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https://quantumcomputing.stackexchange.com/questions/4163/how-scalable-are-quantum-computers-when-measurement-operations-are-considered/4168 | # How scalable are quantum computers when measurement operations are considered?
From a high-level point of view, given a quantum program, typically the last few operations are measurements.
In most cases, in order to extract a useful answer, it is necessary to run multiple times to reach a point where it is possible to estimate the probability distribution of the output q-bits with some level of statistical significance.
Even when ignoring noise, if the number of q-bits of the output increases, the number of required runs will increase too. In particular, for some output probability distributions it is likely that the number of required samples can be very high to obtain some sensible statistical power.
While increasing the number of q-bits seems to indicate an exponential growth, the number of measurements required (each run taking x nanoseconds) seem to limit how will quantum computation do scale.
Is this argument sensible or there is some critical concept that I am missing? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8992127180099487, "perplexity": 484.74265626603693}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00274.warc.gz"} |
https://arxiv.org/abs/1501.00855 | gr-qc
# Title:Closure constraints for hyperbolic tetrahedra
Abstract: We investigate the generalization of loop gravity's twisted geometries to a q-deformed gauge group. In the standard undeformed case, loop gravity is a formulation of general relativity as a diffeomorphism-invariant SU(2) gauge theory. Its classical states are graphs provided with algebraic data. In particular closure constraints at every node of the graph ensure their interpretation as twisted geometries. Dual to each node, one has a polyhedron embedded in flat space R^3. One then glues them allowing for both curvature and torsion. It was recently conjectured that q-deforming the gauge group SU(2) would allow to account for a non-vanishing cosmological constant Lambda, and in particular that deforming the loop gravity phase space with real parameter q>0 would lead to a generalization of twisted geometries to a hyperbolic curvature. Following this insight, we look for generalization of the closure constraints to the hyperbolic case. In particular, we introduce two new closure constraints for hyperbolic tetrahedra. One is compact and expressed in terms of normal rotations (group elements in SU(2) associated to the triangles) and the second is non-compact and expressed in terms of triangular matrices (group elements in SB(2,C)). We show that these closure constraints both define a unique dual tetrahedron (up to global translations on the three-dimensional one-sheet hyperboloid) and are thus ultimately equivalent.
Comments: 24 pages Subjects: General Relativity and Quantum Cosmology (gr-qc); Mathematical Physics (math-ph) Journal reference: Class.Quant.Grav. 32 (2015) 13, 135003 DOI: 10.1088/0264-9381/32/13/135003 Cite as: arXiv:1501.00855 [gr-qc] (or arXiv:1501.00855v1 [gr-qc] for this version)
## Submission history
From: Christoph Charles [view email]
[v1] Mon, 5 Jan 2015 13:34:13 UTC (154 KB) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8453622460365295, "perplexity": 1442.2975086334702}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572289.5/warc/CC-MAIN-20190915195146-20190915221146-00193.warc.gz"} |
http://isiarticles.com/article/25963 | دانلود مقاله ISI انگلیسی شماره 25963
عنوان فارسی مقاله
مرزهای دقیق برای تجزیه و تحلیل حساسیت سازه با پارامترهای نامشخص ولی محدود شده
کد مقاله سال انتشار مقاله انگلیسی ترجمه فارسی تعداد کلمات
25963 2008 15 صفحه PDF سفارش دهید محاسبه نشده
خرید مقاله
پس از پرداخت، فوراً می توانید مقاله را دانلود فرمایید.
عنوان انگلیسی
Exact bounds for the sensitivity analysis of structures with uncertain-but-bounded parameters
منبع
Publisher : Elsevier - Science Direct (الزویر - ساینس دایرکت)
Journal : Applied Mathematical Modelling, Volume 32, Issue 6, June 2008, Pages 1143–1157
کلمات کلیدی
حساسیت - عدم قطعیت - روش تجزیه و تحلیل فاصله - مقادیر ویژه سازه -
پیش نمایش مقاله
چکیده انگلیسی
Based on interval mathematical theory, the interval analysis method for the sensitivity analysis of the structure is advanced in this paper. The interval analysis method deals with the upper and lower bounds on eigenvalues of structures with uncertain-but-bounded (or interval) parameters. The stiffness matrix and the mass matrix of the structure, whose elements have the initial errors, are unknown except for the fact that they belong to given bounded matrix sets. The set of possible matrices can be described by the interval matrix. In terms of structural parameters, the stiffness matrix and the mass matrix take the non-negative decomposition. By means of interval extension, the generalized interval eigenvalue problem of structures with uncertain-but-bounded parameters can be divided into two generalized eigenvalue problems of a pair of real symmetric matrix pair by the real analysis method. Unlike normal sensitivity analysis method, the interval analysis method obtains informations on the response of structures with structural parameters (or design variables) changing and without any partial differential operation. Low computational effort and wide application rang are the characteristic of the proposed method. Two illustrative numerical examples illustrate the efficiency of the interval analysis.
مقدمه انگلیسی
The purpose of sensitivity analysis is to work out the structure response or the variety of performance through the transformation of parameters or designing variables [1]: equation(1) u=u(b1,b2,…,bn),u=u(b1,b2,…,bn), Turn MathJax on where b1,b2,…,bnb1,b2,…,bn are structure parameters or designing variables. Thus, via partial differential operation and bring b0=(b10,b20,…,bn0)Tb0=(b10,b20,…,bn0)T into Eq. (1), we have equation(2) View the MathML source∂u∂bb0=∂u(b10,b20,…,bn0)∂b. Turn MathJax on The absolute value of Eq. (2) denotes the sensitivity degree of structure responses or capability to structure parameters. On condition that View the MathML source∂u∂bb0>0 the structure response or performance u is monotone increased around the parameter b0=(b10,b20,…,bn0)Tb0=(b10,b20,…,bn0)T. If View the MathML source∂u∂bb0<0 the structure response or performance u is monotone degressive around the parameter b0=(b10,b20,…,bn0)Tb0=(b10,b20,…,bn0)T. Bear in mind Eq. (2), we also gain the transformation of the structure response or performance as equation(3) View the MathML sourceδu=∂u∂bδb. Turn MathJax on In engineering practice, very often difference operation methods are used instead of differential operation methods, but the results are often unreliable. However, the above-mentioned normal sensitivity analysis method has many problems: Case I. The mathematical foundation of the normal sensitivity analysis method is the differential calculus of real analysis. In terms of differential calculus principle, based on partial differential, the sensitivity analysis result is only local information. Namely, in this local bound the structure response or performance is most sensitive to this parameter; but in another local bound, the structure response or performance is likely least sensitive to this parameter. In the same way, the structure response or performance is monotone increased to this parameter in this local bound; but in another local bound, the structure response or performance is likely monotone degressive to this parameter. However, in practice analysis and designing, people were concerned with the sensitivity information of the structure response or performance in a certain large bound. The sensitivity analysis which based on differential calculus cannot satisfy the requirement of such global information. Although we can process normal sensitivity analysis many times and receive the sensitivity information in a certain large bound, the efficiency of the calculation will decrease seriously. Case II. For most practical engineering, it is impossible to present the parse expressions of structure response or performance in virtue of complexity and legion dimensions. So, usually we use difference or perturbation analysis instead of differential calculus. However, in engineering practice, the variety of parameters or designing variables oversize or undersize will all impact the precision of the sensitivity analysis and present complete incorrect information. As shown in Fig. 1, we have equation(4) View the MathML sourceδu1δb1=u(b1)-u(b0)b1-b0=u1-u0b1-b0>0 Turn MathJax on and equation(5) View the MathML sourceδu2δb2=u(b2)-u(b0)b2-b0=u2-u0b2-b0<0. Turn MathJax on It presents an opposite sensitivity information. So, sometimes the results of the difference sensitivity analysis method are without confidence to the distinct nonlinearity structures. Full-size image (7 K) Fig. 1. Distinct nonlinearity function. Figure options Case III . Having found the results of the sensitivity analysis, how can we work out the variation of structure responses or performances through the variation of parameters or designing variables? The normal sensitivity analysis calculate the variation by Eq. (3), but it is only present the local information and δb should not be too large, otherwise the result will be without confidence. For instance, as shown in Fig. 1, we process a calculation according to δu=(∂u/∂b)δbδu=(∂u/∂b)δb as following: equation(6) View the MathML sourceu1=u0+δu1δb1δb1. Turn MathJax on From Eq. (6) we get that u 1 is increased when compared with u 0. Also, we process another calculation according to δu=(∂u/∂b)δbδu=(∂u/∂b)δb as following: equation(7) View the MathML sourceu2=u0+δu2δb2δb2. Turn MathJax on From Eq. (7) we work out that u2 is decreased when compared with u0. Case IV. In the difference calculation of sensitivity, the result that calculated by δu is the distance u(b)-u(b0)u(b)-u(b0) of structure responses or parameters between b and b 0 rather than the distance umax-uminumax-umin of structure responses or parameters between b and b 0. Where u(b)≠umaxu(b)≠umax and u(b0)≠uminu(b0)≠umin will be correct with distinct nonlinearity problem. Therefore, in engineering practice the variation of parameters or design variables δb=b-b0δb=b-b0 oversize or undersize will all impact the precision of sensitivity analysis and present complete incorrect information indeed. Case V. If the structure parameters in the expressions of structure responses, performances or mathematical calculations are uncertain, especially in the large-scale structure, the result of the sensitivity analysis is often without confidence due to the cumulation of the uncertainty. If we use interval mathematics to process sensitivity analysis, the above problem of normal sensitivity analysis will be completely solved.
نتیجه گیری انگلیسی
In this paper, considering the properties of the stiffness matrix and the mass matrix in structural engineering, making use of the structural parameters and the non-negative decomposition of a matrix, we present a new method to determine the lower and upper bounds on the sensitivity due to uncertain-but-bounded parameters for the interval sensitivity analysis problem. Without any partial differential operations, the interval analysis method obtained the information on the response of the structure with the structural parameters (or design variables) changing. The effectiveness and correctness of the algorithm was illustrated by two numerical examples. For large-scale structures with interval parameters, a fast computing technique for obtaining the approximate sensitivity and the corresponding errors is desirable.
خرید مقاله
پس از پرداخت، فوراً می توانید مقاله را دانلود فرمایید. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8568086624145508, "perplexity": 1817.4525644678658}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607786.59/warc/CC-MAIN-20170524035700-20170524055700-00057.warc.gz"} |
https://arxiv.org/abs/1210.3857 | Skip to main content
# Mathematics > Analysis of PDEs
arXiv:1210.3857 (math)
# Title:Regularity criterion for 3D Navier-Stokes Equations in Besov spaces
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Abstract: Several regularity criterions of Leray-Hopf weak solutions $u$ to the 3D Navier-Stokes equations are obtained. The results show that a weak solution $u$ becomes regular if the gradient of velocity component $\nabla_{h}{u}$ (or $\nabla{u_3}$) satisfies the additional conditions in the class of $L^{q}(0,T; \dot{B}_{p,r}^{s}(\mathbb{R}^{3}))$, where $\nabla_{h}=(\partial_{x_{1}},\partial_{x_{2}})$ is the horizontal gradient operator. Besides, we also consider the anisotropic regularity criterion for the weak solution of Navier-Stokes equations in $\mathbb{R}^3$. Finally, we also get a further regularity criterion, when give the sufficient condition on $\partial_3u_3$.
Comments: arXiv admin note: text overlap with arXiv:1005.4463 by other authors Subjects: Analysis of PDEs (math.AP) Cite as: arXiv:1210.3857 [math.AP] (or arXiv:1210.3857v1 [math.AP] for this version)
## Submission history
From: Daoyuan Fang [view email]
[v1] Sun, 14 Oct 2012 23:32:49 UTC (17 KB)
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https://math.libretexts.org/Bookshelves/PreAlgebra/Book%3A_Prealgebra_(OpenStax)/10%3A_Polynomials | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 10: Polynomials
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Expressions known as polynomials are used widely in algebra. Applications of these expressions are essential to many careers, including economists, engineers, and scientists. In this chapter, we will find out what polynomials are and how to manipulate them through basic mathematical operations.
• 10.1: Add and Subtract Polynomials
In this section, we will work with polynomials that have only one variable in each term. The degree of a polynomial and the degree of its terms are determined by the exponents of the variable. Working with polynomials is easier when you list the terms in descending order of degrees. When a polynomial is written this way, it is said to be in standard form. Adding and subtracting polynomials can be thought of as just adding and subtracting like terms.
• 10.2: Use Multiplication Properties of Exponents (Part 1)
In this section, we will begin working with variable expressions containing exponents. Remember that an exponent indicates repeated multiplication of the same quantity. You have seen that when you combine like terms by adding and subtracting, you need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different, too. We’ll derive the properties of exponents by looking for patterns in several examples.
• 10.3: Use Multiplication Properties of Exponents (Part 2)
All the exponent properties hold true for any real numbers, but right now we will only use whole number exponents. The product property of exponents allows us to multiply expressions with like bases by adding their exponents together. The power property of exponents states that to raise a power to a power, multiply the exponents. Finally, the product to a power property of exponents describes how raising a product to a power is accomplished by raising each factor to that power.
• 10.4: Multiply Polynomials (Part 1)
In this section, we will begin multiplying polynomials with degree one, two, and/or three. Just like there are different ways to represent multiplication of numbers, there are several methods that can be used to multiply a polynomial by another polynomial. The Distributive Property is the first method that you have already encountered and used to find the product of any two polynomials.
• 10.5: Multiply Polynomials (Part 2)
The FOIL method is usually the quickest method for multiplying two binomials, but it works only for binomials. When you multiply a binomial by a binomial you get four terms. Sometimes you can combine like terms to get a trinomial, but sometimes there are no like terms to combine. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers.
• 10.6: Divide Monomials (Part 1)
In this section, we will look at the exponent properties for division. A special case of the Quotient Property is when the exponents of the numerator and denominator are equal. It leads us to the definition of the zero exponent, which states that if a is a non-zero number, then a^0 = 1. Any nonzero number raised to the zero power is 1. The quotient to a power property of exponents states that to raise a fraction to a power, you raise the numerator and denominator to that power.
• 10.7: Divide Monomials (Part 2)
We have now seen all the properties of exponents. We'll use them to divide monomials. Later, you'll use them to divide polynomials. When we divide monomials with more than one variable, we write one fraction for each variable. Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.
• 10.8: Integer Exponents and Scientific Notation (Part 1)
The negative exponent tells us to re-write the expression by taking the reciprocal of the base and then changing the sign of the exponent. Any expression that has negative exponents is not considered to be in simplest form. We will use the definition of a negative exponent and other properties of exponents to write an expression with only positive exponents.
• 10.9: Integer Exponents and Scientific Notation (Part 2)
When a number is written as a product of two numbers, where the first factor is a number greater than or equal to one but less than 10, and the second factor is a power of 10 written in exponential form, it is said to be in scientific notation. It is customary to use × as the multiplication sign, even though we avoid using this sign elsewhere in algebra. Scientific notation is a useful way of writing very large or very small numbers. It is used often in the sciences to make calculations easier.
• 10.E: Polynomials (Exercises)
• 10.S: Polynomials (Summary)
• 10.10: Introduction to Factoring Polynomials
Earlier we multiplied factors together to get a product. Now, we will be reversing this process; we will start with a product and then break it down into its factors. Splitting a product into factors is called factoring. In The Language of Algebra we factored numbers to find the least common multiple (LCM) of two or more numbers. Now we will factor expressions and find the greatest common factor of two or more expressions. The method we use is similar to what we used to find the LCM.
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http://mathhelpforum.com/algebra/176029-simulatenous-equations-quadratic-linear.html | # Math Help - Simulatenous equations (Quadratic + linear)
1. ## Simulatenous equations (Quadratic + linear)
If (2,1) is a solution fo the simultaneous equations,
x^2 + xy + ay = b
2ax + 3y = b
A) Find the value of a and of b. [a=1, b =7]
B) Find also the other solution [x=-7 and y = 7)
I have solved both parts, but I face a problem.
After part a), I substituted values of a and b into the equation to obtain
x^2 + xy + y = 7 and 2x + 3y = 7
Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
I obtain the quadratic expression y^2 - 9y + 14 = 0
Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2
Why do I reject y =2 and x = 1/2 combination ?
I know that substituting the above into 2x + 3y = 7 satisfies the equation
But it doesn't satisfy x^2 + xy + y = 7. How would I know?
Thanks
2. Originally Posted by Drdj
If (2,1) is a solution fo the simultaneous equations,
x^2 + xy + ay = b
2ax + 3y = b
A) Find the value of a and of b. [a=1, b =7]
B) Find also the other solution [x=-7 and y = 7)
I have solved both parts, but I face a problem.
After part a), I substituted values of a and b into the equation to obtain
x^2 + xy + y = 7 and 2x + 3y = 7
Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
I obtain the quadratic expression y^2 - 9y + 14 = 0
Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2
Why do I reject y =2 and x = 1/2 combination ?
I know that substituting the above into 2x + 3y = 7 satisfies the equation
But it doesn't satisfy x^2 + xy + y = 7. How would I know?
Thanks
you don't ... extraneous solutions show up when you sub them back into the original equations
for example ...
$\sqrt{x+2} = x$
$x+2 = x^2$
$0 = x^2 - x - 2$
$0 = (x - 2)(x + 1)$
$x = 2$ , $x = -1$
$x = -1$ is an extraneous solution caused by the resulting quadratic ... $\sqrt{whatever} \ge 0$
3. Originally Posted by Drdj
If (2,1) is a solution fo the simultaneous equations,
x^2 + xy + ay = b
2ax + 3y = b
A) Find the value of a and of b. [a=1, b =7]
B) Find also the other solution [x=-7 and y = 7)
I have solved both parts, but I face a problem.
After part a), I substituted values of a and b into the equation to obtain
x^2 + xy + y = 7 and 2x + 3y = 7
Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
I obtain the quadratic expression y^2 - 9y + 14 = 0
Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2
Why do I reject y =2 and x = 1/2 combination ?
I know that substituting the above into 2x + 3y = 7 satisfies the equation
But it doesn't satisfy x^2 + xy + y = 7. How would I know?
Thanks
There are no extraneous solutions. Your quadratic is incorrect. I get $y^2 - 8y + 7 = 0$
-Dan
4. Thanks skeeter ! So, for all quadratic equations, I have to substitute the final answers back, in order to check if both sets of equations are valid! Many thanks for your quick reply !
Originally Posted by skeeter
you don't ... extraneous solutions show up when you sub them back into the original equations
for example ...
$\sqrt{x+2} = x$
$x+2 = x^2$
$0 = x^2 - x - 2$
$0 = (x - 2)(x + 1)$
$x = 2$ , $x = -1$
$x = -1$ is an extraneous solution caused by the resulting quadratic ... $\sqrt{whatever} \ge 0$
5. Hello, Drdj!
A simple algebra error . . .
$\text{If }(2,1)\text{ is a solution of the simultaneous equations:}$
. . $\begin{array}{c} x^2 + xy + ay \;=\;b \\ 2ax + 3y \;=\;b \end{array}$
$\text{(A) Find the values of }a\text{ and }b.\;\;[a=1,\:b =7]$
$\text{(B) Find also the other solution. }\;\;[x=-7,\:y = 7)$
$\text{I have solved both parts, but I face a problem. }$
$\text{After part (A), I substituted values of }a\text{ and }b\text{ into the equations}$
. . $\text{to obtain: }\:x^2 + xy + y \:=\: 7\:\text{ and }\:2x + 3y \:=\: 7$ . Yes!
