text:
# Search by Topic

#### Resources tagged with Working systematically similar to Spot the Card:

Filter by: Content type:
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### There are 127 results

Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically

### Spot the Card

##### Age 14 to 16 Challenge Level:

It is possible to identify a particular card out of a pack of 15 with the use of some mathematical reasoning. What is this reasoning and can it be applied to other numbers of cards?

### Bent Out of Shape

##### Age 14 to 18 Challenge Level:

An introduction to bond angle geometry.

### Problem Solving, Using and Applying and Functional Mathematics

##### Age 5 to 18 Challenge Level:

Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.

### Counting on Letters

##### Age 11 to 14 Challenge Level:

The letters of the word ABACUS have been arranged in the shape of a triangle. How many different ways can you find to read the word ABACUS from this triangular pattern?

### Instant Insanity

##### Age 11 to 18 Challenge Level:

Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

### Tea Cups

##### Age 7 to 14 Challenge Level:

Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.

### Teddy Town

##### Age 5 to 14 Challenge Level:

There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?

##### Age 11 to 14 Challenge Level:

Rather than using the numbers 1-9, this sudoku uses the nine different letters used to make the words "Advent Calendar".

### Factor Lines

##### Age 7 to 14 Challenge Level:

Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.

### Weights

##### Age 11 to 14 Challenge Level:

Different combinations of the weights available allow you to make different totals. Which totals can you make?

### Ratio Sudoku 3

##### Age 11 to 16 Challenge Level:

A Sudoku with clues as ratios or fractions.

### Wallpaper Sudoku

##### Age 11 to 16 Challenge Level:

A Sudoku that uses transformations as supporting clues.

### Ratio Sudoku 1

##### Age 11 to 16 Challenge Level:

A Sudoku with clues as ratios.

### Seasonal Twin Sudokus

##### Age 11 to 16 Challenge Level:

This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?

### LOGO Challenge - Triangles-squares-stars

##### Age 11 to 16 Challenge Level:

Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential.

##### Age 11 to 14 Challenge Level:

A few extra challenges set by some young NRICH members.

### Corresponding Sudokus

##### Age 11 to 18

This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule.

### Colour in the Square

##### Age 7 to 16 Challenge Level:

Can you put the 25 coloured tiles into the 5 x 5 square so that no column, no row and no diagonal line have tiles of the same colour in them?

##### Age 11 to 14 Challenge Level:

You need to find the values of the stars before you can apply normal Sudoku rules.

### Intersection Sudoku 1

##### Age 11 to 16 Challenge Level:

A Sudoku with a twist.

##### Age 11 to 16 Challenge Level:

Four small numbers give the clue to the contents of the four surrounding cells.

### Magic Potting Sheds

##### Age 11 to 14 Challenge Level:

Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?

### Ratio Sudoku 2

##### Age 11 to 16 Challenge Level:

A Sudoku with clues as ratios.

### Fence It

##### Age 11 to 14 Challenge Level:

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

### Isosceles Triangles

##### Age 11 to 14 Challenge Level:

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

### Twin Corresponding Sudokus II

##### Age 11 to 16 Challenge Level:

Two sudokus in one. Challenge yourself to make the necessary connections.

### More on Mazes

##### Age 7 to 14

There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.

##### Age 11 to 16 Challenge Level:

Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.

### Colour Islands Sudoku

##### Age 11 to 14 Challenge Level:

An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.

### Multiples Sudoku

##### Age 11 to 14 Challenge Level:

Each clue in this Sudoku is the product of the two numbers in adjacent cells.

### Twin Corresponding Sudoku III

##### Age 11 to 16 Challenge Level:

Two sudokus in one. Challenge yourself to make the necessary connections.

### Inky Cube

##### Age 7 to 14 Challenge Level:

This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken?

### Troublesome Triangles

##### Age 7 to 14 Challenge Level:

Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .

### Olympic Logic

##### Age 11 to 16 Challenge Level:

Can you use your powers of logic and deduction to work out the missing information in these sporty situations?

### Latin Lilies

##### Age 7 to 16 Challenge Level:

Have a go at this game which has been inspired by the Big Internet Math-Off 2019. Can you gain more columns of lily pads than your opponent?

### More Magic Potting Sheds

##### Age 11 to 14 Challenge Level:

The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?

### Magic W

##### Age 14 to 16 Challenge Level:

Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.

### First Connect Three

##### Age 7 to 14 Challenge Level:

Add or subtract the two numbers on the spinners and try to complete a row of three. Are there some numbers that are good to aim for?

### Number Daisy

##### Age 11 to 14 Challenge Level:

Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?

### Window Frames

##### Age 5 to 14 Challenge Level:

This task encourages you to investigate the number of edging pieces and panes in different sized windows.

### Pole Star Sudoku

##### Age 14 to 18 Challenge Level:

A Sudoku based on clues that give the differences between adjacent cells.

### Tetrahedra Tester

##### Age 11 to 14 Challenge Level:

An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?

### Number Sandwiches

##### Age 7 to 14 Challenge Level:

Can you arrange the digits 1, 1, 2, 2, 3 and 3 to make a Number Sandwich?

### Olympic Magic

##### Age 14 to 16 Challenge Level:

in how many ways can you place the numbers 1, 2, 3 … 9 in the nine regions of the Olympic Emblem (5 overlapping circles) so that the amount in each ring is the same?

### Twinkle Twinkle

##### Age 7 to 14 Challenge Level:

A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.

### Cayley

##### Age 11 to 14 Challenge Level:

The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?

### Intersection Sudoku 2

##### Age 11 to 16 Challenge Level:

A Sudoku with a twist.

### One Out One Under

##### Age 14 to 16 Challenge Level:

Imagine a stack of numbered cards with one on top. Discard the top, put the next card to the bottom and repeat continuously. Can you predict the last card?

### Twin Corresponding Sudoku

##### Age 11 to 18 Challenge Level:

This sudoku requires you to have "double vision" - two Sudoku's for the price of one

### Football Sum

##### Age 11 to 14 Challenge Level:

Find the values of the nine letters in the sum: FOOT + BALL = GAME


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# Conflict between Conditional Independence and Conditional Dependence

In "Causal Inference in Statistics" (Pearl et al.), Conditional Independence is defined as (page 10):

Later on, in the context of colliders, Conditional Dependence is defined as (page 41):

and an example is provided for that:

Consider a simultaneous (independent) toss of two fair coins and a bell that rings whenever at least one of the coins lands on heads. Let the outcomes of the two coins be denoted X and Y, respectively, and let Z stand for the state of the bell, with Z = 1 representing ringing, and Z = 0 representing silence.

+ an extra condition is later added (page 43)

Suppose we do not hear the bell directly, but instead rely on a witness who is somewhat unreliable; whenever the bell does not ring, there is 50% chance that our witness will falsely report that it did. Letting W stand for the witness’s report

The authors claim and prove that conditioning on W makes X and Y dependent:

(full probability table provided at the bottom)

I understand that adding Y="Heads" to the 2nd expression impacts the probability thus showing that the two variables X and Y are now dependent given W.

But, we also then get that P(X = "Heads"| Y="Heads", W = 1) is equal to P(X = "Heads"| Y="Heads") (=0.5) which is equation 1.2 of conditional independence... so how can it be? that the variables are both conditionally independent and dependent?

The probability table:

The difference is in which variables are independent.

• X is independent of W, conditional on Y.
• X is dependent on Y, conditional on W.

These statements express different relationships. There is no contradiction between them.

I will explain a little further to address your comment on Dilip Sarwate's answer. You ask, "would you say {X="Heads"|Y="Heads", W=1} is dependent or independent?" But that expression isn't something that can be dependent or independent. (I'm not even sure how to interpret it, because when we talk about events outside the context of probability statements, we can't condition on events, we can only conjoin them. For example you could write {X="Heads" & Y="Heads" & W=1}, and that would make sense.)

There are three entities involved in conditional (in)dependences: the two variables that are (in)dependent - call them A and B - and the set of variables conditioned on - call it C.

To know whether A and B are dependent or independent conditional on C, you test the equality between two probability statements:

P(A|B, C) =?= P(A|C)

Both probability statements are relevant; you are testing whether they are equal. If you change the statement on the right to P(A|B) then you are testing a different equality, a different conditional independence, and you can get a different answer.

$P(X|Y, W) \neq P(X|W)$, showing that X is dependent on Y given W;
$P(X|Y, W) = P(X|Y)$, showing that X is independent of W given Y.

format: plain


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# Persistence of a number

In mathematics, the persistence of a number is the number of times one must apply a given operation to an integer before reaching a fixed point at which the operation no longer alters the number.

Usually, this involves additive or multiplicative persistence of a non-negative integer, which is how often one has to replace the number by the sum or product of its digits until one reaches a single digit. Because the numbers are broken down into their digits, the additive or multiplicative persistence depends on the radix. In the remainder of this article, base ten is assumed.

The single-digit final state reached in the process of calculating an integer's additive persistence is its digital root. Put another way, a number's additive persistence counts how many times we must sum its digits to arrive at its digital root.

## Examples

The additive persistence of 2718 is 2: first we find that 2 + 7 + 1 + 8 = 18, and then that 1 + 8 = 9. The multiplicative persistence of 39 is 3, because it takes three steps to reduce 39 to a single digit: 39 → 27 → 14 → 4. Also, 39 is the smallest number of multiplicative persistence 3.

## Smallest numbers of a given multiplicative persistence

For a radix of 10, there is thought to be no number with a multiplicative persistence > 11: this is known to be true for numbers up to 1020000.[1][2] The smallest numbers with persistence 0, 1, ... are:

0, 10, 25, 39, 77, 679, 6788, 68889, 2677889, 26888999, 3778888999, 277777788888899. (sequence A003001 in the OEIS)

The search for these numbers can be sped up by using additional properties of the decimal digits of these record-breaking numbers. These digits must be sorted, and, except for the first two digits, all digits must be 7, 8, or 9. There are also additional restrictions on the first two digits. Based on these restrictions, the number of candidates for n-digit numbers with record-breaking persistence is only proportional to the square of n, a tiny fraction of all possible n-digit numbers. However, any number that is missing from the sequence above would have multiplicative persistence > 11; such numbers are believed not to exist, and would need to have over 20,000 digits if they do exist.[1]

• The additive persistence of a number is smaller or equal to the number itself, with equality only when the number is zero.
• For base ${\displaystyle b}$  and natural numbers ${\displaystyle k}$  and ${\displaystyle n>9}$  the numbers ${\displaystyle n}$  and ${\displaystyle n*b^{k}}$  have the same additive persistence.

