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# The Golden Ratio

(Difference between revisions)
 Revision as of 09:48, 9 July 2012 (edit)← Previous diff Current revision (03:00, 12 December 2012) (edit) (undo) (6 intermediate revisions not shown.) Line 1: Line 1: {{Image Description Ready {{Image Description Ready |ImageName=The Golden Ratio |ImageName=The Golden Ratio - |Image=Goldenratioapplet1.jpg + |Image=Goldenratio.jpg - |ImageIntro=The '''golden number,''' often denoted by lowercase Greek letter "phi", is
${\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399.... + |ImageIntro=The '''golden number,''' often denoted by lowercase Greek letter "phi", is - The term '''golden ratio''' refers to any ratio which has the value phi. The image to the right is a warped representation of dividing and subdividing a rectangle into the golden ratio. The result is [[Field:Fractals|fractal-like]]. This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number. + :[itex] \frac{a+b}{a} = \frac{a}{b} \equiv \varphi,$ - |ImageDescElem=The golden number, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. [http://en.wikipedia.org/wiki/Golden_ratio "Golden ratio"], Retrieved on 20 June 2012. + where the Greek letter [[Phi (letter)|phi]] (\varphi) represents the golden ratio. Its value is: -
+ - Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa use the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids.
+ - However, such claims have been criticized in scholarly journals as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.[http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf "Misconceptions about the Golden Ratio"], Retrieved on 24 June 2012. + + ${\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399...$.
+ 1/1=1.000000
+ 2/1=2.000000
+ 3/2=1.500000
+ 5/3=1.666666
+ 8/5=1.600000
+ 13/8=1.625000
+ 21/13=1.615385
+ 34/21=1.619048
+ 55/34=1.617647
+ 89/55=1.618182
+ 144/89=1.717978
+ 233/144=1.618056
+ 377/233=1.618026
+ 610/377=1.618037
+ 987/610=1.618033
- ==A Geometric Representation== - ===The Golden Ratio in a Line Segment=== + ==The Golden Ratio in Nature== - [[Image:Golden_segment.jpg|400px]][[Image:Animation2.gif]] + - + ===Spirals & Phyllotaxis=== - The golden number can be defined using a line segment divided into two sections of lengths ''a'' and ''b''. If ''a'' and ''b'' are appropriately chosen, the ratio of ''a'' to ''b'' is the same as the ratio of ''a'' + ''b'' to ''a'' and both ratios are equal to $\varphi$. The line segment above (left) exhibits the golden proportion. The line segments above (right) are also examples of the golden ratio. In each case, + [[Image:Sunflower head.jpeg|px200]] - $\frac{{\color{Red}\mathrm{red}}+\color{Blue}\mathrm{blue}}{{\color{Blue}\mathrm{blue}} }= \frac{{\color{Blue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} }= \varphi . + Spirals are abundant concerning phyllotaxis, which describes the way leaves are arranged on a plant stem. In 92 percent of Norway spruce cones, the spirals were found to appear in rows five and eight rows. They appeared in rows of four and seven in six percent, and four and six in four percent. In addition, the number of right-handed spirals appears to be equal to the number of left-handed spirals. The arrangements of the spirals in these spruce cones are found as the following pairs of rows: 2/3,3/5,5/8,8/13,13/21,21/34,34/55,55/89, and 89/144. These numbers are all numbers that belong to the Fibonacci Series. + [[Image:Pinecone3.gif]] - + A similar phenomenon can be observed in sunflower heads. A sunflower head has both clockwise and counterclockwise spirals. The numbers of the spirals in a sunflower usually depend on the size of the sunflower. However, the ratios of the spirals that usually occur are 89/55,144/89, and even 233/144. Once again, all of the numbers in these ratios are consecutive Fibonacci numbers. - ===The Golden Rectangle=== + - A '''golden rectangle''' is any rectangle where the ratio between the sides is equal to phi. When the sides lengths are proportioned in the golden ratio, the rectangle is said to possess the '''golden proportions.''' A golden rectangle has sides of length [itex]\varphi \times r and [itex]1 \times r where [itex]r$ can be any constant. Remarkably, when a square with side length equal to the shorter side of the rectangle is cut off from one side of the golden rectangle, the remaining rectangle also exhibits the golden proportions. This continuing pattern is visible in the golden rectangle below. + - :[[Image:Coloredfinalrectangle1.jpg]] + - + |AuthorName=Joyce Han - + - ===Triangles=== + - [[Image:1byrrectangle1.jpg|500px]][[Image:Pentagon_final.jpg|300px]] + - + - The golden number, $\varphi$, is used to construct the '''golden triangle,''' an isoceles triangle that has legs of length $\varphi \times r$ and base length of $1 \times r$ where $r$ can be any constant. It is above and to the left. Similarly, the '''golden gnomon''' has base $\varphi \times r$ and legs of length $1 \times r$. It is shown above and to the right. These triangles can be used to form regular pentagons (pictured above) and pentagrams. + - + - The pentgram below, generated by the golden triangle and the golden gnomon, has many side lengths proportioned in the golden ratio. + - :[[Image:Star1.jpg]] + - :::$\frac{{\color{SkyBlue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} } = \frac{{\color{Red}\mathrm{red}} }{{\color{Green}\mathrm{green}} } = \frac{{\color{Green}\mathrm{green}} }{{\color{Magenta}\mathrm{pink}} } = \varphi .$ + - + - These triangles can be used to form [[Field:Fractals| fractals]] and are one of the only ways to tile a plane using '''pentagonal symmetry'''. Pentagonal symmetry is best explained through example. Below, we have two fractal examples of pentagonal symmetry. Images that exhibit pentagonal symmetry have five symmetry axes. This means that we can draw five lines from the image's center, and all resulting divisions are identical. + - + - :[[Image:Penta1.jpg|400px]] + - :[[Image:Pent111.jpg|400px]] + - |ImageDesc==An Algebraic Derivation of Phi= + - + - {{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText=How can we derive the value of $\varphi$ from its characteristics as a ratio? We may algebraically solve for the ratio ($\varphi$) by observing that ratio satisfies the following property by definition: + - :$\frac{b}{a} = \frac{a+b}{b} = \varphi$|FullText= + - Let $r$ denote the ratio : + - :$r=\frac{a}{b}=\frac{a+b}{a}$. + - + - So + - :$r=\frac{a+b}{a}=1+\frac{b}{a}$ which can be rewritten as + - + - :$1+\cfrac{1}{a/b}=1+\frac{1}{r}$ thus, + - + - :$r=1+\frac{1}{r}$ + - + - Multiplying both sides by $r$, we get + - + - :${r}^2=r+1$ + - + - which can be written as: + - :$r^2 - r - 1 = 0$. + - + - Applying the quadratic formula + - , we get $r = \frac{1 \pm \sqrt{5}} {2}$. + - + - The ratio must be positive because we can not have negative line segments or side lengths. Because the ratio has to be a positive value, + - + - :$r=\frac{1 + \sqrt{5}}{2} = 1.61803399... =\varphi$. + - |NumChars=500}} + - + - + - + - + - ==Continued Fraction Representation and [[Fibonacci sequence|Fibonacci Sequences]]== + - The golden ratio can also be written as what is called a '''continued fraction,'''a fraction of infinite length whose denominator is a quantity plus a fraction, which latter fraction has a similar denominator, and so on. This is done by using recursion. + - + - {{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText= |FullText=We have already solved for $\varphi$ using the following equation: + - + - ${\varphi}^2-{\varphi}-1=0$. + - + - We can add one to both sides of the equation to get + - + - ${\varphi}^2-{\varphi}=1$. + - + - Factoring this gives + - + - $\varphi(\varphi-1)=1$. + - + - Dividing by $\varphi$ gives us + - + - $\varphi -1= \cfrac{1}{\varphi }$. + - + - Adding 1 to both sides gives + - + - $\varphi =1+ \cfrac{1}{\varphi }$. + - + - Substitute in the entire right side of the equation for $\varphi$ in the bottom of the fraction. + - + - $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }$ + - + - Substituting in again, + - + - $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\varphi}}}$ + - + - + - + - $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}$ + - + - Continuing this substitution forever, the last infinite form is a continued fraction + - + - If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction, replacing $\varphi$ with 1, we produce the ratios between consecutive terms in the [[Fibonacci sequence]]. + - + - $\varphi \approx 1 + \cfrac{1}{1} = 2$ + - + - $\varphi \approx 1 + \cfrac{1}{1+\cfrac{1}{1}} = 3/2$ + - + - $\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1} } } = 5/3$ + - + - $\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cfrac{1}{1}}}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}} =1 + \cfrac{1}{1 + \cfrac{2}{3}} = 8/5$ + - + - Thus we discover that the golden ratio is approximated in the Fibonacci sequence. + - + - $1,1,2,3,5,8,13,21,34,55,89,144...\,$ + - + -
$1/1$$=$$1$
$2/1$$=$$2$
$3/2$$=$$1.5$
$8/5$$=$$1.6$
$13/8$$=$$1.625$
$21/13$$=$$1.61538462...$
$34/21$$=$$1.61904762...$
$55/34$$=$$1.61764706...$
$89/55$$=$$1.61818182...$
+ - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - $\varphi = 1.61803399...\,$ + - + - As you go farther along in the Fibonacci sequence, the ratio between the consecutive terms approaches the golden ratio. Many real-world applications of the golden ratio are related to the Fibonacci sequence. For more real-world applications of the golden ratio [[Fibonacci Numbers|click here!]] + - + - In fact, we can prove that the ratio between terms in the Fibonacci sequence approaches the golden ratio by using mathematical [[Induction]]. + - + - {{Switch|link1=Click to show proof|link2=Click to hide proof|1= |2= + - + - Since we have already shown that + - + - $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}$, + - + - we only need to show that each of the terms in the continued fraction is the ratio of Fibonacci numbers as shown above. + - + - First, let + - :$x_1=1$, + - :$x_2=1+\frac{1}{1}=1+\frac{1}{x_1}$, + - :$x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2}$ and so on so that + - :$x_n=1+\frac{1}{x_{n-1}}$. + - + - These are just the same truncated terms as listed above. Let's also denote the terms of the Fibonacci sequence as + - :$s_n=s_{n-1}+s_{n-2}$ where $s_1=1$,$s_2=1$,$s_3=2$,$s_4=3$ etc. + -
+ - + - We want to show that + - :$x_n=\frac{s_{n+1}}{s_n}$ for all n. + - + - First, we establish our [[Induction|base case]]. We see that + - :$x_1=1=\frac{1}{1}=\frac{s_2}{s_1}$, and so the relationship holds for the base case. + - + - Now we assume that + - :$x_k=\frac{s_{k+1}}{s_{k}}$ for some $1 \leq k < n$ (This step is the [[Induction|inductive hypothesis]]). We will show that this implies that + - :$x_{k+1}=\frac{s_{(k+1)+1}}{s_{k+1}}=\frac{s_{k+2}}{s_{k+1}}$. + - + -

