Nov 28 '19 at 11:49
• @Gerry Myerson If I understand the question correctly. Minimum calculations to generate the desired number of points. – Ivan Triumphov Nov 28 '19 at 12:01
• @Gerry Myerson Maybe I didn’t formulate correctly. Any algorithm to generate a given number of points in a polygon. – Ivan Triumphov Nov 28 '19 at 12:05

To generate points with the same distribution as the Halton sequence in a polygon you can just take a rectangle enclosing the polygon - for example take the convex hull of the polygon (assuming it may be concave ), pick two orthogonal directions, say the coordinate axes, project all points onto them, and take the min and max of the projections in each axis - this gives you 4 lines which define the enclosing rectangle. You can then generate the Halton sequence within this rectangle and throw away all the points which don't lie in your polygon. If you have a highly non-convex or self-intersecting polygon you can first subdivide the polygon into convex pieces using this answer and apply the above method to each convex piece separately.

If efficiency is a major consideration, which it may not be if the Halton sequence generation is fast enough, then you can look at finding enclosing rectangles with minimum area - this can be done using this answer. Note that you can always find an enclosing rectangle around a convex shape with area at most twice the shape area but typically it will be closer to 1 if the shape is reasonably regular - for an ellipse it is $$4/\pi\approx1.3$$.

Generation of Halton Sequence.

$$\text{Rev[D_,b_]:=Reverse[IntegerDigits[D,b]]}$$

$$\text{Halton[n_,b_]:=FromDigits[Rev[n,b]/b^Length[Rev[n,b]],b]}$$

$$\text{Halton2[n_,b1_,b2_]:=Table[{Halton[i,b1],Halton[i,b2]},{i,1,n}]}$$

The above Mathematica function $$\text{Halton2[n,b1,b2]}$$ generates a Halton sequence in $$[0,1]\times [0,1]$$ of length $$n$$ with respect to the bases b1 and b2. Here is a plot of $$\text{Halton2[1000, 2, 3]}$$:

(If you don't know Mathematica, $$\text{IntegerDigits[D,b]}$$ returns an ordered list of digits of $$D$$ in base $$b$$, $$\text{FromDigits[l,b]}$$ generates an integer with digits $$l$$, base $$b$$. $$\text{Reverse[l]}$$ returns the reverse of an ordered list and $$\text{Table}$$ generates an ordered list)

Example

Essentially to generate the $$n$$-th Halton number base b you take the digits of n base b, reverse them and place a decimal point in front. So $$\text{Halton[12,2]} = .0011_2$$ since 12 in base 2 is 1100. You generate a Halton point sequence in 2d by pairing up the Halton sequences for 2 different primes. So the 12-th Halton 2d point for the bases 2 and 3 is $$(\text{Halton}[12,2], \text{Halton}[12,3]) = (.0011_2,.011_3)$$

Filtering out points not in your polygon

If you have an ordered list $$\{(x_1,y_1),(x_2,y_2),\cdots,(x_n,y_n)\}$$ of $$n$$ points comprising your convex polygon in anti-clockwise order then a point $$(px,py)$$ is inside the polygon iff the determinant

$$\begin{vmatrix} px & py & 1 \\ x_i & y_i & 1 \\ x_{i+1} & y_{i+1} & 1 \end{vmatrix}$$

is greater than 0 for all $$0\leq i\leq n$$ where indices are taken mod n.

Figures below show the process of generating the Halton sequence for the enclosing rectangle (Fig 1) then throwing away points outside in the dark region (Fig2).

Fig 1.

Fig 2.