More about the additive persistence of a number can be found here.

## Smallest numbers of a given additive persistence

The additive persistence of a number, however, can become arbitrarily large (proof: For a given number ${\displaystyle n}$ , the persistence of the number consisting of ${\displaystyle n}$  repetitions of the digit 1 is 1 higher than that of ${\displaystyle n}$ ). The smallest numbers of additive persistence 0, 1, ... are:

0, 10, 19, 199, 19999999999999999999999, ... (sequence A006050 in the OEIS)

The next number in the sequence (the smallest number of additive persistence 5) is 2 × 102×(1022 − 1)/9 − 1 (that is, 1 followed by 2222222222222222222222 9's). For any fixed base, the sum of the digits of a number is at most proportional to its logarithm; therefore, the additive persistence is at most proportional to the iterated logarithm.

## Functions with limited persistence

Some functions only allow persistence up to a certain degree.

For example, the function which takes the minimal digit only allows for persistence 0 or 1, as you either start or step to a single-digit number.

## References

1. ^ a b Sloane, N. J. A. (ed.). "Sequence A003001". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
2. ^ Eric W. Weisstein. "Multiplicative Persistence". mathworld.wolfram.com.


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Quick Draw / Quick Sheath Items?

Does the Quick Draw feat allow you to also store the weapon you were wielding as part of the free action? If not, what sort of action is it to 'switch weapons' without dropping one or the other on the floor?

-

The benefits don't mention putting away anything. I have always used this as a draw only feat, as I believe it's been that way pretty much since 3.0. Now, as you mentioned, that doesn't preclude you from using a free/swift action to drop your weapon, and quick draw a different one. But by default, if you want to put your weapon away, you must take a move action to do so.

-
This. It takes you much less time to grab and pull a weapon out of its sheath, but takes significantly longer to put it back into that same sheath. – Pulsehead Jun 13 '12 at 16:33
Indeed. Purely opinion: Were it my game, and I the DM, I would probably rule in favor of the creation of a second feat called "Quick Sheath". That way you could do it, but it'd cost you pretty significantly to be able to do so. I can't imagine how that could be unbalancing. Though that could just be lack of imagination on my part... – Naetuir Jun 13 '12 at 16:50
In my circles we include sheathing because of things like iaijutsu where a character would quickly draw, slash, sheathe the weapon in a fluid motion. (Granted that's in spirit to the original source of the skill). Creating an extra feat to put a weapon away kind of black balls a character's development. – CatLord Jun 14 '12 at 11:33
However, Iaijutsu is about pulling quickly and slashing quickly - and then standing there dramatically waiting to see if you fall over or they do :-P - or else keep slashing. But if you want to put the weapon away in the same turn that's a move action to me. Maybe even one that provokes unless you have quickdraw? Especially if you clean the blade of blood first, as they usually do with a quick air strike: Check out this video so you know what I mean. youtube.com/watch?v=_YdSKZzXOFo – Julix Apr 26 '14 at 4:24
I would absolutely take a Quick Sheath feat for my switch-hitter Ranger if I could, but our GM has decided I can't. – mcv Jan 9 '15 at 13:10

in the APG page 179 under Quick draw Shield it does give some info on the mechanics,and does state:If you have the quick draw feat,you may don or put away a quick draw shield as a free action.It also says If you have a base attack bonus of +1 or higher,you may don or put away a quick draw shield as a swift action combined with a regular move.I'd rule a weapon is easier to sheath then to take off a quick draw shield but that's for you and your GM to decide

-

simple solution, with Quick Draw, make sheathing a Swift Action, tada, solved.

-
Welcome to the site, Utomneian, could you unpack this a bit? Take a look at our FAQ. – Brian Ballsun-Stanton Oct 29 '12 at 8:22
-1 Please explain how to turn sheathing into a swift action. I'll remove my downvote once this answer explains that. If it's just a house ruling, and not possible via RAW, that isn't really helpful (because you could just immediately resolve the entire issue via houserules by saying Quick Draw lets you quickly swap). – doppelgreener Oct 29 '12 at 12:07


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# The domain of $f(x)=\left(\Bigl\lvert\bigl\lfloor |x|-1\bigr\rfloor\Bigr\rvert-5\right)^{-1/2}$

How should I find the domain of the following function?

$$f(x)=\left(\Bigl\lvert\bigl\lfloor |x|-1\bigr\rfloor\Bigr\rvert-5\right)^{-1/2}$$

I am getting something like $6\leq|x|<7$ but not sure whether I'm right. What will be the best approach?

You have tried to find the domain, which is the possible $x$-values. The range is the possible $y$-values.
The function inside the absolute value signs can be any integer greater than or equal to $-1$. So the possible values for $f(x)$ is a list of numbers, one value each for most of those integers.
• $\lfloor| x|\rfloor-1>5$ or $\lfloor |x|\rfloor-1<-5$. The second case can't happen, so you have $\lfloor|x|\rfloor>6$ or $|x|\geq7$ – Empy2 Apr 13 '15 at 18:34


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Simulations of a Non-linear Least Squares fit of an Exponential Decay Process with R Shiny App

SLIDES


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# Página Inicial » Inscrições Científicas » Trabalhos

## Dados do Trabalho

### Título

A KERNEL K-MEANS CLUSTERING ALGORITHM BASED ON AN ADAPTIVE MAHALANOBIS KERNEL

### Resumo

Clustering method are useful tools for exploiting structures in datasets and have been widely used for unsupervised pattern recognition. Clustering means organizing a set of observations (objects, individuals, genes, pixels, etc.) into cluster so that observations belonging to a given cluster have a high degree of similarity, whereas observations belonging to different cluster have a high degree of dissimilarity. Euclidean distance is the most commonly used in clustering methods. However, methods based on this distance have good performance in data whose cluster are approximately hyperspherical and linearly separable. Because of this limitation, several methods capable of dealing with data whose structure is complex have been proposed, among which, kernel based clustering methods whose essence involves realization $\Phi$ arbitrary nonlinear mapping of the original space $p$- dimensional $X\subset\mathbb{R}^p$ for a dimension space higher (possibly infinite), called the feature space, $\mathcal{F}$. The most commonly used kernel function is the Gaussian. Despite its good characteristics, this kernel is based on the Euclidean distance, that is, it assumes that the observations are more likely to be distributed in a hyperspherical region (that is, equal variances and zero covariance). However, examples in two different cluster are more likely to be distributed inside two hyperelipsoid regions diferente. The distance from Mahalanobis, which takes into account the correlations between variables and is invariant in scale, is a better choice for dealing with hyperelipsoid regions. We propose, under the approach kernelization of the metric, a algorithm kernel K-means based on an adaptive Mahalanobis kernel. The effectiveness of the proposed algorithm was demonstrated by experiments with simulated data.

### Palavras-chave

Clustering, Kernel, Mahalanobis.

### Área

Dados Funcionais, Dados em Alta Dimensão e Aprendizado Estatístico de Máquinas

### Autores

Fernanda Florencio Costa, Marcelo Rodrigo Portela Ferreira


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1. jeroendehaas

For the last three years I’ve used the moderncv package to typeset my CV: http://www.ctan.org/pkg/moderncv. Well worth checking out if you’re considering writing your CV in LaTeX.

• Nice, thanks! I was not aware of the package. I will add it to the list. Best, Tom.

• I tried the europecv package quite some time ago and forgot to add it. Thanks for the link! Cheers, Tom.

• Thanks for the links. The latter has a few interesting templates. Best, Tom.

2. henningninneh

excellent. this looks exactly like my word-typeset cv. now i must give it a try in LaTex. thanks for this post.

• Thanks for your comment and have fun typing your CV in LaTeX. Cheers, Tom.

3. It looks good, but I would go with the cv or the moderncv packages. They’re quite easy to use, and yet produce simple but elegant résumés.

• Thanks for your thoughts on that topic. Appreciate it. Best, Tom.

4. Out of completeness, you should also mention the CurVe class (shameless plug). Available on CTAN.

• Very nice, thanks Didier.

Best, Tom.

5. that is not mother tongue it is native language

• Thanks for your remark. I guess both work. It depends on your preference. Thanks, Tom.

6. Bala

beautiful!!!

7. […] Here is an another excellent post discussing   resume. 13.060416 80.249634 Rate this: Sharing is fun:MoreLike this:LikeBe the first to like this post. […]

8. akmkat

• Hi!

Thanks for your comment. Did you use the minipage environment as in the example above? Please provide a minimal example and I’ll try to help.

Best, Tom.

9. […] In any case, what got me started on a resume was a Google+ post that linked to this page: http://texblog.org/2012/04/25/writing-a-cv-in-latex/ […]

10. eiselekd

The layout is nice however there a problems I think:
When using tabular and you have a long list, i.e.
in the “Practical Experience” section, tabular will
not break across multiple pages. Also when you add
footnotes inside tables you need to be an expert. A simpler
2 column layout would be, it misses the vertical ruler however:

\documentclass{scrartcl}
\usepackage{lipsum}
\begin{document}
\begin{labeling}{XXXXXXX}%template for longest item.
\item[2000] \lipsum[1-2]
\item[2001] \lipsum[1-2]
\end{labeling}
\end{document}
• Thanks for the sample code. I agree, one may run into problems, but LaTeX almost always provides a reasonable solution. I wrote about footnotes in tables in one of my previous posts. Concerning the multipage issue, often one can arrange things with margins, font size, line spacing, etc. Another possibility would be to use the longtable package instead, which allows tables to span across multiple pages.