$\text{Substituting }x \:=\:\frac{7-3y}{2}\text{ into the equation }x^2 + xy + y \:=\: 7$
. . $\text{I obtain the quadratic expression: }\:y^2 - 9y + 14 \:=\: 0$ . No
You should have had: . $\left(\dfrac{7-3y}{2}\right)^2 + \left(\dfrac{7-3y}{2}\right)\!y \,+\, y \;=\;7$
. . . . . . . . . . . . . . . $\displaystyle \frac{49-42y + 9y^2}{4} + \frac{7y - 3y^2}{2} + y \;=\;7$
Multiply by 4: . $49 - 42y + 9y^2 + 14y - 6y^2 + 4y \;=\;28$
. . which simplifies to: . $3y^2 - 24y + 21 \:=\:0$
Divide by 3: . $y^2 - 8y + 7 \:=\:0$
Factor: . $(y - 1)(y - 7) \:=\:0$
Hence, we have: . $y \;=\;1,\:7$
. . .which yields: . $x \;=\;2,\:\text{-}7$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 34, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8659614324569702, "perplexity": 525.7272820797175}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394021878262/warc/CC-MAIN-20140305121758-00009-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/problem-involving-moment-components-and-angles.712375/ | # Problem involving moment components and angles
1. Sep 24, 2013
### Loopas
1
Gripper C of the industrial robot is accidentally subjected to a 60 lb side load directed perpendicular to BC (see attachment). The lengths of the robot's links are AB = 22 in and BC = 18 in. By using the moment components method, determine the moment of the force about the center of joint A.
2
M = Fd (d is the lever arm)
3
I'm not sure what angle to use when calculating the components of the 60 lb force. Is the 60 lb force at a 140° angle with the coordinate plane given at point A? Since this is the first step of the problem, I want to make sure I'm doing it correctly.
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2. Sep 24, 2013
### SteamKing
Staff Emeritus
I don't know where you get 140 deg. from. Most right angles are 90 deg., and BC is 55 deg. above the horizontal.
3. Sep 24, 2013
### Loopas
The 140 degrees was referring to the direction of the force (in relation to the xy coordinate frame thats given at point A). I don't know which angles to use to calculate the force components. Would I use 55 degrees?
4. Sep 24, 2013
### SteamKing
Staff Emeritus
You want the moment calculated about A in component form. I suggest you find out what the components of the 60-lb Force are relative to the x-y axis. You know what angle BC makes with the x-axis, and the force is acting a right angles to BC.
5. Sep 24, 2013
### Loopas
So the correct angle would be:
180-90-55 = 35 degrees?
I think I may have just over thought this...
6. Sep 25, 2013
### SteamKing
Staff Emeritus
Take a look at the force vector. Look at the position of the arrow head relative to the opposite end of the vector. What quadrant is the arrow head in. Is that the same quadrant that an angle of 35 degrees would be in?
If these things are confusing, you can always use a protractor to check yourself. They are handy tools for this sort of work.
7. Sep 25, 2013
### Loopas
35 should be the reference angle, which means that 145 would be the real angle, since Fx needs to be negative. Right?
8. Sep 25, 2013
### SteamKing
Staff Emeritus
Draft saved Draft deleted
Similar Discussions: Problem involving moment components and angles | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8109554648399353, "perplexity": 1222.2394318264735}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823731.36/warc/CC-MAIN-20171020044747-20171020064747-00217.warc.gz"} |
http://ethos.bl.uk/OrderDetails.do?uin=uk.bl.ethos.565868 | Use this URL to cite or link to this record in EThOS: http://ethos.bl.uk/OrderDetails.do?uin=uk.bl.ethos.565868
Title: Exceptional Lebesgue densities and random Riemann sums
Author: Grahl, J.
Awarding Body: University College London (University of London)
Current Institution: University College London (University of London)
Date of Award: 2012
Availability of Full Text:
Access through EThOS: Access through Institution:
Abstract:
We will examine two topics in this thesis. Firstly we give a result which improved a bound for a question asking which values the Lebesgue density of a measurable set in the real line must have (joint work with Toby O'Neil and Marianna Csörnyei). We also show how this result relates to the results obtained by others. Secondly, we give several results which indicate when a Lebesgue measurable function has a random Riemann integral which converges, in either the weak and strong sense. A Lebesgue measurable set A, subset of R, has density either 0 or 1 at almost every point. Here the density at some point x refers to the proportion of a small ball around x which belongs to A, in the limit as the size of the ball tends to 0. Suppose that A is not either a nullset, which has density 0 at every single point, or the complement of a nullset, which similarly has density 1 everywhere. Then there are certain restrictions on the range of possible values at those exceptional points where the density is neither 0 nor 1. In particular, it is now known that if δ < 0.268486 ..., where the exact value is the positive root of 8δ^3+8 δ^2+ δ1 = 0, then there must exist a point at which the density of A is between δ and 1 - δ, and that this does not remain true for any larger value of δ. This was proved in a recent paper by Ondrej Kurka. Previous to his work our result given in this thesis was the best known counterexample. We give the background to this, construct the counterexample, and discuss Kurka's proof of the exact bound. The random Riemann integral is defined as follows. Given a Lebesgue measurable function f : [0, 1] \rightarrow R and a partition of [0, 1] into disjoint intervals, we can choose a point belonging to each interval, independently and uniformly with respect to Lebesgue measure. We then use these random points to form a Riemann sum, which is itself a random variable. We are interested in knowing whether or not this random Riemann sum converges in probability to some real number. Convergence in probability to r means that the probability that Riemann sum differs from r by more than \varepsilon, is less than \varepsilon, provided that the maximum length of an interval in the partition is sufficiently small. We have previously shown that this type of convergence does take place provided that f is Lebesgue integrable. In other words, the random Riemann integral, defined as the limit in probability of the random Riemann sums, has at least the power of the Lebesgue integral. Here we prove that the random Riemann integral of f does not converge unless |f|^1-e is integrable for e > 0 arbitrarily small. We also give another, more technical, necessary condition which applies to functions which are not Lebesgue integrable but are improper Riemann integrable. We have also done some work on the question of almost sure convergence. This works slightly differently. We must choose, in advance, a sequence of partitions (Pn)n∞=1, with the size of the intervals of Pn tending to zero. We form a probability space on which we can take random Riemann sums independently on each partition of the sequence. Almost sure convergence means that the sequence of random Riemann sums converges to some (unique) limit with probability 1 in this space. There are two complementary results; firstly that almost sure convergence holds if the function is in Lp and the sequence of partition sizes is in l^p-1 for some p \ge 1. Secondly, we have a partial converse which only applies to nonnegative functions, and if the ratio between the lengths smallest and biggest intervals in each partition is bounded uniformly. This says that if for some p \ge 1 f is not in L^p and the partition sizes are not in l^p-1, then the sequence of Riemann sums diverges with probability 1.
Supervisor: Not available Sponsor: Not available
Qualification Name: Thesis (Ph.D.) Qualification Level: Doctoral
EThOS ID: uk.bl.ethos.565868 DOI: Not available
Share: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9790269732475281, "perplexity": 293.15967435907083}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187820487.5/warc/CC-MAIN-20171016233304-20171017013304-00071.warc.gz"} |
http://mathoverflow.net/questions/108926/the-complete-list-of-continued-fractions-like-the-rogers-ramanujan | The complete list of continued fractions like the Rogers-Ramanujan?
I have two questions about q-continued fractions, but a little intro first. Given Ramanujan's theta function,
$$f(a,b) = \sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}$$
then the following, $$A(q) = q^{1/8} \frac{f(-q,-q^3)}{f(-q^2,-q^2)}$$
$$B(q) = q^{1/5} \frac{f(-q,-q^4)}{f(-q^2,-q^3)}$$
$$C(q) = q^{1/3} \frac{f(-q,-q^5)}{f(-q^3,-q^3)}$$
$$D(q) = q^{1/2} \frac{f(-q,-q^7)}{f(-q^3,-q^5)}$$
$$E(q) = q^{1/1} \frac{f(-q,-q^{11})}{f(-q^5,-q^7)}$$
are q-continued fractions of degree $4,5,6,8,12$, respectively, namely,
$$A(q) = \cfrac{q^{1/8}}{1 + \cfrac{q}{1+q + \cfrac{q^2}{1+q^2 + \ddots}}},\;\;B(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \ddots}}}$$
$$C(q) = \cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \ddots}}},\;\;\;\;D(q) = \cfrac{q^{1/2}}{1 + q +\cfrac{q^2}{1+q^3 + \cfrac{q^4}{1+q^5 + \ddots}}}$$
$$E(q) = \cfrac{q(1-q)}{1-q^3 + \cfrac{q^3(1-q^2)(1-q^4)}{(1-q^3)(1+q^6)+\cfrac{q^3(1-q^8)(1-q^{10})}{(1-q^3)(1+q^{12}) + \ddots}}}$$
The first three are by Ramanujan, the fourth is the Ramanujan-Gollnitz-Gordon cfrac, while the last is by Naika, et al (using an identity by Ramanujan). Let $q = e^{2\pi i \tau}$ where $\tau = \sqrt{-n}$ and these can be simply expressed in terms of the Dedekind eta function $\eta(\tau)$ as,
$$\tfrac{1}{A^4(q)}+16A^4(q) = \left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8$$
$$\tfrac{1}{B(q)}-B(q) = \left(\tfrac{\eta(\tau/5)}{\eta(5\tau)}\right)+1$$
$$\tfrac{1}{C(q)}+4C^2(q) = \left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3$$
$$\tfrac{1}{D(q)}-D(q) = \big(\tfrac{1}{A(q^2)}\big)^2$$
$$E(q) = \;???$$
Question 1: Does anybody know how to express $E(q)$ in terms of $\eta(\tau)$? (It's SO frustrating not to complete this list. I believe there might be a simple relationship between orders 6 and 12, just like there is between 4 and 8.) This cfrac can be found in "On Continued Fraction of Order 12", but the authors do not address this point.
Question 2: Excluding these five and the Heine cfrac which gives $\eta(\tau)/\eta(2\tau)$, are there any other q-continued fractions which yield an algebraic value at imaginary arguments?
-
[Edited again to give a second identity relating $E$ to eta products]
Continued fraction or not, an expression $q^{\frac{(r-s)^2}{8(r+s)}} f(\pm q^r, \pm q^s)$ is a modular form of weight $1/2$ for all integers $r,s$ with $r+s>0$, because it is a sum $\sum_{n=-\infty}^\infty \pm q^{(cn+d)^2}$ with rational $c,d$ and periodic signs. Therefore the quotient of two such expressions is a modular function, and takes algebraic valus at quadratic imaginary values.
The quotient $$E(q) = q - q^2 + q^6 - q^7 + q^8 - q^9 + q^{11} - 2q^{12} + 2q^{13} - 2q^{14} + 2q^{15} \cdots$$ looks like a modular unit $-$ its logarithmic derivative has small coefficients $-$ but not quite an eta product; instead it seems to be a quotient of Klein forms: $$E(q) = q \prod_{n=1}^\infty (1-q^n)^{\chi(n)},$$ where $\chi$ is the Dirichlet character of conductor $12$, given by
$$\chi(n) = \cases{ +1,& if n \equiv \pm 1 \bmod 12; \cr -1,& if n \equiv \pm 5 \bmod 12; \cr 0,& otherwise. }$$ Two identities relating $E$ to $\eta$ products, similar to but somewhat more complicated than the ones you give for $A,B,C,D,$ are $$\frac1{E(q)} - E(q) = \frac{\eta(2\tau)^2 \eta(6\tau)^4}{\eta(\tau)\eta(3\tau)\eta(12\tau)^4},$$ and (a bit simpler) $$\frac1{E(q)} + E(q) = \frac{\eta(4\tau)}{\eta(\tau)} \Bigl(\frac{\eta(3\tau)}{\eta(12\tau)}\Bigr)^3.$$
-
Thanks, Dr. Elkies. It is good to finally know a relation of $E(q)$ to an eta product! However, it may be possible to simplify this considering the other cfracs also have more complicated expressions. For example, in Duke's "Continued Fractions and Modular Functions", we have $\frac{1}{D^2(q)}+D^2(q) = \frac{\eta(4\tau)^2\eta(\tau)^4}{\eta(2\tau)^2\eta(8\tau)^4}+6$. I'll see if I can tweak your relation to find a reasonably simple one between $C(q)$ and $E(q)$. – Tito Piezas III Oct 6 at 15:55 The $E^{-1} - E$ relation might not simplify much, but the one I just added for $E^{-1} + E$ might come closer to what you're hoping for. – Noam D. Elkies Oct 7 at 21:05 [It seems that meanwhile Somos also called attention to $E^{-1} + E$...] – Noam D. Elkies Oct 7 at 21:07
Based on Elkies' answer and an email by Michael Somos, we can give an alternative expression to my Question 1. If a sum is used, instead of a difference,
$$u = \frac{1}{E(q)}+E(q)$$
then,
$$\frac{u(u-4)^3}{(u-1)^3} = \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8$$
Since this eta quotient and $C(q)$ appear in j-function formulas, we can relate the two. Let,
$$v = \frac{u}{2}-1$$ then, $$2\left(\frac{1}{v}+v+2\right) = \left(\frac{1}{C(q)}\right)^3$$
I knew there was a relatively simple relationship, with "relatively" being the operative word.
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http://www.ask.com/question/how-many-miles-over-the-speed-limit-is-considered-reckless-driving-in-il | # How Many Miles over the Speed Limit Is Considered Reckless Driving in Il?
In Illinois, it is considered reckless driving when someone is driving 15 MPH over the speed limit. Reckless driving is described as a disregard for the safety of persons or property.
Q&A Related to "How Many Miles over the Speed Limit Is Considered..."
Tennessee law does not define a http://www.chacha.com/question/how-many-miles-over...
In most states its 15 miles over the speed limit. There may be variations on this such as if you are in a city, or a construction or school zone. http://www.kgbanswers.com/how-many-miles-over-the-...
Any amount over the speed limit is considered reckless driving. The DMV says even if you drive one mile more you can get a ticket. http://answers.yahoo.com/question/index?qid=200711...
16 mph you can be cited for reckless driving in WV. http://www.transportation.wv.gov/dmv/Dri…. http://answers.yahoo.com/question/index?qid=201203...
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The reckless driving depends on the speed limit which varies from place to place. Speaking in context of California, 20 miles in excess of the posted speed limit ...
In the state of California if you are caught speeding 26 miles over the speed limit, the amount of points that it will cost you on your driving record will depend ...
22 miles an hour over the speed limit is typically considered a major speed violation and can sometimes result in a license suspension not to mention the points. ... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9067685008049011, "perplexity": 1383.7883968906715}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00385-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2804580/proving-x3-9-0-has-no-solutions-in-mathbbz-31-mathbbz/2804582 | # Proving $x^3 - 9 = 0$ has no solutions in $\mathbb{Z}/31\mathbb{Z}$
This is from Paolo Aluffi's book "Algebra: Chapter 0". First, find the order of $[9]_{31}$ in the group $( \mathbb{Z}/31\mathbb{Z})^*$. Then, does the equation $x^3 - 9 = 0$ have any solutions in $\mathbb{Z}/31\mathbb{Z}$?
The order of $9$ is $15$: Repeated squaring shows $9^{16} = 9$. The order has to divide the order of the group, which is $30$, so that looks fine.
But I get stuck trying to prove $x^3=9$ does not have a solution. I tried to say something like:
"$9^{16} = 9$ but $3 \nmid 16$ so there is no solution"
... but that does not seem correct. Can I get a hint (not solution) on how to solve this? I am learning Group Theory, and have not yet gotten to Rings, Fields, or Modules.
• What's $9^{10}$? Jun 1 '18 at 17:17
Knowing that the (multiplicative) order of $9$ modulo $31$ is $15$, suppose there is an element $x$ such that $x^3 = 9$. What will its order be? You should find a contradiction.
• This and Lord Shark The Unknown helped out a lot. Since $x^{30} = 9^{10} \neq 1$, but $a^{30} = 1$ for all $a$ in the group, there cannot be a solution for $x$ in this group. Jun 1 '18 at 17:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8742470741271973, "perplexity": 124.96052409745289}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00410.warc.gz"} |
http://blog.computationalcomplexity.org/2010/07/seventh-mil-problem.html | ## Monday, July 26, 2010
### A Seventh Mil. Problem
Richard Lipton had a wonderful post asking for a seventh Millennium Prize now that Poincare's conjecture has been solved. I posted a suggestion on his blog but got no comments. I'll expand on it and see if I get any comments here.
HISTORY: The original proof of VDW's theorem , in 1927, yields INSANE (not primitive recursive ) bounds on the VDW numbers. (Shelah (1988) later got primitive recursive bounds and Gowers (2001) got bounds you can actually write down!) Inspired by VDW's proof Erdos and Turan (1936), made two conjectures:
1. If A is a subset of N of positive upper density then A has arbitrarily long arithmetic sequences. Proven by Szemeredi in 1975 (see here for more.)
2. If &Sigmax ∈ A 1/x diverges then A has arbitrarily long arithmetic sequences. (This conjecture implies the first one.)
A proof of either of these yields a proof of VDW theorem. The hope was that it would lead to a proof with better bounds. Szemeredi's proof of the first conjecture did not yield better bounds; however, Gowers proved the first conjecture a different way that did yield better bounds on the VDW numbers.
The second conjecture is still worthwhile since it may yield even better bounds and because it is interesting in its own right. So, I propose the second conjecture of Erdos-Turan as the 7th Millennium problem. (It might need a snazzier name. The Divergence Conjecture? The k-AP Conjecture? Suggestions are welcome!)
1. Greene and Tao have already shown that the primes have arbitrarily large arithmetic progressions.
2. The work that has gone into Szemeredi's theorem and the Greene-Tao theorem spanned many areas of mathematics. Hence this is not just an isolated problem.
3. The problem has been open since 1936. Hence it is a hard problem.
4. Will more connections to other parts of math be made? Is the problem too hard? A NO answer to both of these would make it not that good a problem.
5. The converse to the conjecture is not true. Note the following set:
A = &cupk&isin N {2^k + i : 0 &le i < k }
The set A has arbitrarily long arithmetic sequences but If &Sigmax ∈ A 1/x converges.
6. Is there a plausible condition that characterizes the sets that have arbitrarily long arithmetic sequences?
7. There is already (I think) a 3000 dollar bounty on the second conjecture. So the Clay Math Institute will have to just give 997,000 dollars.
1. how about finding the probability that nobody will ever find proofs for any one of those conjectures, not now, not ever?
2. I think it's a great problem for
generating public attention,
as it is easy to understand
even for non mathematicians.
As the Tao-Green-Theorem was
widley recognized as a big
breaktrough in mathematics,
i would like to ask, if there are
other sets where a proof of the
special case would be similar
important.
This could be an advantage for
choosing the conjecture, as even
if the main problem is to hard,
many impotant special cases
would benefit from the status of a
millenium problem.
I think this might lower the risk of
working on such a problem and
making no progress whatsoever.
3. Now we need 2 new problems!
And i'll get \$1.200.000!
Just awesome | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9467771649360657, "perplexity": 1192.979490580712}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462751.85/warc/CC-MAIN-20150226074102-00177-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://forum.allaboutcircuits.com/threads/problem.2041/#post-13326 | # problem
Joined Dec 29, 2004
83
Hi,
I am lost with this problem.
A bottle containig air is closed with a watertight yet smoothly moving piston. The bottle with its air has a total mass of 0.30kg. At the surface of a body of water whose temperature is a uniform 285 K throughout, the volume of air contained in the bottle is 1.5L.
Recall that the pressure of water increases with depth below the surface, D, as p=po+ρ *g*D, where po is the surface pressure and ρ=1.0 kg/L.
The bottle is submerged.
a- What is the volume of the air in the bottle as a function of depth?
b- Calculate the buoyant force on the bottle as a function of depth?
There are other questions but I need to understand thoses first.
B
#### CoulombMagician
Joined Jan 10, 2006
37
a) Boyle's law P*V/T = constant. T is constant here so at a depth D the pressure is p and the piston compresses the gas inside until the internal pressure is also p. At this point the new volume V is
V = V0*p0/p = (p0/p)*1.5L
B) The bouyant force is the weight of the displaced water less the weight of the bottle
F = 1.0kg/L * 1.5L * (p0/p) -0.3kg
When Archimedes discovered this principle while he was bathing he was reportedly so excited that he ran naked thorough the city streets.
Joined Dec 29, 2004
83
Originally posted by CoulombMagician@Jan 21 2006, 04:12 AM
a) Boyle's law P*V/T = constant. T is constant here so at a depth D the pressure is p and the piston compresses the gas inside until the internal pressure is also p. At this point the new volume V is
V = V0*p0/p = (p0/p)*1.5L
B) The bouyant force is the weight of the displaced water less the weight of the bottle
F = 1.0kg/L * 1.5L * (p0/p) -0.3kg
When Archimedes discovered this principle while he was bathing he was reportedly so excited that he ran naked thorough the city streets.