+ - + - By our assumptions about ''x1'',''x2''...''xn'', we have + - + - :$x_{k+1}=1+\frac{1}{x_k}$. + - + - By our inductive hypothesis, this is equivalent to + - + - :$x_{k+1}=1+\frac{1}{\frac{s_{k+1}}{s_{k}}}$. + - + - Now we only need to complete some simple algebra to see + - + - :$x_{k+1}=1+\frac{s_k}{s_{k+1}}$ + - + - :$x_{k+1}=\frac{s_{k+1}+s_k}{s_{k+1}}$ + - + - Noting the definition of $s_n=s_{n-1}+s_{n-2}$, we see that we have + - + - $x_{k+1}=\frac{s_{k+2}}{s_{k+1}}$ + - + - So by the principle of mathematical induction, we have shown that the terms in our continued fraction are represented by ratios of consecutive Fibonacci numbers. + - + - The exact continued fraction is + - :$x_{\infty} = \lim_{n\rightarrow \infty}\frac{s_{n+1}}{s_n} =\varphi$. + - + - }}|NumChars=75}} + - + - + - + - ==Proof of the Golden Ratio's Irrationality== + - + - {{SwitchPreview|ShowMessage=Click to expand|HideMessage=Click to hide|PreviewText= |FullText= + - Remarkably, the Golden Ratio is irrational, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers. + - We will use the method of contradiction to prove that the golden ratio is irrational. + - + - Suppose $\varphi$ is rational. Then it can be written as fraction in lowest terms $\varphi = b/a$, where ''a'' and ''b'' are integers. + - + - Our goal is to find a different fraction that is equal to $\varphi$ and is in lower terms. This will be our contradiction that will show that $\varphi$ is irrational. + - + - First note that the definition of $\varphi = \frac{b}{a}=\frac{a+b}{b}$ implies that $b > a$ since clearly $b+a>b$ and the two fractions must be equal. + - + -
+ - + - Now, since we know + - + - :$\frac{b}{a}=\frac{a+b}{b}$ + - + - we see that $b^2=a(a+b)$ by cross multiplication. Foiling this expression gives us $b^2=a^2+ab$. + - + - Rearranging this gives us $b^2-ab=a^2$, which is the same as :$b(b-a)=a^2$. + - + - Dividing both sides of the equation by ''a(b-a)'' gives us + - + - :$\frac{b}{a}=\frac{a}{b-a}$. + - + - Since $\varphi=\frac{b}{a}$, this means $\varphi=\frac{a}{b-a}$. + - + - Since we have assumed that a and b are integers, we know that b-a must also be an integer. Furthermore, since [itex] a - |ToDo=-animation? - - http://www.metaphorical.net/note/on/golden_ratio - http://www.mathopenref.com/rectanglegolden.html - |InProgress=Yes - |HideMME=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra - |InProgress=No - } - |Field=Algebra |InProgress=Yes |InProgress=Yes }} }}

## Current revision

The Golden Ratio
Fields: Algebra and Geometry
Image Created By: Joyce Han

The Golden Ratio

The golden number, often denoted by lowercase Greek letter "phi", is
$\frac{a+b}{a} = \frac{a}{b} \equiv \varphi,$

where the Greek letter phi ($\varphi$) represents the golden ratio. Its value is:

${\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399...$.
The term golden ratio refers to any ratio which has the value phi. The image to the right illustrates dividing and subdividing a rectangle into the golden ratio. This page explores how the Golden Ratio can be observed and found in the arts, mathematics, and nature.

# Basic Description

==The Golden Ratio as an Irrational Number==

## The Golden Ratio in the Arts

### Parthenon

There is an abundance of artists who consciously used the Golden Ratio from as long ago as 400 BCE. Once such example is the Greek sculptor Phidias, who built the Parthenon. The exterior dimensions of the Parthenon form the golden rectangle, and the golden rectangle can also be found in the space between the columns.

### Vitruvian Man

Another instance in which the Golden ratio appears is in Leonardo Da Vinci’s drawing of the Vitruvian Man. Da Vinci’s picture of man’s body fits the approximation of the golden ratio very closely. This picture is considered to a depiction of a perfectly proportioned human body. The Golden Ratio in this picture is the distance from the naval to the top of his head, divided by the distance from the soles of the feet to the top of the head.

### Music

Some people have argued that the Golden Ratio in music produces aesthetically pleasing sounds. Major sixth and the minor sixth chords are considered to be the most pleasing intervals, and subsequently, they are the intervals related to the Golden Ratio. The standard tuning tone used is an A and it vibrates at 440 vibrations per second. A major sixth interval below that would be a C, which has a frequency of 264 vibrations per second. The ratio of the two frequencies reduces to 5/3, which is the ratio of two Fibonacci numbers. Similarly, a minor sixth can be obtained from a high C, which has a frequency of 528 vibrations per second, and an E, which is 330 vibrations per second. This ratio reduces to 8/5, which is also a ratio of two Fibonacci numbers.

The Golden Ratio manifests in the idea that music is proportionally balanced. However, it is an area of debate, because some scholars claim that these appearances are purely coincidental, or a result of number juggling from aficionados. However, many contemporary artists have intentionally used the Golden Ratio in their pieces. Composer, mathematician and teacher Joseph Schillinger believed that music could be based entirely on mathematical formulation. He developed a System of Musical Composition in which successive notes in a melody were followed by Fibonacci intervals when counted in half steps. Half steps are the smallest intervals possible, and it is the closest note that can be played higher or lower. For example, if C were the first note in a composition, the following note would be half a step higher, because one is a Fibonacci number. So, the second note would be a D flat. Schillinger would alternate between moving up and down in intervals. Thus, the third note would be two half steps (because two is the next Fibonacci number), down from the D flat, which would be B flat. The next Fibonacci number after two is three, so the next note would be three half steps higher than the last. Three half steps higher than B flat is E flat. Schillinger believed that these notes convey the same sense of harmony as the phyllotactic ratios found in leaves.