Best, Tom.

11. Daniel

Thanks for a great post and blog!
I’m wondering about how to change the indentation of the dates in the first column. Suppose you want to specify the month e.g. 201203 – 201206. This causes a line-break but it seems there is more space to use to the left. What would you do?

• Hi Daniel,

Thanks for your comment. What you would do is to change the width of the manually defined columns. Try for example 0.25\textwidth for L and 0.6\textwidth for R as in the code below.

\usepackage{array, xcolor}
\definecolor{lightgray}{gray}{0.8}
\newcolumntype{L}{>{\raggedleft}p{0.25\textwidth}}
\newcolumntype{R}{p{0.6\textwidth}}
\newcommand\VRule{\color{lightgray}\vrule width 0.5pt}

Best, Tom.

12. Daniel

Thanks! By the way, is there a way to let the red line come before the first column?

• Hi Daniel,

Do you mean a vertical red line? You could color a vrule with the xcolor package:

\documentclass[10pt]{article}
\usepackage[table]{xcolor}
\usepackage{array, lipsum, bibentry}
\usepackage{tabu}
\usepackage[margin=3cm]{geometry}
\taburulecolor{red}
\title{\bfseries\Huge Tom T. Texblog}
\author{texblog+cv@gmail.com}
\date{}
\definecolor{lightgray}{gray}{0.8}
\newcolumntype{L}{|>{\raggedleft}p{0.14\textwidth}}
\newcolumntype{R}{p{0.8\textwidth}}
\newcommand\VRule{\color{lightgray}\vrule width 0.5pt}
...

Changes are highlighted.

Best, Tom.

13. Daniel

Nice! Yes, that’s what I meant:), (of course I mean the lightgray line which I changed to red, thus the confusion). I changed \tabular to \tabu and then everything worked perfectly.
Cheers, Daniel

• Got it, sorry I misunderstood you initially. I’m happy it worked out though! Best, Tom.

14. crlsmrgf

Thank u very much!
It was also something in my TO-DO list for too long, and it was much ore easy than I thought!

• Great to hear you found my post useful, thanks. Tom.

15. Alex

I’m having trouble getting the publications section to work… Even if I copy your source code exactly, I get and error: “Citation ‘knuth2006art’ undefined”. Any idea how to fix this?

• Hi Alex,

Two things:

• Did you run latex, bibtex, latex, latex?
• Do you have a publication.bib file containing the bib-entries?

In case you use the complete example provided in the post, the file will be created when you typeset the document (latex) the first time.

Hope this helps.
Best, Tom

• Luís Martins

Hi, I was testing your example and the same error appeared. In line 69 – \nobibliography{publication.bib}, I just removed the extension and the error disappeared.

• Changed, thanks! Tom.

16. With the publications list, I get a vertical space in front of each of my \bibentry bits, so that the year is a line above the text. If I add some text or “\hspace{0pt}” ahead of it, it jumps back on the right line. This is an ugly hack, though. Any suggestions?

• Hi Matt,

Unfortunately, I wasn’t able to reproduce what you describe above. Could you provide a minimal working example? Something similar to the code below would do.

Thanks, Tom.

\documentclass{article}
\usepackage{bibentry}
\usepackage{array, xcolor}
\definecolor{lightgray}{gray}{0.8}
\newcolumntype{L}{>{\raggedleft}p{0.14\textwidth}}
\newcolumntype{R}{p{0.8\textwidth}}
\newcommand\VRule{\color{lightgray}\vrule width 0.5pt}
\begin{filecontents}{publication.bib}
@book{knuth2006art,
title={The art of computer programming: Generating all trees: history of combinatorial generation},
author={Knuth, D.E.},
volume={4},
year={2006},
\end{filecontents}
\begin{document}
\bibliographystyle{plain}
\nobibliography{publication.bib}
\section*{Publications}
\begin{tabular}{L!{\VRule}R}
2006&\bibentry{knuth2006art}\vspace{5pt}\\
\end{tabular}
\end{document}
• Hi Tom,

In the process of eliminating things to give a minimal example, I discovered the culprit: the hyperref package. (This is crucial for me if we’re going to be posting the CV online or sending it electronically.) I also managed to find a solution: use the longtable package instead of the ordinary tabular environment.

Best,
Matt

• Thanks for the feedback. I appreciate it. Best, Tom.

• Pablo

Hi Matthew, the hyperref package was causing trouble to me too. Would you mention what you did to have publications AND the hyperref package both work? Thanks!

• Pablo

I don’t understand how longtable can help me to enable href. Do you mean to use it instead of the array package, and that will allow using the href package?

• Yep. Let me know if you run into problems and I will try my best to help. Best, Tom.

17. interesting,thanks for sharing.

• thanks! cheers, tom.

18. Martin Vogel

Hi,
Thanks for your nice CV template.
I just wonder how to modify the code to include properly a photo in the beginning. I am not very familiar with the minipage environment.
Martin

• Thanks for your question, Martin. That’s fairly simple. Copy-paste the code above in the post which has the photo black box and replace:

\rule{width}{height}

with

\includegraphics[width=3.5cm]{path/photo.png}

You’ll have to load the graphicx package to make use of the width option.

Let me know if you have any problems with that and if you provide the code, I’ll most certainly be able to help.

Cheers, Tom.

19. […] Blog’s Writing a CV in LaTeX (where I found out about the bibentry […]

20. Jolyon Bloomfield

Heya,

Just wanted to point out that the use of “\vspace{5pt}\\” to get nice spacing in the tables can backfire if the \vspace will cause a new line to be added to that row (if your text is _just_ the wrong length). Better is to use “\\[5pt]” which doesn’t have this problem.

Best,
Jolyon

• Hi Jolyon,

Thanks for taking the time to point this out. I changed the code accordingly.

Thanks,
Tom.

21. Phil

Hi there.
I’m very new to latex so forgive me naive question.

I noticed that when a section was a bit long it started on a new page leaving sometime a lot of extra white page after the previous section.

Is it something that can be changed easily?

Otherwise, thank for the excellent post.

• Hi Phil,

Thanks for your question. You probably have to play around to find the optimal arrangement. However, in general, I would not expect a section to be automatically moved to the next page if there is blank space left on the current page. Unless you add \clearpage or \newpage of course.

Could you provide a minimal example for illustration?

Thanks, Tom.

• jolyonbloomfield

Hi there,

This happens because LaTeX tries to fit a tabular environment on a single page; it won’t break it across pages. I ran into this same issue recently. The fix is to use the longtable package:

\usepackage{longtable}

You probably also want to use these commands in the preamble (otherwise the spacing gets a little big):

\setlength{\LTpre}{0pt}
\setlength{\LTpost}{0pt}

Then, wherever \begin{tabular} … \end{\tabular} appears, change it to the following.

\begin{longtable}
...
\end{longtable}

I hope this helps!
Jolyon

• Phil

Thank you very much everyone. I’ll try it once I’m back to the office and will let you know how it works!

Thank again,
Phil

• Phil

Just a quick reply to let you it worked like a charm. Thank guys!

• Great, thanks for the feedback! Tom.

22. Ben

Dear All,

Thank you for this template. It works really great.

But I want to add a foot note stating the date and company for wich this Resume is made.

I’ve been searching the net for how to do this, but I can’t figure out how to get a footer on the first page …

Is there any good trick to do this?

Thank you!

• Hi Ben,

Assuming you are using the fancyhdr package and redefined the header/footer according to your needs. Now to get a header/footer on the first page, just add the following line after \maketitle:

\thispagestyle{fancy}

Best, Tom.

23. Johnathan

Just wanted to say this is a great guide, many thanks for this.

Cheers,
Johnathan

24. Rivka

Hi,
I used this as a template to write my CV, thanks!

25. dottag

Thank you for the awesome CV! This was exactly my Word-style but my limited knowledge of LaTeX did not allow me to do the same.

One question: how do I make the second column left-justified? I’d loke to do it because I don’t like that LaTeX stretches out the between-words spaces (it’s not an article, so the right margin looks “ragged” anyway).

• Hello!
Left-align the right side is done similarly to right-align the left side using \raggedright in the column definition. I attached a minimal working example below.

\documentclass{article}
\usepackage{xcolor, blindtext, array}
\definecolor{lightgray}{gray}{0.8}
\newcolumntype{L}{>{\raggedleft}p{0.14\textwidth}}
\newcolumntype{R}{>{\raggedright}p{0.8\textwidth}}
\newcommand\VRule{\color{lightgray}\vrule width 0.5pt}
\begin{document}
\section*{Title}
\begin{tabular}{L!{\VRule}R}
2013&\blindtext\\
\end{tabular}
\end{document}
26. peddrim

Thank you very much! Your CV template is the most beautiful I could find

• Thanks, glad you like it. Best, Tom.

27. Tom Walton

Hi Tom,

Have you seen the LaTeX CV template on http://www.writelatex.com – you can open it for editing in their online editor with one click. Is pretty neat (the original template is from howtotex)

cheers,
Tom

• Yup, saw it. Thanks! Pretty neat indeed. Tom

28. Chaanakya

This post was really nice. Thank you One suggestion/improvement: for the address/other info, you could use this code instead, which would put the address and other info flush against the left/right side respectively:

\begin{minipage}[ht]{0.48\textwidth}
\begin{flushleft}
666 Devil St. \\
Hell
\end{flushleft}
\end{minipage}
\begin{minipage}[ht]{0.48\textwidth}
\begin{flushright}
666-666-6666 \\
satan@gmail.com
\end{flushright}
\end{minipage}
• Hi!

Thanks for the code! Indeed a nice way to place the address. Best, Tom.

29. lazhar

• Hi!

You can download the LaTeX distribution from TUG. I put together a few direct links for different operating systems on my resources page.

HTH,
Tom

• Hi Katia,

Thanks for getting in touch. It seems that similar ideas were used in the CV in question than what I published here. Still, I wouldn’t want to re-engineer the CV in question. Why don’t you ask the person directly for a LaTeX template? That would make it much easier for you. Also, I’d be more than happy to help, should you get stuck preparing your CV with the template.