[post=13305]Quoted post[/post]
Thank you,
I understand your reasonning but how can we find the p0??
#### CoulombMagician
Joined Jan 10, 2006
37
It is one atmosphere, ~15 psi if my memory is working. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8443030118942261, "perplexity": 1616.5253339845876}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00309.warc.gz"} |
https://www.talks.cam.ac.uk/talk/index/170972 | # Homotopical Lagrangian Monodromy
• Noah Porcelli, Cambridge
• Wednesday 09 March 2022, 16:00-17:00
• MR13.
Given a Lagrangian submanifold L in a symplectic manifold X, a natural question to ask is: what diffeomorphisms f:L → L can arise as the restriction of a Hamiltonian diffeomorphism of X? Assuming L is relatively exact, we will extend results of Hu-Lalonde-Leclercq about the action of f on the homology of L, and deduce that f must be homotopic to the identity if L is a sphere or K(\pi, 1). The proof will use various moduli spaces of pseudoholomorphic curves as well as input from string topology. While motivated by HLL ’s Floer-theoretic proof, we will not encounter any Floer theory.
This talk is part of the Differential Geometry and Topology Seminar series. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9598932862281799, "perplexity": 824.032954211317}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570741.21/warc/CC-MAIN-20220808001418-20220808031418-00176.warc.gz"} |
https://ompf2.com/viewtopic.php?p=5132&sid=ed15daf50aa33180ef86a7250b2773e5 | ## Generalizing the Golden Ratio Sequence to Higher Dimensions
Practical and theoretical implementation discussion.
Paleos
Posts: 16
Joined: Tue Apr 08, 2014 1:32 am
### Generalizing the Golden Ratio Sequence to Higher Dimensions
I am trying to figure out a way to generalize the golden ratio sequence to higher dimensions in a way that is extensible(does not make any assumption about the amount of samples that are going to taken), as well as scaling well to 6 or more dimensions.
The golden ratio sequence is desirable because it uses the irrationality of golden ratio to avoid the structured noise and moire artifacts of other sequences, and it is the cheapest low discrepancy sequence known, requiring only an addition and check for overflow.
update: I have tried doing 2 steps for the second, 1 for first, the result is that the diagonal resulting from using the same values for both x and y just repeats twice instead of once.
The original paper https://www.graphics.rwth-aachen.de/pub ... /2/jgt.pdf uses a permutation like the faure sequence to generalize sequence to 2 dimensions,
which assumes that i know how many samples are going to be taken, which I do not want to know.
A later paper http://www.researchgate.net/profile/Col ... 000000.pdf generalizes the sequence using a Hilbert Curve,
which I am not so confident about the performance in 32 bit precision, especially 6 or more dimensions(tell me if I am wrong) or the quality of it mapped to a sphere compared to the original.
One idea for the unit square i have is to take inspiration from the Halton sequence
What i would prefer is one generalization for the unit square and another one specifically for the unit sphere.
Last edited by Paleos on Fri Feb 27, 2015 12:55 am, edited 6 times in total.
hobold
Posts: 56
Joined: Wed Dec 21, 2011 6:08 pm
### Re: Generalizing the Golden Ratio Sequence to Higher Dimensi
I cannot back this up with a citation or even only with plausible reasoning. But my first experiment would be bit interleaving, i.e. Z-order, also known as Morton code. In other words, instead of the Hilbert-Curve, I'd try a cheaper mapping from the unit interval to the unit square. I feel that the golden ratio's "maximal irrationality" should ensure that sample points are spread evenly across the "tiles" of any recursively defined self-similar curve. I would be surprised if that wasn't true on all scales of that fractal curve.
You need high precision arithmetic for the golden ratio constant and its wrap-around addition. If you want a mapping to 64 bit wide X and Y coordinates on the unit square, then 128 bits are required on the unit interval.
Paleos
Posts: 16
Joined: Tue Apr 08, 2014 1:32 am
### Re: Generalizing the Golden Ratio Sequence to Higher Dimensi
hobold
Posts: 56
Joined: Wed Dec 21, 2011 6:08 pm
### Re: Generalizing the Golden Ratio Sequence to Higher Dimensi
I gave golden ratio sampling along a Z-curve a try. It does cover the unit square nicely, but the sample spacing is less regular than true 2D low discrepancy sequences. I suspect that is due to the discontinuities of the Z-curve; the example images with the Hilbert curve looked a bit better in that regard. But overall the difference was not large. The Z-curve might well be good enough, depending on what exactly you need it for. It is certainly much better than purely random sample points; neither clusters nor gaps are nearly as bad.
More random remarks:
- Bit interleaving generalizes to an arbitrary number of dimensions, but the required precision for the golden ratio accumulator variable will grow fairly quickly.
- Independent threads of computation can be decorrelated nicely in the case of golden ratio sampling: just initialize each thread's accumulator independently at random. This is particularly neat for GPU computing, as each thread will continue to run identical code.
- Unfortunately, inverse bit interleaving isn't exactly cheap: http://graphics.stanford.edu/~seander/b ... bleObvious
- Some exotic RISC ISA extensions used to provide support for bit interleaving back in the 1990s. Not sure if any current (i.e. 'x86) machines still do.
- GPUs used to lay out textures in Z order. I don't know if they still do that, and if that address computation is available to programmers.
Paleos
Posts: 16
Joined: Tue Apr 08, 2014 1:32 am
### Re: Generalizing the Golden Ratio Sequence to Higher Dimensi
What I have figured out that I could do is take the Golden point set as described in the original paper and and then assume that a total of, for instance 2^64 will be generated, and then choose the index within the period with the van der corput sequence. The problem is however is given the how second coordinate is selected through a permutation, how to query the permuted index directly? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8993270397186279, "perplexity": 965.9759482128779}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00279.warc.gz"} |
https://www.arxiv-vanity.com/papers/hep-ex/9908054/ | # Uctp-112-99 Recent Charm Results From Fermilab Experiment E791
A.J. SCHWARTZ
Department of Physics, University of Cincinnati,
Cincinnati, Ohio 45221 (USA)
\abstracts
Fermilab experiment E791 studied weak decays of , , and mesons produced in collisions of 500 GeV/ negative pions with Pt and C targets. The experiment collected over 200 000 fully reconstructed charm decays. Four recent results are discussed here: (a) measurement of the form factor ratios , , and in and decays; (b) measurement of the difference in decay widths between the two mass eigenstates; (c) search for rare and forbidden decays to dilepton final states; and (d) search for a “Pentaquark,” a bound state of .
## 1 Introduction
Fermilab E791 is a charm hadroproduction experiment studying the weak decays of charmed mesons and baryons. The experiment took data from September, 1991 to January, 1992, recording over interactions and reconstructing over 200 000 charm decays. This large sample has led to numerous published results. [1] Four recent results are discussed here: (a) measurement of the form factor ratios , , and in and decays; [2] (b) measurement of the difference in decay widths between the two mass eigenstates; [3] (c) search for rare and forbidden decays; [4] and (d) search for a “Pentaquark,” a bound state of . [5, 6]
The E791 collaboration comprises approximately 70 physicists from 17 institutions. 111 C.B.P.F. (Brazil), Tel Aviv (Israel), CINVESTAV (Mexico), Puebla (Mexico), U.C. Santa Cruz, University of Cincinnati, Fermilab, Illinois Institute of Technology, Kansas State, University of Massachusetts, University of Mississippi, Princeton, University of South Carolina, Stanford, Tufts, University of Wisconsin, and Yale. The experiment produced charmed mesons and baryons using a beam of momentum 500 GeV/ incident on five thin target foils (one platinum, four carbon). The foils were separated along the beamline by approximately 1.5 cm such that most charm decays occurred in air rather than in solid material. Immediately downstream of the target was a silicon strip vertex detector consisting of 17 planes of silicon strips oriented along the and directions, where and point from the vertical direction (). Following the vertex detector was a spectrometer consisting of two large-aperture dipole magnets providing kicks of 212 MeV/ and 320 MeV/, and 37 planes of wire drift chambers and proportional chambers. Downstream of the second magnet were two threshold Čerenkov counters used to discriminate among pions, kaons, and protons. Following the Čerenkov counters was a Pb/liquid-scintillator calorimeter used to measure the energy of electrons and photons, and an Fe/plastic-scintillator calorimeter used to measure the energy of hadrons. Downstream of the calorimeters was approximately 1.0 m of iron to range out any remaining hadrons, and following the iron were two stations of plastic scintillator – one -measuring and one -measuring – to identify muons. The experiment used a loose transverse energy trigger ( GeV) that was almost fully efficient for charm decays.
After events were reconstructed, those with evidence of a decay vertex separated from the interaction vertex were retained for further analysis. The analyses presented here selected events using numerous kinematic and quality criteria; the most important of these are listed in Table 1. An especially effective criterion for enhancing signal over background was “SDZ,” the longitudinal distance between the production and decay vertices divided by the total measurement error in this quantity.
## 2 D+ and D+s Form Factors
Semileptonic decays such as and proceed via spectator diagrams. As such, all hadronic effects are parametrized by four Lorentz-invariant form factors: , , , and . Unfortunately, the limited size of current data samples precludes measurement of the dependence, and we assume this dependence to be given by a nearest pole dominance model: , where GeV/ for the vector form factor and 2.5 GeV/ for the axial-vector form factors . Because appears in every term in the differential decay rate, we factor out and measure the ratios , , and . These ratios are insensitive to the total decay rate and to the weak mixing matrix element .
To select and decays, we identify 3-track vertices in which one track is identified as a kaon and one track as a lepton. We cut on the “transverse” mass , where ( ). In this expression, is the momentum of the neutrino transverse to the direction of the as inferred by momentum balance. The distribution of semileptonic decays forms a Jacobian peak with an endpoint at , and thus we require that lie in the range 1.6–2.0 GeV/ (1.7–2.1 GeV/) for the () sample. We also require that either or . The resulting samples contain very little background, and we do a maximum likelihood fit for the form factors using a likelihood function based on three angles: , the polar angle in the () rest frame between the () and the (); , the polar angle in the rest frame between the and the (); and , the azimuthal angle in the () rest frame between the () and decay planes. The results of the fit to the data for the samples combined are: and . We measure from the sample alone. The results for are: and . These results are compared with theoretical predictions in Table 2; the errors in the measurements are smaller than the spread in theoretical predictions.
## 3 D0-¯¯¯¯¯D0 Mixing and ΔΓ
E791 has published a limit on the - mixing rate using semileptonic decays [19] and hadronic and decays. [20] The flavor of the or when produced is determined by combining the with a low momentum pion to reconstruct a or decay. The semileptonic decays yield a 90% C.L. limit % [where , while the hadronic decays yield a 90% C.L. limit %. The latter limit assumes no violation in the mixing and no violation in a doubly-Cabibbo-suppressed (DCS) amplitude which also contributes to the rate. However, violation is allowed in the interference between the mixing and DCS amplitudes. Since the DCS amplitude is in fact substantially larger than that expected from mixing, the presence of “wrong-sign” decays in the hadronic data – while a signature for mixing – is more easily interpreted as evidence for DCS decays. If we assume no mixing, then the numbers of wrong-sign decays observed in our data, corrected for acceptance, imply ratios of DCS decays to Cabibbo-favored decays of % and %.
Since , where and are the differences between the masses and decay widths of the mass eigenstates, the upper limit for implies an upper limit for the difference in widths: ps. E791 has made a direct measurement of using and decays. Since the former results in a -even eigenstate, only the -even component contributes and the lifetime distribution is proportional to . The final state, however, is a admixture and the lifetime distribution is proportional to . [21] Over the range of lifetimes for which the experiment has sensitivity, and thus: .
Our samples of and are shown in Fig. 1a. We bin these events by reduced proper lifetime, which is the distance traveled by the candidate beyond that required to survive our selection criteria, multiplied by mass and divided by momentum. For each bin of reduced lifetime we fit the mass distribution for the number of signal events. Plotting this number (corrected for acceptance) as a function of reduced lifetime gives the distributions shown in Fig. 1b. Fitting these distributions to exponential functions yields ps, ps, and thus ps. This implies ps at 90% C.L., which is more stringent than the constraint resulting from .
## 4 Rare and Forbidden D Decays
E791 has searched for rare and forbidden dilepton decays of the , , and . The decay modes can be classified as follows:
1. flavor-changing neutral current decays and , in which is a pion or kaon;
2. lepton-flavor violating decays , , and , in which the leptons belong to different generations; and
3. lepton-number violating decays , in which the leptons belong to the same generation but have the same sign charge.
Decay modes belonging to (1) occur within the Standard Model via higher-order diagrams, but the branching fractions are estimated [22] to be only to . This is below the sensitivity of current experiments. Decay modes belonging to (2) and (3) do not conserve lepton number and thus are forbidden within the Standard Model. However, a number of theoretical extensions to the Standard Model predict lepton number violation, [23] and the observation of a signal in these modes would indicate new physics. We have searched for 24 different rare and forbidden decay modes and have found no evidence for them. We therefore present upper limits on their branching fractions. Eight of these modes have no previously reported limits, and fourteen are reported with substantial improvements over previously published results.
For this study we used a “blind” analysis technique. Before our selection criteria were finalized, all events having masses within a window around the mass of the , , or were masked so that the presence or absence of potential signal candidates would not bias our choice of selection criteria. All criteria were then chosen by studying signal events generated by Monte Carlo simulation and background events obtained from the data. The background events were chosen from mass windows above and below the signal window . The criteria were chosen to maximize the ratio , where and are the numbers of signal and background events, respectively. Only after this procedure were events within the signal window unmasked. The signal windows used for decay modes containing electrons are asymmetric around to allow for the bremsstrahlung low-energy tail.
We normalize the sensitivity of our search to topologically similar Cabibbo-favored decays. For the modes we use ; for the modes we use ; and for the we use . The upper limit on the branching fraction for decay mode is:
BX=NXNnormεnormεX⋅Bnorm , (1)
where is the upper limit on the mean number of signal events, is the number of normalization events, and and are overall detection efficiencies. The geometric acceptances and reconstruction efficiencies are found from Monte Carlo simulation, and the particle identification efficiencies are measured from data.
The background consists of random combinations of tracks and vertices, and reflections from more copious hadronic decays. The former is essentially flat in the reconstructed invariant mass, and we estimate this background by scaling the level from mass regions above and below the signal region . The hadronic decay background in which a is misidentified as a lepton is explicitly removed via a or invariant mass cut. The hadronic background in which a is misidentified as a lepton cannot be removed in this manner, as the reflected mass and true mass are too close and such a cut would remove a substantial fraction of signal events. We thus estimate this background by multiplying the number of , , or decays falling within the signal region by the rate for double particle misidentification , or . The misidentification rates were measured from data using decays misidentified as . Because the latter samples have substantial feedthrough background from the former (which is Cabibbo-favored), we do not attempt to establish a limit for decays. Rather, we use the observed signals to measure the lepton misidentification rates under the assumption that all decays observed arise from misidentified . Most of our final event samples are shown in Fig. 2, and all results are tabulated in Table 3.
## 5 Search for the Pentaquark P0¯csuud
E791 has searched for a “Pentaquark” , which is a bound state of five quarks having flavor quantum numbers . This state was originally proposed by Lipkin [25] and Gignoux et al.[26] over ten years ago, but no experimental searches have been undertaken. The is predicted to have a mass below the threshold for strong decay ( GeV/) by 10–150 MeV/. The lifetime is expected to be similar to that of the shortest-lived charm meson, 0.4–0.5 ps. We have searched for decays into and final states.
The analysis proceeds by first selecting four-track vertices in which one track is identified as a proton and two opposite-sign tracks are identified as kaons. We require that either or and remove events in which either or the or momentum projects back to the production vertex. We normalize the sensitivity of the search to and decays; these are topologically similar to and (except for the proton) and several systematic errors cancel. After all selection criteria are applied, we observe no excess of events above background in either decay channel. We thus obtain upper limits for the product of production cross section and branching fraction, relative to that for the . The expression used is (here for ):
σP⋅BP→ϕpπσDs⋅BDs→ϕπ = NP→ϕpπNDs→ϕπ εDs→ϕπεP→ϕpπ , (2)
where is the upper limit on the mean number of decays, and is the number of events observed in the normalization channel. All numbers and the resulting limits are listed in Table 4. When calculating acceptance, we assume the lifetime to be 0.4 ps. The limits are given for two possible values of ; the difference in the limits is due mainly to the difference in the numbers of events observed in the mass spectrum around these mass values. Our upper limits are 2–4% of that for the corresponding decay, which is similar to the theoretical estimate ( 1%).
## 6 Summary
We have presented four recent results from Fermilab experiment E791: a measurement of the form factors governing and decays; a measurement of the difference in decay widths between the two mass eigenstates of ; new limits on two dozen rare and forbidden dilepton decays of , , and ; and a limit on for a “Pentaquark” relative to that for the . Almost all measurements and limits are superior to previously published results. In the case of the and eight of the rare and forbidden dilepton decays, our limits are the first such limits reported.
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• [26] C. Gignoux et al., Phys. Lett. B 193, 323 (1987). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9863521456718445, "perplexity": 1594.8540903816072}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00531.warc.gz"} |
https://math.stackexchange.com/questions/4064539/define-a-relation-r-on-mathbbz-by-arb-iff-3a-5b-is-even-prove-r | Define a relation $R$ on $\mathbb{Z}$ by $aRb$ iff $3a - 5b$ is even. Prove $R$ is an equivalence relation and describe equivalence classes.
I think I got most of the proof, but feel free to critique anything you would like :).
First, notice that $$aRa$$ means $$3a - 5a = a(3 - 5) = 2(-a)$$. This is even, thus, $$R$$ is reflexive.
Second we show that $$aRb \implies bRa$$. Well, $$aRb$$ means that $$3a - 5b$$ is even. Then we have $$3a - 5b = 2k$$ for some integer k. Rearrange like so $$$$\begin{split} 3a - 5b &= 2k \\ -5b &= 2k - 3a \\ -5b &= 2(k - a) + (-a) \end{split}$$$$ Since, we stated that the LHS and RHS were both even in the beginning and still are (i think?), this forces $$a$$ and $$b$$ to be even. Thus, $$3b - 5a$$ is even because we are dealing with two even numbers. Therefore $$bRa$$ which means $$R$$ is symmetric.
Now for transitive. We have show that $$(aRb \wedge bRc) \implies aRc$$. Since $$aRb \wedge bRc$$, we have $$3a - 5b$$ even and $$3b - 5c$$ even. Note when we add, the sum is even. Observe $$(3a - 5b) + (3b - 5c) \\ 3a - 2b - 5c \\ 2(-b) + (3a - 5c)$$ $$2(-b)$$ is even and this forces $$3a - 5c$$ to be even. Therefore $$R$$ is transitive. QED.
For the equivalences classes we know that if $$a$$ is even we have that $$b$$ must be even. Similarly, if $$a$$ is odd then $$b$$ is odd. Therefore, the equivalence classes are the set of odd integers and the set of even integers.
My question is on proving that $$R$$ is symmetric. Is it mathematically sound?