# A More Mathematical Explanation

## Fibonacci Numbers

One of the greatest breakthroughs regarding the Golden Ratio came when its rela [...]

## Fibonacci Numbers

One of the greatest breakthroughs regarding the Golden Ratio came when its relation to Fibonacci numbers, also known as the Fibonacci sequence, was discovered. Fibonacci numbers can also be found in the arts, and nature. The Fibonacci sequence is the series of numbers, 0,1,1,2,3,5,8,13,21,34,55,89,144,233… The next term in the Fibonacci sequence, starting from the third, is determined by adding the previous two terms together. The Fibonacci sequence is related to the Golden Ratio because as the sequence grows, the ratio of consecutive terms gradually approaches the Golden Ratio. For example here are the ratios of the successive numbers in the Fibonacci sequence:
1/1=1.000000
2/1=2.000000
3/2=1.500000
5/3=1.666666
8/5=1.600000
13/8=1.625000
21/13=1.615385
34/21=1.619048
55/34=1.617647
89/55=1.618182
144/89=1.717978
233/144=1.618056
377/233=1.618026
610/377=1.618037
987/610=1.618033

## The Golden Ratio in Nature

### Spirals & Phyllotaxis

Spirals are abundant concerning phyllotaxis, which describes the way leaves are arranged on a plant stem. In 92 percent of Norway spruce cones, the spirals were found to appear in rows five and eight rows. They appeared in rows of four and seven in six percent, and four and six in four percent. In addition, the number of right-handed spirals appears to be equal to the number of left-handed spirals. The arrangements of the spirals in these spruce cones are found as the following pairs of rows: 2/3,3/5,5/8,8/13,13/21,21/34,34/55,55/89, and 89/144. These numbers are all numbers that belong to the Fibonacci Series.

A similar phenomenon can be observed in sunflower heads. A sunflower head has both clockwise and counterclockwise spirals. The numbers of the spirals in a sunflower usually depend on the size of the sunflower. However, the ratios of the spirals that usually occur are 89/55,144/89, and even 233/144. Once again, all of the numbers in these ratios are consecutive Fibonacci numbers.


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# Nuclear Science and Techniques

《核技术》(英文版) ISSN 1001-8042 CN 31-1559/TL     2019 Impact factor 1.556

Nuclear Science and Techniques ›› 2013, Vol. 24 ›› Issue (5): 050503

• Special Section on International Workshop on Nuclear Dynamics in Heavy-Ion Reaction (IWND2012) •

### Constraints on symmetry energy and n/p effective mass splitting with improved quantum molecular dynamics model

ZHANG Yingxun1,2,*  LI Zhuxia1  ZHAO Kai1  LIU Hang3  TSANG M B2

1. 1China Institute of Atomic Energy, Beijing102413, China
2National Superconducting Cyclotron Laboratory, Michigan State University, East Lansing, MI 48824, USA
3Texas Advanced Computing Center, University of Texas, Austin, Texas 78758, USA
• Received:2013-06-27
• Contact: ZHANG Yingxun E-mail: zhyx@ciae.ac.cn
• Supported by:

Supported by National Natural Science Foundation of China (NSFC) projects (Nos.11075215, 10875031, 11005022, 11005155 and 11275052), 973 Program of China (No. 2013CB834404) and National Science Foundation (Grants No. PHY-0606007).

PDF ShareIt Export Citation
ZHANG Yingxun, LI Zhuxia, ZHAO Kai, LIU Hang, TSANG M B. Constraints on symmetry energy and n/p effective mass splitting with improved quantum molecular dynamics model.Nuclear Science and Techniques, 2013, 24(5): 050503

Abstract:

A new version of improved quantum molecular dynamics model that includes standard Skyrme interactions has been developed. Based on the new code, four commonly used parameter sets, SLy4, SkI2, SkM* and Gs are adopted in the improved quantum molecular dynamics model and the isospin sensitive observables, namely isospin transport ratios, single and double ratios of the yields of neutrons and protons are investigated. The isospin transport ratios are strongly sensitive to the slope of symmetry energy, and are not very sensitive to the nucleon effective mass splitting. On the other hand, the high energy neutrons and protons yields ratios from reactions at different incident energies provide a good observable to the momentum dependence of nucleon effective mass splitting. By comparing our calculations with the data, we find that the constrained L value (the slope of density dependence of symmetry energy) is about ~46 MeV when the Skyrme type interaction is considered in transport models, and the isospin diffusion data prefer to mn*>mp* , but it is not a strong constraint with deep χ2 minimum.

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# How to represent quantum systems using graphs

Represent quantum systems for optimization, simulation, and other tasks using graphs

The typical purpose of graph objects in Boulder Opal is to represent quantum systems. The graph representation of such a system can be used to perform several tasks, including robust control (for calculating optimized control pulses), simulation (to understand the dynamics of the system in the presence of specific controls and noises), and system identification (to estimate the values of unknown system parameters based on measurements of the system).

Please refer to our topic Understanding graphs in Boulder Opal for context on what graphs are used for and why.

In what follows, we use graphs to define time-dependent Hamiltonians for simulation or optimization.

## Summary workflow

Here we outline the general procedure for representing your quantum systems and the computations performed on them in the form of graph objects.

### 1. Create graph inputs

For an optimization, these are optimizable variables that can be tuned by the optimizer to minimize a cost function. For a simulation, the inputs can be known values for control pulses, or quantities derived from known values.

### 2. Create signals

Signals are scalar-valued functions of time, which may be non-linear, have enforced temporal structure such as time symmetry, or more generally can depend arbitrarily on the inputs. They only contain one numerical value in each period of time, which means their shape is zero-dimensional. Signals represent the time-dependent envelope of the Hamiltonian. You create piecewise-constant (PWC) signals using the graph.pwc_signal operation.

### 3. Create Hamiltonian operators

You create operators or PWC operator-valued (2D) functions of time by multiplying constant matrices (for example Pauli matrices) with signals. Usually these operators represent individual terms in your Hamiltonian. You can also create constant operators to represent static terms in your Hamiltonian. Finally, you sum the individual operators created into a single Hamiltonian operator.

Once you have defined the nodes that describe your quantum system, you can add extra nodes representing the computations that you want to perform on it. For example, Boulder Opal offers time evolution operations, among many others.

Note that while this approach to constructing Hamiltonians is the most common, it is not a requirement. You can use graphs to perform a wide variety of other computations too. For example, Boulder Opal also provides specialized functions for working with trapped ions systems that take advantage of certain approximations to bypass Hamiltonian-level descriptions of the system (see the How to design error-robust Mølmer–Sørensen gates for trapped ions user guide for details).

## Worked example: Two-qubit system with a tunable coupling qubit

Consider a system consisting of three transmon qubits, in which the central qubit acts as a tunable coupler between the other two, as proposed by Yan et al. This system can be approximated by the following two-qubit Hamiltonian: $$\tilde H = \frac{\omega_a}{2} \sigma_{z, a} +\frac{\omega_b}{2} \sigma_{z, b} + \frac{g_a^2}{2 \Delta_a} \sigma_{z,a}+ \frac{g_b^2}{2 \Delta_b} \sigma_{z,b} + \tilde g \left(\sigma_{+, a} \sigma_{-,b} + \sigma_{-,a} \sigma_{+,b} \right) ,$$ where $\omega_k$ are the qubit frequencies, $\Delta_k=\omega_k-\omega_c$ are the detunings of each qubit from the coupling qubit, $g_k$ are the direct couplings between each qubit and the coupling qubit, and $\tilde g$ is the effective coupling between the two qubits.

