text:
# Placing Intermediate and Output Files in Another Folder

I am trying to figure out a way to cleanly separate the TEX files from all of the intermediate and output files. Various approaches I've tried, including the one described by ‘Bundle’ TeX output in a directory don't seem to work. I'm using TexMaker + MikTex (portable). Ideally, I would like my file/directory structure to look something like this:

• texproject
• texfiles
• 1.tex
• 2.tex
• intfiles
• 1.out
• 1.log
• 1.aux
• 2.out
• 2.log
• 2.aux
• pdffiles
• 1.pdf
• 2.pdf

Instead, what I have, is the jumbled-up mess where all of these files live in the same directory. I've tried setting the ouput-directory, for example, on pdflatex but this doesn't work. The pdf file still gets created in the same place but now it doesn't show up in the previewer. This should be pretty easy to solve but I've so far been unsuccessful. How can I set up my environment to use the above structure when compiling?

Update: I believe I have narrowed this down to something specific to TexMaker. If the desired aux and out directories are passed via the command line to pdflatex, everything builds as it should. Yet, I get numerous message boxes popping up saying that files can't be found and the PDF doesn't display in the internal viewer. I can't see any obvious options within TexMaker to change to address this. So, has anyone been able to make the scenario I'm describing work with TexMaker?

• Have you tried -aux_directory=../intfiles and -output-directory=../pdffiles? That should work with MikTeX. – Peter Grill Nov 16 '11 at 16:41
• @PeterGrill Well, it partially works; I've tried passing these parameters to PDFLATEX but there are still things that seem to be looking in the current directory for the output and intermediate files. Although build succeeds, numerous "can't find file" message boxes pop up and the internal pdf viewer can't seem to find the file – 3Sphere Nov 16 '11 at 17:36
• Wondering if that has anything to do with .aux files from previous runs? I will try to test this later today when I am on my PC that has MikTeX. – Peter Grill Nov 16 '11 at 17:48
• @PeterGrill No, even when all aux files everywhere are cleared out it has this problem. Something (TexMaker?) is trying to write to a log file and can't find the file and the internal PDF viewer can't find the PDF; I'm thinking this is something TexMaker-specific because MikTex seems to build the files properly. – 3Sphere Nov 16 '11 at 17:58
• Just tried that as well (using Win7+MikTex+Texmaker). Texmaker invokes pdflatex, which writes all files to the folders specified. Writing works and all files are created as they should. But then Texmaker tries to read and display the log in the integrated Messages/Log viewer and the pdf in the integrated pdf viewer and does not find the files. – matth Nov 16 '11 at 18:27

As suggested by lockstep, I write a short answer: AFAIK, this is currently not possible, because the path cannot be changed for the messages / log viewer and for the pdf viewer from within the Texmaker GUI. There is a feature request on the Texmaker issue list to make the path configurable: http://code.google.com/p/texmaker/issues/detail?id=97 Maybe you are lucky and in a later release this will be available. I personally would also use this new feature (and only backup the input folder, not the output folder).

Since Texmaker 3.4, you can specify the option "Use a build directory for output", which creates a "build" folder in your project folder, and saves everything but the .tex file there. See the manual Now you can also specify the output directory as an option to pdflatex as e.g. -output-directory pdf. But you will also have a copy of the pdf in the build folder no matter what. And I haven't found a straight-forward way to have the output file in the root folder, either.

(And I also have a .gz file in the folders, but that might be an option you can turn off.)

• note that you then need to update your biber / bibtex command as biber build/% (respectively bibtex build/%.aux). Regarding the pdf as output in the root folder, I am using a symbolic link (ln -s build/pdfName.pdf pdfName.pdf). – ebosi Oct 22 '15 at 9:50

For TeXstudio (since a fork from texmaker, it might be similar there):

1. Add the options -aux_directory and -output-directory to the latex pdflatex and other commands, if needed.
2. Change the appropriate lines (160 and 161 in my case) in the texstudio.ini file, in W7 usually located under

C:\Users\username\AppData\Roaming\texstudio

Modify the path for the two varables accordingly.

Tools\Log%20Paths=C:\\\Users\\\username\\\AppData\\\Local\\\Temp

Tools\PDF%20Paths=D:\\\repos\\\pdf

(double-backslashes by intention)

Enjoy!

• Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. A suggestion: Do us a favour and change your username to something more telling than "user1234". – Martin Schröder Jan 22 '14 at 9:40

In Texstudio, if you want to save outputs of latex in a directory (and thereby have the directory that contains your source files uncluttered, set the -output-directory option of pdflatex command in Tools → Configure TextStudio → Commands Change the value of PdfLatex the -output-directory option and output is the directory you need the output files to be stored. It can be any name and in any location, but make sure you give is the full path (or relative path).

pdflatex -synctex=1 -interaction=nonstopmode -output-directory=output %.tex

The Linux version of Texstudio (that I am using) does not support -aux-directory option.

After doing that, if the pdf preview does not work, refer to this solution.

• After doing this, I get an error when building the bibliography: "Process started: bibtex "filename.aux". I couldn't open file name filename.aux'. Process exited with error(s)." And the bibliography is not created. – Erel Segal-Halevi Dec 2 '16 at 13:06

In TeXstudio (2.10.8) we can set a subdirectory, in this example named "output", as follows:

Menu > Options > Configure TeXstudio ...

• Commands [Tab] > PdfLaTeX. Add key/value switch -output-directory=output.

So my command changed from

pdflatex.exe -synctex=1 -interaction=nonstopmode %.tex

... to ...

pdflatex.exe -synctex=1 -interaction=nonstopmode -output-directory=output %.tex

If you are using other build commands aside from "PdfLaTeX" then make analogous changes there. I'm only building against PdfLaTex so that's all I had to change.

• Commands [Tab] > External PDF Viewer. I'm using Foxit Reader, so in my case I changed from ...

 "C:\PROGRAM FILES (X86)\FOXIT SOFTWARE\FOXIT READER\FOXITREADER.EXE" "?am.pdf"

... to ...

 "C:\PROGRAM FILES (X86)\FOXIT SOFTWARE\FOXIT READER\FOXITREADER.EXE"  C:\Users\John\Documents\Sda\Code\Markup\Latex\Examples\output\?rm.pdf"

It would be desirable to only need a relative path, something like \output\?rm.pdf, but so far I've only been able to get it working with the hardcoded absolute path.

• Commands [Tab] > Biber (this is used for bibliographies with the biblatex package).

Change

 biber.exe %

... to ...

 biber.exe -output-directory=output -input-directory=output %

• Commands [Tab] > BibTeX (this is used for bibliographies with the bibtex package).

Change

 bibtex.exe %.aux

... to ...

 bibtex.exe  output/%.aux

• Build [Tab] > Additional Search Paths. Add \output (with the backslash) to "Log File" and "PDF File" fields.

With your source MyExample.tex file loaded in the current window press F5 (TeXstudio Menu > Tools > Build & View). After the build completes the PDF viewer should correctly pickup and display \output\MyExample.pdf.