• Thanks! I was hoping for formulas in the answer. – Ivan Triumphov Nov 29 '19 at 7:50
• @IvanTriumphov I've added additional details concerning the various steps of the process which I hope will help. – Ivan Meir Nov 29 '19 at 19:23


text:
# CAT Quantitative Aptitude Questions | CAT Ratios, Mixtures, Alligations and Averages Questions

###### CAT Questions | Mixtures and Alligation | Alligation

The question is from CAT Ratio, Mixtures and Averages. How would we go about solving this one when we are mixing two things and making certain amount of profit? CAT exam is known to test on basics rather than high funda ideas. A range of CAT questions can be asked from Ratios and Proportions, Mixtures, Alligations and Averages. Make sure you master the topics. 2IIMs CAT questions bank provides you with CAT questions that can help you gear for CAT Exam CAT 2019.

Question 31: What is the ratio in which water should be mixed with a coke concentrate costing Rs 15/litre to make a profit of 30% by selling the resultant drink at Rs 18/litre?

1. 12:1
2. 11:1
3. 1:10
4. 1:12

## Best CAT Coaching in Chennai

### Explanatory Answer

##### Method of solving this CAT Question from Ratio, Mixtures and Averages: When we mix two mixtures in a particular ratio, we get a third mixture. Given the third mixture how does one find the ratio in which they were mixed.

C.P. to get 30% profit at Rs 18/litre = $$frac{18}{1.3}\\$ = $\frac{180}{13}\\$ Therefore, => $\frac{15}{13}\\$ : $\frac{180}{13}\\$ => 1: 12 Note: Make sure you don’t mark 12:1 since in the question, water has been mentioned first. This can be a silly trap to fall into after all the hard work. The question is "What is the ratio in which water should be mixed with a coke concentrate costing Rs 15/litre to make a profit of 30% by selling the resultant drink at Rs 18/litre?" ##### Hence, the answer is "1:12". Choice D is the correct answer. ###### Online CAT CoachingTry it free! Register in 2 easy steps and start learning in 30 seconds! ###### Already have an Account? ###### CAT Coaching in Chennai Next Weekend Batch Starts Sat, Mar 23rd, 2019 ###### Best CAT Coaching in Chennai Sign up for a Free Demo of our CAT Coaching in Chennai ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 10-C, Kalinga Colony, Bobbili Raja Salai K.K.Nagar, Chennai. India. Pin - 600 078 ##### How to reach 2IIM? Phone:$91) 44 4505 8484
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text:
# Method of integral equations in the polytropic theory of stars with axial rotation. II. Polytropes with indices $n>1$

2021;
: pp. 474–485

A new method for finding solutions of the nonlinear equilibrium equations for rotational polytropes was proposed, which is based on a self-consistent description of internal region and periphery using the integral form of equations.  Dependencies of geometrical parameters, surface form, mass, moment of inertia and integration constants on angular velocity were calculated for indices $n=2.5$ and $n=3$.


text:
## Section40.4Online (HTML) Options

These options affect the base conversion to web pages (online, HTML). Many, but not all, will affect subsequent conversions based on HTML, such as a conversion to EPUB or Jupyter notebooks. See Chapter 27 for a more general overview of this conversion, including options that will eventually migrate here.

### Subsection40.4.1HTML Analytics

The

/publication/html/analytics

element can have the following attributes:

• @google-gst: a Google global site tag, and ID you get from Google. Do not copy the identification numbers from another project—be sure to obtain your own for your project.

• @statcounter-project, @statcounter-security: ID numbers you get from StatCounter. Do not copy the identification numbers from another project—be sure to obtain your own for your project.

Setting these attributes to non-empty strings is the signal to add the relevant code to each of the pages of your HTML output. See Section 27.7 for more.

### Subsection40.4.2HTML Base URL

The

/publication/html/baseurl/@href

attribute may be given as a complete URL for the top-level of where HTML output is hosted. Of course, this may be different for different publishers. The value is a directory, and so should end with a slash (path separator). See Subsection 27.1.3 for reasons why you might want to specify this.

### Subsection40.4.3HTML Embedded Calculator

The

/publication/html/calculator

element has the following attributes:

• @model: used to control which calculator is available on every page. Possible values are:

• geogebra-classic

• geogebra-graphing

• geogebra-geometry

• geogebra-3d

• none

See Subsection 27.1.2 for more.