The couplings $g_k,\tilde g$ are given by $$g_k \approx \frac{1}{2}\frac{C_{kc}}{\sqrt{C_kC_c}}\sqrt{\omega_k\omega_c},$$ $$\tilde g = \frac{1}{2}\left[\frac{\omega_c}{2\Delta}\eta - \frac{\omega_c}{2\Sigma}\eta + \eta + 1\right]\frac{C_{ab}}{\sqrt{C_aC_b}}\sqrt{\omega_a\omega_b} ,$$ where $C_k,C_c$ are the capacitances of the qubits, $C_{xy}$ are the qubit-qubit capacitances between each pair of qubits, and the derived quantities are given by $\eta=C_{ac}C_{bc}/C_{ab}C_c$, $1/\Sigma=(1/\Sigma_a+1/\Sigma_b)/2$, $\Sigma_k=\omega_k+\omega_c$, and $1/\Delta=(1/\Delta_a+1/\Delta_b)/2$. We can see that the effective coupling strength $\tilde g$ can be tuned by adjusting the frequency $\omega_c$ of the coupling qubit.

In this example, we will show how you can represent this system as a graph, and calculate its time evolution operators for a particular choice of the $\omega_c$ control. The chosen controls will produce a gate that is, up to single-qubit phases, close to an iSWAP.

We will consider a simple PWC pulse for $\omega_c$ which is then passed through imperfect control lines. These control lines distort the signal that actually reaches the system, and thus the relevant term in the Hamiltonian. See the How to add smoothing and band-limits to optimized controls user guide for more details about incorporating filters into your graphs.

From this filtered $\omega_c$ signal, we will also create signals representing the different terms in the Hamiltonian, such as $g_a^2/(2\Delta_a)$, $g_b^2/(2\Delta_b)$, and $\tilde g$. We will also add constant operators to the Hamiltonian, representing its static terms—in this case the fixed frequency terms for each qubit.

import matplotlib.pyplot as plt
import numpy as np
from qctrlvisualizer import plot_controls

from qctrl import Qctrl

# Start a Boulder Opal session.
qctrl = Qctrl()

# Define system parameters.

# Physical parameters.
C_a = 70  # fF
C_b = 72  # fF
C_ac = 4  # fF
C_bc = 4.2  # fF
C_c = 200  # fF
C_ab = 0.1  # fF
omega_a = 2 * np.pi * 4e9  # Hz
omega_b = 2 * np.pi * 4e9  # Hz
eta = C_ac * C_bc / (C_ab * C_c)
gate_duration = 100e-9  # s

# Create graph.
graph = qctrl.create_graph()

# Standard qubit operators.
iswap = np.array(
[
[1.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 1j, 0.0],
[0.0, 1j, 0.0, 0.0],
[0.0, 0.0, 0.0, 1.0],
]
)

# 2-qubit operators.
sigma_z_a = graph.pauli_kronecker_product([("Z", 0)], 2)
sigma_z_b = graph.pauli_kronecker_product([("Z", 1)], 2)
sigma_plus_minus = graph.pauli_kronecker_product([("P", 0), ("M", 1)], 2)

# Create basic piecewise-constant signal.
omega_c_raw_values = (
(5.4 - 0.34 * np.array([0, 1, 2, 3, 2, 1, 0])) * 2 * np.pi * 1e9
)  # Hz
omega_c_raw = graph.pwc_signal(
duration=gate_duration, values=omega_c_raw_values, name="omega_c_raw"
)

# Apply a sinc filter (and rediscretize) to simulate
# the effect of control line imperfections.
omega_c = graph.utils.filter_and_resample_pwc(
pwc=omega_c_raw,
cutoff_frequency=2 * np.pi * 0.1e9,
segment_count=100,
name="omega_c",
)

# Calculate derived couplings in the Hamiltonian terms.
Delta_a = omega_a - omega_c
Delta_b = omega_b - omega_c
Delta = 2 / (1 / Delta_a + 1 / Delta_b)
Sigma_a = omega_a + omega_c
Sigma_b = omega_b + omega_c
Sigma = 2 / (1 / Sigma_a + 1 / Sigma_b)

g_a = 0.5 * C_ac * graph.sqrt(omega_a * omega_c / (C_a * C_c))
g_b = 0.5 * C_bc * graph.sqrt(omega_b * omega_c / (C_b * C_c))
g_tilde = (
0.5
* (omega_c * eta / (2 * Delta) - omega_c * eta / (2 * Sigma) + eta + 1)
* C_ab
* graph.sqrt(omega_a * omega_b / (C_a * C_b))
)

detuning_a = g_a**2 / (2 * Delta_a)
detuning_b = g_b**2 / (2 * Delta_b)

# Build the system's Hamiltonian from its terms.

detuning_a_term = detuning_a * sigma_z_a
# This is equivalent to using graph.pwc_operator:
# detuning_a_term = graph.pwc_operator(detuning_a, sigma_z_a).

detuning_b_term = detuning_b * sigma_z_b
coupling_term = g_tilde * 2.0 * graph.hermitian_part(sigma_plus_minus)

# Tensors and arrays are considered constant operators.
fixed_frequency_a_term = omega_a * sigma_z_a / 2

# We can also explicity declare constant PWC operators.
fixed_frequency_b_term = graph.constant_pwc_operator(
operator=omega_b * sigma_z_b / 2, duration=gate_duration
)

# Add all operators into full Hamiltonian.
hamiltonian = (
detuning_a_term
+ detuning_b_term
+ coupling_term
+ fixed_frequency_a_term
+ fixed_frequency_b_term
)

# Compute the time evolution operators for the system.
# We provide a name to extract them after you execute the graph.
time_evolution_operators = graph.time_evolution_operators_pwc(
hamiltonian=hamiltonian,
sample_times=np.linspace(0, gate_duration, 10),
name="time_evolution_operators",
)

# Evaluate the time evolution operators for the system, and the raw and filtered input signals defined earlier.
result = qctrl.functions.calculate_graph(
graph=graph,
output_node_names=["time_evolution_operators", "omega_c_raw", "omega_c"],
)

Your task calculate_graph (action_id="1164458") has completed.

plot_controls(
{
"$\omega_c^\mathrm{raw}$": result.output["omega_c_raw"],
"$\omega_c^\mathrm{filtered}$": result.output["omega_c"],
}
)

print("Final time evolution operator:")
print(np.round(result.output["time_evolution_operators"]["value"][-1], 3))

Final time evolution operator:
[[-0.126-0.992j  0.   +0.j     0.   +0.j     0.   +0.j   ]
[ 0.   +0.j    -0.169-0.058j  0.   +0.984j  0.   +0.j   ]
[ 0.   +0.j    -0.   +0.984j -0.169+0.058j  0.   +0.j   ]
[ 0.   +0.j     0.   +0.j     0.   +0.j    -0.126+0.992j]]

This notebook was run using the following package versions. It should also be compatible with newer versions of the Q-CTRL Python package.

Package version
Python 3.9.12
matplotlib 3.5.1
numpy 1.21.5
scipy 1.7.3
qctrl 19.1.0
qctrlcommons 17.1.1
qctrltoolkit 1.5.0
qctrlvisualizer 3.2.1


text:
# Deterministic Sampling of Sparse Trigonometric Polynomials - Mathematics > Numerical Analysis

Deterministic Sampling of Sparse Trigonometric Polynomials - Mathematics > Numerical Analysis - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Abstract: One can recover sparse multivariate trigonometric polynomials from fewrandomly taken samples with high probability as shown by Kunis and Rauhut. Wegive a deterministic sampling of multivariate trigonometric polynomialsinspired by Weil-s exponential sum. Our sampling can produce a deterministicmatrix satisfying the statistical restricted isometry property, and also nearlyoptimal Grassmannian frames. We show that one can exactly reconstruct every\$M\$-sparse multivariate trigonometric polynomial with fixed degree and oflength \$D\$ from the determinant sampling \$X\$, using the orthogonal matchingpursuit, and \$# X\$ is a prime number greater than \$M\log D^2\$. This result isalmost optimal within the \$\log D^2 \$ factor. The simulations show that thedeterministic sampling can offer reconstruction performance similar to therandom sampling.

Autor: Zhiqiang Xu

Fuente: https://arxiv.org/


text:
# Unable to make sense of the output of X gate

Please see the output of the two following Qiskit programs.