• In my TeXstudio 2.12.10 installation, the above works, and relative paths are also accepted everywhere. – Waldir Leoncio Aug 16 '18 at 7:19

I am not an expert in any of these but this is a solution that works for me (I am using texmaker under Linux but I am sure there are similar commands for Windows):

1. Options>Configure Texmaker>Commands mark 'Use a "build" subdirectory for output files'
2. Options>Configure Texmaker>Quick Build mark 'User' and type in the commands that you need for compliling the way you want. (You can use the 'wizard' if you need help.)

|mkdir pdf|cp build/%.pdf pdf

(Copying instead of moving provides the pdf-viewer from not finding the .pdf)

• Welcome to TeX.Stackexchange! A similar approach is also used by ltx2any. Using this tool would have the advantage that it also adjusts the paths in synctex files etc. – user36296 Aug 30 '18 at 13:13

If you are working on Linux, you might use symbolic links as a workaround, e.g. create the directories ../texfiles and ../pdffiles, then run the following script:

#! /bin/bash

for i in *.tex
do
ln -s $(pwd)/$i ../texfiles
done

for i in *.pdf
do
ln -s $(pwd)/$i ../pdffiles
done

The script can be expanded to update the directories dynamically (needs to be called via write18` or maybe a hook in TexMaker?).


text:
# KATEX MATH OP

This is a list of TeX functions supported by KaTeX. It is sorted into logical groups.There is a similar Support Table, sorted alphabetically, that lis

format: plain


text:
# 35. What is 2 as an improper fraction? A-11/4 b-23/4 c-4/23 d-8/4

###### Question:

35. What is 2 as an improper fraction? A-11/4 b-23/4 c-4/23 d-8/4

### Please can you tell me how to solve this. The largest side of a triangle is six more units than the smallest side. The third side is twice the smallest side. If the perimeter of the triangle is 25 units, write and solve an equation to find the lengths of all three sides of the triangle.

Please can you tell me how to solve this. The largest side of a triangle is six more units than the smallest side. The third side is twice the smallest side. If the perimeter of the triangle is 25 units, write and solve an equation to find the lengths of all three sides of the triangle....

### How much did a full set of medival battle equipment weigh

how much did a full set of medival battle equipment weigh...

### Determine whether formulas u and v are logically equivalent (you may use truth tables or properties of logical equivalences). (a) u : (p ⇒ q) ∧ (p ⇒ q) v : p (b) u : p ⇒ q v : q ⇒ p (c) u : p ⇔ q v : q ⇔ p (d) u : (p ⇒ q) ⇒ r v : p ⇒ (q ⇒ r)

Determine whether formulas u and v are logically equivalent (you may use truth tables or properties of logical equivalences). (a) u : (p ⇒ q) ∧ (p ⇒ q) v : p (b) u : p ⇒ q v : q ⇒ p (c) u : p ⇔ q v : q ⇔ p (d) u : (p ⇒ q) ⇒ r v : p ⇒ (q ⇒ r)...

### Sound waves with a large distance between A and B would be sound waves that have a large ________ and produce loud sounds. A) amplitude B) frequency C) pitch D) resonance

Sound waves with a large distance between A and B would be sound waves that have a large ________ and produce loud sounds. A) amplitude B) frequency C) pitch D) resonance...

### What is a composition that is designed so that every element in the composition represents a single idea or theme? A. Abstract B. Allegory C. Chiaroscuro D. Mosaic

What is a composition that is designed so that every element in the composition represents a single idea or theme? A. Abstract B. Allegory C. Chiaroscuro D. Mosaic...

### A cannonball is launched off a cliff 125 m in height (dy) with a HORIZONTAL VELOCITY (Vx) of 33 m/s. How long (flight time) does it take for the cannonball to land?

A cannonball is launched off a cliff 125 m in height (dy) with a HORIZONTAL VELOCITY (Vx) of 33 m/s. How long (flight time) does it take for the cannonball to land?...

### While looking at bacteria under a microscope, you decide to take a break. When you return, the amount of bacteria has doubled in size. Your break probably only took _____. 24 hours 1-3 hours 8 hours 48 hours

While looking at bacteria under a microscope, you decide to take a break. When you return, the amount of bacteria has doubled in size. Your break probably only took _____. 24 hours 1-3 hours 8 hours 48 hours...

### The area of each triangular base is: A = 1/2(b)(h) A = 1/2(4)(3) A = in.2 There are two bases, so total base area is in.2. The lateral area is: A = 8(4) + 8(3) + 8(5) A = in.2 The total surface area is in.2.

The area of each triangular base is: A = 1/2(b)(h) A = 1/2(4)(3) A = in.2 There are two bases, so total base area is in.2. The lateral area is: A = 8(4) + 8(3) + 8(5) A = in.2 The total surface area is in.2....

### What significant event happened in the year of 1807?

What significant event happened in the year of 1807?...

### Simplify LaTeX: \Large\frac{-4^{6} \cdot 4^{2}}{4^{4}}

Simplify LaTeX: \Large\frac{-4^{6} \cdot 4^{2}}{4^{4}}...

### Perceiving pain and assisting in regulating muscle reflexes, breathing, and arousal are all functions of the __________.

Perceiving pain and assisting in regulating muscle reflexes, breathing, and arousal are all functions of the __________....

### Help please I promise to give brainliest

Help please I promise to give brainliest...

### Which is the best approach for avoiding activation of the adductors during a floor bridge? a. The feet should be wider than hip-width apart and straight ahead b. The feet should be together and pointed straight ahead c. The feet should be hip-width apart and straight ahead d. The feet should be hip-width apart and turned out

Which is the best approach for avoiding activation of the adductors during a floor bridge? a. The feet should be wider than hip-width apart and straight ahead b. The feet should be together and pointed straight ahead c. The feet should be hip-width apart and straight ahead d. The feet should be hip-...

### Find the measure of the exterior 21. 60° 7 25

Find the measure of the exterior 21. 60° 7 25...

### Base your answer to the question on the information below and on your knowledge of chemistry. Nitrogen dioxide, NO2, is a dark brown gas that is used to make nitric acid and to bleach flour. Nitrogen dioxide has a boiling point of 294 K at 101.3 kPa. In a rigid cylinder with a movable piston, nitrogen dioxide can be in equilibrium with colorless dinitrogen tetroxide, N2O4. This equilibrium is represented by the equation below. 2NO2(g) ⇌ N2O4(g) + 58 kJ Determine the oxidation state of nitrogen i

Base your answer to the question on the information below and on your knowledge of chemistry. Nitrogen dioxide, NO2, is a dark brown gas that is used to make nitric acid and to bleach flour. Nitrogen dioxide has a boiling point of 294 K at 101.3 kPa. In a rigid cylinder with a movable piston, nitrog...

### What important difference do you observe between the original thesis statement and the revised version (“As these works show, positive thinking is one of the best ways to respond to conflict, as well as an effective path to healing for those who survive p

What important difference do you observe between the original thesis statement and the revised version (“As these works show, positive thinking is one of the best ways to respond to conflict, as well as an effective path to healing for those who survive p...

### Why do scientists believe that light is made of streams of particles?

Why do scientists believe that light is made of streams of particles?...

### Modern cell theory is best defined as being based on

modern cell theory is best defined as being based on...


text:
# Definition:Quotient Group/Historical Note

The idea of a quotient group appeared in the work of Marie Ennemond Camille Jordan in the $1860$s.
However, the modern formulation using cosets did not appear till the work of Otto Ludwig Hölder in $1889$, fairly late on in the history of group theory.


text:
# Transformation Function: Gonzalez and Woods

I have been reading Image Processing from Gonzalez and Woods and in the chapter Image Transformation I have come across this equation

$$T \left(u, v\right) = \sum_{x=0}^{M-1} { \sum_{y=0}^{N-1} { f \left(x, y\right) \, r \left(x, y, u, v\right) } } \,.$$

I am unable to understand this mathematical notation. Any help will be highly appreciated. Thanks

Those $${\sum}\text{-signs}$$ specify sigma notation. They're basically like for loops where you keep summing up the value specified in the body.

The equation, $$T \left(u, v\right) = \sum_{x=0}^{M-1} { \sum_{y=0}^{N-1} { f \left(x, y\right) \, r \left(x, y, u, v\right) } } \,,$$

translates into:

public double T(double u, double v)
{
var sum = 0.0;

for (int x=0; x <= M-1; ++x)
{
for (int y=0; y <= N-1; ++y)
{
sum = sum + (f(x, y) * r(x, y, u, v));
}
}

return sum;
}

In other words, it defines some function, $$T \left(u, v\right) ,$$ that takes two input variables: $$u$$ and $$v .$$ You also need to know two other functions: $$f \left(x, y\right)$$ and $$r\left(x, y, u, v\right) ,$$ which means you'll need at least 2 more equations.