### Subsection40.4.4HTML Index Page

The

/publication/html/index-page

element can have the following attribute:

• @ref: the @xml:id of a division which will be a complete page at whatever level the document is chunked (broken into smaller pages).

An index.html page will be created which redirects immediately to this page. Many webservers will serve this page when a URL stops with the enclosing directory. So the Table of Contents (<book>, <article>) or the <frontmatter> are common choices. See Subsection 27.1.1 for more.

### Subsection40.4.5HTML Knowlization

In a conversion to HTML a wide variety of content can be “born” in a knowl. Publisher switches are specified via multiple attributes of a single

/publication/html/knowl

element, with values of yes (do use a knowl) or no (do not use a knowl, present content normally). For example,

<knowl remark="yes"/>

would make every “remark-like” element occur as a knowl where the content is first introduced. Note that these are distinctly different than cross-reference knowls. For more detail see Section 27.2.

This table gives the various attribute names, the default value, and an indication of the elements affected. Note that some items are automatically born knowled (e.g. footnotes) and so there is no option to control that behavior. Also, items such as <figure>, when included in a <sidebyside> are not influenced by the relevant option and are not born knowled.

### Subsection40.4.6HTML Platforms

The

/publication/html

element can have an attribute @platform with values:

• web: the default, meant for self-hosting with no server configuration, features, or assumptions

• runestone: output meant for hosting on a Runestone server (Chapter 30)

Here platform refers to the server where the HTML output will eventually be hosted. The effect is to create minor variations in the output to take advantage of extra features of the indicated platform.

### Subsection40.4.7HTML Style

The

/publication/html/css

element can have the following attributes:

• @colors: used to construct filename for CSS controlling colors.

• @knowls: used to construct filename for CSS controlling knowls.

• @toc: used to construct filename for CSS controlling the Table of Contents.

• @banner: used to construct filename for CSS controlling the banner (masthead).

In each case the default file is replaced and instead a new filename is formed from the attributes value. For example, the specification

<css colors='bright_red'/>

will cause a CSS file named colors_bright_red.css to replace colors_default.css. The server in use can be configured separately. See Section 27.6 for more.

2019-12-08: at this time we recommend experimenting with only one file at a time.

### Subsection40.4.8HTML Search

The

/publication/html/search

element can have the following attribute:

• @google-cx: a Google cx number, gained from configuring search for a site.

Setting this attributes to a non-empty string is the signal to add the relevant code for a search box in the masthead. See Section 27.8 for more.

### Subsection40.4.9HTML Video Embedding

The

/publication/html/video

element can have the following attribute:

• @privacy: allowed values are yes or no.

Setting this to yes (the default) prevents certain tracking cookies from being used. Currently only supported for videos from YouTube. See Subsection 27.1.4 for more.


text:
# LaTeX2HTML

Author Van Abel
Contributors van-abel
Type Free
Current Version 2.3.7
Last Updated 17 May, 2019
Released 23 Jan, 2013
Active Install 300
Support 0 out of 1

LaTeX2HTML makes you write blog like in LaTeX doc. You can just copy and paste to create a new post in WP from your tex source file, then it will looks like amsart document style. It support a lot of raw latex command including theorem like environment and BibTeX style citation.

### Main Functions

• Enable MathJax to render math formula.
• Translate raw LaTeX (almost all the command of amsart) to HTML+CSS markup.
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Good news, this plugin is free for everyone! Since it’s released under the GPL2, you can use it free of charge on your personal or commercial blog.

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# Criterion for element of alternating group to be real

## Statement

Suppose $n$ is a natural number and $A_n$ is the Alternating group (?) on a set of size $n$. An element $g \in A_n$ is a Real element (?) if and only if the cycle decomposition of $g$ satisfies one of these three conditions:

1. The cycle decomposition has a cycle of even length.
2. The cycle decomposition has two cycles of equal odd length. Note that fixed points are counted as cycles of length 1, so this includes any permutation that has two or more fixed points.
3. All the cycles have distinct odd lengths $r_1,r_2,\dots,r_k$ and $\sum_{i=1}^k (r_i - 1)/2$ is even. In other words, the number of $r_i$s that are congruent to $3$ modulo $4$ is even.