Tom

• Katia Consani

Hi Tom, thanks for the quick reply, I asked the author of that CV for a latex template but I got not reply. I’d appreciate if you can suggest me a basic skeleton to obtain a similar output

• Hi Katia, I will give it a try and post it here…

30. Katia Consani

Thanks Tom, I’ll look forward to your post!

• Hi Katia,

Please find below code for some elements of the CV you liked. You picked the font using the LaTeX font catalogue. LaTeX lets you mix different fonts as it was done in the example CV. I wouldn’t recommend it, there are other ways to highlight certain words or sentences.
I hope with that you’ll be able to prepare your CV.
Cheers, Tom

\documentclass[11pt]{article}

% Font
\usepackage[light,math]{kurier}
\usepackage[T1]{fontenc}

\usepackage{xcolor}
\definecolor{darkred}{rgb}{0.8,0,0}

\usepackage{titlesec}
\newcommand{\lineafter}[1]{#1{} \hrulefill}
\titleformat{\section}{\scshape\huge\centering\color{darkred}}{}{0pt}{\hrulefill{} \lineafter}[]
\titleformat{\subsection}{\scshape\large\color{darkred}}{}{0pt}{\lineafter}[]

% Tabular style for content
\usepackage{array, xcolor}
\newcolumntype{L}{>{\raggedleft}p{0.14\textwidth}}
\newcolumntype{R}{p{0.8\textwidth}}
\newcommand\VRule{\vrule width 0.5pt}

\begin{document}
\begin{minipage}{0.48\textwidth}
Personal information
\end{minipage}
\begin{minipage}{0.48\textwidth}
\end{minipage}

\begin{tabular}{L!{\VRule}R}
2012--13&Some text\\
{\it Something}&Some other text\\
\multicolumn{2}{c}{}\\
2011&Some text\\
{\it Something}&Some other text\\
\end{tabular}

\end{document}
• Katia Consani

Great Tom! Your code will help me a lot in preparing the cv.
Thank you very much for your help.

• You are welcome. Good luck with your CV.

31. Allen

Very good codes. Helps me alot. My question is, can I move the title(my name) a bit upper? Since I wanna keep my CV in one page, the extra blank above title is not that good.

• Hi Allen,

Thanks for your question. I used the geometry package to set the margins to 3cm. You can change or use top, bottom, left, right to set the margins individually. See the package documentation for more details.

Cheers, Tom.

32. lchriste

Hi Tom,

first of all, thanks for a really nice CV template. I’ve tried to tweak it somewhat for my own purposes, but I can’t seem to get footnotes to work properly in the document. Specifically, the the footnote text does not show at the bottom of the page, when I compile the document. I wonder whether this has something to do with that it is overwritten by the tabular environment or something.

Any ideas of how to solve this?

Best,

• Hi there,

From your description, I assume you are trying to place footnote marks within a tabular. You might find this post helpful. If that doesn’t solve your problem, please post a minimal example and I’ll be better able to help.

Cheers, Tom

33. Algore

Best CV made in Latex. I swear, I looked up all other template (moderncv etc.). They are way too overloaded for a proper academic CV. Thanks a lot!

• tom

Thanks for the feedback. Tom.

34. Daniel

Hey Tom,

first of all, thanks for that great code! But I have one question…do you have any idea how to insert a picture that is exactly right of e.g. the “education” section, so that it doesn’t shift the text down?
Thank you very much in advance,

Daniel

• tom

Hi Daniel,

I’d use minipage to split the page into two columns locally. Give the code below a try and change it according to what you had in mind.

Cheers, Tom.

\section*{Education}
\newcolumntype{M}{>{\raggedleft}p{0.28\textwidth}}
\newcolumntype{N}{p{0.66\textwidth}}
\begin{minipage}{0.5\textwidth}
\begin{tabular}{M!{\VRule}N}
2005--2007&{\bf MSc in Computer Science, Great University, Country.}\\[5pt]
2001--2005&BSc in Life Science, Great University, Country.\\[5pt]
1997--2001&Fancy Highschool, Contry.\\
\end{tabular}
\end{minipage}
\begin{minipage}{0.5\textwidth}
\centering\rule{0.35\linewidth}{0.35\linewidth}
\end{minipage}
• Daniel

It works, thanks a lot!

• tom

Great to hear! Thanks, Tom.

35. Hi I get following error while compiling the minimal example.

! Undefined control sequence.
l.2 \abx@aux@sortscheme
{nty}

Am I missing anything?

• tom

It seems you are sorting citations (nty: name, title, year). So you might want to look for the line where you do that. I can’t recall having that in my original code.

Cheers, Tom

36. Shanmukharao

Very very useful. I am happy with this format. Thanks a lot.

• tom

Good to hear, thanks!

• tom

Cool, I wasn’t aware of this. Thanks for the link!


text:
# What’s new in Kwant 1.4¶

This article explains the user-visible changes in Kwant 1.4.0. Subsequently, the user-visible changes for each maintenance release of the 1.4.x series are listed (if there were any).

## Summary: release highlights¶

Backwards-incompatible changes:

## Automatic Peierls phase calculation¶

When defining systems with orbital magnetic fields it is often cumbersome to manually calculate the phases required by the Peierls substitution, and to ensure that the chosen gauge is consistent across the whole system (this is especially true for systems with leads that point in different directions). This release introduces kwant.physics.magnetic_gauge, which calculates the Peierls phases for you:

def hopping(a, b, t, peierls):
return -t * peierls(a, b)

def B_syst(pos):
return np.exp(-np.sum(pos * pos))

syst = make_system(hopping)
syst = syst.finalized()

gauge = kwant.physics.magnetic_gauge(syst)

# B_syst in scattering region, 0 in lead.

kwant.hamiltonian_submatrix(syst, params=params)

Note that the API for this functionality is provisional, and may be revised in a future version of Kwant.

## Conductivity calculations using kwant.kpm.conductivity¶

Kwant now has the ability to calculate conductivities using a combination of the Kernel Polynomial Method and the Kubo-Bastin relation. The following example calculates $$σ_{xy}$$ for various chemical potentials at finite temperature:

syst = make_system().finalized()
sigma_xy = kwant.kpm.conductivity(syst, alpha='x', beta='y')
conductivities = [sigma_xy(mu=mu, temperature=0.1)
for mu in np.linspace(0, 4)]

## Integration with Qsymm library¶

Kwant now contains an integration with the Qsymm library for analysing model symmetries. This functionality is available under kwant.qsymm. Here is an example for extracting the symmetry group of a graphene system:

import numpy as np
import kwant
import kwant.qsymm

s_0 = np.eye(2)

lat = kwant.lattice.honeycomb(norbs=[1, 1])
sym = kwant.TranslationalSymmetry(lat.vec((1, 0)), lat.vec((0, 1)))

graphene = kwant.Builder(sym)
graphene[[lat.a(0, 0), lat.b(0, 0)]] = 0
graphene[lat.neighbors()] = 1

symmetry_generators = kwant.qsymm.find_builder_symmetries(graphene)

# Let's find what the chiral symmetry looks like

def is_chiral(g):
return g.antisymmetry and not g.conjugate and np.allclose(g.R, s_0)

print(next(g for g in symmetry_generators if is_chiral(g)))

kwant.qsymm also contains functionality for converting Qsymm models to Kwant Builders, and vice versa, and for working with continuum Hamiltonians (such as would be used with kwant.continuum). This integration requires separately installing Qsymm, which is available on the Python Package Index.

## System parameter substitution¶

After the introduction of Builder.fill it has become possible to construct Kwant systems by first creating a “model” system with high symmetry and then filling a lower symmetry system with this model. Often, however, one wants to use different parameter values in different parts of a system. In previous versions of Kwant this was difficult to achieve.

Builders now have a method substituted that makes it easy to substitute different names for parameters. For example if a builder model has a parameter V, and one wishes to have different values for V in the scattering region and leads, one could do the following:

syst = kwant.Builder()
syst.fill(model.substituted(V='V_dot'), ...))

syst = syst.finalized()

## System parameters can now be inspected¶

In modern Kwant the preferred way to pass arguments to your models is by name, using a dictionary and params:

def onsite(site, magnetic_field, voltage):
return magnetic_field * sigma_z + voltage * sigma_0

syst = make_system(onsite)
syst = syst.finalized()

# naming the arguments makes things clear!
kwant.smatrix(syst, params=dict(magnetic_field=0.5, voltage=1,

This is a much clearer and less error prone than passing arguments by position using args, as was required in older versions of Kwant. In this version of Kwant we introduce the parameters attribute of finalized systems, which allows inspection of the names of the parameters that the system (and its leads) expects:

>>> syst.parameters
frozenset({'magnetic_field', 'voltage'})

This is a provisional API that may be changed in a future version of Kwant.

## Passing system arguments via args is deprecated in favor of params¶

It is now deprecated to pass arguments to systems by providing the args parameter (in kwant.smatrix and elsewhere). Passing arguments via args is error prone and requires that all value functions take the same formal parameters, even if they do not depend on all of them. The preferred way of passing parameters to Kwant systems is by passing a dictionary using params:

def onsite(site, magnetic_field, voltage):
return magnetic_field * sigma_z + voltage * sigma_0

syst = make_system(onsite).finalized()

kwant.smatrix(syst, params=dict(magnetic_field=0.5, voltage=0.2))

# Compare this to the deprecated 'args'
kwant.smatrix(syst, args=(0.5, 0.2))

Providing args will be removed in a future Kwant version.

## Interpolated density plots¶

A new function, kwant.plotter.density, has been added that can be used to visualize a density defined over the sites of a Kwant system. This convolves the “discrete” density (defined over the system sites) with a “bump” function in realspace. The output of density can be more informative that map when plotting systems with many sites, where it is not important to see the individual contribution from each site.

## Configurable maximum velocity in stream plots¶

The function streamplot has got a new option vmax. Note that this option is not available in current. In order to use it, one has to call streamplot directly as shown in the docstring of current.