• I see. Question @JMoravitz, how did you go from $3b - 8b + 8b - 5a +8a - 8a$ to $(5a - 3b) - 8a + 8b = 2(k- 4a + 4b)$? That is, it seems that you simply changed the sign of $3b$ and $5a$ to make it from $3b - 5a \to 5a - 3b$. – Owen Mar 16 at 20:22
• We started with $3b-5a$ since this is what we want to show is even. We then "added zero" twice which is always allowed and doesn't change anything, zero here being $-8b+8b$ and $8a-8a$ since something minus itself is zero. We then used part of those expressions for zero to be grouped with the initial expression to make it look like $3a-5b$ since we know that is even and grouped the remaining portions of the expression, and factoring a $2$ out of everything at the end... showing that $3b-5a$ is also $2$ times an integer given that $3a-5b$ was as well. – JMoravitz Mar 16 at 20:24
• and I seem to have a typo in the above, I meant of course $(3a-5b)-8a+8b$, not $(5a-3b)-8a+8b$. The point though still holds... you should be showing that $3b-5a$, the expression you are interested in, is equal to $2$ times an integer. – JMoravitz Mar 16 at 20:26
• @JMoravitz, Ah, yes, that makes sense! That is where I was a little confused! Thank you for the explanation! – Owen Mar 16 at 20:27
• It boils down to $3a - 5b = a -b + 2(a-2b)$ is even iff $a-b$ is even iff $a,b$ are both odd or both even. – lhf Mar 16 at 21:08
Suppose that $$a\sim b$$. That is to say, $$3a-5b$$ is even. That is to say, $$3a-5b$$ is equal to $$2$$ times some integer, we'll call it $$k$$ so $$3a-5b=2k$$
We ask whether or not this implies that $$b\sim a$$, that is if $$3b-5a$$ can be written as $$2$$ times an integer as well (note, not necessarily the same integer as before)
$$\begin{array}{l|l}~~~~3b-5a&\text{original}\\=3b+0-5a+0&\text{add zero twice}\\=3b+(-8b+8b)-5a+(8a-8a)&\text{replace zeroes}\\=(3b-8b)+8b+(-5a+8a)-8a&\text{adjust parentheses}\\=(3a-5b)-8a+8b&\text{simplify and rearrange}\\=2k-8a+8b&\text{use hypothesis}\\=2(k-4a+4b)&\text{factor out two}\end{array}$$
This shows that $$3b-5a$$ can also be written as $$2$$ times an integer and so is also even.
Similarly, for transitivity, we suppose $$3a-5b$$ is even and $$3b-5c$$ is even and we ask about $$3a-5c$$.
$$3a-5c = 3a+0-5c=3a-5b+3b+2b-5c = (3a-5b)+(3b-5c)+2b$$ is the sum of three even numbers and thus even as well.
3a-5b is even implie 3a=5b[2]
• a$$R$$a implie $$3a=5a[2]\Rightarrow -2a=0[2]$$ so R is reflexive
• a$$R$$b$$\Rightarrow 3a=5b[2]\Rightarrow 9a=15b[2] \Rightarrow 5a=3b[2]\Rightarrow$$b$$R$$a
(because $$9a=5a[2]$$ and $$15b=3b[2]$$)
So R is symmetry
• a$$R$$b$$\Rightarrow 3a=5b[2]$$
and
b$$R$$c$$\Rightarrow 3b=5c[2]$$ that implies $$9b=15c[2]\Rightarrow 5b=5c[2]$$
So $$3a=5c[2]\Rightarrow$$ a$$R$$c
So a$$R$$b and b$$R$$c $$\Rightarrow$$a$$R$$c
So R is Transitive
Finally After (reflexivity, symmetry, transitivity) we Kan see R is equivalence relation | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 60, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9888927340507507, "perplexity": 196.49973911447174}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00254.warc.gz"} |
https://unapologetic.wordpress.com/2007/12/18/ | # The Unapologetic Mathematician
## Movie news
I just heard this:
• MGM and New Line will co-finance and co-distribute two films, The Hobbit and a sequel to The Hobbit. New Line will distribute in North America and MGM will distribute internationally.
• Peter Jackson and Fran Walsh will serve as Executive Producers of two films based on The Hobbit. New Line will manage the production of the films, which will be shot simultaneously.
• Peter Jackson and New Line have settled all litigation relating to the Lord of the Rings Trilogy.
So great. We’re going to finally have a Hobbit movie. And a seq…what?
Look, I love Tolkien as much as the next guy, and in an intellectual (opp. fantasy fanboy) way. I grew up with it, I’ve dabbled in Quenya and Sindarin, I’ve read the archives of Vinyar Tengwar, and I wasn’t horribly disappointed by the LotR trilogy. But honestly, people, there’s just not that much there in the Hobbit. It won’t really support two movies on its own. So either they’re reeeeeeeeeally stretching the script to squeeze the money out; they’re bringing in a lot of stuff from Unfinished Tales, or maybe even HoME (unlikely given how much of LotR proper was cut); or they’re creating new Hobbit material out of whole cloth.
A friend of mine says that he trusts Jackson’s vision. I trusted Lucas’ vision, and see where that got us. This may be good, but buckle up just in case.
December 18, 2007 Posted by | Uncategorized | 5 Comments
## Real-Valued Functions of a Single Real Variable
At long last we can really start getting into one of the most basic kinds of functions: those which take a real number in and spit a real number out. Quite a lot of mathematics is based on a good understanding of how to take these functions apart and put them together in different ways — to analyze them. And so we have the topic of “real analysis”. At our disposal we have a toolbox with various methods for calculating and dealing with these sorts of functions, which we call “calculus”. Really, all calculus is is a collection of techniques for understanding what makes these functions tick.
Sitting behind everything else, we have the real number system $\mathbb{R}$ — the unique ordered topological field which is big enough to contain limits of all Cauchy sequences (so it’s a complete uniform space) and least upper bounds for all nonempty subsets which have any upper bounds at all (so the order is Dedekind complete), and yet small enough to exclude infinitesimals and infinites (so it’s Archimedean).
Because the properties that make the real numbers do their thing are all wrapped up in the topology, it’s no surprise that we’re really interested in continuous functions, and we have quite a lot of them. At the most basic, the constant function $f(x)=1$ for all real numbers $x$ is continuous, as is the identity function $f(x)=x$.
We also have ways of combining continuous functions, many of which are essentially inherited from the field structure on $\mathbb{R}$. We can add and multiply functions just by adding and multiplying their values, and we can multiply a function by a real number too.
• $\left[f+g\right](x)=f(x)+g(x)$
• $\left[fg\right](x)=f(x)g(x)$
• $\left[cf\right](x)=cf(x)$
Since all the nice properties of these algebraic constructions carry over from $\mathbb{R}$, this makes the collection of continuous functions into an algebra over the field of real numbers. We get additive inverses as usual in a module by multiplying by $-1$, so we have an $\mathbb{R}$-module using addition and scalar multiplication. We have a bilinear multiplication because of the distributive law holding in the ring $\mathbb{R}$ where our functions take their values. We also have a unit for multiplication — the constant function $1$ — and a commutative law for multiplication. I’ll leave you to verify that all these operations give back continuous functions when we start with continuous functions.
What we don’t have is division. Multiplicative inverses are tough because we can’t invert any function which takes the value zero anywhere. Even the identity function $f(x)=\frac{1}{x}$ is very much not continuous at $x=0$. In fact, it’s not even defined there! So how can we deal with this?
Well, the answer is sitting right there. The function $\frac{1}{x}$ is not continuous at that point. We have two definitions (by neighborhood systems and by nets) of what it means for a function between two topological spaces to be continuous at one point or another, and we said a function is continuous if it’s continuous at every point in its domain. So we can throw out some points and restrict our attention to a subspace where the function is continuous. Here, for instance, we can define a function $f:\mathbb{R}\setminus\{0\}\rightarrow\mathbb{R}$ by $f(x)=\frac{1}{x}$, and this function is continuous at each point in its domain.
So what we should really be considering is this: for each subspace $X\subseteq\mathbb{R}$ we have a collection $C^0(X)$ of those real-valued functions which are continuous on $X$. Each of these is a commutative $\mathbb{R}$-algebra, just like we saw for the collection of functions continuous on all of $\mathbb{R}$.
But we may come up with two functions over different domains that we want to work with. How do we deal with them together? Well, let’s say we have a function $f\in C^0(X)$ and another one $g\in C^0(Y)$, where $Y\subseteq X$. We may not be able to work with $g$ at the points in $X$ that aren’t in $Y$, but we can certainly work with $f$ at just those points of $X$ that happen to be in $Y$. That is, we can restrict the function $f$ to the function $f|_Y$. It’s the exact same function, except it’s only defined on $Y$ instead of all of $X$. This gives us a homomorphism of $\mathbb{R}$-algebras $\underline{\hphantom{X}}|_Y:C^0(X)\rightarrow C^0(Y)$. (If you’ve been reading along for a while, how would a category theorist say this?)
As an example, we have the identity function $f(x)=x$ in $C^0(\mathbb{R})$ and the reciprocal function $g(x)=\frac{1}{x}$ in $C^0(\mathbb{R}\setminus\{0\})$. We can restrict the identity function by forgetting that it has a value at ${0}$ to get another function $f|_{\mathbb{R}\setminus\{0\}}$, which we will also denote by $x$. Then we can multiply $f|_{\mathbb{R}\setminus\{0\}}g$ to get the function $1\in C^0(\mathbb{R}\setminus\{0\})$. Notice that the resulting function we get is not the constant function on $\mathbb{R}$ because it’s not defined at ${0}$.
Now as far as language goes, we usually drop all mention of domains and assume by default that the domain is “wherever the function makes sense”. That is, whenever we see $\frac{1}{x}$ we automatically restrict to nonzero real numbers, and whenever we combine two functions on different domains we automatically restrict to the intersection of their domains, all without explicit comment.
We do have to be a bit careful here, though, because when we see $\frac{x}{x}$, we also restrict to nonzero real numbers. This is not the constant function $1:\mathbb{R}\rightarrow\mathbb{R}$ because as it stands it’s not defined for $x=0$. Clearly, this is a little nutty and pedantic, so tomorrow we’ll come back and see how to cope with it.
December 18, 2007 Posted by | Analysis, Calculus | 2 Comments
## The Orbit Method
Over at Not Even Wrong, there’s a discussion of David Vogan’s talks at Columbia about the “orbit method” or “orbit philosophy”. This is the view that there is — or at least there should be — a correspondence between unitary irreps of a Lie group $G$ and the orbits of a certain action of $G$. As Woit puts it
This is described as a “method” or “philosophy” rather than a theorem because it doesn’t always work, and remains poorly understood in some cases, while at the same time having shown itself to be a powerful source of inspiration in representation theory.
What he doesn’t say in so many words (but which I’m just rude enough to) is that the same statement applies to a lot of theoretical physics. Path integrals are, as they currently stand, prima facie nonsense. In some cases we’ve figured out how to make sense of them, and to give real meaning to the conceptual framework of what should happen. And this isn’t a bad thing. Path integrals have proven to be a powerful source of inspiration, and a lot of actual, solid mathematics and physics has come out of trying to determine what the hell they’re supposed to mean.
Where this becomes a problem is when people take the conceptual framework as literal truth rather than as the inspirational jumping-off point it properly is.
December 18, 2007 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 55, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8169996738433838, "perplexity": 297.5160438087805}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860121618.46/warc/CC-MAIN-20160428161521-00179-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://mathhelpforum.com/statistics/143277-how-did-they-get.html | # Math Help - How did they get this?
1. ## How did they get this?
My textbook makes a calculation and just says "with some simplications", and then gets the answer.
We have L0=(1/sqrt(2pi))^n*(e^(-1/2*sum(xi-m0)^2)...i.e. this the max likelihood function for a normal with variance 1 and mean m0. Sum is from 1 to n. We have L1 is the same except replace m0 with m1.
Then they compute L0/L1 and get
e^((n/2)*(m1^2-m0^2)+(m0-m1)*sum(xi))
I do not see where this step comes from. Any ideas?
Thank you.
2. Originally Posted by zhupolongjoe
My textbook makes a calculation and just says "with some simplications", and then gets the answer.
We have L0=(1/sqrt(2pi))^n*(e^(-1/2*sum(xi-m0)^2)...i.e. this the max likelihood function for a normal with variance 1 and mean m0. Sum is from 1 to n. We have L1 is the same except replace m0 with m1.
Then they compute L0/L1 and get
e^((n/2)*(m1^2-m0^2)+(m0-m1)*sum(xi))
I do not see where this step comes from. Any ideas?
Thank you.
This is tedious to type out, but I literally have it sitting in my notes...
What part are you stuck on?
3. Just how they get that form of Lo/L1...I understand the analysis that follows using the Neyman Pearson lemma, but I just don't understand this, and maybe it's an algebraic thing I'm missing, but I don't see how they are getting the e^((n/2)*(m1^2-m0^2)+(m0-m1)*sum(xi)) from the two likelihood functions.....Maybe there is some property of sums I am missing, but I am not sure.
4. $\frac{f_0}{f_1} = \frac{\exp{[-\frac{1}{2\sigma^2}\sum^n (X_i - \mu_0)^2]}}{\exp{[-\frac{1}{2\sigma^2}\sum^n (X_i - \mu_1)^2]}}$ Apply the property $\frac{e^a}{e^b} = e^{a-b}$ and cancel.
$= \exp\{-\frac{1}{2\sigma^2}\sum^n[(X_i - \mu_0)^2 - (X_i - \mu_1)^2)]\}$ Now expand this.
$= \exp{[2\bar X(\mu_0 - \mu_1) + \mu_0^2 - \mu_1^2]}$
Let me know if you have any specific questions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9259684085845947, "perplexity": 712.6947006987433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701168076.20/warc/CC-MAIN-20160205193928-00290-ip-10-236-182-209.ec2.internal.warc.gz"} |
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[1] Sanjiban Santra. On the Lazer-McKenna conjecture and its applications. Proc. Amer. Math. Soc. Abstract, references, and article information View Article: PDF [2] Alessia E. Kogoj. A Liouville-type Theorem on half-spaces for sub-Laplacians. Proc. Amer. Math. Soc. 143 (2015) 239-248. Abstract, references, and article information View Article: PDF [3] Amir Moradifam. Sharp counterexamples related to the De Giorgi conjecture in dimensions $4\leq n \leq 8$. Proc. Amer. Math. Soc. 142 (2014) 199-203. Abstract, references, and article information View Article: PDF [4] Luciano Mari and Daniele Valtorta. On the equivalence of stochastic completeness and Liouville and Khas'minskii conditions in linear and nonlinear settings. Trans. Amer. Math. Soc. 365 (2013) 4699-4727. Abstract, references, and article information View Article: PDF [5] Roberta Filippucci. A Liouville result on a half space. Contemporary Mathematics 595 (2013) 237-252. Book volume table of contents View Article: PDF [6] Lorenzo D’Ambrosio and Enzo Mitidieri. An application of Kato's inequality to quasilinear elliptic problems. Contemporary Mathematics 595 (2013) 205-218. Book volume table of contents View Article: PDF [7] A. I. Nazarov and N. N. Ural′tseva. The Harnack inequality and related properties for solutions of elliptic and parabolic equations with divergence-free lower-order coefficients. St. Petersburg Math. J. 23 (2012) 93-115. MR 2760150. Abstract, references, and article information View Article: PDF [8] Alberto Farina, Yannick Sire and Enrico Valdinoci. Stable solutions of elliptic equations on Riemannian manifolds with Euclidean coverings. Proc. Amer. Math. Soc. 140 (2012) 927-930. MR 2869076. Abstract, references, and article information View Article: PDF [9] Xiaobao Zhu. Hamilton’s gradient estimates and Liouville theorems for fast diffusion equations on noncompact Riemannian manifolds. Proc. Amer. Math. Soc. 139 (2011) 1637-1644. MR 2763753. Abstract, references, and article information View Article: PDF
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https://grocid.net/2017/02/14/bsides17-delphi/ | # BSides’17 – Delphi
In this challenge, we are given a server which accepts encrypted commands and returns the resulting output. First we define our oracle `go(cmd)`.
```import urllib2
def go(cmd):
```
This simply return the status from the server. It is common for this kind of CTF challenges to use some block-cipher variant such as some of the AES modes.
The first guess I had was that AES-CBC was being used. That would mean that if we try to flip some bit in a block somewhere in the middle of the ciphertext, the first decrypted part would remain intact, whilst the trailing blocks would get scrambled.
Assume that we have four ciphertext blocks $C_0, C_1, C_2, C_3$ and the decryption is $\textsf{dec}_k : C_0\| C_1\| C_2\| C_3 \mapsto P_0\|P_1\|P_2\|P_3$. Now, we flip a bit in $C_1$ so that we get $C_1'$, then we have $\textsf{dec}_k : C_0\| C'_1\| C_2\| C_3 \mapsto P_0\|P'_1\|P'_2\|P'_3$. (This is not true, thanks to hellman for pointing that out in the comments).
Turns out this is not the case. In fact, the error did only propagate one block and not further, i.e.,$\textsf{dec}_k : C_0\| C'_1\| C_2\| C_3 \mapsto P_0\|P'_1\|P'_2\|P_3$. Having a look at the Wikipedia page, I found that this is how AES-CFB/(CBC) would behave (image from Wikipedia):
Since $\textsf{dec}_k(C_0) \oplus C_1 = P_1$, we can inject some data into the decrypted ciphertext! Assume that we want $P'_1 = Q$. Then, we can set $C'_1 = C_1 \oplus P_1 \oplus Q$, since then $\textsf{dec}_k(C_0) \oplus C'_1 = P_1\oplus P_1 \oplus Q = Q$. Embodying the above in Python, we might get something like
```def xor(a, b):
return ''.join(chr(ord(x) ^ ord(y))
for x, y in zip(a, b))
response = ' to test multiple-block patterns' # the block we attack
split_blocks = [cmd[i * 32: i * 32 + 32]
for i in range(len(cmd) / 32)]
block = 3 # this is somewhat arbitrary
# get command and pad it with blank space
append_cmd = ' some command'
append_cmd = append_cmd + '\x20' * (16 - len(append_cmd))
new_block = xor(split_blocks[block].decode("hex"),
response).encode('hex')
new_block = xor(new_block.decode("hex"),
append_cmd).encode('hex')
split_blocks[block] = new_block
cmd = ''.join(split_blocks)
#print cmd
print go(cmd)
```
We can verify that this works. Running the server, we get
`This is a longer string th\x8a\r\xe4\xd9.\n\xde\x86\xb6\xbd*\xde\xf8X\x15I some command e-block patterns\n`
OK, so the server accepts it. Nice. Can we exploit this? Obviously — yes. We can guess that the server does something like
```echo "{input string}";
```
First, we break off the echo statement. Then we try to `cat` the flag and comment out the rest. We can do this in one block! Here is how:
```append_cmd = '\"; cat f* #'
```
Then, the server code becomes
```echo "{partial + garbage}"; cat f* #{more string}";
```
The server gives the following response:
```This is a longer string th:\xd7\xb1\xe8\xc2Q\xd7\xe8*\x02\xe8\xe8\x9c\xa6\xf71\n
FLAG:a1cf81c5e0872a7e0a4aec2e8e9f74c3\n```
Indeed, this is the flag. So, we are done!
## 2 thoughts on “BSides’17 – Delphi”
1. Nice writeup!
In CBC decryption also, the modified block is “scrambled”, the next block is simply bitflipped and all the following blocks are untouched too. Since the attack is basically the same (bit flips), are you sure it was CFB actually? 🙂
1. You are of course right, an error does not propagate in decryption of CBC either — only in encryption. Fixed! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 10, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8884464502334595, "perplexity": 1197.7099377094128}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00464.warc.gz"} |
https://www.quantumdiaries.org/2009/11/30/visiting-damtp-cambridge/ | ## View Blog | Read Bio
### Visiting DAMTP, Cambridge.
It’s been long since I have visited Cambridge before, probably it was already 4 years ago. I lived in Cambridge for half a year, as a visiting researcher then at DAMTP (Department of Applied Math and Theoretical Physics), universtiy of Cambridge. I have good memories on my days at Cambridge.
Last time when I was here at Cambridge, I couldn’t imagine that I would be working on nuclear physics. The talk I gave here 4 years ago was on ADHM construction of instantons and its D-brane realization, and the motivation was purely mathematical physics. But this time, again in my talk I told about ADHM construction, but now applied to nucleon nucleon interaction realized and computed in string theory. It is interesting that, although the mathematical tools are quite similar to each other (and almost the same), the motivations were totally different.
Discussions with my friends at DAMTP were joyful, and are always insightful. And, one more thing — winter in Cambridge, it is beautiful. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8924689292907715, "perplexity": 1342.459316041499}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989812.47/warc/CC-MAIN-20210515035645-20210515065645-00463.warc.gz"} |
https://brilliant.org/practice/si-fluid-measure/ | ×
Classical Mechanics
# SI Fluid Measure
If a cylindrical container has a base area of $$12 \text{ cm}^2$$ and a height of $$6 \text{ cm}$$, how many mL of water can it hold?
Liquefied natural gas is filled in a horizontal cylindrical tank, as shown above. The diameter of the circular base and the length of the cylindrical tank are $$d=6\text{ m}$$ and $$l=14\text{ m},$$ respectively, and the height of the liquefied natural gas is $$h=3\text{ m}.$$ Find the volume of the liquefied natural gas in the tank.
$$18 0\text{ mL}$$ of water is poured into a cylindrical cup with base area $$30\text{ cm}^2$$ and height $$15 \text{ cm}.$$ What is the height of the water inside the cup?