Note that the permutations which satisfy condition (1) or (2) are precisely those whose conjugacy class is unsplit from the symmetric group, where every element is real. (see splitting criterion for conjugacy classes in the alternating group). (3) is the case of a conjugacy class that splits in $A_n$ but each element still remains with its inverse.


text:
## Precalculus (10th Edition)

$10$
An absolute maximum is a function value that is larger than or equal to all function values in the whole area of the graph. By this, the absolute maximum is $10$.


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# The Scale Space Theory Understanding

In the scale-space theory the scale-space representation of the signal $f(x), x = (x_1, ..., x_d)$, (in case of image $d = 2$) is given as: $L(x, y; t) = g(x, y; t) * f(x, y)$ where $g(x, y; t)$ is a gaussian kernel with parameter $t$ and $*$ is a convolution. By changing the $t$ parameter we receive a more or less smoothed image. As the result coarser representation (parameter $t$) will not contain small objects or noise.

The main point is to find a way of scale-invariant feature detection, right? So that for some image at it's reduced in size copy the features like keypoints will be detected correctly, even if size is different, without finding other noise keypoints.

1. In the paper they are using the $\gamma$-normalized derivatives. $\delta_{\xi, \gamma-norm} = t^{\gamma / 2} \delta_x$. What is the meaning of using the $\gamma$-normalized derivative, how does it helps in scale-invariancy?

2. From this image we can see that at near the same positions the different keypoints found (different in size). How is that possible?

If you can explain the step-by-step algorithm of scale-invariant feature detection, this would be great. What is actually done? The derivatives can be taken by $x, y$ or $t$. Blob can be detected by taking the the derivative of $L$ by $(x, y)$ variables. How is the derivative by $t$ is helping here?

The paper I was reading is: Feature detection with automatic scale selection

1. It really has been a long time since I have read Lindeberg's papers, so the notation looks a bit strange. As a result, my initial answer was wrong. $\gamma$ is not a scale level. It seems to be a parameter of some sort that can be tuned. It is true that you need to multiply the derivative by the appropriate power of $t$. $t$ itself corresponds to a scale level, and the power depends on the order of the derivative.

2. You can find keypoints at multiple scales in the same location. That is because you look for the local maxima over scales. Here's the intuition: think of an image of a face. At a fine scale you get a blob corresponding to the nose. At a course scale you get a blob corresponding to the whole face. The two blobs are centered at the same point, but have different scales.

3. Here is the whole algorithm:

• Decide which image features you are interested in (e. g. blobs, corners, edges)
• Define a corresponding "detector function" in terms of derivatives, e. g. a Laplacian for blobs.
• Compute derivatives that you need for your detector function at a range of scales.
• Multiply the derivative responses by $t^{m \gamma / 2}$, where $m$ is the order of the derivative, to compensate for magnitude decrease.
• Compute the detector function over the whole scale space.
• Find local maxima of the detector function over $x, y, t$.
• These are your interest points, or keypoints.

## Edit:

1. Lindeberg proves in the paper that $t^{\gamma / 2}$ is the appropriate factor for normalization of derivatives. I don't think I can reproduce the proof here.
2. You do not take derivatives with respect to $t$. You only compute derivatives with respect to $x$ and $y$, but you compute them at a range of scales. One way to think about this is you generate a Gaussian scale space first, by repeatedly blurring the image with a Gaussian filter of some variance $t$. Then compute derivatives with respect to $x$ and $y$ at each scale level.
3. You want to find local maxima over scales because you may have image features of different size at the same location. Think of an image of concentric circles, like a bulls-eye. It will give you high responses of a Laplacian at several scales. Or think of an image of a real human eye filtered by a Laplacian at a range of scales. You will get a high response at a fine scale for the pupil, a high response an some medium scale for the iris, and a high response at a coarse scale for the whole eye.