Then:

1. Start with a starting total value of $$0 .$$

2. Set $$x$$ to $$0 .$$

3. Set $$y$$ to $$0 .$$

4. Calculate $$f\left(x, y\right) r\left(x, y, u, v\right) ,$$ then add this value to the total.

• On the first evaluation, $$x=0$$ and $$y=0 ,$$ so that's $$f\left(0, 0\right) r\left(0, 0, u, v\right) .$$
5. Increment $$y .$$

• If it's still less-than-or-equal-to $$N - 1 ,$$ then go back to Step (4).
6. Increment $$x .$$

• If it's still less-than-or-equal-to $$M - 1 ,$$ then go back to Step (3).
7. The final total is the calculated value for $$T \left(u, v\right) .$$

• What is r(x,y,u,v) referring to? what are those extra parameters u and v? – Turing101 Dec 17 '19 at 13:42
• Are we multiplying each pixel intensity alone with that function? any explanation showing how this works on a matrix level by taking 2 matrices will be of much help – Turing101 Dec 17 '19 at 13:44
• @HIRAKMONDAL I'm not sure. The equation is very generic, so it's hard to guess what the variables might refer to without seeing the book. $u$ and $v$ may be coordinate positions for a pixel. You'll probably have to read where the book declares its notation to figure out what the symbols match up to. – Nat Dec 17 '19 at 13:49


text:
Verify that thefunction

Q1(t) = EoC ( 1 –e-t/(RC))           (1)

Is asolution to the initial value problem

dQ1/dt = - 1/(RC) * Q1 + 1/R *E0,   Q1 (0) =0              (2)

where R, C, and E0 are constants. That is,by plugging t = 0 into the formula (1)

show that the initial conditionis satisfied, and then by differentiating the formula

(1) and comparing with theright-hand side of the differential equation show that

Q1(t)satisfies the differential equation. (In other words, do not try tofind the

solution of the initial-valueproblem, but rather just check that the given function

solves the problem.)

Verify that

Q2(t) = E0 /(R(?2 + (120*pi)2) [ 120*pi ( e-?t – cos(120*pi*t)) +?*sin(120*pi*t)]       (3)

Satisfies the initial-valueproblem

dQ2 / dt = -(1/(RC) *Q2 + 1/R * E0sin(120*pi*t),    Q2(0) =0                      (4)

where theconstant = 1/(RC) has units ofHz=sec−1. What do theseinitial

conditions represent forthe system at the time the switch is closed?


text:
This function exports all subsets in the specified av_state to individual output files. All output files are subsequently packed in a .tar file. Data sets with less than 80 columns are stored inline in the .sas script, otherwise, the data is stored in separate files.

store_file(av_state, filename, inline_data, file_type = c("SPSS", "STATA"))

## Arguments

av_state an object of class av_state the name that the exported files should be saved under. Do not include a path. When this argument is missing, the filename of the input file (used in load_file) is substituted. boolean argument to determine whether data should be stored inline in the .sas script. This argument is optional, and defaults to TRUE if the data set has less than 80 columns, and to FALSE otherwise. sets the type of file export that should be used. Currently, only 'SPSS' is supported.

## Examples

# NOT RUN {
av_state <- load_file("../data/input/ID68 basisbestand.sav",log_level=3)
av_state <- add_trend(av_state)
av_state <- set_timestamps(av_state,date_of_first_measurement="2012-07-12",
measurements_per_day=3,log_level=3)
store_file(av_state)
# }


text:
Compute operator norm by image on orthonormal basis

Let $e_n$ a orthonormal basis for a Hilbert space and $T$ a bounded linear operator. Is the following correct?

$$\lVert T \lVert^2 \leq \sup_{n \in \mathbb{N}} \sum_{k \in \mathbb{N}} |\langle e_n,Te_k\rangle|^2$$

-
Your inequality is not homogeneous in $T$ (a square is missing in the LHS). The $\max$ should be a $\sup$ and note that by Bessel's inequality, the RHS is $\sup_n\lVert T^*e_n\rVert^2$. –  Davide Giraudo Jun 15 '12 at 10:51
Since by Bessel's inequality $$\sum_{k\in\Bbb N}|\langle e_n,Te_k\rangle|^2=\sum_{k\in\Bbb N}|\langle T^*e_n,e_k\rangle|^2=\lVert T^*e_n\rVert^2$$ and $\lVert T\rVert=\lVert T^*\rVert$, the question is: given a linear bounded operator $T\colon H\to H$, do we have $\lVert T\rVert^2=\sup_{n\in\Bbb N}\lVert Te_n\rVert^2$? (the reverse inequality is quite obvious)
The answer is no. Indeed, consider $T\colon \ell^2(\Bbb N)\to \ell^2(\Bbb N)$ given by $T(x)(n)=0$ for $n\geq 1$ and $T(x)(0)=\sum_{j=1}^{+\infty}\frac{x(j)}j$ (well-defined by Cauchy-Schwarz inequality). Such an operator is linear an bounded. It's norm is atteind only for a multiple of the sequence $\{\frac 1j\}_{j\geq 1}$. Denote $e^{(n)}$ the sequence whose $n$-th term is $1$, the other are $0$ (this gives an orthonormal basis). We have $\lVert T(e^{(n)})\rVert=\frac 1n\leq 1<\lVert T\rVert=\frac{\pi}{\sqrt 6}$.
Thank you for your answer. Do you know if there is a different way to estimate the norm in terms of $<e_n, T e_k>$? –  Rejnik Jun 15 '12 at 18:28


text:
0th

Percentile

##### Drawing radar chart (a.k.a. spider plot)

Drawing the radar chart with several lines from a data frame, which must be composed of more than 3 variables as axes and the rows indicate cases as series.

##### Usage
radarchart(df, axistype, seg, pty, pcol, plty, plwd, pdensity, pangle, pfcol,
cglty, cglwd, cglcol, axislabcol, title, maxmin, na.itp, centerzero,
vlabels, vlcex, caxislabels, calcex, paxislabels, palcex, ...)
##### Arguments
df

The data frame to be used to draw radarchart. If maxmin is TRUE, this must include maximum values as row 1 and minimum values as row 2 for each variables, and actual data should be given as row 3 and lower rows. The number of columns (variables) must be more than 2.

axistype

The type of axes, specified by any of 0:5. 0 means no axis label. 1 means center axis label only. 2 means around-the-chart label only. 3 means both center and around-the-chart (peripheral) labels. 4 is *.** format of 1, 5 is *.** format of 3. Default is 0.

seg

The number of segments for each axis (default 4).

pty

A vector to specify point symbol: Default 16 (closed circle), if you don't plot data points, it should be 32. This is repeatedly used for data series.

pcol

A vector of color codes for plot data: Default 1:8, which are repeatedly used.

plty

A vector of line types for plot data: Default 1:6, which are repeatedly used.

plwd

A vector of line widths for plot data: Default 1, which is repeatedly used.

pdensity

A vector of filling density of polygons: Default NULL, which is repeatedly used.

pangle

A vector of the angles of lines used as filling polygons: Default 45, which is repeatedly used.

pfcol

A vector of color codes for filling polygons: Default NA, which is repeatedly usd.

cglty

Line type for radar grids: Default 3, which means dotted line.

cglwd

Line width for radar grids: Default 1, which means thinnest line.

cglcol

Line color for radar grids: Default "navy"

axislabcol

Color of axis label and numbers: Default "blue"

title

if any, title should be typed.

maxmin

Logical. If true, data frame includes possible maximum values as row 1 and possible minimum values as row 2. If false, the maximum and minimum values for each axis will be calculated as actual maximum and minimum of the data. Default TRUE.

na.itp

Logical. If true, items with NA values are interpolated from nearest neighbor items and connect them. If false, items with NA are treated as the origin (but not pointed, only connected with lines). Default FALSE.

centerzero

Logical. If true, this function draws charts with scaling originated from (0,0). If false, charts originated from (1/segments). Default FALSE.

vlabels

Character vector for the names for variables. If NULL, the names of the variables as colnames(df) are used. Default NULL.

vlcex

Font size magnification for vlabels. If NULL, the font size is fixed at text()'s default. Default NULL.

caxislabels

Character vector for center axis labels, overwriting values specified in axistype option. If NULL, the values specified by axistype option are used. Default is NULL.

calcex

Font size magnification for caxislabels. If NULL, the font size is fixed at text()'s default. Default NULL.

paxislabels

Character vector for around-the-chart (peripheral) labels, overwriting values specified in axistype option. If NULL, the values specified by axistype option are used. Default is NULL.

palcex

Font size magnification for paxislabels. If NULL, the font size is fixed at text()'s default. Default NULL.