In the figure above, a water reservoir has a rectangular base floor with dimensions $$3\text{ km}\times 2\text{ km}$$ and a depth of $$8 \text{ m}.$$ If this reservoir is half-full, what is the volume of the water in liters?
What is $$8\text{ L}$$ in $$\text{m}^3?$$
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.942249596118927, "perplexity": 281.84112236712303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818690376.61/warc/CC-MAIN-20170925074036-20170925094036-00265.warc.gz"} |
https://physicshelpforum.com/threads/are-the-resistors-in-series-vs-in-parallel.12286/ | # Are the resistors in series vs in parallel?
#### SophiaL
Nov 2016
3
0
1. The problem statement, all variables and given/known data A metal wire of resistance R is cut into two pieces of equal length. The two pieces are connected together side by side. What is the resistance of the two connected wires?
2. Relevant equations RSeries = R1 + R2 + ... RParallel = (1/R1 + 1/R2 + ...)-1
3. The attempt at a solution Initially I thought that, being connected side by side, the same current would be passing through both halves of the wire thus making them resistors in series. However, the answer turns out to be R/4 as a because they are apparently in parallel. Can someone explain how we know?
Last edited:
#### HallsofIvy
Aug 2010
434
174
You say the wires are "connected together side by side". Are we to assume they are insulated so this does not just form one short wire? If so then, yes, the two wires are parallel so this is the same as to resistors "in parallel".
You say "the same current would be passing through both halves of the wire thus making them resistors in series." Where did you get the idea that "the same current" means they are "in series"?
Reactions: 1 person | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9026190042495728, "perplexity": 476.60077002024343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00044.warc.gz"} |
http://mymathforum.com/real-analysis/343727-following-functions-differentiable-x-1-a.html | My Math Forum Which of the following functions are differentiable at $x = 1$ ?
Real Analysis Real Analysis Math Forum
March 23rd, 2018, 07:32 PM #1 Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Which of the following functions are differentiable at $x = 1$ ? Which of the following functions are differentiable at $x = 1$ ? a) $f(x) = ||x| - 1/2|$ b) $f(x) = max${$|x-1|, |x+1|$} c) $f(x) =|x-1| + e^x$ d) $f(x) = |e^x -1|$ I have tried solving above functions but I didnt get any of them differentiable. I have doubt in solving option (a) incorrectly. Someone help on this!!! Thanks
March 23rd, 2018, 07:59 PM #2 Senior Member Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics Answer is (d). Now can you see why ?
March 23rd, 2018, 08:01 PM #3 Senior Member Joined: Aug 2012 Posts: 2,193 Thanks: 645 When you graph (a) what does it do at x = 1?
March 23rd, 2018, 08:04 PM #4 Math Team Joined: Dec 2013 From: Colombia Posts: 7,616 Thanks: 2605 Math Focus: Mainly analysis and algebra The absolute value function $|x|$ has a derivative everywhere except where it is equal to zero. So you should be looking to find out whether any of the four functions contain an absolute value function that has value zero at $x=1$.The inner absolute value function has derivative except at $x=0$, so it's smooth for $x \gt 0$, where $\big| |x| - \frac12 \big| = \big| x - \frac12 \big|$. Where does this function not have a derivative? $\max{(|x-1|, |x+1|)} = |x+1|$ for all $x \gt 0$. And $|x+1| = x+1$ for all $x \gt -1$. Treat each half of the function separately. When does $e^x-1 = 0$? Last edited by v8archie; March 23rd, 2018 at 08:06 PM.
March 23rd, 2018, 09:17 PM #5
Senior Member
Joined: Nov 2015
Posts: 232
Thanks: 2
Quote:
Originally Posted by SDK Answer is (d). Now can you see why ?
I assume $|e^x - 1|$is same as$|x|$. So I took $-x$ (i.e.,$-e^x+1$) when $x \leq 1$ and $x$(i.e., $e^x-1$) when $x \geq 1$. In this way they are neither differentiable nor continuous.
Let me know how to do it 😢
March 24th, 2018, 06:00 AM #6
Math Team
Joined: Jan 2015
From: Alabama
Posts: 3,264
Thanks: 902
Quote:
Originally Posted by Lalitha183 I assume $|e^x - 1|$is same as$|x|$. So I took $-x$ (i.e.,$-e^x+1$) when $x \leq 1$ and $x$(i.e., $e^x-1$) when $x \geq 1$. In this way they are neither differentiable nor continuous. Let me know how to do it 😢
If you mean, by "is the same as |x|", that you can treat it in the same way, yes.
For any positive x, $e^x> 1$ so $e^x- 1> 0$ and $|e^x- 1|= e^x- 1$ which is differentiable. This problem would be more complicated if the question were about x= 0 rather than x= 1.
March 27th, 2018, 08:05 AM #7
Senior Member
Joined: Nov 2015
Posts: 232
Thanks: 2
Quote:
Originally Posted by Country Boy If you mean, by "is the same as |x|", that you can treat it in the same way, yes. For any positive x, $e^x> 1$ so $e^x- 1> 0$ and $|e^x- 1|= e^x- 1$ which is differentiable. This problem would be more complicated if the question were about x= 0 rather than x= 1.
I have finally solved all 4 problems. And the results are like this :
1. L.H.L = R.H.L = $\frac{1}{2}$
L.H.D = R.H.D = $1$ and $f(1) = \frac{1}{2}$
2. L.H.L = R.H.L = $2$
L.H.D = R.H.D = $0$ and $f(1) = 2$
3. L.H.L = R.H.L = $e$
L.H.D = R.H.D = $0$ and $f(1) = e > 1$
4. L.H.L = R.H.L = $e - 1$
L.H.D = R.H.D = 0 and $f(1) = e-1$
In this, $e-1 >0$ then how will it be equal to $0$ to say $|e^x-1|$ is differentiable at $x = 1$
March 29th, 2018, 07:38 AM #8 Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 You say you have solved all four problems but the original problem was to determine whether or not each function is differentiable at x= 1 and nowhere do you answer that question! a)f(x)= ||x|- 1/2|. For x larger than 1/2, |x|= x and x- 1/2> 0 so ||x|- 1/2|= x- 1/2 for all x> 1/2. Near x= 1, ||x|- 1/2|= x- 1/2 which is differentiable (and has derivative 1) at x= 1. b)f(x)= max(|x- 1|, |x+1|). For x larger than 1, both x- 1 and x+ 1 are positive so |x- 1|= x- 1 and |x+ 1|= x+1. Obviously x+1 is larger than x- 1 so for x> 1, f(x)= x+ 1. If x< 1 then x- 1 is negative so |x- 1|= -(x- 1)= 1- x. So for x< 1, but close to 1, the function is max(1- x, x+ 1)= x+ 1 also. The "difference quotient" is (f(1+ h)- f(1))/h= h/h= 1 for x>1 or x< 1. The derivative at x= 1 is the limit of that as h goes to 0, 1. The function is differentiable at x= 1 (and has derivative 1). c) f(x)= |x- 1|+ e^x. For x larger than 1, |x- 1|= x- 1, for x less than 1, |x- 1|= -(x- 1)= 1- x. The difference quotient, (f(1+h)- f(1))/h is, for h positive, so that 1+h> 1, (1+ h- 1+ e^{1+h}- e)/h= 1+ (e^{1+ h}- e)/h. As h goes to 0, that goes to 1+ 1= 2. For h negative, so that 1+h< 1, the difference quotient is (1- (1+h)+ e^{1+h}- e)/h= -1+ (e^{1+h}- e)/h. As h goes to 0, that goes to -1+ 1= 0. The two limits are not the same so this function is not differentiable at x= 1. d) f(x)= |e^x- 1|. For x close to 1, e^x is close to e and e- 1> 0 so for x close to 1, f(x)= e^x- 1. That is differentiable so f(x) is differentiable for x= 1 (and the derivative is e- 1).
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Contact - Home - Forums - Cryptocurrency Forum - Top | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8929449915885925, "perplexity": 2458.2896112781114}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202003.56/warc/CC-MAIN-20190319163636-20190319185636-00351.warc.gz"} |
https://sinews.siam.org/Details-Page/the-jungle-of-stochastic-optimization | SIAM News Blog
SIAM News
# The Jungle of Stochastic Optimization
There is a vast range of problems that fall under the broad umbrella of making sequential decisions under uncertainty. While there is widespread acceptance of basic modeling frameworks for deterministic versions of these problems from the fields of math programming and optimal control, sequential stochastic problems are another matter.
Motivated by a wide range of applications, entire fields have emerged with names such as dynamic programming (Markov decision processes, approximate/adaptive dynamic programming, reinforcement learning), stochastic optimal control, stochastic programming, model predictive control, decision trees, robust optimization, simulation optimization, stochastic search, model predictive control, and online computation. Problems may be solved offline (requiring computer simulation) or online in the field, which opens the door to the communities working on multi-armed bandit problems. Each of these fields has developed its own style of modeling, often with different notation ($$x$$ or $$S$$ for state, $$x/a/u$$ for decision/action/control), and different objectives (minimizing expectations, risk, or stability). Perhaps most difficult is appreciating the differences in the underlying application. A matrix $$K$$ could be $$5 \times 5$$ in one class of problems, or $$50,000 \times 50,000$$ in another (it still looks like $$K$$ on paper). But what really stands out is how each community makes a decision.
Despite these differences, it is possible to pull them together in a common framework that recognizes that the most important (albeit not the only) difference is the nature of the policy being used to make decisions over time (we emphasize that we are only talking about sequential problems, consisting of decision, information, decision, information, …). We start by writing the most basic canonical form as
\begin{align} & {\min\nolimits_{\pi \in \Pi} \mathbb{E}^{\pi} \sum_{t=0}^T C(S_t, U_t^{\pi}(S_t))} \: \: \: \: \: \: \: \: \: (1)\\ & \text{where }S_{t+1} = S^M(S_t, u_t, W_{t+1}). \nonumber \end{align}
Here we have adopted the notational system where $$S_t$$ is the state (physical state, as well as the state of information, and state of knowledge), and $$u_t$$ is a decision/action/control (alternatives are $$x_t$$, popular in operations research, or $$a_t$$, popular in operations research as well as computer science). We let $$U_t^{\pi}(S_t)$$ be the decision function, or policy, which is one member in a set $$\Pi$$ where $$\pi$$ specifies both the type of function, as well as any tunable parameters $$\theta \in \Theta^\pi$$. The function $$S^M(S_t, u_t, W_{t+1})$$ is known as the transition function (or system model, state model, plant model, or simply “model”). Finally, we let $$W_{t+1}$$ be the information that first becomes known at time $$t+1$$ (control theorists would call this $$W_t$$, which is random at time $$t)$$.
Important problem variations include different operators to handle uncertainty; we can use an expectation in $$(1)$$, a risk measure, or worst case (robust optimization), as well as a metric capturing system stability. We can assume we know the probability law behind $$W_t$$, or we may just observe $$S_{t+1}$$ given $$S_t$$ and $$u_t$$ (model-free dynamic programming).
While equation $$(1)$$ is well-recognized in certain communities (some will describe it as “obvious”), it is actually quite rare to see $$(1)$$ stated as the objective function with anything close to the automatic writing of objective functions for deterministic problems in math programming or optimal control. We would argue that the reason is that there is no clear path to computation. While we have powerful algorithms to solve over real-valued vector spaces (as required in deterministic optimization), equation $$(1)$$ requires that we search over spaces of functions (policies).
Lacking tools for performing this search, we make the argument that all the different fields of stochastic optimization can actually be described in terms of different classes of policies. In fact, we have identified four fundamental (meta) classes, which are the following:
1. Policy function approximations (PFAs). These are analytical functions that map states to actions. PFAs may come in the form of lookup tables, parametric, or non-parametric functions. A simple example might be
$U^{\pi}(S_t\,|\,\theta) = \sum_{f\in F} \theta_f \phi_f(S_t) \: \: \: \: \: \: \: \: \: (2)$
where $$F$$ is a set of features, and $$\bigl(\phi_f(S_t)\bigr), f \in F$$ are sometimes called basis functions.
2. Cost function approximations (CFAs). Here we are going to design a parametric cost function, or parametrically modified constraints, producing a policy that we might write as
$U^{\pi}(S_t\,|\,\theta) = \arg \min\nolimits_{u \in \mathrm{U}_t^{\pi}(\theta)} C_t^{\pi} (S_t, u\,|\,\theta) \: \: \: \: \: \: \: \: \: (3)$
where $$C_t^{\pi} (S_t, u\,|\,\theta)$$ is a parametrically modified set of costs (think of including bonuses and penalties to handle uncertainty), while $$U_t^{\pi}(\theta)$$ might be a parametrically modified set of constraints (think of including schedule slack in an airline schedule, or a buffer stock).
3. Policies based on value function approximations (VFAS). These are the policies most familiar under the umbrella of dynamic programming and reinforcement learning. These might be written as
\begin{align} & U^{\pi}_t(S_t\,|\,\theta) = \arg \min\nolimits_{u \in \mathrm{U}_t^{\pi}(\theta)} C(S_t, u)+{} \nonumber\\ & \quad \mathbb{E}\bigl\{\overline{V}_{t+1}^{\pi}(S^M(S_t, u, W_{t+1})\,|\,\theta)\,|\,S_t\bigr\} \end{align} \: \: \: \: \: \: \: \: \: (4)
where $$\overline{V}_{t+1}^{\pi}(S_{t+1})$$ is an approximation of the value of being in state $$S_{t+1} = S^M(S_t, u, W_{t+1})$$, where $$\pi$$ captures the structure of the approximation and $$\theta \in \Theta^\pi$$ represents any tunable parameters.
\begin{align} & U^{\pi}_t(S_t\,|\,\theta) = \arg \min\nolimits_u\Biggl(C(S_t, u) + \min\nolimits_{\pi \in \Pi} \nonumber \\ & \qquad \mathbb{E}^{\pi}\Biggl\{\sum_{t' = t + 1}^T C(S_{t'}, U_{t'}^{\pi}(S_{t'}))\,|\,S_t, u\Biggr\}\Biggr) \end{align}\: \: \: \: \: \: \: \: \: (5)
The problem is that the second term in $$(5)$$ is not computable (if this were not the case, we could have solved the objective function in $$(1)$$ directly). For this reason, we create a lookahead model which is an approximation of the real problem. Common approximations are to limit the horizon (e.g. from $$T$$, which might be quite long, to $$t+H$$ for some appropriately chosen horizon $$H$$), and (most important) to replace the original stochastic information process with something simpler. The most obvious is a deterministic approximation, which we can write as
$U_t^{\pi}(S_t|\theta)=arg~min_{u_t, \tilde{u}_{t,t+1} ,\ldots,\tilde{u}_{t,t+H}}\ \Biggl(C(S_t,u_t)+\sum^{t+H}_{t'=t+1}C(\tilde{S}_{tt'},\tilde{U}_{tt'})\Biggr).\: \: \: \: \: \: \: \: \: (6)$
To make the distinction from our original base model in $$(1)$$, we put tildes on all our variables (other than those at time $$t$$), and we also index the variables by $$t$$ (to indicate that we are solving a problem at time $$t$$), and $$t’$$ (which is the point in time within the lookahead model).
A widely-used approach in industry is to start with $$(6)$$ and then introduce modifications (often to the constraints) so that the decisions made now are more robust to uncertain outcomes that occur later. This would be a form of (hybrid) cost function approximation.
We may instead use a stochastic lookahead model. For example, the stochastic programming community most often uses
${U}_t^{\pi}(S_t|\theta)=arg~min_{u_t, \tilde{u}_{t,t+1} ,\ldots,\tilde{u}_{t,t+H}}\\ \Biggl(C(S_t,u_t)+\sum_{\omega\in\tilde{\Omega_t}}p(\omega)\sum^{t+H}_{t'=t+1}C(\tilde{S}_{tt'}(\omega),\tilde{U}_{tt'}(\omega))\Biggr).\: \: \: \: \: \: \: \: \: (7)$
Here, we would let $$\theta$$ capture parameters such as the planning horizon, and the logic for constructing $$\hat\Omega_t$$.
Other variations include a robust objective (which minimizes over the worst outcome rather than the expected outcome), or a chance-constrained formulation, which approximates the costs over all the uncertain outcomes using simple penalties for violating constraints.
All of these policies involve tunable parameters, given by $$\theta$$. We would represent the policy $$\pi$$ as the policy class $$f\in F$$, and the parameters $$\theta \in \Theta^f$$. Thus, the search over policies $$\pi$$ in equation $$(1)$$ can now be thought of as the search over policy classes $$f\in F$$, and then over the tunable parameters $$\theta \in \Theta^f$$.
No, this is not easy. But with this simple bit of notation, all of the different communities working on sequential stochastic optimization problems can be represented in a common framework.
Why is this useful? First, a common vocabulary facilitates communication and the sharing of ideas. Second, it is possible to show that each of the four classes of policies can work best on the same problem, if we are allowed to tweak the data. And finally, it is possible to combine the classes into hybrids that work even better than a pure class.
And maybe some day, mathematicians will figure out how to search over function spaces, just as Dantzig taught us to search over vector spaces.
Warren B. Powell is a faculty member of the Department of Operations Research and Financial Engineering at Princeton University. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8152292966842651, "perplexity": 651.3496980428753}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794866938.68/warc/CC-MAIN-20180525024404-20180525044404-00188.warc.gz"} |
http://cms.math.ca/cmb/msc/31A05?fromjnl=cmb&jnl=CMB | location: Publications → journals
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Search: MSC category 31A05 ( Harmonic, subharmonic, superharmonic functions )
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1. CMB 2003 (vol 46 pp. 373)
Laugesen, Richard S.; Pritsker, Igor E.
Potential Theory of the Farthest-Point Distance Function We study the farthest-point distance function, which measures the distance from $z \in \mathbb{C}$ to the farthest point or points of a given compact set $E$ in the plane. The logarithm of this distance is subharmonic as a function of $z$, and equals the logarithmic potential of a unique probability measure with unbounded support. This measure $\sigma_E$ has many interesting properties that reflect the topology and geometry of the compact set $E$. We prove $\sigma_E(E) \leq \frac12$ for polygons inscribed in a circle, with equality if and only if $E$ is a regular $n$-gon for some odd $n$. Also we show $\sigma_E(E) = \frac12$ for smooth convex sets of constant width. We conjecture $\sigma_E(E) \leq \frac12$ for all~$E$. Keywords:distance function, farthest points, subharmonic function, representing measure, convex bodies of constant widthCategories:31A05, 52A10, 52A40
2. CMB 2002 (vol 45 pp. 154)
Weitsman, Allen
On the Poisson Integral of Step Functions and Minimal Surfaces Applications of minimal surface methods are made to obtain information about univalent harmonic mappings. In the case where the mapping arises as the Poisson integral of a step function, lower bounds for the number of zeros of the dilatation are obtained in terms of the geometry of the image. Keywords:harmonic mappings, dilatation, minimal surfacesCategories:30C62, 31A05, 31A20, 49Q05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9802815318107605, "perplexity": 617.282961996021}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00285-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.usgs.gov/publications/rainfall-effects-rare-annual-plants | # Rainfall effects on rare annual plants
January 1, 2008
1. Variation in climate is predicted to increase over much of the planet this century. Forecasting species persistence with climate change thus requires understanding of how populations respond to climate variability, and the mechanisms underlying this response. Variable rainfall is well known to drive fluctuations in annual plant populations, yet the degree to which population response is driven by between-year variation in germination cueing, water limitation or competitive suppression is poorly understood.
2. We used demographic monitoring and population models to examine how three seed banking, rare annual plants of the California Channel Islands respond to natural variation in precipitation and their competitive environments. Island plants are particularly threatened by climate change because their current ranges are unlikely to overlap regions that are climatically favourable in the future.
3. Species showed 9 to 100-fold between-year variation in plant density over the 5–12 years of censusing, including a severe drought and a wet El Niño year. During the drought, population sizes were low for all species. However, even in non-drought years, population sizes and per capita growth rates showed considerable temporal variation, variation that was uncorrelated with total rainfall. These population fluctuations were instead correlated with the temperature after the first major storm event of the season, a germination cue for annual plants.