The whole point is that you do not know at what scale the features of interest might be ahead of time. So you look at all scales.

• 1. How do we provide that the $t^{\gamma / 2}$ is enough multiplier to make the derivative responce actually normalized? 2. I thought that the derivative by $t$ is taken for finding the parameter $t$ where maximum is reached. In order to find the best scale. So as you are writing it seems that I have to compute derivatives by $t$ at a whole selected range. And as the result I will find (by taking derivatives by $x,y$) the features at different scales. Right? – maximus Nov 2 '11 at 1:08
• And as you wrote in 3: Find local maxima of the detector function over $x,y,t$ Why do we need to find local maxima by t? Can you please describe that step in details? Actually how does the derivative by $t$ is used here? – maximus Nov 2 '11 at 1:09
• @maximus Please see the edit to the answer. – Dima Nov 2 '11 at 14:05
• @maximus, I got it wrong before. gamma is not the scale level. I've fixed the answer. – Dima Nov 2 '11 at 17:54


text:
### Multiple Imputation Efficiency

The relative efficiency (RE) of using the finite m imputation estimator, rather than using an infinite number for the fully efficient imputation, in units of variance, is approximately a function of m and (Rubin, 1987, p. 114):

Table 61.7 shows relative efficiencies with different values of m and .

Table 61.7: Relative Efficiencies

m

10%

20%

30%

50%

70%

3

0.9677

0.9375

0.9091

0.8571

0.8108

5

0.9804

0.9615

0.9434

0.9091

0.8772

10

0.9901

0.9804

0.9709

0.9524

0.9346

20

0.9950

0.9901

0.9852

0.9756

0.9662

The table shows that for situations with little missing information, only a small number of imputations are necessary. In practice, the number of imputations needed can be informally verified by replicating sets of m imputations and checking whether the estimates are stable between sets (Horton and Lipsitz, 2001, p. 246).


text:
johnsonscience Posts: 3, Reputation: 1 New Member #1 Oct 5, 2010, 05:09 AM
How much internal energy is converted to Physical Energy?
An 80kg man is standing on a 140kg pontoon boat. The system is motionless. The man dives off. His dive is perfectly horizontal (no PE change). The man goes 7 m/s to the left and the pontoon goes 4 m/s to the right. How much internal energy did the man convert to KE? The KE to the left would be 1960J and the KE to the right would be 1140J. It seems to me that the 820J difference would be dissipated energy due to friction between the boat and the water. So the total overall kinetic energy would be 1960J x 2 = 3920. Is this amount all from the internal energy of the diver? Or is it half of that because the pontoon is experiencing a reaction force? Also, if this could be counted as an explosive collision how would the formula for kinetic energy read? I thought 0 + 0 ----> 1960J + 820J + 1140J but I am not sure this is correct
 ebaines Posts: 12,132, Reputation: 1307 Expert #2 Oct 6, 2010, 05:46 AM

You seem to think that the energy "to the left" must equal the energy "to the right," but this is not true. Kinetic energy is a scalar, not a vector, so it has no direction. The total KE is simply is the KE of the man (1960 J) plus the KE of the boat (1120 J), or 3080 J total.

BTW, check your math for the KE of the boat - I get
$
KE_{Boat}\ = \ \frac 1 2 \times 140\ Kg \times (4 m/s)^2 = 1120\ J
$
 johnsonscience Posts: 3, Reputation: 1 New Member #3 Oct 7, 2010, 05:16 AM
Comment on ebaines's post
Okay! No, I meant to type 1120J!
Is there any dissipated energy?
 ebaines Posts: 12,132, Reputation: 1307 Expert #4 Oct 7, 2010, 05:38 AM

You asked about "dissipated energy." It seems pretty clear that in the real world the boat and the man will slow down due to friction with the water molecules which creates an opoosing force to their movement. Some of their kinetic energy will be transferred to the water (ceating waves and eddies) and some will turn into heat. However, at the instant that the boat is moving at 4 m/s and the man is movig at 7 m/s, their total KE is as shown in the earlier post.