Program 1:

qc = QuantumCircuit(3, 3)

from qiskit.providers.aer import AerSimulator
sim = AerSimulator() # make new simulator object

# Create quantum circuit with 6 qubits and 6 classical bits:
qc = QuantumCircuit(6, 6)

# BELOW LINE IS COMMENTED
#qc.x([3,4,5]) # Perform X-gates on qubits 3,4,5

# measure qubits 0,1,2,3,4,5 to classical bits 0,1,2,3,4,5 respectively
qc.measure([0,1,2,3,4,5],[0,1,2,3,4,5])
qc.draw() # returns a drawing of the circuit

job = sim.run(qc) # run the experiment
result = job.result() # get the results
result.get_counts() # interpret the results as a "counts" dictionary

OUTPUT: {'000000': 1024}

Program 2:

from qiskit import QuantumCircuit
qc = QuantumCircuit(3, 3)

from qiskit.providers.aer import AerSimulator
sim = AerSimulator() # make new simulator object

# Create quantum circuit with 6 qubits and 6 classical bits:
qc = QuantumCircuit(6, 6)

qc.x([3,4,5]) # Perform X-gates on qubits 3,4,5

# measure qubits 0,1,2,3,4,5 to classical bits 0,1,2,3,4,5 respectively
qc.measure([0,1,2,3,4,5],[0,1,2,3,4,5])
qc.draw() # returns a drawing of the circuit

job = sim.run(qc) # run the experiment
result = job.result() # get the results
result.get_counts() # interpret the results as a "counts" dictionary

OUTPUT: {'111000': 1024}

I am not able to make sense of the output. It's as if the output is mapped to the wrong bits. I found this example is on https://learn.qiskit.org/course/introduction/the-atoms-of-computation

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Feb 16 at 19:24


text:
# EDU 623 Week 5 DQ 1 Philosophical Perspectives

This paperwork of EDU 623 Week 5 Discussion Question 1 Philosophical Perspectives comprises:

Philosophical perspectives play an important role in what is taught in the classroom. In a minimum of 300 words, explain the views of progressivism, essentialism, and perennialism. Which philosophical perspective is most closely related to your philosophy? Explain. Respond to at least two of your classmates

• $5.19 ANSWER Tutor has posted answer for$5.19. See answer's preview

*** *** **** 5 DQ 1 ************* ************


text:
# Kunneth theorem

Let $X$, $Y$ be topological spaces. One can ask a question: how homology of $X\times Y$ are related to homologies of $X$ and $Y$. The answer to this question depends on the homology theory we’re talking about and also the coefficients ring. On the other hand, it is well known, that all homology theories are isomorphic on CW-complexes. Thus we may restrict to CW-complexes. Nevertheless the following theorem is more general:

Theorem. (Kunneth) Assume, that $X$, $Y$ are topological spaces and $R$ is a principal ideal domain. Denote by $H_{*}(X,R)$ the singular homology with coefficients in $R$. Then, for any $k>0$ there exists following short exact sequence in the category of $R$-modules:

 $0\to\bigoplus_{i+j=k}H_{i}(X,R)\otimes_{R}H_{j}(Y,R)\to H_{k}(X\times Y,R)\to% \bigoplus_{i+j=k-1}\mathrm{Tor}_{1}^{R}(H_{i}(X,R),H_{j}(Y,R))\to 0,$

where $\mathrm{Tor}$ denotes the Tor functor. Furthermore this sequence splits, i.e. the middle term is a direct sum (up to an isomorphism) of left and right term.

It should be mentioned, that if $R=\mathbb{F}$ is a field, then the Tor functor is always trivial (i.e. $\mathrm{Tor}^{\mathbb{F}}_{1}(M,N)=0$ for all vector spaces $M,N$ over $\mathbb{F}$) and in this case Kunneth formula can be stated as

 $H_{k}(X\times Y,\mathbb{F})\simeq\bigoplus_{i+j=k}H_{i}(X,\mathbb{F})\otimes_{% \mathbb{F}}H_{j}(Y,\mathbb{F})$

for any $k>0$.

Title Kunneth theorem KunnethTheorem 2013-03-22 19:13:57 2013-03-22 19:13:57 joking (16130) joking (16130) 7 joking (16130) Theorem msc 55N10 Kunneth formula Kunneth’s formula Kunneth’s theorem Kunneth theorem


text:
# LGOCV caret package R

i am learning data mining through book . During classification chapters about Neural Networks the authors have below code. I have below questions:

## pre2008 <- 1:nrow(training) ## training is a dataset that has training data
ctrl <- trainControl(method = "LGOCV",
summaryFunction = twoClassSummary,
classProbs = TRUE,
index = list(TrainSet = pre2008),
savePredictions = TRUE)
nnetGrid <- expand.grid(.size = 1:10, .decay = c(0, .1, 1, 2))
maxSize <- max(nnetGrid$.size) set.seed(476) nnetFit <- train(x = training[,reducedSet], y = training$Class,
method = "nnet",
metric = "ROC",
preProc = c("center", "scale"),
tuneGrid = nnetGrid,
trace = FALSE,
maxit = 2000,
MaxNWts = 1*(maxSize * (length(reducedSet) + 1) + maxSize + 1),
trControl = ctrl)

LGOCV - when do we use it? I read the post, but still not clear. the post says that it is a variant of LOOCV for hierarchical data. but my Y variable is not hierarchical :(

twoClassSummary - can it be used only when we have two classes? can i used it for say Iris data?

LGOCV is also known as Monte-Carlo Cross Validation. More details are available here.

• A quick Google establishes "leave-group-out cross validation" as the answer to your first question. Other questions are all focused on software/programming and arguably off-topic here. – Nick Cox Aug 27 '13 at 17:30
• the post that i have mentioned in my email says that LGOCV means "leave-group-out cross validation" , but i would like to know more details of it. for example: when we should use it and does it need Y to be hierarchical .. – user2543622 Aug 27 '13 at 17:56
• OK, but you said "what does it stand for?" and I answered that. Perhaps you should make your question clearer. – Nick Cox Aug 27 '13 at 17:59

From the book: "Repeated training/test splits is also known as 'leave-group-out cross- validation' or 'Monte Carlo cross-validation.'". It is illustrated in Figure 4.7 on page 72.

> LGOCV - when do we use it?

It depends. It has good variance properties if you do a good number of resamples and the bias is really dependent on what percentage of the training data gets left out. If you have a lot of computing power, this might be the preferred method.

> my Y variable is not hierarchical

Not sure what you mean.

Note that we call this LGOCV but we are only holding out a single sample (see the discussion in section 12.1). We needed to call it something in code.

> twoClassSummary - can it be used only when we have two classes?

Yes.

Max

• thanks, i looked at the section 12.1. But didnt see any material describing LGOCV. any particular page that you are referring to? if we are holding only one sample then isn't it same as LOOCV? i looked at the code and a bit curious. why do we have to say "index = list(TrainSet = pre2008)"? i tried changing it with "index = list(pre2008)" but got errors :( – user2543622 Aug 29 '13 at 17:57


text:
In an arbitrary abelian category, does chain complex homology commute with coproduct? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:28:28Z http://mathoverflow.net/feeds/question/75795 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75795/in-an-arbitrary-abelian-category-does-chain-complex-homology-commute-with-coprod In an arbitrary abelian category, does chain complex homology commute with coproduct? Daniel Barter 2011-09-18T22:53:22Z 2011-09-19T01:41:44Z
On page 55 of Weibel's Introduction to homological algebra the following passage appears:

Here are two consequences that use the fact that homology commutes with arbitrary direct sums of chain complexes

I understand why homology commutes with arbitrary direct sums when the direct sum of a collection of monics is a monic (i.e the direct sum functor is exact) but I was under the impression that there were abelian categories where the direct sum functor is not exact. After a bit of thought, I realised that I don't know an example of an abelian category in which the coproduct functor is not exact.

Sheaves of abelian groups on a fixed topological space give an example of an abelian category in which the product functor is not exact.

Question 1: Is the passage from Weibel's book correct? If so, then why?

Question 2: Is there an example of an abelian category where the direct sum functor is not exact?

 http://mathoverflow.net/questions/75795/in-an-arbitrary-abelian-category-does-chain-complex-homology-commute-with-coprod/75809#75809 Answer by Sam Gunningham for In an arbitrary abelian category, does chain complex homology commute with coproduct? Sam Gunningham 2011-09-19T01:41:44Z 2011-09-19T01:41:44Z
I couldn't think of a natural example of an abelian category in which direct sums are not exact (I think this is called axiom AB4). For example, sheaves of abelian groups and R-modules both have this property. However there are natural examples of abelian categories where direct products are not exact (i.e. not satisfying AB4*), for example, the category of abelian sheaves on a space.