4. Temporal variation in the density of the focal species was uncorrelated with the total vegetative cover in the surrounding community, suggesting that variation in competitive environments does not strongly determine population fluctuations. At the same time, the uncorrelated responses of the focal species and their competitors to environmental variation may favour persistence via the storage effect.
5. Population growth rate analyses suggested differential endangerment of the focal annuals. Elasticity analyses and life table response experiments indicated that variation in germination has the same potential as the seeds produced per germinant to drive variation in population growth rates, but only the former was clearly related to rainfall.
6. Synthesis. Our work suggests that future changes in the timing and temperatures associated with the first major rains, acting through germination, may more strongly affect population persistence than changes in season-long rainfall. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8166307806968689, "perplexity": 4294.06775766574}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00522.warc.gz"} |
https://www.physicsforums.com/threads/lattice-field-theory.5678/ | # Lattice Field Theory
#### gnl
21
0
Hi everyone! I would like to post a new thread, related to my research work: QFT on a lattice, i.e. on computers! Is anyone interested?
#### jcsd
Gold Member
2,074
11
Sure, I only know the basic theory behind quantum computing rather than the practicalities, how is the problem of decoherence being overcome?
#### gnl
21
0
lattice
This is what I am talking about. Take some QFT. Write a Euclidean space-time discrete version of the action and then use numerical methods to evaluate Green´s functions. The inverse lattice spacing serves as a momentum cutoff...
#### jcsd
Gold Member
2,074
11
Staff Emeritus
Gold Member
Dearly Missed
6,691
5
This is certainly a very interesting and hot topic, and the opportunity to get some info from the horse's mouth is not to be missed. Fire away, gnl!
#### gnl
21
0
Lattice QCD
One of the most interesting field theories to be studied on the lattice is QCD. QCD is a very complicated theory, with many non-perturbative aspects. The lattice offers a way to investigate, from first principles, such aspects. In the low-energy regime, the QCD coupling becomes too large for any perturbative expansion to make sense. Confinement and hadron structure are among the things one can study in Lattice QCD: hadron masses (QCD spectroscopy in general, including glueballs), hadronic matrix elements.
A good intro can be found in:
hep-lat/9807028
Agreement with experiment has been striking in many cases.
Staff Emeritus
Gold Member
Dearly Missed
6,691
5
I am working my way through the tutorial, and I wondered, gnl what is your topic? And are you going to be doing monte carlo estimations of path integrals like it says?
#### gnl
21
0
my field
My field of research, so far, has been lattice QCD. I have done works on hadron spectroscopy and on the study of leptonic anc semileptonic decays. These decays involve some non-perturbative quantity, like decay constants or form factors.
These objects are calculated as MC estimates (numerical path integral!) of time-ordered products of fields. For example, given the operator that creates a meson with given quantum numbers from the vacuum, one that creates another meson , and a current, lots of things can be calculated.
Lattice QCD needs BIG CPUS!!! However, lots of interesting physics can still be explored with scalar models. The Higgs boson, after all, is such a field!
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• Solo and co-op problem solving | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8599304556846619, "perplexity": 2395.924060448368}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912205597.84/warc/CC-MAIN-20190326160044-20190326182044-00014.warc.gz"} |
https://www.semanticscholar.org/author/Yonglin-Cao/1781136 | # Yonglin Cao
• Yonglin Cao
• 2011
For R a Galois ring and m 1, . . . , m l positive integers, a generalized quasi-cyclic (GQC) code over R of block lengths (m 1, m 2, . . . , m l ) and length $${\sum_{i=1}^lm_i}$$ is an R[x]-submodule of $${R[x]/(x^{m_1}-1)\times\cdots \times R[x]/(x^{m_l}-1)}$$ . Suppose m 1, . . . , m l are all coprime to the characteristic of R and let {g 1, . . . , g t(More)
• 2
• Yonglin Cao
• 2012
Let R be an arbitrary commutative finite chain ring with $$1\ne 0$$ . 1-generator quasi-cyclic (QC) codes over R are considered in this paper. Let $$\gamma$$ be a fixed generator of the maximal ideal of R, $$F=R/\langle \gamma \rangle$$ and $$|F|=q$$ . For any positive integers m, n satisfying $$\mathrm{gcd}(q,n)=1$$ , let \mathcal{R}_n=R[x]/\langle(More)
• 1
• 2015
Keywords: Additive cyclic code Galois ring Linear code Dual code Trace inner product Self-dual code Quasi-cyclic code a b s t r a c t Let R = GR(p ϵ , l) be a Galois ring of characteristic p ϵ and cardinality p ϵl , where p and l are prime integers. First, we give a canonical form decomposition for additive cyclic codes over R. This decomposition is used to(More)
• 2015
In this paper, we study the construction of cyclic DNA codes by cyclic codes over the finite chain ring 2 4 1 F u u . First, we establish a 1-1 correspondence between DNA pairs and the 16 elements of the ring 2 4 1 F u u . Considering the biology features of DNA codes, we investigate the structure and properties of self-reciprocal(More)
• 2015
Let $\mathbb{F}_{p^m}$ be a finite field of cardinality $p^m$ and $R=\mathbb{F}_{p^m}[u]/\langle u^2\rangle=\mathbb{F}_{p^m}+u\mathbb{F}_{p^m}$ $(u^2=0)$, where $p$ is a prime and $m$ is a positive integer. For any $\lambda\in \mathbb{F}_{p^m}^{\times}$, an explicit representation for all distinct $\lambda$-constacyclic codes over $R$ of length $p^sn$ is(More) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9686315059661865, "perplexity": 714.6627973804013}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818691977.66/warc/CC-MAIN-20170925145232-20170925165232-00536.warc.gz"} |
https://dml.cz/handle/10338.dmlcz/108519 | # Article
Full entry | PDF (0.6 MB)
References:
[1] Atkinson V. V.: On second-order non-linear oscillations. Pacific J. Math. 5 (1955), 643-647. MR 0072316 | Zbl 0065.32001
[2] Belohorec S.: Oscillatory solutions of certain nonlinear differential equations of the second order. (Slovak), Mat.-Fyz. Casopis Sloven. Akad. Vied. 11 (1961) 4, 250-255.
[3] Belohorec S.: On some properties of the equation $y" + f(x) y(x) = 0$, $0 < a < 1$. Mat. Časopis Sloven. Akad. Vied. 17 (1967) 1, 10-19. MR 0214854
[4] Coffman C. V., Ullrich D. F.: On the continuation of solutions of a certain nonlinear differential equation. Monatsh, Math., 71 (1967), 385-392. MR 0227494
[5] Gollwitzer H. E.: On nonlinear oscillations for a second order delay equation. J. Math. Anal. Appl., 26 (1969), 385-389. MR 0239224 | Zbl 0169.11401
[6] Gollwitzer H. E.: Nonlinear second order differential equations and Stieltjes integrals. University of Tennessee Report, 1969.
[7] Hastings S. P.: Boundary value problems in one differential equation with a discontinuity. J. of Differential Equations, 1 (1965), 346-369. MR 0180723 | Zbl 0142.06303
124
[8] Heidel J. W: A short proof of Atkinson's oscillation theorem. SI AM Review, in press.
[9] Heidel J. W: Rate of growth of non-oscillatory solutions of $y" + q(t)y^\gamma = 0$, $0 < y < 1$. to appear.
[10] Izjumova D. V., Kiguradze I. T: Certain remarks on solutions of the equation $u" + + a(t)f(u) = 0$. Differencialnye Uravenija 4 (1968), 589-605 (Russian). MR 0227544
[11] Moore R. A., Nehari Z.: Non-oscillation theorems for a class of nonlinear differential equations. Trans. Amer. Math. Soc. 93 (1959), 30-52. MR 0111897
[12] Nehari Z.: On a class of nonlinear second order differential equations. Trans. Amer. Math. Soc. 95 (I960), 101-123. MR 0111898 | Zbl 0097.29501
[13] Wong J. S. W: On second order nonlinear oscillation. Funkcialaj Ekvacioj, 11 (1968), 207-234. MR 0245915 | Zbl 0157.14802
Partner of | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9235695600509644, "perplexity": 4015.963235233367}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00728.warc.gz"} |
https://www.physicsforums.com/threads/arithmetic-series-and-triangular-numbers.185566/ | # Arithmetic Series and Triangular Numbers
1. Sep 18, 2007
### ramsey2879
Re: Arithmetic Series and Factors of Triangular Numbers
A+C*n, B+F*n, A+E*n, B+D*n are all arithmetic series which I define below
It is still not clear to me as why
(A+C*n)*(B+F*n) and (A+E*n)*(B+D*n) are both triangular numbers for all integers n
Can someone please visit my blog and explain it?
Perhaps someone can show somehow that
$$8*(A+C*n)*(B+F*n) + 1$$ and $$8*(A+E*n)*(B+D*n) + 1$$ are both perfect squares for all n using Mathematica or some other program.
--- In [email protected], "ramsey2879"
<ramseykk2@...> wrote:
>
> I found a previously unknow property of Triangular Numbers AFAIK.
> Given that T is a triangular number having exactly M distinct ways
to
> pair the product into two factors, A*B with A </= B. For each of
these
> M distinct pairs, there are two coprime pairs of integers (C,E) and
> (E,D)where C*D = the perfect square of integral part of the square
root
> of 2*T and E*F = the next higher perfect square and each of the
> products (A+Cn)*(B+Fn) and (A+En)*(B+Dn) are triangular numbers for
all
> integer values of n.
> As an example.
> Let T = 666 which can be factored into 6 distinct pairs A,B. The
six
> sets (A,B,C,D,E,F) are as follows
>
> 1. (1,666,1,1369,2,648)
> 2. (2,333,1,1369,8,162)
> 3. (3,222,1,1369,18,72)
> 4. (6,111,1,1369,72,18)
> 5. (9,74,1,1369,162,8)
> 6. (18,37,1,1369,648,2)
>
> I don't have a proof of the general result but am working on it.
>
Still no proof, however, let ab = T(r) = r(r+1)/2, gcd(n,m) = the
greatest common divisor of m and n, then the formula for C,D,E and F
as a function of A and B is
Case 1, r is even
C = (gcd(A,r+1))^2, F = 2*(gcd(B,r))^2
E = 2*(gcd(A,r))^2, D = (gcd(B,r+1))^2
The determinant
|C F|
|E D| = (r+1)^2 - r^2 = 2r+1
Case 2 r is odd
C = 2*(gcd(A,r+1))^2, F = (gcd(B,r))^2
E = (gcd(A,r))^2, D = 2*(gcd(B,r+1))^2
The determinant
|C F|
|E D| = (r+1)^2 - r^2 = 2r+1
Can you offer guidance or do you also need help?
Draft saved Draft deleted
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https://brilliant.org/problems/quadratic-binomial-fusion-2/ | # Quadratic Binomial Fusion
Algebra Level 5
$\displaystyle 1 + \sum_{r = 1}^{10} \left [ 3^r \binom{10}{r} + r \binom{10}{r} \right ] = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$
Consider the above summation, where $\alpha, \beta \in \mathbb N$ and $f(x) = x^2 - 2x -k^2 + 1$.
If $\alpha, \beta$ lies strictly in between the roots of $f(x) = 0$, then find the smallest positive integral value of $k$.
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9913963079452515, "perplexity": 416.37315108307}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153899.14/warc/CC-MAIN-20210729234313-20210730024313-00598.warc.gz"} |
http://math.stackexchange.com/questions/201823/show-p-1pn-1-equiv-1-mod-pn-for-n-in-mathbb-n | Show $[(p-1)!]^{p^{n-1}} \equiv -1$ (mod $p^n$) for n $\in \mathbb N$
Show $[(p-1)!]^{p^{n-1}} \equiv -1$ (mod $p^n$) for n $\in \mathbb N$, by induction. p a prime and p>2.
I can't seem to prove the inductive step for this. Would appreciate help.
My approach has been:
n=1 is just from Wilson.
Assume true for n=m: $[(p-1)!]^{p^{m-1}} \equiv -1$ (mod $p^m$)
Then,
$[(p-1)!]^{p^{(m+1)-1}} = ([(p-1)!]^{p^{m-1}})^p \equiv (-1)^p \equiv -1$ (mod $p^m$)
But, how do I get this to say anything in terms of mod $p^{m+1}$? Since I need the RHS to end up as: -1 (mod $p^{m+1}$).
One thing I could draw from this congruence is that $[(p-1)!]^{p^{(m+1)-1}}$ is not a multiple of p, since multiples of p must be greater than -1 apart from each other.
Hence, GCD($[(p-1)!]^{p^{(m+1)-1}}, p^{m+1})=1.$ I wasn't sure how I might use this.
Alternatively, I could express it as: $[(p-1)!]^{p^{(m+1)-1}} = [(p-1)!]^{p^m}$. But this didn't seem the right way to go about it, since by cancelling the -1+1, there doesn't seem to be any way to use the inductive hypothesis/assumption above.
Another useful result might be that: GCD((p-1)!,$p^{m+1}$)=1.
-
Hint: If $x\equiv -1 \pmod {p^{m-1}}$ show that $x^p\equiv -1 \pmod {p^m}$ when $p$ is an odd prime, when $m>1$ – Thomas Andrews Sep 24 '12 at 20:25
Hint: More generally, if $x\equiv -1\pmod {p^{m-1}}$ then $x^p \equiv -1 \pmod {p^m}$ for $m>1$ and $p$ an odd prime.
-
Thanks @Thomas. how do you derive this result? – confused Sep 24 '12 at 20:48
Showing this is roughly the same as Martini's answer, namely, that if $x=pk-1$ then using the binomial theorem. – Thomas Andrews Sep 24 '12 at 20:53
Am I able to do it without the binomial though? The notes that I saw this problem in don't touch on binomial expansion. – confused Sep 24 '12 at 20:55
Alternatively, you could write $x^p+1 = (x+1)(x^{p-1}-x^{p-2}+x^{p-3}...)$ and note that the terms of $x^{p-1}-x^{p-2}+...$ have to add up to something divisible by $p$. – Thomas Andrews Sep 24 '12 at 20:55
In this case, couldn't (x+1) be divisible by p instead? – jack Sep 24 '12 at 21:04
$(p-1)!^{p^{n-1}} = kp^n - 1$ for some $k\in \mathbb Z$ by induction hypothesis, so \begin{align*} (p-1)!^{p^n} &= \bigl((p-1)!^{p^{n-1}}\bigr)^p\\ &= (kp^n - 1)^p\\ &= \sum_{l=0}^p \binom{p}l (kp^n)^l(-1)^{p-l}\\ &= (-1)^p + p \cdot kp^n(-1)^{p-1} + k^2p^{2n}\sum_{l=2}^p \binom pl (kp^n)^{l-2}(-1)^{p-l}\\ &\equiv (-1)^p + 0 + 0\\ &= -1 \pmod{p^{n+1}} \end{align*} and we are done.
-
I don't really understand the use of combinations like this. – confused Sep 24 '12 at 20:48
Better now?${}{}$ – martini Sep 24 '12 at 21:01
Thanks. @martini I can follow it better now. In the term, $k^2p^{2n}\sum_{l=2}^p \binom pl (kp^n)^{l-2}(-1)^{p-l}$, wouldn't some of the terms of this sum be non-integer fractions, since $\binom pl$ can be a non-integer fraction? Won't this create a problem when working modulo $p^{n+1}$? – confused Sep 24 '12 at 21:36
$\binom{n}{k}$ is always an integer, since it is the number of ways to choose a committee of $k$ people from a group of $n$ people. – André Nicolas Sep 24 '12 at 21:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9675956964492798, "perplexity": 485.67573705653626}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398445080.12/warc/CC-MAIN-20151124205405-00048-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.studypug.com/ca/differential-equations/step-functions | # Step functions
## Step Functions
What is a step function?
The main characteristic of a step function, and the reason why it truly looks like a staircase doodle, is that this function happens to be constant in intervals. These intervals do not have the same value and we end up with a function which "jumps" from one value to the next (following its own conditions) in a certain pattern. Every time the function jumps to a new interval it has a new constant value and we can observe the "steps" on its graphic representation as horizontal lines. Take a look at the graphic representation below for a typical step function where is plain to see how the name of the function came about.
Notice from the figure above that we can also define a step function as a piecewise function of constant steps, meaning that the function is broken down in a finite number of pieces, each piece having a constant value, in other words, each piece is a constant function.
When talking about step functions we cannot forget to talk about the Heaviside function also known as the unit step function. This is a function that is defined to have a constant value of zero up to a certain point on t (the horizontal axis) at which it jumps to a value of 1. This particular value at which the function jumps (or switches from zero to one) is usually taken as the origin in the typical coordinate system representation, but it can be any value c in the horizontal axis.
Consequently, the Heaviside step function is defined as:
Where c represents the point at t in which the function goes from a 0 value to a value of 1. So, if we want to write down the mathematical expression of the heaviside function depicted in figure 3, we would write it as: u3(t). Notice how we only wrote the math expression using the first type of notation found in equation 1, this is because that notation happens to be the most commonly used and is also the one we will continue to use throughout our lesson. Is still important though, that you know about the other notations and keep them in mind just in case you find them throughout your studies.
From its definition, we can understand why the Heaviside function is also called the "unit step" function. As it can be observed, a Heaviside function can only have values of 0 and 1, in other words, the function is always equal to zero before arriving to a certain value t=c at which it "turns on" and jumps directly into having a value of 1, and so, it jumps in a step size of one unit.
Notice how we have used the phrase "turning on" to describe the process of the unit step function to jump from a zero value to a unit value, this is a very common way to refer to Heaviside step functions' behavior and interestingly enough, it is all due to their real life usage and comes from the reason why they were invented. Historically, physicist, self-taught engineer and mathematician Oliver Heaviside, invented Heaviside step functions in order to describe the behaviour of a current signal when you click the switch on of an electric circuit, thus, allowing you to calculate the magnitude of the current from zero when the circuit is off, to a certain value when is tuned on.
There is an important note to clarify here. Electric current does not magically jump from a zero value to a high value. When having a constant current through a circuit, we know that this constant value obviously started off from zero from when the circuit was off and then it arrived until a certain value after gradually increasing in time, the thing is that electric current travels at a very high speed, and so it is impossible for us (in a simple everyday life setting) to see this gradual increase of the current from zero to its final value since it happens in a very short instant of time, and so, we take it as a "jump" from one value to the next and describe it accordingly in graphic representations.
## Heaviside function properties
Although the Heaviside function itself can only have the values of 0 or 1 as mentioned before, this does not mean we cannot obtain a graphic representation of a higher jump using Heaviside step functions. It is actually a very simple task to obtain a higher value since you just need to multiply it for any constant value that you want to have as the size jump, in other words, if you have a step function written as 3u5(t) in here you have a step function which has a value of zero until it gets to a value of t=5, and which point, the function has a final value of 3. This can be seen in the figure below:
One of the greatest properties of Heaviside step functions is that they allow us to model certain scenarios (like the one described for current in an on/off circuit) and mathematically be able to solve for important information on these scenarios. Such cases tend to require the use of differential equations and so here we have yet another tool to solve them.
Being this a course on differential equations, the most important point of this lesson is to give an introduction to a function which will aid in the solution of certain differential equations, such tool will be used along others seen before, such as the Laplace transform. This will serve to come up with important formulas to be used and to be prepared for the next lesson in which you will be solving differential equations with step functions.
We will talk a bit more about this on the last section of this lesson, meanwhile for a review on the definition of the unit step function and a list of a few of its identities, we recommend the next Heaviside step function article.
## Heaviside step function examples
Let us take a look into an example in which you will have to write all of the necessary unit step functions in order to completely describe the graphic representation found in figure 5.
#### Example 1
Write the mathematical expression for the following graph in terms of the Heaviside step function:
Let's divide this in parts so we can see how the functions of the graph behave at each different value given. And so, we will write a separate expression for each of the next pieces: for t < 3, for t=3 to t=4, for t=4 to t=6 and for t>6.
• For t < 3: Notice that this is a regular Heaviside function in which c=0, multiplied by a 2 in order to obtain the jump of 2 units in size. This would fit the requirement of the function being zero for all negative values of t, and then to have a value of 2 for values of t from 0 to 3. And so, the expression is: 2uo(t)=2.