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# 22.4 Magnetic field strength: force on a moving charge in a magnetic

 Page 1 / 7

## Learning objectives

By the end of this section, you will be able to:

• Describe the effects of magnetic fields on moving charges.
• Use the right-hand rule 1 to determine the velocity of a charge, the direction of the magnetic field, and the direction of magnetic force on a moving charge.
• Calculate the magnetic force on a moving charge.

The information presented in this section supports the following AP® learning objectives and science practices:

• 2.D.1.1 The student is able to apply mathematical routines to express the force exerted on a moving charged object by a magnetic field. (S.P. 2.2)
• 3.C.3.1 The student is able to use right-hand rules to analyze a situation involving a current-carrying conductor and a moving electrically charged object to determine the direction of the magnetic force exerted on the charged object due to the magnetic field created by the current-carrying conductor. (S.P. 1.4)

What is the mechanism by which one magnet exerts a force on another? The answer is related to the fact that all magnetism is caused by current, the flow of charge. Magnetic fields exert forces on moving charges , and so they exert forces on other magnets, all of which have moving charges.

## Right hand rule 1

The magnetic force on a moving charge is one of the most fundamental known. Magnetic force is as important as the electrostatic or Coulomb force. Yet the magnetic force is more complex, in both the number of factors that affects it and in its direction, than the relatively simple Coulomb force. The magnitude of the magnetic force     $F$ on a charge $q$ moving at a speed $v$ in a magnetic field of strength $B$ is given by

$F=\text{qvB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \text{,}$

where $\theta$ is the angle between the directions of $\mathbf{\text{v}}$ and $\mathbf{\text{B}}.$ This force is often called the Lorentz force    . In fact, this is how we define the magnetic field strength $B$ —in terms of the force on a charged particle moving in a magnetic field. The SI unit for magnetic field strength $B$ is called the tesla    (T) after the eccentric but brilliant inventor Nikola Tesla (1856–1943). To determine how the tesla relates to other SI units, we solve $F=\text{qvB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ for $B$ .

$B=\frac{F}{\text{qv}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta }$

Because $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ is unitless, the tesla is

$\text{1 T}=\frac{\text{1 N}}{C\cdot \text{m/s}}=\frac{\text{1 N}}{A\cdot m}$

(note that C/s = A).

Another smaller unit, called the gauss    (G), where $1 G={\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}T$ , is sometimes used. The strongest permanent magnets have fields near 2 T; superconducting electromagnets may attain 10 T or more. The Earth’s magnetic field on its surface is only about $5×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}T$ , or 0.5 G.

The direction of the magnetic force $\mathbf{\text{F}}$ is perpendicular to the plane formed by $\mathbf{\text{v}}$ and $\mathbf{\text{B}}$ , as determined by the right hand rule 1 (or RHR-1), which is illustrated in [link] . RHR-1 states that, to determine the direction of the magnetic force on a positive moving charge, you point the thumb of the right hand in the direction of $\mathbf{\text{v}}$ , the fingers in the direction of $\mathbf{\text{B}}$ , and a perpendicular to the palm points in the direction of $\mathbf{\text{F}}$ . One way to remember this is that there is one velocity, and so the thumb represents it. There are many field lines, and so the fingers represent them. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Roofs are sometimes pushed off vertically during a tropical cyclone, and buildings sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena.
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Determine the total force and the absolute pressure on the bottom of a swimming pool 28.0m by 8.5m whose uniform depth is 1 .8m.
how solve this problem?
Foday
P(pressure)=density ×depth×acceleration due to gravity Force =P×Area(28.0x8.5)
Fomukom
for the answer to complete, the units need specified why
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have?
how would I work this problem
Alexia
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
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what is angular velocity
angular velocity can be defined as the rate of change in radian over seconds.
Fidelis
Why does earth exert only a tiny downward pull?
hello
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Why is light bright?
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Kintu
physics can be defined as the natural science that deals with the study of motion through space,time along with its related concepts which are energy and force
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a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole


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# Kris's Research Notes

## November 15, 2011

### Some Droplet Statistics

Filed under: GaAs Simulations — Kris Reyes @ 5:50 am

In this note, we examine droplet width and linear density as a function of temperature $T$ and the amount of Ga deposited $\theta$.