...

Miscellaneous arguments to be given for plot.default().

##### Value

No value is returned.

##### Examples
# NOT RUN {
# Data must be given as the data frame, where the first cases show maximum.
maxmin <- data.frame(
total=c(5, 1),
phys=c(15, 3),
psycho=c(3, 0),
social=c(5, 1),
env=c(5, 1))
# data for radarchart function version 1 series, minimum value must be omitted from above.
RNGkind("Mersenne-Twister")
set.seed(123)
dat <- data.frame(
total=runif(3, 1, 5),
phys=rnorm(3, 10, 2),
psycho=c(0.5, NA, 3),
social=runif(3, 1, 5),
env=c(5, 2.5, 4))
dat <- rbind(maxmin,dat)
op <- par(mar=c(1, 2, 2, 1),mfrow=c(2, 2))
radarchart(dat, axistype=1, seg=5, plty=1, vlabels=c("Total\nQOL", "Physical\naspects",
"Phychological\naspects", "Social\naspects", "Environmental\naspects"),
title="(axis=1, 5 segments, with specified vlabels)", vlcex=0.5)
radarchart(dat, axistype=2, pcol=topo.colors(3), plty=1, pdensity=c(5, 10, 30),
pangle=c(10, 45, 120), pfcol=topo.colors(3),
title="(topo.colors, fill, axis=2)")
radarchart(dat, axistype=3, pty=32, plty=1, axislabcol="grey", na.itp=FALSE,
title="(no points, axis=3, na.itp=FALSE)")


text:
# aiida.orm.nodes.data.remote.stash package¶

Module with data plugins that represent files of completed calculations jobs that have been stashed.

class aiida.orm.nodes.data.remote.stash.RemoteStashData(stash_mode: aiida.common.datastructures.StashMode, **kwargs)[source]

Data plugin that models an archived folder on a remote computer.

A stashed folder is essentially an instance of RemoteData that has been archived. Archiving in this context can simply mean copying the content of the folder to another location on the same or another filesystem as long as it is on the same machine. In addition, the folder may have been compressed into a single file for efficiency or even written to tape. The stash_mode attribute will distinguish how the folder was stashed which will allow the implementation to also unstash it and transform it back into a RemoteData such that it can be used as an input for new CalcJobs.

This class is a non-storable base class that merely registers the stash_mode attribute. Only its subclasses, that actually implement a certain stash mode, can be instantiated and therefore stored. The reason for this design is that because the behavior of the class can change significantly based on the mode employed to stash the files and implementing all these variants in the same class will lead to an unintuitive interface where certain properties or methods of the class will only be available or function properly based on the stash_mode.

__abstractmethods__ = frozenset({})
__init__(stash_mode: aiida.common.datastructures.StashMode, **kwargs)[source]

Construct a new instance

Parameters

stash_mode – the stashing mode with which the data was stashed on the remote.

__module__ = 'aiida.orm.nodes.data.remote.stash.base'
_abc_impl = <_abc_data object>
_logger: Optional[logging.Logger] = <Logger aiida.orm.nodes.data.remote.stash.base.RemoteStashData (REPORT)>
_plugin_type_string = 'data.remote.stash.RemoteStashData.'
_query_type_string = 'data.remote.stash.'
_storable = False
property stash_mode

Return the mode with which the data was stashed on the remote.

Returns

the stash mode.

class aiida.orm.nodes.data.remote.stash.RemoteStashFolderData(stash_mode: aiida.common.datastructures.StashMode, target_basepath: str, source_list: List, **kwargs)[source]

Data plugin that models a folder with files of a completed calculation job that has been stashed through a copy.

This data plugin can and should be used to stash files if and only if the stash mode is StashMode.COPY.

__abstractmethods__ = frozenset({})
__init__(stash_mode: aiida.common.datastructures.StashMode, target_basepath: str, source_list: List, **kwargs)[source]

Construct a new instance

Parameters
• stash_mode – the stashing mode with which the data was stashed on the remote.

• target_basepath – the target basepath.

• source_list – the list of source files.

__module__ = 'aiida.orm.nodes.data.remote.stash.folder'
_abc_impl = <_abc_data object>
_logger: Optional[logging.Logger] = <Logger aiida.orm.nodes.data.remote.stash.folder.RemoteStashFolderData (REPORT)>
_plugin_type_string = 'data.remote.stash.folder.RemoteStashFolderData.'
_query_type_string = 'data.remote.stash.folder.'
_storable = True
property source_list

Return the list of source files that were stashed.

Returns

the list of source files.

property target_basepath

Return the target basepath.

Returns

the target basepath.

## Submodules¶

Data plugin that models an archived folder on a remote computer.

class aiida.orm.nodes.data.remote.stash.base.RemoteStashData(stash_mode: aiida.common.datastructures.StashMode, **kwargs)[source]

Data plugin that models an archived folder on a remote computer.

A stashed folder is essentially an instance of RemoteData that has been archived. Archiving in this context can simply mean copying the content of the folder to another location on the same or another filesystem as long as it is on the same machine. In addition, the folder may have been compressed into a single file for efficiency or even written to tape. The stash_mode attribute will distinguish how the folder was stashed which will allow the implementation to also unstash it and transform it back into a RemoteData such that it can be used as an input for new CalcJobs.

This class is a non-storable base class that merely registers the stash_mode attribute. Only its subclasses, that actually implement a certain stash mode, can be instantiated and therefore stored. The reason for this design is that because the behavior of the class can change significantly based on the mode employed to stash the files and implementing all these variants in the same class will lead to an unintuitive interface where certain properties or methods of the class will only be available or function properly based on the stash_mode.

__abstractmethods__ = frozenset({})
__init__(stash_mode: aiida.common.datastructures.StashMode, **kwargs)[source]

Construct a new instance

Parameters

stash_mode – the stashing mode with which the data was stashed on the remote.

__module__ = 'aiida.orm.nodes.data.remote.stash.base'
_abc_impl = <_abc_data object>
_logger: Optional[logging.Logger] = <Logger aiida.orm.nodes.data.remote.stash.base.RemoteStashData (REPORT)>
_plugin_type_string = 'data.remote.stash.RemoteStashData.'
_query_type_string = 'data.remote.stash.'
_storable = False
property stash_mode

Return the mode with which the data was stashed on the remote.

Returns

the stash mode.

Data plugin that models a stashed folder on a remote computer.

class aiida.orm.nodes.data.remote.stash.folder.RemoteStashFolderData(stash_mode: aiida.common.datastructures.StashMode, target_basepath: str, source_list: List, **kwargs)[source]

Data plugin that models a folder with files of a completed calculation job that has been stashed through a copy.

This data plugin can and should be used to stash files if and only if the stash mode is StashMode.COPY.

__abstractmethods__ = frozenset({})
__init__(stash_mode: aiida.common.datastructures.StashMode, target_basepath: str, source_list: List, **kwargs)[source]

Construct a new instance

Parameters
• stash_mode – the stashing mode with which the data was stashed on the remote.

• target_basepath – the target basepath.

• source_list – the list of source files.