Taking the opposite category of such a category will then give an example of a category not satisfying AB4 (albeit, not a very nice one).

Once you have such an example, homology of chain complexes in this category will not commute with direct sum:

if $A_i \to B_i$ is a sequence of monos such that $\bigoplus (f_i :A_i \to B_i)$ is not a mono, then consider the sequence of two-term complexes

$A_i \to B_i$.

$H^0$ of each of these complexes is zero, but $H^0$ of the direct sum is the kernel of $\bigoplus f_i$.

Here is one way to see that Sh(X) does not satisfy AB4* (probably not the easiest). Assume for simplicity X = [0,1]. Take a finite open cover, $\mathcal U_i$ of X by balls of radius $1/i$. Let $A_i$ be the sheaf

$\prod _{U \in \mathcal U_i} j_{U!} \mathbb Z_U$.

This has an epimorphism to $\mathbb Z_X$, but the direct product of all of them together is not epimorphic: taking sections over any open set $V$ will kill off any $A_i$ when no $1/i$-ball contains $V$.

I hope this is correct!


text:
# Question #334b6

Oct 26, 2017

$B = \frac{40}{3}$

$R = \frac{52}{3}$

#### Explanation:

• $B =$ Becky's Age
• $R =$Rob's Age

So

$B = 2 R - 4$

$B + R = 48$

Start by substituting. We know that

$48 - R = B$

so we can substitute that into the first equation and get

$48 - R = 2 R - 4$

Add $R$ and $4$ to both sides and you get

$52 = 3 R$

So

$R = \frac{52}{3}$

and that can be plugged back into the equation as

$B = 2 \left(\frac{52}{3}\right) - 4$

which becomes

$B = \frac{40}{3}$


text:
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# Mr. and Mrs. Wiley have a child every J years. Their oldest

Author Message
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Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]  06 Feb 2010, 12:17
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Question Stats:

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Mr. and Mrs. Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total?

(A) \frac{T+2}{J} + 1

(B) JT + 1

(C) \frac{J}{T} + \frac{1}{T}

(D) TJ - 1

(E) \frac{T+J}{J}

Source: Manhattan Guide

[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Nov 2012, 04:41, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Mr. and Mrs Wiley, VIC [#permalink]  06 Feb 2010, 12:35
jeeteshsingh wrote:
Need the solution using Algebra....

Mr. & Mrs Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total?

(A) \frac{T+2}{J} + 1

(B) JT + 1

(C) \frac{J}{T} + \frac{1}{T}

(D) TJ - 1

(E) \frac{T+J}{J}

[Reveal] Spoiler: OA:
(A)

Source: Manhattan Guide

Lets take it this way,

They had first child today, after J years they will have 2nd child, and again after J years they will have 3rd child..

And now according to the Q stem in T + 2 years they will have J kids

Hence number kids in T + 2 years = T+2 /J

We will have to add 1 (their first/oldest Kid)

Hence Total number of kids = (T+2 / J) + 1

cheers
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Re: Mr. and Mrs Wiley, VIC [#permalink]  06 Feb 2010, 15:05
Their oldest kid is T y/o.
From the question stem, their youngest kid alive is J-2 y/o.
We need to know how many kids are there between T and J-2 spaced exactly J years apart. This is an arithmetic progression.

(J-2) + (n-1)J = T, inclusive. This gives n = (T+2)/J. Question stem also states that they are going to have a baby after 2 yrs from now, so we have to add that kid to n to get the total number of kids the Wiley family will have.
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Re: Mr. and Mrs Wiley, VIC [#permalink]  13 Feb 2010, 00:21
BarneyStinson wrote:
Their oldest kid is T y/o.
From the question stem, their youngest kid alive is J-2 y/o.
We need to know how many kids are there between T and J-2 spaced exactly J years apart. This is an arithmetic progression.

(J-2) + (n-1)J = T, inclusive. This gives n = (T+2)/J. Question stem also states that they are going to have a baby after 2 yrs from now, so we have to add that kid to n to get the total number of kids the Wiley family will have.

Hi Barney,

Your solution makes sense as its simple AP you applied, but I failed to comprehend that how you get J-2 as first element.

As Q says that they'll have kids after every J years and their oldest kid is T years old and the next kid will be due after two years from now but we don't know that how many years back first kid was born....may be I didn't get rightly.

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Re: Mr. and Mrs Wiley, VIC [#permalink]  13 Feb 2010, 01:03
Ans A

their first child was born after J years...

thus 1 child ---> j years

=> thus after another J years his age = J

thus his age is J --> after 2J years and 2j after 3j years

his present age is T which is after T years.

thus total time after 2years will be T+2
since after every J year they have a child after T+2 they will have \frac{(T+2)}{J} + 1 ( +1 is for the oldest)

thus A
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Re: Mr. and Mrs Wiley, VIC [#permalink]  21 Nov 2012, 07:46
jeeteshsingh wrote:
Need the solution using Algebra....

Mr. & Mrs Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total?

(A) \frac{T+2}{J} + 1

(B) JT + 1

(C) \frac{J}{T} + \frac{1}{T}

(D) TJ - 1

(E) \frac{T+J}{J}

[Reveal] Spoiler: OA:
(A)

Source: Manhattan Guide

Bunuel - would really appreciate you providing your bit on solving the original problem above algebraically. The problem and various explanations remain confusing. Should we think of it as a progression or some other way? Please share your take. Thank you.
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Re: Mr. and Mrs Wiley, VIC [#permalink]  21 Nov 2012, 09:45
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Expert's post
jeeteshsingh wrote:
Need the solution using Algebra....

Mr. & Mrs Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total?

(A) \frac{T+2}{J} + 1

(B) JT + 1

(C) \frac{J}{T} + \frac{1}{T}

(D) TJ - 1

(E) \frac{T+J}{J}

[Reveal] Spoiler: OA:
(A)

Source: Manhattan Guide

Think of it as an Arithmetic Progression where every subsequent term (child) has a difference of J yrs from the previous term (child).

1st child, 2nd child, 3rd child, ....... nth child (to be born after 2 yrs)

What is the difference between first and last terms (children)? (T + 2) yrs

What is the common difference (age difference between two consecutive kids)? J yrs

What is the number of terms (children)? (T + 2)/J + 1
(Number of terms of an AP is n = (Last term - First term)/Common Difference + 1. )
_________________

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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Re: Mr. and Mrs Wiley, VIC [#permalink] 21 Nov 2012, 09:45 Similar topics Replies Last post Similar Topics: 1 Mr. and Mrs. O’Leary (SC) 5 08 Jul 2012, 07:15 Mr. INVESTOR invested a total of$12,000 for a one-year 4 30 Mar 2007, 09:24
2 Mr. and Mrs. Wiley have a child every J years. Their oldest 7 19 Feb 2007, 11:40
Mr.kevincan 6 16 Aug 2006, 12:26
PS: Mr. & Mrs. Smith 2 06 Dec 2005, 00:03
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text:
# Introduction to Crosstable

## Crosstable

Crosstable is a package centered on a single function, crosstable(), which easily computes descriptive statistics on datasets.

Before starting this vignette, here are a few points:

• If you are not familiar with dplyr and pipes (%>%, Ctrl+Shift+M in RStudio), I warmly recommend you to read the vignette, or, if you can read French, Julien Barnier’s awesome tidyverse tutorial. Nevertheless, this vignette should still be easy to understand otherwise, as crosstable is perfectly usable with base R.

## Example dataset: modified mtcars

First, since crosstable() uses the power of the label attribute, let’s start by building a labelled dataset.

In this vignette, we will use a modified version of the mtcars famous dataset, which comprises 11 aspects of design and performance for 32 automobiles. Let’s modify it to add textual categories, make some numeric variables factors, and add labels from a table using import_labels.