We follow the experimental procedure outlined here. All simulations were performed on a domain of width 1024 atoms (about 292 nm). Droplet statistics were calculated by examining the final atom configuration of each trial. Droplet widths were calculated using the autocorrelation of the resulting height profile and with a simple thresholding scheme. Thresholding was done by ignoring those atoms below a specified threshold height above the minimum height of the configuration, as illustrated below:

### $\theta$ series ($T = 456$K)

We varied $\theta$ in:

$\displaystyle \theta = \left\{2.7, 2.9, 3.7, 4.2, 4.7\right\}$ monolayers of Ga.

Temperature was fixed at $T = 456$K. Statistics were obtained by ensemble averaging 128 independent trials, and are summarized below:

 $\theta$ # Droplets Width – Thresholding (#atoms) Width – Autocorrelation (#atoms) 2.7 8.5234 32.4204 29.7435 2.9 8.6250 35.2875 30.4465 3.7 8.3672 54.2442 32.8690 4.2 7.7891 69.3675 34.9988 4.7 7.3047 84.8789 35.9573

Here are plots of the data along with our experimental results:

Droplet Width (nm) vs Thickness

Linear Density (#/μm) vs. Thickness

### Temperature Series $\theta = 2.9 ML$

Here, we vary temperature $T \in \left\{456, 464, 723\right\}$K and fix $\theta 2.9$ monolayers Ga. Simulation statistics was obtained by ensemble averaging 128 independent trials.

 $T$(K) # Droplets Width – Thresholding (#atoms) Width – Autocorrelation (#atoms) 456 8.5625 35.4258 31.1199 464 8.1667 35.2945 31.1053 472 8.3402 34.1838 29.5492

Here is the data plotted with experimental results:

Droplet Width (nm) vs. Temperature (°C)

Linear Density (#/μm) vs. Temperature (°C)

We observe that simulation results are essentially constant while there is a clear trend in experimental data. This is readily explained by the fact that in our simulations, low temperatures yield rough surfaces (exacerbated by the fact that simulation results in 1+1 dimension are often rougher than 2+1). Hence, there is a significant amount of noise in simulation data. This can be addressed by both increasing the amount of Ga deposited (yielding larger droplets and hence delineating surface roughness from genuine droplets) and increasing temperature (making the surface smoother).

Indeed, if we consider $\theta = 3.5$ monolayers Ga and a larger for temperature (including higher temperature cases) we get similar trends to the data. For example, if we consider temperature in

$\displaystyle T \in \left\{ 450, 470, 490, 510, ..., 590 \right\}$ K,

and fix $\theta = 3.5$ monolayer Ga, we obtain the following plots:

Droplet Width (nm) vs. Temperature (°C)

Linear Density (#/μm) vs. Temperature (°C)

It appears that simulation data is too noisy for cases $T < 220^\circ$C, but correct qualitative trends appear past this point.

N.B. For these trials, statistics were obtained from 16 independent trials for each choice of $T$.


text:
Synopsis: Pinpointing Planck’s Constant with GPS

Using the orbiting network of GPS satellites, researchers have placed new limits on how much Planck’s constant varies with respect to relativistic changes.

GPS is helping drivers find their way and parents track their kids and pets. But now a pair of researchers—reporting in Physical Review Letters—has used the same technology to put new limits on variations in Planck’s constant.