## Improved heuristic for colorscale limits in kwant.plotter.map¶

Previously map would set the limits for the color scale to the extrema of the data being plotted when vmin and vmax were not provided. This is the behaviour of matplotlib.imshow. When the data to be plotted has very sharp and high peaks this would mean that most of the data would appear near the bottom of the color scale, and all of the features would be washed out by the presence of the peak. Now map employs a heuristic for setting the colorscale when there are outliers, and will emit a warning when this is detected.

## Sites from different families are plotted in different colors by default¶

Previously kwant.plotter.plot would plot all sites in black. Now sites from different families are plotted in different colors, which improves the default plotting style. You can still customize the site coloring using the site_color parameter, as before.

## kwant.physics.Bands can optionally return eigenvectors and velocities¶

kwant.physics.Bands now takes extra parameters that allow it to return the mode eigenvectors, and also the derivatives of the dispersion relation (up to second order) using the Hellman-Feynman relation:

syst = make_system().finalized()

bands = kwant.physics.Bands(syst)
(energies, velocities, vectors) = bands(k=0, derivative_order=1,
return_eigenvectors=True)

## Finalized Builders keep track of which sites were added when attaching leads¶

When attaching leads to an irregularly shaped scattering region, Kwant adds sites in order to make the interface with the leads “smooth”. Previously, the information of which sites were added was not inspectable after finalization. Now the sites that were added from each lead are available in the lead_paddings attribute. See the documentation for FiniteSystem for details.

## kwant.continuum.discretize can be used with rectangular lattices¶

Previously the discretizer could only be used with lattices with the same lattice constant in all directions. Now it is possible to pass rectangular lattices to the discretizer:

kwant.continuum.discretize(
'k_x**2 + k_y**2',
grid=kwant.lattice.general([(1, 0), (0, 2]),
)

This is useful when you need a finer discretization step in some spatial directions, and a coarser one in others.

## Restrictions on value functions when named parameters are given¶

New restrictions apply to how value functions may accept arguments, when named parameters are given through params. (Nothing changes when the now deprcated args mechanism is used). The restrictions follow the principle that each value function must take a clearly specified set of named parameters. This allows to make the parameter handling less error-prone and faster.

In particular, when params is used, it is no longer possible for value functions to - take *args or **kwargs, - take keyword-only parameters, - have default parameters for arguments.

As an example, the following snippet no longer works because it uses default values:

syst = kwant.Builder()

# Parameter 't' has a default value of 1
def onsite(site, V, t=1):
return V = 2 * t

def hopping(site_a, site_b, t=1):
return -t

syst[...] = onsite
syst[...] = hopping

# Raises ValueError
syst = syst.finalized()

As a solution, simply remove the default values and always provide t. To deal with many parameters, the following idiom may be useful:

defaults = dict(a=0, b=1, c=2, d=3)
...
smatrix = kwant.smatrix(syst, E, params=dict(defaults, d=4, e=5))

Note that this allows to override defaults as well as to add additional parameters.

## Installation on Microsoft Windows is available via Conda¶

Kwant is now packaged for the Conda package manager on Windows, and using Conda is the preferred method for installing Kwant on that platform. Please refer to the installation section of the Kwant website for details. Currently the MUMPS solver is not available for the Windows version of the Conda package; we hope to include MUMPS support in a later patch release.

## Minimum required versions for some dependencies have increased¶

Kwant now requires at least the following versions:

• Python 3.5

• numpy 0.11.0

• scipy 0.17.0

• matplotlib 1.5.1

These versions (or newer) are available in the latest stable releases of Ubuntu and Debian GNU/Linux.

## Changes in Kwant 1.4.1¶

• The list of user-visible changes was rearranged to emphasize backwards-incompatible changes by moving them to the top of the list.

• Restrictions on value functions no longer apply when the old args mechanism is used, this restores most of the backwards compatibility with previous Kwant versions: Restrictions on value functions when named parameters are given.

• The args parameter passing mechanism works again with wraparound-treated systems. Some restriction continue to appply, notably it is not possible to use wraparound with value functions that take *args or *kwargs.

• Kwant no longer requires the existence of a parameters attribute for low-level systems.

• A note about an API change that affects low-level systems that ocurred in Kwant 1.3 was added.

## Changes in Kwant 1.4.2¶

• Due to bugs in the Qsymm library that required modifying its API, the version of Qsymm that Kwant depends on had to be increased from 1.1.2 to 1.2.6. SInce Qsymm is an optional dependency of Kwant, this backwards-incompatible change only affects Kwant users who use it.

• Minor bugs were fixed in the KPM module.

## Changes in Kwant 1.4.3¶

• Builder properly handles Hermitian conjugation of array-like values. Before, if the value was provided not as a true array but, for example, as a list of lists, it was not conjugated, which could lead to generation of a non-Hermitian Hamiltonian matrix without emitting any error.

• Various maintenance.


text:
+0

# I need help with this problem.

-5
131
1
+-18

I need help with this problem.

Mar 13, 2019
edited by GAMEMASTERX40  Mar 15, 2019

#1
+102804
+2

I think that you have already been given the answers to many of these.

So it is time for you to start learning how to do them.

I will give you some examples that will help.

$$5^{1/2}=\sqrt5\\ 7^{-3}=\frac{1}{7^{+3}}\\ \\~\\ 8^{3/2}=(8^3)^{1/2}=(8^{1/2})^3$$

Now think about it and you tell me what  these are equal to.

$$1) \;\; 2^{-1/2}\\ 2) \;\; 2^{-3/2}\\ and \\ 3) \;\;b^{1/2}$$


text:
# Common Review Quotes

In my reviews a couple of points come up again and again (and again). It's like somebody is teaching bad practice out there.

As a result a lot of my reviews look very similar (or get terse with my annoyance). It would be nice if there was an area we could have canned quotes that we could link to had all the text we need for a point. Then we could peer review these canned quotes to make sure they are accurate. Update them with wisdom etc. Then when doing the review, just link to the canned quote.

Do we support anything like that? On Stack Overflow, they had added the tag to handle this. Could we introduce something similar on this site? A tag we use to curate common mistakes and the perfect review answer for them.

• IIRC there's a userscript for this, I'll try to find it. – Riker Mar 6 '17 at 18:58
• I have my own long list of "frequently made review suggestions" but it would be nice to be able to have a list instead. I like this idea a lot! – Edward Mar 6 '17 at 18:59
• That's probably something we could maintain on Github / Gist as community, but I don't think the site's meta is a good place for this, since all languages/frameworks/etc. have their own common issues. Thoughts? We can throw some ideas around in chat if you like. – Phrancis Mar 6 '17 at 20:24
• Interestingly enough, another user and I were working on some stuff for a website a while back, but it kind of died: github.com/Hosch250/FrequentlyPostedReviews – user34073 Mar 6 '17 at 21:40
• I got a couple of half-baked ideas (pastebin). I don't know how I should post them, or if they're any good, so I'll leave them here. – Peilonrayz Mar 7 '17 at 0:29
• – 200_success Mar 7 '17 at 0:38
• How about linking to something in Stack Overflow Documentation instead? – 200_success Mar 7 '17 at 0:48
• @200_success That's probably not a good idea. Documentation still has a lot of problems that need fixing. Maybe once it's stabilized, but that hasn't happened yet, and probably won't for, oh, 6-8 weeks or so. – Fund Monica's Lawsuit Mar 7 '17 at 3:41
• I like the idea of using the tag wiki. But if it is just one long page it becomes less useful as you force the user to scroll through a whole bunch of irelavant stuff. In the tag Wiki; is there a way to add anchors to the text so we can link to a specific section? – Martin York Mar 7 '17 at 18:56
• @200_success I don't think documentation is the correct place. Documentation shows you the functionality. But does not have common mistakes or common themes. – Martin York Mar 7 '17 at 18:57
• I would prefer not to use an offsite store. Git/pastbin/gist etc. There are a whole bunch of ownership issues. – Martin York Mar 7 '17 at 19:00

Experiment one did not really work.

I have rolled back the C++ tag wiki page.
With the minor addition of a couple of lines at the top:

As an experiment I am starting to collect snippets of common Code Review things that happen repeatedly here: Code Review

I have started a github repository. CodeReview

If you want to help fill out this repository (especially for other languages) I would love the help. But as a starting point you need to have reached the level of being able to edit the tag wiki for the language you want to contribute for.

Yes this is an arbitrary rule so just using it as a starting point and we can see if it goes anywhere.

One possible way we could approach this would be to ask (and answer) questions on Stack Overflow like this one: https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice

It also has the advantage that if there are different views or tradeoffs, these are likely to be more fully discussed (in the form of multiple answers) than in any one person's answer.

• Yes we should definitely link to these articles. But when I give this link I also try and explain that why its std and not standard (so you only have to type 5 characters and that's not a big deal). – Martin York Mar 7 '17 at 19:15
• Also your text on return 0 is a couple of links and some explaining text. I would like to capture that in a single place to re-use verbatim. – Martin York Mar 7 '17 at 19:16
• If we went this route, it'd be nice to be able to categorise these questions. Since we probably shouldn't use Stack Overflow tags to do this, do you have any ideas to achieve this? – Peilonrayz Mar 25 '17 at 23:09
• I've added my return 0 text as an answer on StackOverflow: stackoverflow.com/questions/204476/… – Edward Apr 22 '17 at 11:39

OK I don't think we have a real consensus.
But as an experiment I have started editing the tag wiki for C++.

• It's on site.
• The previous content was an outdated copy of what was already on SO (so no loss)
• People with the appropriate privileges can edit it (and help)


text:
# Amplitude's time dependence in Heisenberg representation

1. Apr 6, 2013

### LayMuon

$$A = \langle q_f(t) \mid q_i(t) \rangle = \langle q_{f,H} \mid e^{iH(t_0-t)} e^{-iH(t-t_0)} \mid q_{i,H} \rangle = \langle q_{f,H} \mid q_{i,H} \rangle$$

This means that A is time-independent, and depends only on the reference point $t_0$. How is it possibly? From Schoedinger picture it does depend on time! Thanks.