For convenience, this dataset is already packed into crosstable so don’t bother re-creating it for your own tests.

library(crosstable)
library(dplyr)
name label
mpg  'Miles/(US) gallon'
cyl  'Number of cylinders'
disp 'Displacement (cu.in.)'
hp   'Gross horsepower'
drat 'Rear axle ratio'
wt   'Weight (1000 lbs)'
qsec '1/4 mile time'
vs   'Engine'
am   'Transmission'
gear 'Number of forward gears'
carb 'Number of carburetors'
")
mtcars2 = mtcars %>%
mutate(vs=ifelse(vs==0, "vshaped", "straight"),
am=ifelse(am==0, "auto", "manual")) %>%
mutate_at(c("cyl", "gear"), factor) %>%
import_labels(mtcars_labels, name_from="name", label_from="label")
#I also could have used Hmisc::label() or expss::apply_labels() to add labels

## First overview

As a first example, let’s describe the columns mpg and cyl, grouping by the column am (the transmission).

As tables are not very readable in the console, let’s also use as_flextable() to turn the resulting crosstable into a beautiful, ready-to-print HTML table. This table will be automatically displayed in the Viewer pane if your are using RStudio.

crosstable(mtcars2, c(mpg, cyl), by=am) %>%
as_flextable(keep_id=TRUE)
 .id label variable Transmission auto manual mpg Miles/(US) gallon Min / Max 10.4 / 24.4 15.0 / 33.9 Med [IQR] 17.3 [14.9;19.2] 22.8 [21.0;30.4] Mean (std) 17.1 (3.8) 24.4 (6.2) N (NA) 19 (0) 13 (0) cyl Number of cylinders 4 3 (27.27%) 8 (72.73%) 6 4 (57.14%) 3 (42.86%) 8 12 (85.71%) 2 (14.29%)

By default, numeric variables (like mpg and disp) are described with min/max, median/IQR, mean/sd and number of observations/missing, while categorical (factor/character) variables (like cyl) are described with levels counts and fractions. All of this is fully customizable, as you will see hereafter.

There are many ways to select variables: with names, character vector, tidyselect helpers, formula… This is described in details in vignette("crosstable-selection").

The by column is usually a factor, character or logical vector. If it is a numeric vector, then only numeric vectors can be described and correlation coefficients will be displayed. While it is possible to apply several variables (by=c(am, vs)), I will use only one variable here for clarity.

In this vignette, I will often set keep_id=TRUE so you can see the variable name, but in practice you usually omit it. See vignette("crosstable-report") for more about as_flextable() and on how to integrate crosstables in MS Word document (using {officer}) and Rmarkdown. On the other hand, you can set label=FALSE if you don’t want them to appear.

## Totals

To display totals, use the total argument as one of c("none", "row", "column", "both").

#of course, the total of a "column" in only meaningful for categorical variables.
crosstable(mtcars2, c(am, mpg), by=vs, total="both") %>%
as_flextable(keep_id=TRUE)
 .id label variable Engine Total straight vshaped am Transmission auto 7 (36.84%) 12 (63.16%) 19 (59.38%) manual 7 (53.85%) 6 (46.15%) 13 (40.62%) Total 14 (43.75%) 18 (56.25%) 32 (100.00%) mpg Miles/(US) gallon Min / Max 17.8 / 33.9 10.4 / 26.0 10.4 / 33.9 Med [IQR] 22.8 [21.4;29.6] 15.6 [14.8;19.1] 19.2 [15.4;22.8] Mean (std) 24.6 (5.4) 16.6 (3.9) 20.1 (6.0) N (NA) 14 (0) 18 (0) 32 (0)

Note that totals always take missing values into account. Therefore, be aware that if showNA="no", totals may be higher than the sum of the values inside the table.

## Arguments to control variables description

The crosstable() function comes with a lot of arguments to add control on how it will describe your dataset. Some arguments are tied to a type of variable and will only apply for the descriptions of those.

### Categorical variables

Categorical variables are described using counts and percentages. A numeric variable is considered categorical if its number of levels is lesser than the unique_numeric argument (default=3).

If by is set, you can use the margin argument to control whether percentages should be calculated by row (default), column, or cell. You can also use percent_digits (default=2) to control how many decimals should be displayed.

Use showNA (one of c("ifany", "always", "no")) to control when missing values should be displayed.

mtcars3 = mtcars2
mtcars3$cyl[1:5] = NA crosstable(mtcars3, c(am, cyl), by=vs, margin=c("column","row"), percent_digits=0, showNA="always") %>% as_flextable(keep_id=TRUE)  .id label variable straight vshaped NA am Transmission auto 7 (50% / 37%) 12 (67% / 63%) 0 manual 7 (50% / 54%) 6 (33% / 46%) 0 NA 0 0 0 cyl Number of cylinders 4 9 (75% / 90%) 1 (7% / 10%) 0 6 3 (25% / 75%) 1 (7% / 25%) 0 8 0 (0% / 0%) 13 (87% / 100%) 0 NA 2 3 0 You can see that missing values are never taken into account when calculating percentage calculation in R. You can change this behaviour by using tidyr::replace_na() or forcats::fct_explicit_na() on your dataset before applying crosstable(). ### Numeric variables Numeric variables are described with summary functions. By default, it outputs min/max, median/IQR, mean/sd and number of observations/missing. However, this might not be the information you need, so you can use the funs argument to apply the set of functions of your choice: crosstable(mtcars2, c(mpg, wt), funs=c(median, mean, "std dev"=sd)) %>% as_flextable(keep_id=TRUE)  .id label variable value mpg Miles/(US) gallon median 19.2 mean 20.1 std dev 6.0 wt Weight (1000 lbs) median 3.3 mean 3.2 std dev 1.0 As you can see, you can name your functions. For more advanced use cases, note that you can also use anonymous and lambda functions: crosstable(mtcars2, c(mpg, wt), funs=c("mean square"=function(xx) mean(xx^2), "mean cube"= ~mean(.x^3))). You might want to use crosstable’s convenience functions such as meansd(), meanCI(), mediqr(), minmax(), or nna(). For additional arguments, you can use the funs_arg argument. For instance, you could write crosstable(mtcars2, c(disp, hp), funs=c(quantile), funs_arg=list(probs=c(0.25,0.75))). Numbers are formatted to have the same number of decimal places. You can use funs_arg=list(dig=3) to customize the number of decimals. You might also want to take a look at ?summaryFunctions and ?format_fixed. On the other hand, if by refers to a numeric variable, correlation coefficients will be calculated. library(survival) crosstable(mtcars2, where(is.numeric), by=mpg) %>% as_flextable(keep_id=TRUE)  .id label variable Miles/(US) gallon disp Displacement (cu.in.) pearson -0.85 95%CI [-0.92;-0.71] hp Gross horsepower pearson -0.78 95%CI [-0.89;-0.59] drat Rear axle ratio pearson 0.68 95%CI [0.44;0.83] wt Weight (1000 lbs) pearson -0.87 95%CI [-0.93;-0.74] qsec 1/4 mile time pearson 0.42 95%CI [0.08;0.67] carb Number of carburetors pearson -0.55 95%CI [-0.75;-0.25] You can use the cor_method argument to choose which coefficient to calculate ("pearson", "kendall", or "spearman"). ### Other variables #### Survival data Crosstable is also able to describe survival data. Use times to set the specific times of interest and followup to compute the median followup. For each time, you will get the survival, the number of events before this time, and the number of patients at risk, as per survival:::summary.survfit(). library(survival) aml$surv = Surv(aml$time, aml$status)
crosstable(aml, surv, by=x, times=c(0,50,150), followup=TRUE) %>%
as_flextable(keep_id=TRUE)
 .id label variable x Maintained Nonmaintained surv surv t=0 1.00 (0/11) 1.00 (0/12) t=50 0.18 (7/1) 0 (11/0) t=150 0.18 (0/1) 0 (0/0) Median follow up [min ; max] 103 [13 ; 161] NA [16 ; 45] Median survival 31 23

Note that, using the formula interface, you could declare the Surv object directly inside the crosstable function: crosstable(aml, Surv(time, status) ~ x).