• For t=3 to t=4:
In this range we have to cancel the unit step function that we had before, that means we need a negative unit step function in here but this one will start to be applied at t=3 and will have to be multiplied by 2 again in order to cancel the value of the previous expression. Thus, our expression for this part of the function is: -2u3(t).
• For t=4 to t=6:
If we weren't to add any function at t=4, the value of the function would remain as zero to infinity since the second function cancelled the first one, but instead, we see in the graph that we have a diagonal line increasing one unit step size for each one unit of distance traveled on t.
Thus, since the function is increasing at the same rate as t, we could easily multiply a new unit step function which starts at 4 by t and be over with, this would produce a diagonal line following the same behavior. The problem is that just multiplying u4(t) by t would produce a line that would come out of the origin instead of t=4, and for that, we need to multiply the unit step function by (t-4) so the function starts at the same time as the unit step function will be applied (which is at t=4). And so the expression is: (t-4)u4(t).
Notice (t-4)u4(t) produces the values for y of: 0 (when t=4), 1 (when t=5) and 2 (when t=6), which is what the graph requires.
• For t > 6:
For this last piece of the graph should be already easy that we just need to cancel our last function in order to have a value of zero back to the graph. For that we use a negative unit step function which starts at t=6, and which should be multiplied again by (t-4). And so, the expression to cancel our last one and completes the graph is: -(t-4)6(t).
We add all of the four pieces of function we found to produce the expression that represents the whole graph shown in figure 5:
Due to this particular example having multiple step function samples, we continue onto the next section to work on more complicated problems.
If you would like to continue practicing how to write down Heaviside step functions, we recommend you to visit these notes of step functions for more Heaviside function examples along with a little introduction. Notice these notes also introduce the topic for our next section: unit step function Laplace transforms and the overall use of the Laplace transform when associated to Heaviside functions.
# Laplace transform of Heaviside function
You have already had an introduction to the Laplace transform in recent past lessons, still, at this time we do recommend you to give the topic a review if you think it appropriate or necessary. The lesson on calculating Laplace transforms is of special use for you to be prepared for this section.
Let us continue with the Heaviside step function and how we will use it along the Laplace transform. The Laplace transform will help us find the value of y(t) for a function that will be represented using the unit step function, so far we have talked about step functions in which the value is a constant (just a jump from zero to a constant value, producing a straight horizontal line in a graph) but we can have any type of function to start at any given point in time (which is what we represent with t mostly).
What are we saying? Well, think on the graphic representation of a function, you can have any function, with any shape, but this special case comes from the fact that the function will be zero through time, until a certain point in which the signal will turn on, and then it will "jump" into this "any shape" function behavior. This is what we call a "shifted function" and this is because we can think of any regular function graphed, and then saying "oh but we want this function to start happening later" and we "switch" it to start at a later point in time (at t=c).
Since these shifted functions will be equal to zero until at certain point c in which they will "turn on", they can be represented as:
The shifted function then is defined as:
Now let's see what happens when we take the Laplace transform of a shifted function!
First, remember that the mathematical definition of the Laplace transform is:
Therefore the Laplace transform for the shifted function is:
Notice how the Laplace transform gets simplified from an improper integral to a regular integral where you have dropped the unit step function. The reason for this is that although the range of the whole Laplace transform integral is from 0 to infinity, before c (whatever value c has) the unit step function is equal to zero, and therefore the whole integral goes to zero. After c, the unit step function has a value of 1, and thus, we can just take it as a constant value of 1 multiplying the rest of the integral, which range is not from c to infinity.
Continuing with the simplification of the Laplace transform of the shifted function, we set x = t - c, which means that t = x+c and so the transformation looks as follows
By making the integral to be in relation to one single variable rather that a t-c term, we have simplified this transformation so we could obtain a quickly manageable formula we can readily use with future problems and our already known table of Laplace transforms from past lessons.
Having solved equation 5 makes it easier to obtain another important formula: the unit step function Laplace transform. Notice, not the transform for a shifted function, but the Laplace transform of the unit step function (Heaviside function) itself and alone.
If you notice, equation 5 was useful while obtaining equation 6 because taking the Laplace transformation of the Heaviside function by itself can be taken as having a shifted function in which the f(t-c) part equals to 1, and so you end up with the Laplace transform of a unit step function times 1, which results in the simple and very useful formula found in equation 6.
Now let us finish this lesson by working on some transformation of a unit step function examples.
#### Example 2
Find the Laplace transform of each of the following step functions:
• Applying the Laplace transform to the function and using equations 5 and 6 we obtain:
• Using equation 5, we set x=t-c (for this case x=t-5) and work through the transformation on the second term of the last equation:
Notice that in order to solve the last term, we used the method of comparison with the table of Laplace transforms. You can find such table on the past lessons related to the Laplace transform.
• Now let us solve the third transformation (remember we set x = t-c, which in this case is x=t-7):
• Now we put everything together to form the final answer to the problem:
#### Example 3
Find the Laplace transform for each of the following functions in g(t):
• Now we use equation 5 and set up x=t-c to solve the Laplace transform:
• We separate the two terms found on the right hand side of the equation, and solve for the first one (for this case x=t-) using the trigonometric function: sin(a+b)=sin(a)cos(b)+cos(a)sin(b).
Therefore:
• Now let us solve the second term, where x = t-4, therefore, t=x+4:
• Putting the whole result together:
And now we are ready for our next section where we will be solving differential equations with what we learned today. If you would like to see some extra notes on the Heaviside function and its relation with another function which we will study in later lessons, the Dirac delta function, visit the link provided.
### Step functions
#### Lessons
A Heaviside Step Function (also just called a "Step Function") is a function that has a value of 0 from 0 to some constant, and then at that constant switches to 1.
The Heaviside Step Function is defined as,
The Laplace Transform of the Step Function:
$L${$u_{c}(t)$ $f(t - c)$} = $e^{-sc}$$L${$f(t)$}
$L${$u_{c}(t)$} = $\frac{e^{-sc}}{s}$
These Formulae might be necessary for shifting functions:
$\sin{(a + b)} = \sin(a)\cos(b) + \cos(a)\sin(b)$
$\cos{(a + b)} = \cos(a)\cos(b) - \sin(a)\sin(b)$
$(a + b)^{2} = a^{2} + 2ab +b^{2}$
• Introduction
a)
What is the Heaviside Step Function?
b)
What are some uses of the Heaviside Step Function and what is the Laplace Transform of a Heaviside Step Function?
• 1.
Determining Heaviside Step Functions
Write the following graph in terms of a Heaviside Step Function
• 2.
Determining the Laplace Transform of a Heaviside Step Function
Find the Laplace Transform of each of the following Step Functions:
a)
$f(t) = 6u_{3}(t) - e^{3t - 15}u_{5}(t) + 3(t - 7)^{2}u_{7}(t)$
b)
$g(t) = -\sin{(t)}u_{\pi}(t) + 2t^{2}u_{4}(t)$
• 3.
Determining the Inverse Laplace Transform of a Heaviside Step Function
Find the inverse Laplace Transform of the following function:
$F(s) = \frac{4e^{-3s}}{(s - 2)(s + 3)}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8628491163253784, "perplexity": 239.58458368028974}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655890105.39/warc/CC-MAIN-20200706042111-20200706072111-00087.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_24&diff=prev&oldid=33127 | # Difference between revisions of "2008 AMC 12B Problems/Problem 24"
## Problem 24
Let . Distinct points lie on the -axis, and distinct points lie on the graph of . For every positive integer is an equilateral triangle. What is the least for which the length ?
## Solution
Let . We need to rewrite the recursion into something manageable. The two strange conditions, 's lie on the graph of and is an equilateral triangle, can be compacted as follows: which uses , where is the height of the equilateral triangle and therefore times its base.
The relation above holds for and for , so Or, This implies that each segment of a successive triangle is more than the last triangle. To find , we merely have to plug in into the aforementioned recursion and we have . Knowing that is , we can deduce that .Thus, , so . We want to find so that . is our answer. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9638452529907227, "perplexity": 499.9830878608061}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570741.21/warc/CC-MAIN-20220808001418-20220808031418-00586.warc.gz"} |
http://lists.gnu.org/archive/html/lilypond-user/2006-10/msg00318.html | lilypond-user
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## error in manual?
From: Julian Peterson Subject: error in manual? Date: Thu, 19 Oct 2006 14:36:15 -0400 User-agent: Thunderbird 1.5.0.7 (Windows/20060909)
I think I found an error in the lilypond documentation for 2.8.x, although as I'm still quite new to lilypond there is a good chance that it's my own misunderstanding...
```
Section 4.5 contains this snippet:
```
```
```
Which seems to contain an extra # preceding \$padding. I had to remove it to get this example to work, and further examples seem to be consistent with using only a \$.
```
Thanks,
Julian Peterson
``` | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8418446779251099, "perplexity": 2832.9282685017456}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917126538.54/warc/CC-MAIN-20170423031206-00453-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://asmedigitalcollection.asme.org/appliedmechanics/article-abstract/51/3/636/423229/Thermoelastic-Displacements-and-Stresses-Due-to-a?redirectedFrom=PDF | A solution is given for the surface displacement and stresses due to a line heat source that moves at constant speed over the surface of an elastic half plane. The solution is obtained by integration of previous results for the instantaneous point source. The final results are expressed in terms of Bessel functions for which numerically efficient series and asymptotic expressions are given.
This content is only available via PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9342133402824402, "perplexity": 203.2964497105308}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00580.warc.gz"} |
http://www.wias-berlin.de/publications/wias-publ/run.jsp?template=abstract&type=Preprint&year=1997&number=326 | WIAS Preprint No. 326, (1997)
# Twice degenerate equations in the space of vector-valued functions
Authors
• Kloeden, Peter E.
• Krasnosel´skii, Alexander M.
2010 Mathematics Subject Classification
• 47H11 47H30
Keywords
• index at infinity, asymptotically linear and asymptotically homogeneous vector fields, periodic problems, system of two nonlinear first-order ODE, nonlinear second-order ODE
Abstract
New results are suggested which allow to calculate an index at infinity for asymptotically linear and asymptotically homogeneous vector fields in spaces of vector-valued functions. The case is considered where both linear approximation at infinity and "linear + homogeneous" approximation are degenerate. Applications are given to the 2π-periodic problem for a system of two nonlinear first order ODE's and to the two-point BVP for a system of two nonlinear second order ODE's. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9143341183662415, "perplexity": 1507.6458668558853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540503656.42/warc/CC-MAIN-20191207233943-20191208021943-00180.warc.gz"} |
https://datascience.stackexchange.com/questions/29695/batch-norm-why-the-initial-normalization | # Batch norm: why the initial normalization?
I'm a beginner in NNs and the first thing I don't understand with batch norm is the following 2 steps:
• First we normalize the batch data on a z parameter to Mu=0, sigma^2=1
• Then we change z via the coefficients of Mu, sigma^2 (usu. alpha, beta) by updating them as learnable parameters.
I don't understand why the first step is necessary if we change the distribution in the second step anyway. Could someone explain please? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.950867772102356, "perplexity": 1428.5015449507548}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00528.warc.gz"} |
https://www.physicsforums.com/threads/how-to-reduce-fermi-dirac-to-maxwell-boltzmann-in-a-solid.388486/ | # How to reduce fermi-dirac to maxwell-boltzmann in a solid?
1. Mar 21, 2010
### kof9595995
For indistinguishable particles we use fermi-dirac(FD) or bose-einstein(BE), and for distinguishable we use maxwell-boltzmann(MB).For the distinguishable case our prof gave us the example of atoms in solid, because the positions of the atoms are fixed, so they are distinguishable, thus satisfy MB statistics.
But the so called "fixed" i think is just extremely narrow wave packets with very small overlaps, so in the strict sense atoms in solid are still indistinguishable, then how can we reduce FD or BE to MB, from the condition "wave packets are narrow with small overlaps".
Like in the treatment of dilute gas, when we assume the occupation number is much smaller than the number of degeneracies for each energy level, from the math FD and BE reduce to MB nicely. So it puzzled me whether for the atoms in solid we can find a nice way to reduce to MB.
2. Apr 22, 2010
### kof9595995
Well, no reply for a long time...Maybe I did not state the question clear enough, let me try again:
For atoms in solid, the overlaps between wavefunctions are so small that we can treat them as distinguishable, hence the MB statistics.
But let's take a alternative point of view: atoms in solid are actually indistinguishable, so it's always correct to use FD statistics (let say they're fermions), but we said we could use MB because overlaps are small, which means that $${\Omega _{MB}} \approx {\Omega _{FD}}$$ when overlaps are small. So how do we get "$${\Omega _{MB}} \approx {\Omega _{FD}}$$" from "overlaps are small"
For reference:
MB:$${\Omega _{MB}} = \prod\limits_{j = 1}^n {\frac{{g_j^{{N_j}}}}{{{N_j}!}}}$$
FD:$${\Omega _{FD}} = \prod\limits_{j = 1}^n {\frac{{{g_j}!}}{{{N_j}!\left( {{g_j} - {N_j}} \right)!}}}$$
Where gj is the degeneracy for different energies and Nj is the occupation number | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8958027362823486, "perplexity": 1049.6381151041412}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267155924.5/warc/CC-MAIN-20180919043926-20180919063926-00556.warc.gz"} |
https://www.codecademy.com/learn/math-ds-probability/modules/math-ds-rules-of-probability/cheatsheet | Rules of Probability
Union
The union of two sets encompasses any element that exists in either one or both of them. We can represent this visually as a venn diagram as shown. Union is often represented as:
$(A\ or\ B)$
Intersection
The intersection between two sets encompasses any element that exists in BOTH sets and is often written out as:
$(A\ and\ B)$
If there are two events, A and B, the addition rule states that the probability of event A or B occurring is the sum of the probability of each event minus the probability of the intersection:
$P(A\ or\ B) = P(A) + P(B) - P(A\ and\ B)$
If the events are mutually exclusive, this formula simplifies to:
$P(A\ or\ B) = P(A) + P(B)$
Multiplication Rule
The multiplication rule is used to find the probability of two events, A and B, happening simultaneously. The general formula is:
$P(A \text{ and } B) = P(A) \cdot P(B \mid A)$
For independent events, this formula simplifies to:
$P(A \text{ and } B) = P(A) \cdot P(B)$
This is because the following is true for independent events:
$P(B \mid A) = P(B)$
The tree diagram shown displays an example of the multiplication rule for independent events.
Complement
The complement of a set consists of all possible outcomes outside of the set.
Let’s say set A is rolling an odd number with a 6-sided die: {1, 3, 5}. The complement of this set would be rolling an even number: {2, 4, 6}.
We can write the complement of set A as AC. One key feature of complements is that a set and its complement cover the entire sample space. In this die roll example, the set of even numbers and odd numbers would cover all possible rolls: {1, 2, 3, 4, 5, 6}.
Independent Events
Two events are independent if the occurrence of one event does not affect the probability of the other one occurring.
Let’s say we have a bag of five marbles: three are red and two are blue. If we select two marbles out of the bag WITH replacement, the probability of selecting a blue marble second is independent of the outcome of the first event.
The diagram below outlines the independent nature of these events. Whether a red marble or a blue marble is chosen randomly first, the chance of selecting a blue marble second is always 2 in 5.
Dependent Events
Two events are dependent if the occurrence of one event does affect the probability of the other one occurring.
Let’s say we have a bag of five marbles: three are red and two are blue. If we select two marbles out of the bag WITHOUT replacement, the probability of selecting a blue marble second depends on the outcome of the first event.
The diagram below outlines this dependency. If a red marble is randomly selected first, the chance of selecting a blue marble second is 2 in 4. Meanwhile, if a blue marble is randomly selected first, the chance of selecting a blue marble second is 1 in 4.
Mutually Exclusive Events
Two events are considered mutually exclusive if they cannot occur at the same time. For example, consider a single coin flip: the events “tails” and “heads” are mutually exclusive because we cannot get both tails and heads on a single flip.
We can visualize two mutually exclusive events as a pair of non-overlapping circles. They do not overlap because there is no outcome for one event that is also in the sample space for the other.
Conditional Probability
Conditional probability is the probability of one event occurring, given that another one has already occurred. We can represent this with the following notation:
\begin{aligned} \text{Probability of event A occurring given event B has occurred} \\ P(A \mid B) \\ \end{aligned}
For independent events, the following is true for events A and B:
\begin{aligned} P(A \mid B) = P(A) \\ \text{and} \\ P(B \mid A) = P(B) \\ \end{aligned}
Bayes’ Theorem
Bayes’ theorem is a useful tool to find the probability of an event based on prior knowledge. The formula for Bayes’ theorem is:
$P(B \mid A) = \frac{P(A \mid B) \cdot P(B)}{P(A)}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9782246351242065, "perplexity": 221.66015876726047}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00746.warc.gz"} |
http://mathhelpforum.com/pre-calculus/153838-graphing-parabola.html | # Math Help - graphing this parabola
1. ## graphing this parabola
4y^2-4y-4x+5=0
this is what i did:
4y^2-4y=4x-5
4y^2-4y=4x-5+4
4^2-4y+4=4x-5+4
4(y^2-2y+2)=4x-1
4(y-2)^2=4x-1
i don't really know how to move to the next step in order to get something like (y-k)^2=4p(x-h) or (x-h)^2=4p(y-k)
it doesn't make sense cause i have the 4 outside which seems like its part of the 4p but (y-2) is squared.... and i can't complete the square on the right side since no x^2 value, only x.
2. Originally Posted by Frenchie
4y^2-4y-4x+5=0
this is what i did:
4y^2-4y=4x-5
4y^2-4y=4x-5+4
4^2-4y+4=4x-5+4
4(y^2-2y+2)=4x-1
4(y-2)^2=4x-1
i don't really know how to move to the next step in order to get something like (y-k)^2=4p(x-h) or (x-h)^2=4p(y-k)
it doesn't make sense cause i have the 4 outside which seems like its part of the 4p but (y-2) is squared.... and i can't complete the square on the right side since no x^2 value, only x.
Sorry, this is wrong. If you are going to complete the square, the coefficient of the square term must be 1.
$4y^2-4y-4x+5=0$
$4y^2 - 4y + 5 = 4x$
$4\left(y^2 - y + \frac{5}{4}\right) = 4x$
$4\left[y^2 - y + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{5}{4}\right] = 4x$
$4\left[\left(y-\frac{1}{2}\right)^2 + 1\right] = 4x$
$\left(y - \frac{1}{2}\right)^2 + 1 = x$.
You should be able to graph this now...
3. thank you! I have another question, is that equivalent to this:
(y-1/2)^2=(x-1)
see it doesn't make sense because whats P? there is nothing infront of (x-1) such as 8(x-1) where in that case we could say 4p(x-1), p=2
it doesn't look like this general form (y-k)^2=4p(x-h) =/
4. ah nvm i think its (y-1/2)^2=1/4(x-1)... or is it (y-1/2^2)=4(x-1)... eh w.e good enough
thank you
5. Originally Posted by Frenchie
thank you! I have another question, is that equivalent to this:
(y-1/2)^2=(x-1)
see it doesn't make sense because whats P? there is nothing infront of (x-1) such as 8(x-1) where in that case we could say 4p(x-1), p=2
"no" coefficient means the coefficient is 1. If 4p= 1, then p= 1/4.
it doesn't look like this general form (y-k)^2=4p(x-h) =/
But it is.
$(y- 1/2)^2= x- 1$ is the same as $(y- 1/2)^2= 4(1/4)(x- 1)$. p= 1/4, k= 1/2, and h= 1.
6. ah thank you all this is helpful | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8409505486488342, "perplexity": 1139.9345034111261}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398460982.68/warc/CC-MAIN-20151124205420-00008-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://www.ehow.com/video_12222323_rate-chance-calculus.html | # Rate of Chance in Calculus
Found This Helpful
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One-Sided Limits in Calculus....5
Rate of chance is a calculus term that typically applies to one of two things. Find out about rate of chance in calculus with help from an experienced math tutor in this free video clip.