2. Apr 6, 2013

### vanhees71

You have to be careful about the meaning of your vectors. From what you write, I conclude that $|q(t) \rangle$ are the (generalized) eigenvectors of some operator representing an observable in the Heisenberg picture. Then, from the Heisenberg equation of motion
$$\mathrm{d}_t \hat{O}=\frac{1}{\mathrm{i}}[\hat{O},\hat{H}].$$
Assuming we have time independent Hamiltonian and that implies
$$\hat{O}(t)=\exp(\mathrm{i} t \hat{H}) \hat{O}(0) \exp(-\mathrm{i} t \hat{H}).$$ For sake of simplicity, I've set $t_0=0$.

Now by definition $|q(t) \rangle$ is the eigenvector of $\hat{O}(t)$ with the fixed eigenvalue $q$:
$$\forall t: \quad \hat{O}(t) |q(t) \rangle=q |q(t) \rangle.$$
This implies that
$$|q(t) \rangle=\exp(\mathrm{i} t \hat{H}) |q(0) \rangle.$$
So up to a sign this implies your formula.

Of course, because the time evolution is unitary, the scalar product is time independent
$$\langle q(t)|q'(t) \rangle=\langle q(0) | q'(0) \rangle=\delta(q,q'),$$
where the $\delta$ stands for a Kronecker $\delta$ (discrete eigenvalues) or a Dirac $\delta$ distribution (continuous eigenvalues).


text:
### Top 10 Arxiv Papers Today in Probability

##### #1. Stochastic quantization associated with the $\exp(Φ)_2$-quantum field model driven by space-time white noise on the torus
###### Masato Hoshino, Hiroshi Kawabi, Seiichiro Kusuoka
We consider a quantum field model with exponential interactions on the two-dimensional torus, which is called the $\exp (\Phi)_{2}$-quantum field model or H{\o}egh-Krohn's model. In the present paper, we study the stochastic quantization of this model by singular stochastic partial differential equations, which is recently developed. By the method, we construct a unique time-global solution and the invariant probability measure of the corresponding stochastic quantization equation, and identify with an infinite-dimensional diffusion process, which has been constructed by the Dirichlet form approach.
more | pdf | html
None.
###### Tweets
MathPHYPapers: Stochastic quantization associated with the $\exp(\Phi)_2$-quantum field model driven by space-time white noise on the torus. https://t.co/LFxv1WEE2D
mathPRb: Masato Hoshino, Hiroshi Kawabi, Seiichiro Kusuoka : Stochastic quantization associated with the $\exp(Φ)_2$-quantum field model driven by space-time white noise on the torus https://t.co/53RsT7ojh2 https://t.co/6nms0jANYc
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##### #2. Moderate deviations for the size of the giant component in a random hypergraph
###### Jingjia Liu, Matthias Löwe
We prove a moderate deviations principles for the size of the largest connected component in a random $d$-uniform hypergraph. The key tool is a version of the exploration process, that is also used to investigate the giant component of an Erd\"os-R\'enyi graph, a moderate deviations principle for the martingale associated with this exploration process, and exponential estimates.
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###### Tweets
jkbren: @aliceschwarze @_fernando_rosas This is a good thread, in part because now I’m just gonna indiscriminately post papers that seem to be about hypergraphs here :) https://t.co/YUQqrpMW1v
mathPRb: Jingjia Liu, Matthias Löwe : Moderate deviations for the size of the giant component in a random hypergraph https://t.co/RqiTLNa0xx https://t.co/g96mgQPyXf
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##### #3. Asymptotic behavior of solution and non-existence of global solution to a class of conformable time-fractional stochastic equation
###### Erkan Nane, Eze R. Nwaeze, McSylvester Ejighikeme Omaba
Consider the following class of conformable time-fractional stochastic equation $$T_{\alpha,t}^a u(x,t)=\lambda\sigma(u(x,t))\dot{W}_t,\,\,\,\,x\in\mathbb{R},\,t\in[a,T], T<\infty,\,\,0<\alpha<1,$$ with a non-random initial condition $u(x,0)=u_0(x),\,x\in\mathbb{R}$ assumed to be non-negative and bounded, $T_{\alpha,t}^a$ is a conformable time - fractional derivative, $\sigma:\mathbb{R}\rightarrow\mathbb{R}$ is globally Lipschitz continuous, $\dot{W}_t$ a generalized derivative of Wiener process and $\lambda>0$ is the noise level. Given some precise and suitable conditions on the non-random initial function, we study the asymptotic behaviour of the solution with respect to the time parameter $t$ and the noise level parameter $\lambda$. We also show that when the non-linear term $\sigma$ grows faster than linear, the energy of the solution fails to have global existence at all times $t$ for all $\alpha\in (0,\frac{1}{2})\cup(\frac{1}{2},1)$ and the solution grows to infinity (fails to exist) even with linear condition on $\sigma$...
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mathPRb: Erkan Nane, Eze R. Nwaeze, McSylvester Ejighikeme Omaba : Asymptotic behavior of solution and non-existence of global solution to a class of conformable time-fractional stochastic equation https://t.co/V7ciE1SMbK https://t.co/62yO2qk55z
MathPHYPapers: Asymptotic behavior of solution and non-existence of global solution to a class of conformable time-fractional stochastic equation. https://t.co/x9nE6ZUWea
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##### #4. Polynomial localization of the 2D-Vertex Reinforced Jump Process
###### Christophe Sabot
We prove polynomial decay of the mixing field of the Vertex Reinforced Jump Process (VRJP) on $\Bbb{Z}^2$ with bounded conductances. Using [17] we deduce that the VRJP on $\Bbb{Z}^2$ with any constant conductances is almost surely recurrent. It gives a counterpart of the result of Merkl, Rolles [14] and Sabot, Zeng [17] for the 2-dimensional Edge Reinforced Random Walk.
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mathPRb: Christophe Sabot : Polynomial localization of the 2D-Vertex Reinforced Jump Process https://t.co/BBDYDeh7um https://t.co/wIXlGTMgHd
MathPHYPapers: Polynomial localization of the 2D-Vertex Reinforced Jump Process. https://t.co/MH2TvtWSAS
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##### #5. On the cut-off phenomenon for the maximum of a sampling of Ornstein-Uhlenbeck processes
###### Gerardo Barrera
In the article we study the so-called cut-off phenomenon in the total variation distance when $n$ tends to infinity for the family of continuous-time stochastic processes indexed by $n\in \mathbb{N}$, $\left( \mathcal{Z}^{(n)}_t= \max\limits_{j\in \{1,\ldots,n\}}{X^{(j)}_t}:t\geq 0\right),$ where $X^{(1)},\ldots,X^{(n)}$ is a sampling of $n$ ergodic Ornstein-Uhlenbeck processes driven by an $\alpha$-stable processes. It is not hard to see that for each $n\in \mathbb{N}$, the random variable $\mathcal{Z}^{(n)}_t$ converges in the total variation distance to a limiting distribution $\mathcal{Z}^{(n)}_\infty$ as $t$ goes by. Using the asymptotic theory of extremes; in the Gaussian case we prove that the total variation distance between $\mathcal{Z}^{(n)}_t$ and $\mathcal{Z}^{(n)}_\infty$ converges to a universal function in a constant time window around the cut-off time, a fact known as profile cut-off in the context of stochastic processes. On the other hand, in the heavy-tailed case we prove that there is not cut-off.
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mathPRb: Gerardo Barrera : On the cut-off phenomenon for the maximum of a sampling of Ornstein-Uhlenbeck processes https://t.co/cEOALu1z63 https://t.co/btVYDqjrcP
StatsPapers: On the cut-off phenomenon for the maximum of a sampling of Ornstein-Uhlenbeck processes. https://t.co/dtaw9bwvqw
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##### #6. Three-Parametric Marcenko-Pastur Density
###### Taiki Endo, Makoto Katori
The complex Wishart ensemble is the statistical ensemble of $M \times N$ complex random matrices with $M \geq N$ such that the real and imaginary parts of each element are given by independent standard normal variables. The Marcenko--Pastur (MP) density $\rho(x; r), x \geq 0$ describes the distribution for squares of the singular values of the random matrices in this ensemble in the scaling limit $N \to \infty$, $M \to \infty$ with a fixed rectangularity $r=N/M \in (0, 1]$. The dynamical extension of the squared-singular-value distribution is realized by the noncolliding squared Bessel process, and its hydrodynamic limit provides the two-parametric MP density $\rho(x; r, t)$ with time $t \geq 0$, whose initial distribution is $\delta(x)$. Recently, Blaizot, Nowak, and Warchol studied the time-dependent complex Wishart ensemble with an external source and introduced the three-parametric MP density $\rho(x; r, t, a)$ by analyzing the hydrodynamic limit of the process starting from $\delta(x-a), a > 0$. In the present paper, we give...
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mathPRb: Taiki Endo, Makoto Katori : Three-Parametric Marcenko-Pastur Density https://t.co/YcQEX8gMvE https://t.co/xmq00dIpte
MathPHYPapers: Three-Parametric Marcenko-Pastur Density. https://t.co/G0xqHw0seK
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##### #7. Isolation by distance patterns arising from short range and long range dispersal -- a forwards in time approach
###### Raphaël Forien
In this paper, we consider a mathematical model for the evolution of neutral genetic diversity in a spatial continuum including mutations, genetic drift and either short range or long range dispersal. The model we consider is the spatial $\Lambda$-Fleming-Viot process introduced by Barton, Etheridge and V\'eber, which describes the state of the population at any time by a measure on $\mathbb{R}^d \times [0,1]$, where $\mathbb{R}^d$ is the geographical space and $[0,1]$ is the space of genetic types. In both cases (short range and long range dispersal), we prove a functional central limit theorem for the process as the population density becomes large and under some space-time rescaling. We then deduce from these two central limit theorems a formula for the asymptotic probability of identity of two individuals picked at random from two given spatial locations. In the case of short range dispersal, we recover the classical Wright-Mal\'ecot formula, which is widely used in demographic inference for spatially structured...
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mathPRb: Raphaël Forien : Isolation by distance patterns arising from short range and long range dispersal -- a forwards in time approach https://t.co/KocblYEJJ7 https://t.co/P79QVhx5iZ
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##### #8. Brownian bridge with random length and pinning point for modelling of financial information
###### Mohammed Louriki
Developed countries are increasingly relying on gas storage to ensure security of supply. In this article we consider an approach to gas storage valuation in which the information about the time at which the holder of a gas storage contract should choose to inject or withdraw gas is modelled using a Brownian bridge that starts at zero and is conditioned to equal a constant x in the time of injection and a constant y in the time of withdrawal. This enables to catch some empirical facts on the behavior of gas storage markets: when the Brownian bridge process is away from the boundaries x and y, the holder of the gas storage contract can be relatively sure that the best decision is to do nothing. However, when the bridge information process absorbs y, the holder of the contract should take the decision of withdrawing gas on the other hand, they should take the decision to inject gas when the process absorbs x. In this sense the Brownian bridge information process leaks information concerning the time at which the holder of a storage...
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mathPRb: Mohammed Louriki : Brownian bridge with random length and pinning point for modelling of financial information https://t.co/aO65pYqWi1 https://t.co/gfC6Wa7SNY
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##### #9. Optimization Of Quasi-convex Function Over Product Measure Sets
###### Jerome Stenger, Fabrice Gamboa, Merlin Keller
We consider a generalization of the Bauer maximum principle. We work with tensorial products of convex measures sets, that are non necessarily compact but generated by their extreme points. We show that the maximum of a quasi-convex lower semicontinuous function on this product space is reached on the tensorial product of finite mixtures of extreme points. Our work is an extension of the Bauer maximum principle in three different aspects. First, we only assume that the objective functional is quasi-convex. Secondly, the optimization is performed over a space built as a product of measures sets. Finally, the usual compactness assumption is replaced with the existence of an integral representation on the extreme points. We focus on product of two different types of measures sets, called the moment class and the unimodal moment class. The elements of these classes are probability measures (respectively unimodal probability measures) satisfying generalized moment constraints. We show that an integral representation on the extreme...
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mathPRb: Jerome Stenger (EDF R\D PRISME, GdR MASCOT-NUM), Fabrice Gamboa (IMT), Merlin Keller (EDF R\D PRISME) : Optimization Of Quasi-convex Function Over Product Measure Sets https://t.co/WktmGjKgP6 https://t.co/b4dBjSAZut
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##### #10. Renewal Theory for Transient Markov Chains with Asymptotically Zero Drift
###### Denis Denisov, Dmitry Korshunov, Vitali Wachtel
We solve the problem of asymptotic behaviour of the renewal measure (Green function) generated by a transient Lamperti's Markov chain $X_n$ in $\mathbf R$, that is, when the drift of the chain tends to zero at infinity. Under this setting, the average time spent by $X_n$ in the interval $(x,x+1]$ is roughly speaking the reciprocal of the drift and tends to infinity as $x$ grows. For the first time we present a general approach relying in a diffusion approximation to prove renewal theorems for Markov chains. We apply a martingale type technique and show that the asymptotic behaviour of the renewal measure heavily depends on the rate at which the drift vanishes. The two main cases are distinguished, either the drift of the chain decreases as $1/x$ or much slower than that, say as $1/x^\alpha$ for some $\alpha\in(0,1)$. The intuition behind how the renewal measure behaves in these two cases is totally different. While in the first case $X_n^2/n$ converges weakly to a $\Gamma$-distribution and there is no law of large numbers...
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###### Tweets
mathPRb: Denis Denisov, Dmitry Korshunov, Vitali Wachtel : Renewal Theory for Transient Markov Chains with Asymptotically Zero Drift https://t.co/pw9rTyL4Oa https://t.co/qNYgpwiAIR
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text:
# zbMATH — the first resource for mathematics