Certain theories allow physical constants, such as the speed of light or the gravitational constant, to vary, and some astronomical observations have been interpreted as suggesting the electromagnetic coupling was different in the past. Testing these hypotheses often requires sophisticated instruments. But James Kentosh and Makan Mohageg of California State University, Northridge, have found a way to use the ubiquitous global positioning system, or GPS, to evaluate the constancy of Planck’s constant, $h$.

GPS relies on atomic clocks, which are sensitive to Planck’s constant through their ticking frequency, $f=E/h$, where $E$ is the energy of a specific atomic transition. For a clock orbiting in one of the $32$ GPS satellites, this frequency can shift with respect to ground-based clocks because of well-known relativistic effects. The GPS system keeps track of this frequency drift and broadcasts a clock correction with its signal.

Kentosh and Mohageg looked through a year’s worth of GPS data and found that the corrections depended in an unexpected way on a satellite’s distance above the Earth. This small discrepancy could be due to atmospheric effects or random errors, but it could also arise from a position-dependent Planck’s constant. Assuming the latter, the authors derive an upper limit on Planck variation. – Michael Schirber

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text:
# Markdown cells not rendering directives

Hello!

I am trying to render some {math} and {note} directives in my notebook but I can’t get them to render. The markdown is the following:

Single Dollar sign directive

$\hat{Y} = \hat{\beta}_{0} + \sum \limits _{j=1} ^{p} X_{j}\hat{\beta}_{j}$

-------------------------------

Double dollar sign:

$$w_{t+1} = (21 + r_{t+1}) s(w_t) + y_{t+1}$$

--------------------------------

Math directive

{math}

w_{t+1} = (12 + r_{t+1}) s(w_t) + y_{t+1}

-------------------------------

Inline math directive:

{math}x_{hey}=it+is^{math}

Since Pythagoras, we know that {math}a^2 + b^2 = c^2

-------------------------------

Latex math

$$\int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15}$$

-------------------------------

Note directive

note
this is supposed to be a note

I would expect the note and math formulas to render, but instead here is what I get:

Any ideas why only the Latex and dollar math are rendering, and the rest isn’t? I am not sure how to debug this as I am pretty new to Jupyter. (Note: this happens in both Jupyter Notebook as well as Jupyter Lab).

Troubleshoot info:

||jupyter-client 7.3.4|
||jupyter-core 4.11.1|
||jupyter-server 1.18.1|
||jupyterlab 3.4.5|
||jupyterlab-mathjax3 4.3.0|
||jupyterlab-pygments 0.2.2|
||jupyterlab-server 2.15.0|

I think that this is not a valid Markdown syntax. Plain JupyterLab uses marked.js which is compatible with CommonMark, GitHub Flavored Markdown and the original markdown.pl.

Adding syntax highlighting in CommonMark and GitHub Flavored Markdown is performed via text after backticks without the braces, see examples: CommonMark 142 and GFM 112, but I am not aware of standard syntax other than $ and $\$ which would typeset LaTeX in markdown. If you have a reference for such, please do provide!

What you are trying to do appears to be RMarkdown flavour. Unfortunately, contrary to universally accepted flavours, I could not find a formal specification for RMarkdown; it would make it easier to re-use if there was one making it a proper open standard. There are however already three options if you would like to use that format:

Personal opinion: I really wish that RStudio went for closer collaboration with existing standards (including Jupyter) rather than reinventing a wheel, but here we are.

2 Likes

This also looks a lot like MyST, although I’ve never seen a bare :::note directive (vs :::{note}). For MyST, as @krassowski notes, we do have a JupyterLab renderer: jupyterlab-myst · PyPI

You could try this latest pre-release (pip install --pre jupyterlab-myst) and see if it works for your use case.

2 Likes

Thanks for the replies. Seems like my confusion came from me reading the documentation of Jupyter Book thinking it was Jupyter Notebook. I wonder if there should be some warning on that documentation to disambiguate the two.

I tried the jupyterlab-myst · PyPI and it worked! So I think I will go with that one.

Thanks again!

1 Like