__module__ = 'aiida.orm.nodes.data.remote.stash.folder'
_abc_impl = <_abc_data object>
_logger: Optional[logging.Logger] = <Logger aiida.orm.nodes.data.remote.stash.folder.RemoteStashFolderData (REPORT)>
_plugin_type_string = 'data.remote.stash.folder.RemoteStashFolderData.'
_query_type_string = 'data.remote.stash.folder.'
_storable = True
property source_list

Return the list of source files that were stashed.

Returns

the list of source files.

property target_basepath

Return the target basepath.

Returns

the target basepath.


text:
# nLab vector field

### Context

#### Differential geometry

differential geometry

synthetic differential geometry

# Contents

## Definition

A vector field is a section of a vector bundle.

More specifically, a vector field is a tangent vector field which is a section of a tangent bundle. Equivalently this is a rank $(1,0)$-tensor field on $X$

## References

For instance section 1.3 and 1.4 of

Revised on January 6, 2013 17:35:05 by Urs Schreiber (89.204.138.182)


text:
# Quickstart¶

To launch a fault-tolerant job, run the following on all nodes.

torchrun
--nnodes=NUM_NODES
--nproc-per-node=TRAINERS_PER_NODE
--max-restarts=NUM_ALLOWED_FAILURES
--rdzv-id=JOB_ID
--rdzv-backend=c10d
YOUR_TRAINING_SCRIPT.py (--arg1 ... train script args...)

To launch an elastic job, run the following on at least MIN_SIZE nodes and at most MAX_SIZE nodes.

torchrun
--nnodes=MIN_SIZE:MAX_SIZE
--nproc-per-node=TRAINERS_PER_NODE
--max-restarts=NUM_ALLOWED_FAILURES_OR_MEMBERSHIP_CHANGES
--rdzv-id=JOB_ID
--rdzv-backend=c10d
YOUR_TRAINING_SCRIPT.py (--arg1 ... train script args...)

Note

TorchElastic models failures as membership changes. When a node fails, this is treated as a “scale down” event. When the failed node is replaced by the scheduler, it is a “scale up” event. Hence for both fault tolerant and elastic jobs, --max-restarts is used to control the total number of restarts before giving up, regardless of whether the restart was caused due to a failure or a scaling event.

HOST_NODE_ADDR, in form <host>[:<port>] (e.g. node1.example.com:29400), specifies the node and the port on which the C10d rendezvous backend should be instantiated and hosted. It can be any node in your training cluster, but ideally you should pick a node that has a high bandwidth.

Note

If no port number is specified HOST_NODE_ADDR defaults to 29400.

Note

The --standalone option can be passed to launch a single node job with a sidecar rendezvous backend. You don’t have to pass --rdzv-id, --rdzv-endpoint, and --rdzv-backend when the --standalone option is used.

Note

If torchrun does not meet your requirements you may use our APIs directly for more powerful customization. Start by taking a look at the elastic agent API.


text:
# Specific Heat capacity

Classical Mechanics Level 2

How many specific heat capacity are there for a particular gas?

×


text:
# Relation Problem

• February 20th 2008, 06:38 PM
Edbaseball17
Relation Problem
Let (x1,y1)~(x2,y2) if (y1-y2)=x1^2-x2^2.

is this reflexive, Symmetrical, Transitive and an Equiv. Relation? if so y?

The above problem is a homework problem that I need to figure out. Any help would be appreciated. Thanks, Ed.
• February 20th 2008, 06:57 PM
Jhevon
Quote:

Originally Posted by Edbaseball17
Let (x1,y1)~(x2,y2) if (y1-y2)=x1^2-x2^2.

is this reflexive, Symmetrical, Transitive and an Equiv. Relation? if so y?

The above problem is a homework problem that I need to figure out. Any help would be appreciated. Thanks, Ed.

please refrain from using chat talk, that is, type out "why" as opposed to typing "y". this is for two reasons: (1) this is a math forum, we usually use single letters to represent variables or things like that, so it may cause confusion, (2) this is a prestigious math site :D i'm pushing the envelope enough just by not capitalizing words that should be capitalized and neglecting the proper use of punctuations.

we cannot give you a full solution to homework problems. so you have to be satisfied with hints.

do you know what each of the terms you typed mean? what does it mean to be reflexive? symmetric? transitive? what does it mean to be an equivalence relation? answer those and we'll take it from there
• February 20th 2008, 07:02 PM
Edbaseball17
i know that reflexive is a,a and symmetric is b,a and trans. is a,b . . b,c . . . a,c. And that if it is all three, then it is and equiv. relation.
• February 20th 2008, 07:15 PM
Jhevon
Quote:

Originally Posted by Edbaseball17
i know that reflexive is a,a and symmetric is b,a and trans. is a,b . . b,c . . . a,c. And that if it is all three, then it is and equiv. relation.

if that's how you define them, then you're going to have a hard time proving these.

A relation $\sim$ on a set $A$ is:

reflexive if $a \sim a$ for all $a \in A$

symmetric if $a \sim b \implies b \sim a$ for all $a,b \in A$

transitive if $a \sim b$ and $b \sim c$ implies $a \sim c$ for all $a,b,c \in A$

and you are correct, the relation is an equivalence relation if it is reflexive, symmetric and transitive

let $A$ be the set the relation is on. we pick some $(x_1,y_1) \in A$. does this point relate to itself? that is, can you show whether or not it satisfies the condition mentioned?
• February 21st 2008, 11:32 AM
Edbaseball17
It should be relative because it's a point on a graph, right?

edit:

At least it should be if the points are equal. (1,1) or (2,2). I wouldnt be if the points where (1,3) or (3,6). Is what i'm getting after reading through my book.
• February 21st 2008, 07:02 PM
Jhevon
Quote:

Originally Posted by Edbaseball17
It should be relative because it's a point on a graph, right?

edit:

At least it should be if the points are equal. (1,1) or (2,2). I wouldnt be if the points where (1,3) or (3,6). Is what i'm getting after reading through my book.

no

you must follow the condition they gave you. you can't be guessing randomly like that.

let me do reflexive, you do the others.

we are told that $(x_1,y_1) \sim (x_2,y_2)$ if $\color{red}y_1 - y_2 = x_1^2 - x_2^2$

does this work for a single point? in other words, we want to know if $(x_1,y_1) \sim (x_1,y_1)$.

Let $(x_1,y_1) \in A$

Since $y_1 - y_1 = 0 = x_1^2 - x_1^2$, we have $(x_1,y_1) \sim (x_1,y_1)$. thus, $\sim$ is reflexive


text:
Department of

# Mathematics

Seminar Calendar
for events the day of Monday, December 9, 2019.

.
events for the
events containing

Questions regarding events or the calendar should be directed to Tori Corkery.
    November 2019          December 2019           January 2020
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
1  2    1  2  3  4  5  6  7             1  2  3  4
3  4  5  6  7  8  9    8  9 10 11 12 13 14    5  6  7  8  9 10 11
10 11 12 13 14 15 16   15 16 17 18 19 20 21   12 13 14 15 16 17 18
17 18 19 20 21 22 23   22 23 24 25 26 27 28   19 20 21 22 23 24 25
24 25 26 27 28 29 30   29 30 31               26 27 28 29 30 31

Monday, December 9, 2019

3:00 pm in 441 Altgeld Hall,Monday, December 9, 2019

#### The Gross-Hopkins duality (part 2)

###### Ningchuan Zhang (UIUC Math)

Abstract: Following last week’s talk, we will sketch the proof of the Gross-Hopkins duality following Neil Strickland’s paper. In addition, we will explain how the Gross-Hopkins period map is related to the duality phenomenon.

5:00 pm in 241 Altgeld Hall,Monday, December 9, 2019

#### The Dixmier trace

###### Haojian Li (University of Illinois at Urbana-Champaign)

Abstract: 1. Banach algebra of modulated functions. 2. Extension of the non-commutative residue.


text:
# When Probability Meets Combinatorics: One Problem, Two Approaches

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th March, 2021

Now, we’d like to take a look at an Official GMAT Probability problem to pull everything together. The following is a good example for two reasons:

1. It illustrates a quirky case that is difficult more conceptually than mathematically, and thus is better for the GMAT.