Second order Hamiltonian vector fields on tangent bundles. (English) Zbl 0822.58021
Let $$M$$ be a differentiable manifold. A vector field $$\Gamma$$ of $$TM$$ which corresponds to a system of second order ordinary differential equations on $$M$$ is called a second order Hamiltonian vector field if it is the Hamiltonian field of a function $$F \in C^ \infty(TM)$$ with respect to a Poisson structure $$P$$ of $$TM$$. We formulate the direct problem as that of finding $$\Gamma$$ if $$P$$ is given, and the inverse problem as that of finding $$P$$ if $$\Gamma$$ is given. We show the solution of the direct problem if $$P$$ is a symplectic structure such that the fibers of $$TM$$ are Lagrangian submanifolds. For the inverse problem we generalize Henneaux’ method of looking for a solution by studying a corresponding system of linear algebraic restrictions.
Reviewer: I.Vaisman (Haifa)

##### MSC:
 37J99 Dynamical aspects of finite-dimensional Hamiltonian and Lagrangian systems
Full Text:
##### References:
 [1] Coste, A.; Dazord, P.; Weinstein, A., Groupoïdes symplectiques, Dept. math. univ. Lyon 2/A, 1-62, (1987) · Zbl 0668.58017 [2] Dazord, P.; Lichnerowicz, A.; Marle, Ch.-M., Structures locale des variétés de Jacobi, J. math. pures et appl., 70, 101-152, (1991) · Zbl 0659.53033 [3] Henneaux, M., On the inverse problem of the calculus of variations, J. physics A, 15, L93-98, (1992) [4] Kosmann-Schwarzbach, Y.; Magri, F., Poisson-Nijenhuis structures, Ann. inst. H. Poincaré série A (physique théorique), 53, 35-81, (1990) · Zbl 0707.58048 [5] Lichnerowicz, A., LES variétés de Poisson et leurs algèbres de Lie associées, J. diff. geometry, 12, 253-300, (1977) · Zbl 0405.53024 [6] MacKenzie, K., Lie groupoids and Lie algebroids in differential geometry, () · Zbl 0683.53029 [7] Morandi, G.; Ferrario, C.; Lo Vecchio, G.; Marmo, G.; Rubano, C., The inverse problem in the calculus of variations and the geometry of the tangent bundle, Physics reports, 188, 3-4, (1990) · Zbl 1211.58008 [8] Vaisman, I., Cohomology and differential forms, (1973), M. Dekker New York [9] Vaisman, I., The Poisson-Nijenhuis manifolds revisited, Rendiconti sem. mat. Torino, 52, 377-394, (1994) · Zbl 0852.58042 [10] Weinstein, A., The local structure of Poisson manifolds, J. diff. geometry, 18, 523-557, (1983) · Zbl 0524.58011 [11] Weinstein, A., Lagrangian mechanics and groupoids, (1992), Univ. of Calif Berkeley, Preprint [12] Yano, K.; Ishihara, S., Tangent and cotangent bundles, (1973), M. Dekker New York · Zbl 0262.53024
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text:
## On the James and Hilton-Milnor splittings, and the metastable EHP sequence

##### Doc. Math. 26, 1423-1464 (2021)
DOI: 10.25537/dm.2021v26.1423-1464

### Summary

This note provides modern proofs of some classical results in algebraic topology, such as the James Splitting, the Hilton-Milnor Splitting, and the metastable EHP sequence. We prove fundamental splitting results $\Sigma \Omega \Sigma X \simeq \Sigma X \vee (X\wedge \Sigma\Omega \Sigma X)$ and $\Omega(X \vee Y) \simeq \Omega X\times \Omega Y\times \Omega \Sigma(\Omega X \wedge \Omega Y)$ in the maximal generality of an $\infty$-category with finite limits and pushouts in which pushouts squares remain pushouts after basechange along an arbitrary morphism (i.e., Mather's Second Cube Lemma holds). For connected objects, these imply the classical James and Hilton-Milnor splittings. Moreover, working in this generality shows that the James and Hilton-Milnor splittings hold in many new contexts, for example in: elementary $\infty$-topoi, profinite spaces, and motivic spaces over arbitrary base schemes. The splitting results in this last context extend Wickelgren and Williams' splitting result for motivic spaces over a perfect field [\textit{K. Wickelgren} and \textit{B. Williams}, Geom. Topol. 23, No. 4, 1691--1777 (2019; Zbl 1428.14032)]. We also give two proofs of the metastable EHP sequence in the setting of $\infty$-topoi: the first is a new, non-computational proof that only utilizes basic connectedness estimates involving the James filtration and the Blakers-Massey Theorem, while the second reduces to the classical computational proof.

### Mathematics Subject Classification

55P35, 55P40, 55P99, 55Q20, 18N60, 14F42

### Keywords/Phrases

infinity categories

### References

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### Affiliation

Devalapurkar, Sanath
Harvard University, 1 Oxford Street, Cambridge, MA 02138, USA
Haine, Peter
Department of Mathematics, University of California, Berkeley, 753 Evans Hall, Berkeley, CA 94720, USA


text:
Fiducial, total and differential cross-section measurements of $t$-channel single top-quark production in $pp$ collisions at 8 TeV using data collected by the ATLAS detector

Abstract : Detailed measurements of $t$-channel single top-quark production are presented. They use 20.2 fb$^{-1}$ of data collected by the ATLAS experiment in proton-proton collisions at a centre-of-mass energy of 8 TeV at the LHC. Total, fiducial and differential cross-sections are measured for both top-quark and top-antiquark production. The fiducial cross-section is measured with a precision of 5.8 % (top quark) and 7.8 % (top antiquark), respectively. The total cross-sections are measured to be $\sigma_{\mathrm{tot}}(tq) = 56.7^{+4.3}_{-3.8}\;$pb for top-quark production and $\sigma_{\mathrm{tot}}(\bar{t}q) = 32.9^{+3.0}_{-2.7}\;$pb for top-antiquark production. In addition, the ratio of top-quark to top-antiquark production cross-sections is determined to be $R_t=1.72 \pm 0.09$, with an improved relative precision of 4.9 % since several systematic uncertainties cancel in the ratio. The differential cross-sections as a function of the transverse momentum and rapidity of both the top quark and the top antiquark are measured at both the parton and particle levels. The transverse momentum and rapidity differential cross-sections of the accompanying jet from the $t$-channel scattering are measured at particle level. All measurements are compared to various Monte Carlo predictions as well as to fixed-order QCD calculations where available.
Type de document :
Article dans une revue
European Physical Journal C: Particles and Fields, Springer Verlag (Germany), 2017, 77, pp.531. 〈https://link.springer.com/article/10.1140%2Fepjc%2Fs10052-017-5061-9〉. 〈10.1140/epjc/s10052-017-5061-9〉

http://hal.in2p3.fr/in2p3-01464170
Contributeur : Claudine Bombar <>
Soumis le : vendredi 10 février 2017 - 09:33:52
Dernière modification le : mardi 24 juillet 2018 - 15:46:14

Citation

M. Aaboud, N. Berger, A.M. Burger, M. Delmastro, L. Di Ciaccio, et al.. Fiducial, total and differential cross-section measurements of $t$-channel single top-quark production in $pp$ collisions at 8 TeV using data collected by the ATLAS detector. European Physical Journal C: Particles and Fields, Springer Verlag (Germany), 2017, 77, pp.531. 〈https://link.springer.com/article/10.1140%2Fepjc%2Fs10052-017-5061-9〉. 〈10.1140/epjc/s10052-017-5061-9〉. 〈in2p3-01464170〉

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text:
# how do i work out the area of a circle without a calculator?