#### Dates

Although less usual, you can describe variables of call Date or POSIXt with crosstable(). Use date_format to apply a specific format (see ?strptime for formats). Beside this, they are considered as numeric variables.

mtcars2$x_date = as.Date(mtcars2$hp , origin="2010-01-01") %>% set_label("Date")
mtcars2$x_posix = as.POSIXct(mtcars2$qsec*3600*24 , origin="2010-01-01") %>% set_label("Date+time")
crosstable(mtcars2, c(x_date, x_posix), date_format="%d/%m/%Y") %>%
as_flextable(keep_id=TRUE)
 .id label variable value x_date Date Min / Max 22/02/2010 - 02/12/2010 Med [IQR] 04/05/2010 [06/04/2010;30/06/2010] Mean (std) 27/05/2010 (2.3 months) N (NA) 32 (0) x_posix Date+time Min / Max 15/01/2010 - 23/01/2010 Med [IQR] 18/01/2010 [17/01/2010;19/01/2010] Mean (std) 18/01/2010 (1.8 days) N (NA) 32 (0)

For the standard deviation to be readable, date unit is chosen automatically among ["seconds", "minutes", "hours", "days", "months", "years"]. If you don’t want two groups to have different date unit, you can set it globally using the date_unit key in funs_arg:

crosstable(mtcars2, c(x_date, x_posix), funs=meansd, funs_arg=list(date_unit="days")) %>%
as_flextable(keep_id=TRUE)
 .id label variable value x_date Date meansd 2010-05-27 (68.56 days) x_posix Date+time meansd 2010-01-18 21:22:12 (1.79 days)

## Effects

If there is onely one by variable with only 2 levels, it is possible to automatically compute an effect-size, most often along with its confidence interval.

library(flextable)
crosstable(mtcars2, c(vs, qsec), by=am, funs=mean, effect=TRUE) %>%
as_flextable(keep_id=TRUE)
 .id label variable Transmission effect auto manual vs Engine straight 7 (50.00%) 7 (50.00%) Odds ratio [95% Wald CI], ref='manual vs auto'vshaped vs straight: 0.50 [0.11 to 2.08] vshaped 12 (66.67%) 6 (33.33%) qsec 1/4 mile time mean 18.2 17.4 Difference in means (t-test CI), ref='auto'manual minus auto: -0.82 [-2.12 to 0.48]

Type of effect (method, bootstrap, …) are also chosen depending on the characteristics of the crossed variables (class, size, distribution, …). See ?crosstable_effect_args for more details on the effect choice algorithm and how to customize it.

## Tests

It is also possible to perform statistical tests automatically.

library(flextable)
crosstable(mtcars2, c(vs, qsec), by=am, funs=mean, test=TRUE) %>%
as_flextable(keep_id=TRUE)
 .id label variable Transmission test auto manual vs Engine straight 7 (50.00%) 7 (50.00%) p value: 0.3409 (Pearson's Chi-squared test) vshaped 12 (66.67%) 6 (33.33%) qsec 1/4 mile time mean 18.2 17.4 p value: 0.2057 (Two Sample t-test)

Of course, this should only be done in an exploratory context, as it would cause extensive alpha inflation otherwise.

Tests are chosen depending on the characteristics of the crossed variables (class, size, distribution, …). See ?crosstable_test_args for more details on the test choice algorithm.

## Acknowledgement

crosstable is a rewrite of the awesome biostat2 package written by David Hajage. The user interface is quite different but the concept is the same. Thanks David!


text:
## Dynamics Seminar

These seminars are generally on results related to dynamical systems.   This includes ergodic theory, invariant manifold theory, KAM theory, complex dynamics, topological dynamics, and many more.

Results in related fields, e.g. probability theory, theory of computations, statistical mechanics (to name a few), are also welcome.

Previous Dynamics Seminar listings can be found at the following link: http://www.math.toronto.edu/dynamics/

### A proof of Pilgrim's conjecture

by Nikita Selinger | Stony Brook University
Time: 11:00  (Wednesday, Nov. 23, 2011)
Location: BA2135, Bahen Center, 40 St. George St.
Abstract:
Let $f$ be a postcritically finite branched self-cover of a 2-dimensional topological sphere. Such a map induces an analytic self-map $\sigma_f$ of a finite-dimensional Teichm\"uller space. We prove that this map extends continuously to the augmented Teichm\"uller space and give an explicit construction for this extension. This allows us to characterize the dynamics of Thurston's pullback map near invariant strata of the boundary of the augmented Teichm\"uller space. The resulting classification of invariant boundary strata is used to prove a conjecture by Pilgrim.

#### Dates in this series

· Monday, Feb. 08, 2010: Dynamics of Lagrangian particles beyond the shocks (Konstantin Khanin)
· Monday, Mar. 01, 2010: On the rigidity of circle maps with break singularities (Sasha Kocic)
· Monday, Mar. 08, 2010: Thurston mappings and Galois actions (Sylvain Bonnot)
· Monday, Mar. 15, 2010: Rate of convergence and Kolmogorov Complexity in the Ergodic Theorem (Cristobal Rojas)
· Monday, Mar. 22, 2010: Dynamical charts for irrationally indifferent fixed points (Mitsuhiro Shishikura)
· Monday, Sep. 13, 2010: Thurston equivalence is decidable (joint work with S. Bonnot and M. Braverman) (Michael Yampolsky)
· Monday, Sep. 20, 2010: Decomposition of Thurston maps (Sylvain Bonnot)
· Monday, Sep. 27, 2010: Ergodicity of some open systems with particle-disk interactions (Tanya Yarmola)
· Monday, Oct. 04, 2010: A mechanism for generic Arnold diffusion (Ke Zhang)
· Monday, Oct. 18, 2010: Pentagram map (Fedor Soloviev)
· Monday, Oct. 25, 2010: Semicircular outer billiards (Dimitry Dolgopyat)
· Monday, Nov. 15, 2010: Singular perturbation analysis of the Boltzmann equation (Alexander Grigo)
· Monday, Nov. 22, 2010: Normally Hyperbolic foliations by circles (Pablo Carrasco)
· Monday, Dec. 06, 2010: Rigidity of circle diffeomorphisms with breaks for almost all rotation numbers (Sasa Kocic)
· Monday, Feb. 14, 2011: Subshift factors of the beta-shift are intrinsically ergodic (Dan Thompson)
· Monday, Feb. 28, 2011: Kowalevski-type integrable systems and discriminantly separable polynomials (Vladimir Dragovic)
· Monday, Mar. 07, 2011: Exponentially small splitting of separatrices for one and a half degrees of freedom Hamiltonian Systems close to a resonance (Marcel Guardia)
· Monday, Mar. 28, 2011: Existence and non-existence of KAM tori near an invariant torus of a Hamiltonian system with Diophantine frequency (Bassam Fayad)
· Monday, Sep. 19, 2011: Multi-dimensional Fatou bifurcation (Igor Gorbovickis)
· Wednesday, Sep. 28, 2011: KAM theory for dissipative systems (Renato Calleja)
· Monday, Oct. 03, 2011: Hyperbolicity of the minimizer for some randomly kicked Hamiltonian systems (Ke Zhang)
· Monday, Oct. 31, 2011: Thermodynamics with small obstructions (Vaughn Climenhaga)
· Monday, Nov. 07, 2011: The Consequences of Quasi-Isometry (Andy Hammerlindl)
· Monday, Nov. 14, 2011: Global well-posedness of a barely supercritical surface quasi-geostrophic equation (Michael Dabkowski)
· Wednesday, Nov. 23, 2011: A proof of Pilgrim's conjecture (Nikita Selinger)
· Friday, Nov. 25, 2011: Critical Locus for Complex Hénon Maps (Tanya Firsova)
· Monday, Jan. 16, 2012: Convergence of renormalizations and rigidity theory for circle diffeomorphisms with a break (Sasa Kocic)
· Monday, Sep. 17, 2012: Perturbing Misiurewicz parameters in the exponential family (Neil Dobbs)
· Monday, Oct. 01, 2012: Homogenization of convex Hamiltonians on arbitrary manifolds (Renato Iturriaga)
· Wednesday, Oct. 24, 2012: Computability and computational complexity of Julia sets (Michael Yampolsky)
· Friday, Oct. 26, 2012: Limit theorems for toral translations (Dmitry Dolgopyat)
· Monday, Oct. 29, 2012: A method to estimate the Lyapunov exponents for some quasi-periodic Schr\"{o}dinger cocycles (Kristian Bjerklov)
· Monday, Jan. 14, 2013: Renormalization and rigidity theory for circle diffeomorphisms with breaks (Sasha Kocic)
· Monday, Jan. 21, 2013: Dynamics of parabolic germs via Resurgence Theory (Artem Dudko)
· Monday, Feb. 04, 2013: Renormalization for Lorenz maps (Denis Gaydashev)