2. It can be tackled either through straight probability or through a combination of probability and combinatorics.

Here’s the question:

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

A) 1/24
B)
1/8
C) 1/4
D) 1/3
E) 3/8

First, as always, give the problem a shot before reading on for the explanation. If possible, see if you can tackle it with both methods (pure probability and probability w/ combinatorics).

#### Explanation #1:

First, we’ll tackle pure probability. Let’s label the letters A, B, C, and D, and let’s say that A is the letter we wish to match with its correct envelope. The other three will be matched with incorrect envelopes. We now must examine the individual probabilities of the following events happening (green for correct, red for incorrect):

_A_   _B_   _C_   _D_

For the above, each slot represents a letter matched with an envelope. There are four envelopes and only one is correct for letter A. That means Tanya has a 1/4 chance of placing letter A in its correct envelope:

_1/4__   _B_   _C_   _D_

We now desire letter B to be placed in an incorrect envelope. Two of the remaining three envelopes display incorrect addresses, so there is a 2/3 chance of that happening:

_1/4__   _2/3_   _C_   _D_

We then desire letter C to also be placed in an incorrect envelope. Only one of the remaining two envelopes displays an incorrect address, so there is a 1/2 chance of that happening:

_1/4__   _2/3_   _1/2_   _D_

At that point, the only remaining option is to place the last remaining letter in the last remaining envelope (i.e. a 100% chance, so we place a 1 in the final slot):

_1/4__   _2/3_   _1/2_   _1_

Multiplying the fractions, we can hopefully see that some cancelling will occur:

¼ x ⅔ x ½ x 1

= 1 x 2 x 1
———–
4 x 3 x 2

= 1/12

But lo and behold, 1/12 is not in our answer choices. Did you figure out why?

We can’t treat letter A as the only possible correct letter. Any of the four letters could possibly be the correct one. However, the good news is that in any of the four cases, the math will be exactly the same. So all we have to do is take the original 1/12 we just calculated and multiply it by 4 to get the final answer: 4 x 1/12 = 4/12 = 1/3. The correct answer is D.

#### Explanation #2:

So what about a combinatorics approach?

As we’ve discussed in our previous GMAT probability posts, all probability can be boiled down to Desired Outcomes / Total Possible Outcomes. And as we discussed in our posts on GMAT combinatorics, we can use factorials to figure out the total possible outcomes in a situation such as this, which is actually a simple PERMUTATION. There are four envelopes, so for the denominator of our fraction (total possible outcomes), we can create a slot for each envelope and place a number representing the letters in each slot to get:

_4_  _3_  _2_  _1_  =  4! = 24  possible outcomes

This lets us know that if we were to put the four letters into the four envelopes at random, as the problem says, there would be 24 permutations, giving us the denominator of our fraction (total possible outcomes).

So what about the desired outcomes? How many of those 24 involve exactly one correctly placed letter? Well, let’s again treat letter A as the correctly placed letter. Once it’s placed, there are three slots (envelopes) left:

___  ___  ___

But the catch is: the next envelope has only two letters that could go into it. Remember, one of the letters correctly matches the envelope in address, and we want a mismatch:

_2_  ___  ___

Likewise, that would leave two letters available for the next envelope, but only one of them would have the wrong address:

_2_  _1_  ___

And finally, there would be only one choice left for the final envelope:

_2_  _1_  _1_

That would mean for the correctly-placed A letter, there are only two permutations in which each of the other letters is placed incorrectly:

_2_ x  _1_ x  _1_ = 2 possible outcomes.

But as before, we must consider that any of the four letters could be the correct letter, not just letter A. So we must multiply the 2 possible outcomes by four to get 8 desired outcomes involving exactly one letter being placed in its correct envelope. That gives us our numerator of Desired Outcomes. Our denominator, remember, was 24 total possible outcomes. So our final answer, once again, is 8/24 = 1/3.

This is a great example of how GMAT combinatorics can intersect with probability.

To tide you over until next time, give this Official GMAT problem a try. It will also give a nice segue into Number Theory, which we’ll begin to talk more about going forward. Explanation next time…

If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6


text:
# Greedy algorithm

Prove or disprove that the greedy algorithm for making change always uses the fewest coins possible when the denominations available are pennies (1-cent coins), nickels (5-cent coins), and quarters (25-cent coins).

Does anyone know how to solve this?

## 2 Answers

Since the each coin divides the face value of every larger coin, a single larger coin will always represent an integer multiple of smaller coins.

After giving out the maximal number of quarters, there will be $0-24$ cents remaining. Then there will be at most 4 nickels to give out. After giving out nickels greedily, there will be $0-4$ cents remaining, so there will be at most 4 pennies to give out. Now, can you prove that we cannot rearrange our change to use fewer coins?


text:
# zbMATH — the first resource for mathematics

Twistor construction of Einstein metrics. (English) Zbl 0615.53057
Global Riemannian geometry, Proc. Symp., Durham/Engl. 1982, 115-125 (1984).
[For the entire collection see Zbl 0614.00017.]
One of the most intriguing and important consequences of Yau’s proof of the Calabi conjecture [S. T. Yau, Proc. Natl. Acad. Sci. USA 74, 1798-1799 (1977; Zbl 0355.32028)] is the following theorem:
Theorem 1.1 (Yau). Let X be a Kählerian K3 surface, and $$\alpha \in H^{1,1}(X,{\mathbb{R}})$$ a positive cohomology class. Then there exists a unique Kähler metric on X such that the Ricci tensor vanishes, and $$\alpha$$ is the class defined by the Kähler form.
The proof of this theorem consists of an existence theorem for certain non-linear partial differential equations, the complex Monge-Ampère equations, and the end result is of importance not only to differential geometers, but also to those who work in algebraic geometry and in general relativity [D. Page, Phys. Lett. B 80, 55-57 (1978)]. The question Yau’s theorem provokes is clear:
Problem. Find the K3 metric explicitly.
The aim of this lecture is not to solve this problem (would that it were!) but first to describe how Penrose’s twistor theory gies a method of attacking it, and secondly to show how that method may be used to provide explicit approximations to the K3 metric. The existence of one of these approximations was deduced heuristically by D. Page [Phys. Lett. B 100, 313-315 (1981)] as a degeneration of the K3 metric.

##### MSC:
 53C55 Global differential geometry of Hermitian and Kählerian manifolds 32Q99 Complex manifolds 53C80 Applications of global differential geometry to the sciences 14M99 Special varieties


text:
activation

Every zone has particular demand which translates into the probability of picking the n-th scooter in a give time frame. We look at a proper distribution to estimate a zone characteristic parameter for each zone.

activation probability

Depending on the demand of a zone we can define a probability for each ordinal number of scooters. Probability will of course drops for any new scooter added to the same zone.

activation probability for every n-th scooter in the area for a give timeframe

We sum the number of used scooters inside an area for a given time frame. Since our scope is to optimize the deployments we define the

def sumProb(used,left,max_occ=20):
"""ordinal probability from a row of positive events: [4,3,5,6,...]"""
prob = np.zeros(max_occ)
for i,j in zip(used,left):
## summing used, left over, unknown
p = [1. for x in range(i)] + [0. for x in range(j)] + [np.nan for x in range(max_occ-j-i)]
p = s_s.interpMissing(p)
prob = prob + np.array(p[:max_occ])
prob = prob/len(used)
return prob

empirical distribution of activation probability per n-th scooter

We therefore use a Fermi-Dirac distribution which describes the occupation of particles for a given energy state which in our case translates into the zone demand.