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how do i work out the area of a circle without a calculator?

thanx

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pen and paper, basic arithmetic would be my choice

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use approxamations?

pi = 3 etc?

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2pi x the Radius Squared (or something like that).

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Use 3.142 for Pi.

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can anyone give me an example???

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For a circle of 2m radius:

Area = Pi x r2

Area = 3.142 x 4

Area = 12.6m2

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2pi = 3.142 x 2 thus the 1st bit is 6.284

whats the Radius? lets assume its 9cms so whats 9 Squared? thats 81

81 x 6.284 = 509 cms

Hmmm... I think I just worked out a Sphere?

nevermind, I said my maths was crap!

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I think a good approximate for pi is 3.1416 since the next after the five there's a 9.

Memorizing pi is probably the most idiotic thing to do, but for some strange reason I've memorized up to 35 decimals. I wonder what's the current world record...

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Why dko you need to approximate anyway? the area of a circle is pi*r^2, you shpould always keep the pi in, unless for some reason you need to know where it approximately lies on the real number line, but generally there's no need to know.

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Why do you need to approximate, eh? How about pi being an irrational number.

Btw, World Record: Goto, Hiroyuki | Japan | 42195 digits, World Rec since -95

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Why do you need to approximate' date=' eh? How about pi being an irrational number.

Btw, World Record: Goto, Hiroyuki | Japan | 42195 digits, World Rec since -95[/quote']

You should never approximate unless you have to, the decimals are only one representation of the real numbers.

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If you didn't notice, the thread is about Ice wanting to know about counting the area of a circle without a calculator; therefore, you need to approximate quite a bit.

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If your asked to find the raidus of a circle of say radius 9m, then you should alewyas express your answer as 81*pi m^2, of course if for some reason (I relaly can't think of any reaosn though) you need to know wherre it approximatley lies on the real number line then you would approximate pi.

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If you didn't notice, the thread is about Ice wanting to know about counting the area of a circle without a calculator; therefore, you need to approximate quite a bit.

If he can't multiply numbers just because they're on the wrong side of a decimal place I'd have concerns about what his maths teacher is doing all day.

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I would just multiply 3.14 by the Radius squared

3.14®®

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when i have to calculate the area of a circle without a calculator, I usually leave it in terms of pi.

pi*r^2=A

r=2

2^2=4

A=4pi

Ask whoever is telling you to do this if they want an approximate or exact value. If they want exact you must leave it in terms of pi. However, if you are just doing this on your own, pick one.

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I don't think I've used a calculator to work out areas of things approximately for quite some time now. Leaving it in terms of pi is a lot better.

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He could of corse leave it with Pi without turning it into numbers. This can be useful where there is a more complex question with more than 1 circle. I remember doing quesitions where pi canceled out, this is the most likely solution if it crops up on a non calculator paper.

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Hey there,

I recently did found myself wanting to work out the area of a circle by hand, and here was what I came up with...

Tools:

A compass, a pen/pencil/stylus, a straightedge, paper or otherwise neat drawing surface, and oodles of free time which may or may not drive you into a crushing torpor.

Method:

Take the compass and make for yourself a nice circle of some arbitrary radius (make it big enough to really see everything -- about hand-sized). Be sure to fix this radius in the compass (perhaps you've a nice one with a screw-lock or something else handy). Do not swallow your compass.

Now taking your compass, carefully stick the pointy end on an arbitrary point somewhere on the circle (perhaps your favorite spot?). Trace-out your second circle here and be sure to note how it should cross directly over the center of the first circle you made. It also should intersect the first circle at exactly two points. Choose your favorite of these two and stick the center of a third circle (same radius you've been using again) on that point. It should intersect at two points on the original circle, and you can just pick up your compass and move around that way again. After doing this carefully for a while, you'll notice that you've made a total of five lovely circles--one mother circle in the middle and four clones radiating around her.

Take a step back and examine what you've done. You should definitely notice how these four circles intersect each other outside and inside the mother circle in two places for every pair of circles. There should be four intersections outside the momma circle, and four inside (all at the exact center of the momma). Connect these intersection points by making them straight through the center (normal to the mother circle).

At this time, you should have exactly five circles and two straight lines drawn (via-straightedge). These lines should be perpendicular to each other and make a cross.

If you will notice, these lines make new intersection points on the circle....you may continue to divide the circle this way (by powers of two) if you repeat the above method by placing your compass on these new points and trace around the momma circle again until all new points on the circle have become home to a new circle center (same radii of course).

Now, while that is interesting certainly, it is only a mere step (necessary of course) in the right direction to answering your original question..."how do i work out the area of a circle without a calculator?"

So here we go....

Please notice your original two lines making the cross intersect the mother circle in four places that are perfectly (at least in your head they are perfect...because this is all an exercise to emphasize the validity of mathematical ideas here) and evenly spaced around the circular perimeter/circumference. Please draw a square inside this circle (a circumscription, you might say) with vertices at those four points.

Now, I hope you've made your circle big enough to really see all of what comes next.

Focus your attention on your favorite quadrant of the circle....I'll use the second just because I'm weird like that. You should see a very nice 45-45-90 triangle there...one fourth of the area of the square (which happens to be the biggest square that can fit).

The legs of this triangle are each a magnitude R of the radius you randomly picked, making the hypotenuse a line of magnitude [(2RR)^(1/2)]. This simplifies to R times the square root of 2.

Now notice the line you drew before (unless you stopped after making the cross...if so, keep dividing the circle's angles in half as described in the beginning) cuts this triangle in half, bisecting the hypotenuse and running through the center of the momma circle. That line should have intersected the mother circle 45 degrees above the quadrant you're looking at....in other words, it should have split the cross's 90 degree angle in half there. Draw a new line from the point where that splitting line intersects the mother circle to one of the two vertices of the 45-45-90 triangle that is NOT at the center of the mother circle.

This should make for a new right triangle...one that has a long leg and a very short leg. You will notice that you could repeat this for all the quadrants and you'd have made for yourself a nice little octagon. By continuing to divide the circle's angles in half and making these new little triangles on top of the hypotenuses of the old right triangles, you can successfully construct polygons with powers of two as their side number (square-->2^2 : octogon-->2^3 : 16-sides-->2^4 etc.).

You know the area of a square is base times height.....in this case R squared divided by two gets our triangle, and 4 times that for the entire polygon.

Area =(( (2^2)(R^2) )/ 2) + ...next polygons...

The base of the triangle sitting on top of the triangle that is part of the big square is just half of the old hypotenuse we just found a second ago....R times the square root of 2.

So side B for this triangle is R times the square root of 2 all divided by 2.

We'll refer to that as B.

The other leg of this new triangle, leg A, is at present unstated. Notice, however, that there is another right triangle that has half the area in the original 45-45-90 triangle. It is kinda flipped with it's hypotenuse as the radius of the circle. Well, we already know that it's hypotenuse is R, and one of it's sides is just half of the 45-45-90's hypotenuse...this is side B of the new triangle, too. Now we can find the other leg (actually it is obvious in this special case....but a couple polygons down the line, it would not have been so obvious because everything becomes a radical mess!). The other leg is the square root of all of R squared minus B squared. Now, this is the missing leg of that triangle, but as for the smallest leg of the new triangle--the one whose hypotenuse and side A are unknown--we can say that R minus the square root of all of R squared minus B squared is that side length....side A for the new triangle (the one piggy-backing on the 45-45-90).

With this knowledge, we can find the hypotenuse of that triangle, and simply multiply A and B and divide by 2 to get the area of the triangle. Multiply by the next power of two (with respect to whatever polygon that hypotenuse is a part of) and you've an area that is even closer to that of a circle...

Area ==( (2)(R^2) ) + ( (2)(R^2)(-1+sqrt(2)) + ...next polygons...

You can repeat this method to get closer and closer to the area of a circle, and if you assume R is one, closer and closer to a hand estimation of the elusive Pi constant.

Hope this helped!

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well obviously the easiest way to approximate the are is to fill the circle with tiny squares

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2pi = 3.142 x 2 thus the 1st bit is 6.284

whats the Radius? lets assume its 9cms so whats 9 Squared? thats 81

81 x 6.284 = 509 cms

Hmmm... I think I just worked out a Sphere?

nevermind' date=' I said my maths was crap! [/quote']

You appear to have magically input a '2' into the equation. A = pi r^2, not 2 pi r ^2.

The volume of a sphere is 4/3 pi r^3, the surface area is 4 pi r ^2.

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well obviously the easiest way to approximate the are is to fill the circle with tiny squares

It's the SIMPLEST. In no way is it the easiest.

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You guys are cracking me up with this pi R squared stuff.

Pi are round --- cornbread r square.

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Never post again.