Fermi distribution:

$$n_i = \frac{1}{e^{ (\epsilon_i - \mu)/k_BT} + 1}$$

Where μ is called the chemical potential (how favorable is to add a new particle to the system) and kBT is the thermal noise which we connect with the normalized temperature of that day and hour

p0 = [0.75,thermal_noise]
x0 = [1.3,thermal_noise]
t = np.array(list(range(len(y))))
def ser_residual(x,t,y,p):
"""residuals of distribution"""
return (y-dis_Fermi(x,t,p))
res = least_squares(ser_residual,x0,args=(t,y,p0))
x0 = res.x
y1 = dis_Fermi(x0,t,p0)

example of fitting some empirical distributions

We allow a maximum number of 20 scooters to have a similar distribution across all zones. We iterate the fit across all locations.

we fit all the empirical distributions with a Fermi-Dirac law and compute the chemical potential

A better fitting is obtained optimizing both the chemical potential and the thermal noise but we see that if we keep the thermal noise constant (at least for similar weather) the error is not much larger and the chemical potential is well defined

distribution of chemical potential, noise and error with or without a fix noise parameter

We can visualize on the map the areas with most revenue

areas with most revenues

Actually most of the revenue comes from the fact that we mostly deployed in that region

areas where most of scooters were deployed

We than calculate the average revenue per scooter and see that outskirts have a high potential

areas with most revenues per scooter

We than calculate the chemical potential and the potential revenue per area depending on the shift and the weekday

geohash weekday shift chem n urev pot
8a63a06995affff 4 1 7.09 66 9.39 61.97
8a63a06988dffff 6 1 6.94 130 11.84 76.33
8a63a069958ffff 3 2 6.89 69 6.19 39.64
8a63a0686367fff 5 2 6.89 52 7.54 48.22
8a63a068a8effff 6 1 6.88 70 11.37 72.65

Displaying the chemical potential tells us how many scooters should be deployed in an area

chemical potential for single area

We can than theoretically know the potential revenue per location, weekday and shift

potential of that area for a given weekday and shift number

We iteratively group by different zooms of geohash to have at least an estimate in areas where we don’t have an accurate measurment of the activation probabilty and we can see that there is the indication that on same areas we would have good revenues

We calculate the error of the model per zone and obtain a median error of 18% with no significant correlation with the number of scooters analyzed

distribution of error per zone, the smaller error is where we have a larger history

    conn_string = "dbname='{}' port='{}' host='{}' user='{}' password='{}'".format(cred['DATABASE'],cred['PORT'],cred['HOST'],cred['USER'],cred['PASSWORD'])
con = psycopg2.connect(conn_string)
sql = "select event.city, event.properties, event.time, event.device_id, event.user_id  from fivetran_mixpanel.event"
sql += " where name = 'appStart' and time > '%s 00:00:00' and time < '%s 23:59:59' and lower(city) = 'berlin' " % (day,day)
cur = con.cursor()
cur.execute(sql)
con.commit()
resq = cur.fetchall()
time.sleep(10)

distribution of error per zone, the smaller error is where we have a larger history


text:
# GFDL - Geophysical Fluid Dynamics Laboratory

Some feedbacks in AR4 models, from Held and Shell 2012.  The three red columns on the right provide the traditional perspective:  the “Planck feedback”– the response to uniform warming of surface and troposphere with fixed specific humidity ($\lambda_T$), the lapse rate feedback at fixed specific humidity ($\lambda_L$), and the water vapor feedback ($\lambda_Q$).  The three blue columns on the left provide an alternative perspective — with the fixed relative humidity uniform warming feedback ($\tilde{\lambda}_T$), the fixed relative humidity lapse rate feedback, ($\tilde{\lambda}_L$), and the relative humidity feedback ($\tilde{\lambda}_H$).  The sum of the three terms, shown in the middle black column, is the same from either perspective.  Surface albedo and cloud feedbacks are omitted. Each model is a dot.


text:
# bit64 v0.9-7

0

0th

Percentile

## A S3 Class for Vectors of 64bit Integers

Package 'bit64' provides serializable S3 atomic 64bit (signed) integers. These are useful for handling database keys and exact counting in +-2^63. WARNING: do not use them as replacement for 32bit integers, integer64 are not supported for subscripting by R-core and they have different semantics when combined with double, e.g. integer64 + double => integer64. Class integer64 can be used in vectors, matrices, arrays and data.frames. Methods are available for coercion from and to logicals, integers, doubles, characters and factors as well as many elementwise and summary functions. Many fast algorithmic operations such as 'match' and 'order' support inter- active data exploration and manipulation and optionally leverage caching.


text:
# Intersection of curves on projective toric surface and some enumerative questions

Reading on the tropical approach to enumerative geometry I have come across the claim: given a projective toric surface from a polygon P, we can consider a tautological bundle of algebraic / holomorphic functions with newton polygon contained in P (i.e. take functions that in the dense torus can be written as a combination of monomial corresponding only to points in P).

The claim I've come across (in Mikhalkin's paper [1] (section on severi varieties) or in Shustin's book [2] (proof of lemma 2.42 page 59-61) is that if we take the zeroes of such a function from our bundle, then in the general case it is a curve that is transversal to the boundary divisors of the toric surface. Moreover, "Any curve (which is not a toric divisor) crosses at least two toric divisors; take the Newton polygon of a curve equation, then its intersections with the toric divisors are as follows: for each side of that Newton polygon equipped with the outer normal, take the parallel supporting line for the polygon P defining the toric surface, if the intersection is a vertex, then the curve passes through the corresponding intersection point of toric divisors, if the intersection is a side, then the curve crosses the corresponding toric divisor; for a Newton segment (binomial curve), we take it twice - with both normals." [I quote an explanation given to me on what are 'rules' of intersection]

A) I could not find any proof or explanation why it is true. I am currently less interested in a fully rigorous proof but more in an explanation why this is true.

B) Another question come to my mind, does any other curve on the toric surface must come from zeroes of a function from the tautological bundle? if not, why do authors focus so much on curves that are zeroes of this bundle in the context of counting curves via n points? (i.e. GW invariants)?

Thanks!

And i apologize if this was a bit cumbersome way of asking my question, when things are not clear enough, even asking properly is somewhat difficult :)

References

-

The answer to your question B is yes and no :) You see: replacing $P$ by $kP$ for any $k \geq 1$ gives rise to the same toric surface (the latter is precisely the $k$-uple Veronese embedding of the former). And, the defining polynomial of any curve will fit (up to translation) in $kP$ for a sufficiently large $P$.
For the phenomenon in your first question, I can give an answer in the case that the underlying field $\mathbb{k}$ is algebraically closed. Here it goes: let $f$ be the Laurent polynomial defining the curve $C$ and $S$ is an edge of the Newton Polygon $N$ of $f$. Let $\nu := (p, q)$ be an outward pointing (with respect to $N$) normal to $S$. For simplicity assume $q > 0$. Then there is a branch of $C$ with degree-wise Puisuex series of the form: $\gamma(t) = (t^p, \sum_{k = 0}^\infty a_k t^{q_k})$, where $q = q_0 > q_1 > \cdots$ are rational numbers with bounded denominators. Now let $\psi_P: X_P \to \mathbb{P}^N$ (where $N := |P \cap \mathbb{Z}^n| - 1$) be the embedding of the toric variety defined by the monomials in $P$, i.e. the restriction of $\psi_P$ to $(\mathbb{k}^*)^n$ is given by: $\psi_P(x) := [x^{\alpha_0}: \cdots : x^{\alpha_N}]$, where $P \cap \mathbb{Z}^n = \lbrace \alpha_0, \ldots \alpha_N \rbrace$. Let $Q$ be the face (i.e. edge or vertex) of $P$ such that $\nu$ is an outer normal to $Q$. W.l.o.g. assume that $Q \cap \mathbb{Z}^n = \lbrace \alpha_0, \ldots, \alpha_q \rbrace$, $q < N$. Then precisely the first $q+1$ coordinates of $x := \lim_{t \to \infty} \psi_P(\gamma(t))$ are non-zero, i.e. $x$ belongs to the subvariety of $X_P$ determined by $Q$.
Remark: Usually in the books on toric varieties, inner normals are used instead of outer normals. That gives rise to a usual Puiseux series (the exponents being increasing , as opposed to the one in the preceding paragraph). But then the point at infinity (on the curve) is approached as $t \to 0$. I prefer that one approaches the point at infinity as $t$ approaches infinity as well.


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# Math Help - Find eigenvalues and eigenvectors of matrices

1. ## Find eigenvalues and eigenvectors of matrices

Find eigenvalues and eigenvectors of this matrix

I found the eigenvalue λ1=-3 and λ2=6
When I added λ1=-3 to the matrix, I have the equation system like this

x1-2x2+2x3=0
-2x1+4x2-4x3=0
2x1-4x2+4x3=0

I don't know how to solve this equation system to figure out x1,x2,x3 and eigenvector. Please help me!

2. ## Re: Find eigenvalues and eigenvectors of matrices

Hey math88.

You should eliminate two rows (since they are all linearly dependent) and you will get x1 - 2x2 + 2x3 = 0.

Now you need to convert that to a vector with a free parameter (call it t) by letting one variable equal to t and solving for the other parameters (in terms of t) and your vector will be (x1(t),x2(t),x3(t)) and this is your eigenvector. You can then normalize this to have length 1 (this is typically done for eigenvectors).

3. ## Re: Find eigenvalues and eigenvectors of matrices

I'm not sure, but
if x3=t and x2=-t so x1= -4t so the eigenvectors is t(-4,-1,1)
or if I choose x1=t, x2=2t so x3=3/2 t -> the eigenvectors is t(1,2,3/2)
I don't know which one is correct.

If x1=t -> x2-x3=t/2
What should I do next to find x2, x3 in term of t?

4. ## Re: Find eigenvalues and eigenvectors of matrices

Originally Posted by math88
Find eigenvalues and eigenvectors of this matrix

I found the eigenvalue λ1=-3 and λ2=6
When I added λ1=-3 to the matrix, I have the equation system like this

x1-2x2+2x3=0
-2x1+4x2-4x3=0
2x1-4x2+4x3=0

I don't know how to solve this equation system to figure out x1,x2,x3 and eigenvector. Please help me!
An nXn matrix always has n eigenvalues. But some of them may be repeated. What are your three eigenvalues?

5. ## Re: Find eigenvalues and eigenvectors of matrices

There are, as I am sure you know, to solve three equations in three variables. Here, an obvious solution is $x_1= x_2= x_3= 0$. Since you used an eigenvalue to get them, there must be another, in fact, infinitely many, non-zero ("non-trivial") solutions.

Here, you can immediately see that the third equation is just 2 times the first. And the second equation is -1 times the third. That is, we really only have one equation here. From the first equation, $x_1- 2x_2+ 2x_3= 0$ we can solve for x_1= 2x_2- 2x_3. Then any eigenvector can be written in the form [tex]<x_1, x_2, x_3>= <2x_2- 2x_3, x_2, x_3>= <2x_2, x_2, 0>+ <-2x_3, 0, x_3>= x_2<2, 1, 0>+ x_3<2, 0, 1>. That is, this "eigenspace" the space of all eigenvectors corresponding to eigenvalue -3 is two-dimensional and is spanned by <2, 1, 0> and <2, 0, 1>.


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Lefschetz pencils on a certain hypersurface in positive characteristic.(English)Zbl 1388.14127

Oguiso, Keiji (ed.) et al., Higher dimensional algebraic geometry. In honour of Professor Yujiro Kawamata’s sixtieth birthday. Proceedings of the conference, Tokyo, Japan, January 7–11, 2013. Tokyo: Mathematical Society of Japan (MSJ) (ISBN 978-4-86497-046-4/hbk). Advanced Studies in Pure Mathematics 74, 265-278 (2017).
Summary: We examine Lefschetz pencils of a certain hypersurface in $$\mathbb{P}$$ over an algebraically closed field of characteristic $$p>2$$, and determine the group structure of sections of the fiber spaces derived from the pencils. Using the structure of a Lefschetz pencil, we give a geometric proof of the unirationality of Fermat surfaces of degree $$p^a+1$$ with a positive integer $$a$$ which was first poved by T. Shioda [Math. Ann. 211, 233–236 (1974; Zbl 0276.14018)].
As byproducts, we also see that on the hypersurface there exists a $$(q^3+q^2+q+1)_{q+1}$$-symmetric configuration (resp. a $$((q^3+ 1)(q^2+ 1)_{q+1}$$, $$(q^3+ 1)(q+1)_{q^2+1})$$-configuration) made up of the rational points over $$\mathbb{F}_q$$ (resp. over $$\mathbb{F}_{q^2}$$) and the lines over $$\mathbb{F}_q$$ (resp. over $$\mathbb{F}_{q^2}$$) with $$q=p^a$$.
For the entire collection see [Zbl 1388.14012].

MSC:

 14J70 Hypersurfaces and algebraic geometry 14G17 Positive characteristic ground fields in algebraic geometry 14J25 Special surfaces 14C21 Pencils, nets, webs in algebraic geometry

Zbl 0276.14018
Full Text:


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We have moved to a new Sailfish OS Forum. Please start new discussions there.

# Run script on certain time/date

Is there a way to make the OS execute a script on a certain time/date? I know that Sailfish doesn't support cronjobs, but are there any alternative way to establish the same functionality?

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Very true! This would be awesome. There is quite a similar question that should be cross referenced here: https://together.jolla.com/question/1770/alarm-could-trigger-any-shell-command/

( 2014-04-04 21:48:19 +0300 )edit

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You might wanna have a closer look to "fcron" from OpenRepos.net. Maybe it will suits your needs.

more

Yes, it seems like a nifty program. However I cannot manage to make it work. When I try fcrontab -e (to edit current users fcrontab) I get:

Could not authenticate user using PAM (7): Authentication failure

Doesn't matter if I run as root or not

( 2014-04-07 23:52:04 +0300 )edit

i have not tried it but sailfish os was changed (in update 1.0.3.8) to introduce a security model. have a look at the file /usr/share/mapplauncherd/privileges

might work if you edit it?

( 2014-04-08 00:55:58 +0300 )edit

I compiled it from source instead, and it works like a charm. Thanks for the tips.

EDIT: seems like fcron misses deadlines if the phone isn't active.

( 2014-04-12 01:09:02 +0300 )edit

If you are using fcron package from openrepos the /etc/pam.d/fcron file is not a pam config file.

create a file /etc/pam.d/fcrontab like this:

auth       sufficient   pam_rootok.so
auth       include      system-auth
account    include      system-auth
session    include      system-auth

and maybe /etc/pam.d/fcron the same.

And you can use fcrontab -e as root.

( 2014-04-22 20:32:50 +0300 )edit

Thanks for this advice, it worked. So far I've only done the first step (creating the fcrontab file). The /etc/pam.d/fcron seems to be the same as /etc/fcron.conf (which I suppose is wrong for PAM). But are you saying that Ican safely replace the contents of this file too with the same content as that of the /etc/pam.d/fcrontab file?

( 2014-04-24 10:59:53 +0300 )edit

Systemd supports this. Not sure from which version and up though and i have no idea which version is included in sailfish.

more

Hmm, I've looked in to systemd and it seems like this could really solve my problem. However, I don't get it to work. Either I've done something wrong or Sailfish doesn't support timers

EDIT: though it seems like someone has done it before: https://www.wolf-kissendorf.de/wordpress/?p=179

( 2014-04-04 19:55:44 +0300 )edit

i don't think systemd on the phone supports this (it is old. only supports monotonic timers). newer version supports calendar timers, which is what you are looking for. this is available from v197 onwards. so we are waiting for jolla to upgrade systemd.

( 2014-04-07 14:00:35 +0300 )edit

Yes, I've seen the first link you posted, but since came over that guide I thought that maybe timers worked anyway. Also, I can see that there are some .timer-files in /usr/lib/systemd/system/ and /lib/systemd/system/

( 2014-04-07 23:04:57 +0300 )edit