Part of the Video Series: Calculus Explained
## Video Transcript
Hi there. This is Ryan Malloy here at the Worldwide Center of Mathematics. In this video, we're going to discuss the concept of the rate of change as it applies to calculus. So when you hear the term rate of change in calculus, there's typically one of two meanings that's being talked about. Let's say we've got some function f of x, we won't define it for now. There's two things we might be asked about it. The average rate of change over an interval, which is sometimes called the AROC, or average rate of change, and we have the instantaneous rate of change at a value. Sometimes called the IROC. So how do we express the average rate of change over an interval? If we are looking at the average rate of change on the interval from a to b where a and b are simply two numbers that are within the domain of our function. The average rate of change can be expressed quite simply as value of the function at a minus the value of the function at b divided by a minus b. And it's just that simply. The instantaneous rate of change is a little bit more complicated and it uses some more advanced techniques. So let's say that we want instantaneous rate of change at some value a. This is given by a limit. The limit as h approaches zero where h is just some arbitrary variable of f of a plus h minus f of a divided by h. Over here we can't simply plug in h directly since there'll be a zero in the denominator. But typically this limit is not very difficult to compute, and as a result, there are a number of rules and properties that have been well established as to how to do it quickly. For example, if we have f of x equals let's say x cubed. Instead of computing this limit for x cubed, we can simply use what's known as the power rule. And so if we want to find the instantaneous rate of change of f of a, sometimes indicated by f prime a. Well we'd simply take three x squared at a. Which just gives us three a squared. So for example if a were two, we get two squared is four times three is 12. My name is Ryan Malloy and we've just discussed the rate of change in calculus. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9363283514976501, "perplexity": 213.80012482436788}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375094501.77/warc/CC-MAIN-20150627031814-00192-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://formulasearchengine.com/wiki/Operator_(physics) | # Operator (physics)
In physics, an operator is a function over the space of physical states. As a result of its application on a physical state, another physical state is obtained, very often along with some extra relevant information.
The simplest example of the utility of operators is the study of symmetry. Because of this, they are a very useful tool in classical mechanics. In quantum mechanics, on the other hand, they are an intrinsic part of the formulation of the theory.
## Operators in classical mechanics
In classical mechanics, the movement of a particle (or system of particles) is completely determined by the Lagrangian ${\displaystyle L(q,{\dot {q}},t)}$ or equivalently the Hamiltonian ${\displaystyle H(q,p,t)}$, a function of the generalized coordinates q, generalized velocities ${\displaystyle {\dot {q}}=\mathrm {d} q/\mathrm {d} t}$ and its conjugate momenta:
${\displaystyle p={\frac {\partial L}{\partial {\dot {q}}}}}$
If either L or H are independent of a generalized coordinate q, meaning the L and H do not change when q is changed, which in turn means the dynamics of the particle are still the same even when q changes, the corresponding momenta conjugate to those coordinates will be conserved (this is part of Noether's theorem, and the invariance of motion with respect to the coordinate q is a symmetry). Operators in classical mechanics are related to these symmetries.
More technically, when H is invariant under the action of a certain group of transformations G:
${\displaystyle S\in G,H(S(q,p))=H(q,p)}$.
the elements of G are physical operators, which map physical states among themselves.
### Table of classical mechanics operators
where ${\displaystyle R({\hat {\boldsymbol {n}}},\theta )}$ is the rotation matrix about an axis defined by the unit vector ${\displaystyle {\hat {\boldsymbol {n}}}}$ and angle θ.
## Concept of generator
If the transformation is infinitesimal, the operator action should be of the form
${\displaystyle I+\epsilon A}$
where ${\displaystyle I}$ is the identity operator, ${\displaystyle \epsilon }$ is a parameter with a small value, and ${\displaystyle A}$ will depend on the transformation at hand, and is called a generator of the group. Again, as a simple example, we will derive the generator of the space translations on 1D functions.
As it was stated, ${\displaystyle T_{a}f(x)=f(x-a)}$. If ${\displaystyle a=\epsilon }$ is infinitesimal, then we may write
${\displaystyle T_{\epsilon }f(x)=f(x-\epsilon )\approx f(x)-\epsilon f'(x).}$
This formula may be rewritten as
${\displaystyle T_{\epsilon }f(x)=(I-\epsilon D)f(x)}$
where ${\displaystyle D}$ is the generator of the translation group, which in this case happens to be the derivative operator. Thus, it is said that the generator of translations is the derivative.
## The exponential map
The whole group may be recovered, under normal circumstances, from the generators, via the exponential map. In the case of the translations the idea works like this.
The translation for a finite value of ${\displaystyle a}$ may be obtained by repeated application of the infinitesimal translation:
${\displaystyle T_{a}f(x)=\lim _{N\to \infty }T_{a/N}\cdots T_{a/N}f(x)}$
with the ${\displaystyle \cdots }$ standing for the application ${\displaystyle N}$ times. If ${\displaystyle N}$ is large, each of the factors may be considered to be infinitesimal:
${\displaystyle T_{a}f(x)=\lim _{N\to \infty }(I-(a/N)D)^{N}f(x).}$
But this limit may be rewritten as an exponential:
${\displaystyle T_{a}f(x)=\exp(-aD)f(x).}$
To be convinced of the validity of this formal expression, we may expand the exponential in a power series:
${\displaystyle T_{a}f(x)=\left(I-aD+{a^{2}D^{2} \over 2!}-{a^{3}D^{3} \over 3!}+\cdots \right)f(x).}$
The right-hand side may be rewritten as
${\displaystyle f(x)-af'(x)+{a^{2} \over 2!}f''(x)-{a^{3} \over 3!}f'''(x)+\cdots }$
which is just the Taylor expansion of ${\displaystyle f(x-a)}$, which was our original value for ${\displaystyle T_{a}f(x)}$.
The mathematical properties of physical operators are a topic of great importance in itself. For further information, see C*-algebra and Gelfand-Naimark theorem.
## Operators in quantum mechanics
The mathematical formulation of quantum mechanics (QM) is built upon the concept of an operator.
The wavefunction represents the probability amplitude of finding the system in that state. The terms "wavefunction" and "state" in QM context are usually used interchangeably.
Physical pure states in quantum mechanics are represented as unit-norm vectors (probabilities are normalized to one) in a special complex vector space: a Hilbert space. Time evolution in this vector space is given by the application of the evolution operator.
Any observable, i.e., any quantity which can be measured in a physical experiment, should be associated with a self-adjoint linear operator. The operators must yield real eigenvalues, since they are values which may come up as the result of the experiment. Mathematically this means the operators must be Hermitian.[1] The probability of each eigenvalue is related to the projection of the physical state on the subspace related to that eigenvalue. See below for mathematical details.
In the wave mechanics formulation of QM, the wavefunction varies with space and time, or equivalently momentum and time (see position and momentum space for details), so observables are differential operators.
In the matrix mechanics formulation, the norm of the physical state should stay fixed, so the evolution operator should be unitary, and the operators can be represented as matrices. Any other symmetry, mapping a physical state into another, should keep this restriction.
### Wavefunction
{{#invoke:main|main}}
The wavefunction must be square-integrable (see Lp spaces), meaning:
${\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }|\psi ({\mathbf {r}})|^{2}{\rm {d}}^{3}{\mathbf {r}}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\psi ({\mathbf {r}})^{*}\psi ({\mathbf {r}}){\rm {d}}^{3}{\mathbf {r}}<\infty }$
and normalizable, so that:
${\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }|\psi ({\mathbf {r}})|^{2}{\rm {d}}^{3}{\mathbf {r}}=1}$
Two cases of eigenstates (and eigenvalues) are:
${\displaystyle |\psi \rangle =\sum _{i}c_{i}|\phi _{i}\rangle }$
where ci are complex numbers such that |ci|2 = ci*ci = probability of measuring the state ${\displaystyle |\phi _{i}\rangle }$, and has the corresponding set of eigenvalues ai is also discrete - either finite or countably infinite,
${\displaystyle |\psi \rangle =\int c(\phi ){\rm {d}}\phi |\phi _{i}\rangle }$
where c(φ) is a complex function such that |c(φ)|2 = c(φ)*c(φ) = probability of measuring the state ${\displaystyle |\phi \rangle }$, there is an uncountably infinite set of eigenvalues a.
### Linear operators in wave mechanics
{{#invoke:main|main}}
Let ψ be the wavefunction for a quantum system, and ${\displaystyle {\hat {A}}}$ be any linear operator for some observable A (such as position, momentum, energy, angular momentum etc.), then
${\displaystyle {\hat {A}}\psi =a\psi ,}$
where:
If ψ is an eigenfunction of a given operator A, then a definite quantity (the eigenvalue a) will be observed if a measurement of the observable A is made on the state ψ. Conversely, if ψ is not an eigenfunction of A, then it has no eigenvalue for A, and the observable does not have a single definite value in that case. Instead, measurements of the observable A will yield each eigenvalue with a certain probability (related to the decomposition of ψ relative to the orthonormal eigenbasis of A).
In bra–ket notation the above can be written;
{\displaystyle {\begin{aligned}&{\hat {A}}\psi ={\hat {A}}\psi (\mathbf {r} )={\hat {A}}\left\langle \mathbf {r} \mid \psi \right\rangle =\left\langle \mathbf {r} \mid {\hat {A}}\mid \psi \right\rangle \\&a\psi =a\psi (\mathbf {r} )=a\left\langle \mathbf {r} \mid \psi \right\rangle =\left\langle \mathbf {r} \mid a\mid \psi \right\rangle \\\end{aligned}}}
Due to linearity, vectors can be defined in any number of dimensions, as each component of the vector acts on the function separately. One mathematical example is the del operator, which is itself a vector (useful in momentum-related quantum operators, in the table below).
An operator in n-dimensional space can be written:
${\displaystyle \mathbf {\hat {A}} =\sum _{j=1}^{n}\mathbf {e} _{j}{\hat {A}}_{j}}$
where ej are basis vectors corresponding to each component operator Aj. Each component will yield a corresponding eigenvalue. Acting this on the wave function ψ:
${\displaystyle \mathbf {\hat {A}} \psi =\left(\sum _{j=1}^{n}\mathbf {e} _{j}{\hat {A}}_{j}\right)\psi =\sum _{j=1}^{n}\left(\mathbf {e} _{j}{\hat {A}}_{j}\psi \right)=\sum _{j=1}^{n}\left(\mathbf {e} _{j}a_{j}\psi \right)}$
in which
${\displaystyle {\hat {A}}_{j}\psi =a_{j}\psi .}$
In bra–ket notation:
{\displaystyle {\begin{aligned}&\mathbf {\hat {A}} \psi =\mathbf {\hat {A}} \psi (\mathbf {r} )=\mathbf {\hat {A}} \left\langle \mathbf {r} \mid \psi \right\rangle =\left\langle \mathbf {r} \mid \mathbf {\hat {A}} \mid \psi \right\rangle \\&\left(\sum _{j=1}^{n}\mathbf {e} _{j}{\hat {A}}_{j}\right)\psi =\left(\sum _{j=1}^{n}\mathbf {e} _{j}{\hat {A}}_{j}\right)\psi (\mathbf {r} )=\left(\sum _{j=1}^{n}\mathbf {e} _{j}{\hat {A}}_{j}\right)\left\langle \mathbf {r} \mid \psi \right\rangle =\left\langle \mathbf {r} \mid \sum _{j=1}^{n}\mathbf {e} _{j}{\hat {A}}_{j}\mid \psi \right\rangle \\\end{aligned}}\,\!}
### Commutation of operators on Ψ
{{#invoke:main|main}}
If two observables A and B have linear operators ${\displaystyle {\hat {A}}}$ and ${\displaystyle {\hat {B}}}$, the commutator is defined by,
${\displaystyle \left[{\hat {A}},{\hat {B}}\right]={\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}}$
The commutator is itself a (composite) operator. Acting the commutator on ψ gives:
${\displaystyle \left[{\hat {A}},{\hat {B}}\right]\psi ={\hat {A}}{\hat {B}}\psi -{\hat {B}}{\hat {A}}\psi .}$
If ψ is an eigenfunction with eigenvalues a and b for observables A and B respectively, and if the operators commute:
${\displaystyle \left[{\hat {A}},{\hat {B}}\right]\psi =0,}$
then the observables A and B can be measured simultaneously with infinite precision i.e. uncertainties ${\displaystyle \Delta A=0}$, ${\displaystyle \Delta B=0}$ simultaneously. ψ is then said to be the simultaneous eigenfunction of A and B. To illustrate this:
{\displaystyle {\begin{aligned}\left[{\hat {A}},{\hat {B}}\right]\psi &={\hat {A}}{\hat {B}}\psi -{\hat {B}}{\hat {A}}\psi \\&=a(b\psi )-b(a\psi )\\&=0.\\\end{aligned}}}
It shows that measurement of A and B does not cause any shift of state i.e. initial and final states are same (no disturbance due to measurement). Suppose we measure A to get value a. We then measure B to get the value b. We measure A again. We still get the same value a. Clearly the state (ψ) of the system is not destroyed and so we are able to measure A and B simultaneously with infinite precision.
If the operators do not commute:
${\displaystyle \left[{\hat {A}},{\hat {B}}\right]\psi \neq 0,}$
they can't be prepared simultaneously to arbitrary precision, and there is an uncertainty relation between the observables,
${\displaystyle \Delta A\Delta B\geq {\frac {\hbar }{2}}}$
even if ψ is an eigenfunction the above relation holds.. Notable pairs are position and momentum, and energy and time - uncertainty relations, and the angular momenta (spin, orbital and total) about any two orthogonal axes (such as Lx and Ly, or sy and sz etc.).[2]
### Expectation values of operators on Ψ
The expectation value (equivalently the average or mean value) is the average measurement of an observable, for particle in region R. The expectation value ${\displaystyle \langle {\hat {A}}\rangle }$ of the operator ${\displaystyle {\hat {A}}}$ is calculated from:[3]
${\displaystyle \langle {\hat {A}}\rangle =\int _{R}\psi ^{*}\left(\mathbf {r} \right){\hat {A}}\psi \left(\mathbf {r} \right)\mathrm {d} ^{3}\mathbf {r} =\langle \psi |{\hat {A}}|\psi \rangle .}$
This can be generalized to any function F of an operator:
${\displaystyle \langle F({\hat {A}})\rangle =\int _{R}\psi (\mathbf {r} )^{*}\left[F({\hat {A}})\psi (\mathbf {r} )\right]\mathrm {d} ^{3}\mathbf {r} =\langle \psi |F({\hat {A}})|\psi \rangle ,}$
An example of F is the 2-fold action of A on ψ, i.e. squaring an operator or doing it twice:
{\displaystyle {\begin{aligned}&F({\hat {A}})={\hat {A}}^{2}\\&\Rightarrow \langle {\hat {A}}^{2}\rangle =\int _{R}\psi ^{*}\left(\mathbf {r} \right){\hat {A}}^{2}\psi \left(\mathbf {r} \right)\mathrm {d} ^{3}\mathbf {r} =\langle \psi \vert {\hat {A}}^{2}\vert \psi \rangle \\\end{aligned}}\,\!}
### Hermitian operators
{{#invoke:main|main}}
The definition of a Hermitian operator is:[1]
${\displaystyle {\hat {A}}={\hat {A}}^{\dagger }}$
Following from this, in bra–ket notation:
${\displaystyle \langle \phi _{i}|{\hat {A}}|\phi _{j}\rangle =\langle \phi _{j}|{\hat {A}}|\phi _{i}\rangle ^{*}.}$
Important properties of Hermitian operators include:
• real eigenvalues,
• eigenvectors with different eigenvalues are orthogonal,
• eigenvectors can be chosen to be a complete orthonormal basis,
### Operators in Matrix mechanics
An operator can be written in matrix form to map one basis vector to another. Since the operators and basis vectors are linear, the matrix is a linear transformation (aka transition matrix) between bases. Each basis element ${\displaystyle \phi _{j}}$ can be connected to another,[3] by the expression:
${\displaystyle A_{ij}=\langle \phi _{i}|{\hat {A}}|\phi _{j}\rangle ,}$
which is a matrix element:
${\displaystyle {\hat {A}}={\begin{pmatrix}A_{11}&A_{12}&\cdots &A_{1n}\\A_{21}&A_{22}&\cdots &A_{2n}\\\vdots &\vdots &\ddots &\vdots \\A_{n1}&A_{n2}&\cdots &A_{nn}\\\end{pmatrix}}}$
A further property of a Hermitian operator is that eigenfunctions corresponding to different eigenvalues are orthogonal.[1] In matrix form, operators allow real eigenvalues to be found, corresponding to measurements. Orthogonality allows a suitable basis set of vectors to represent the state of the quantum system. The eigenvalues of the operator are also evaluated in the same way as for the square matrix, by solving the characteristic polynomial:
${\displaystyle \det \left({\hat {A}}-a{\hat {I}}\right)=0,}$
where I is the n × n identity matrix, as an operator it corresponds to the identity operator. For a discrete basis:
${\displaystyle {\hat {I}}=\sum _{i}|\phi _{i}\rangle \langle \phi _{i}|}$
while for a continuous basis:
${\displaystyle {\hat {I}}=\int |\phi \rangle \langle \phi |d\phi }$
### Inverse of an operator
A non-singular operator ${\displaystyle {\hat {A}}}$ has an inverse ${\displaystyle {\hat {A}}^{-1}}$ defined by:
${\displaystyle {\hat {A}}{\hat {A}}^{-1}={\hat {A}}^{-1}{\hat {A}}={\hat {I}}}$
If an operator has no inverse, it is a singular operator. In a finite-dimensional space, the determinant of a non-singular operator is non-zero:
${\displaystyle \det({\hat {A}})\neq 0}$
and hence it is zero for a singular operator.
### Table of QM operators
The operators used in quantum mechanics are collected in the table below (see for example,[1][4]). The bold-face vectors with circumflexes are not unit vectors, they are 3-vector operators; all three spatial components taken together.
### Examples of applying quantum operators
The procedure for extracting information from a wave function is as follows. Consider the momentum p of a particle as an example. The momentum operator in one dimension is:
${\displaystyle {\hat {p}}=-i\hbar {\frac {\partial }{\partial x}}}$
Letting this act on ψ we obtain:
${\displaystyle {\hat {p}}\psi =-i\hbar {\frac {\partial }{\partial x}}\psi ,}$
if ψ is an eigenfunction of ${\displaystyle {\hat {p}}}$, then the momentum eigenvalue p is the value of the particle's momentum, found by:
${\displaystyle -i\hbar {\frac {\partial }{\partial x}}\psi =p\psi .}$
For three dimensions the momentum operator uses the nabla operator to become:
${\displaystyle \mathbf {\hat {p}} =-i\hbar \nabla .}$
In Cartesian coordinates (using the standard Cartesian basis vectors ex, ey, ez) this can be written;
${\displaystyle \mathbf {e} _{\mathrm {x} }{\hat {p}}_{x}+\mathbf {e} _{\mathrm {y} }{\hat {p}}_{y}+\mathbf {e} _{\mathrm {z} }{\hat {p}}_{z}=-i\hbar \left(\mathbf {e} _{\mathrm {x} }{\frac {\partial }{\partial x}}+\mathbf {e} _{\mathrm {y} }{\frac {\partial }{\partial y}}+\mathbf {e} _{\mathrm {z} }{\frac {\partial }{\partial z}}\right),}$
that is:
${\displaystyle {\hat {p}}_{x}=-i\hbar {\frac {\partial }{\partial x}},\quad {\hat {p}}_{y}=-i\hbar {\frac {\partial }{\partial y}},\quad {\hat {p}}_{z}=-i\hbar {\frac {\partial }{\partial z}}\,\!}$
The process of finding eigenvalues is the same. Since this is a vector and operator equation, if ψ is an eigenfunction, then each component of the momentum operator will have an eigenvalue corresponding to that component of momentum. Acting ${\displaystyle \mathbf {\hat {p}} }$ on ψ obtains:
{\displaystyle {\begin{aligned}{\hat {p}}_{x}\psi &=-i\hbar {\frac {\partial }{\partial x}}\psi =p_{x}\psi \\{\hat {p}}_{y}\psi &=-i\hbar {\frac {\partial }{\partial y}}\psi =p_{y}\psi \\{\hat {p}}_{z}\psi &=-i\hbar {\frac {\partial }{\partial z}}\psi =p_{z}\psi \\\end{aligned}}\,\!} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 125, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9863354563713074, "perplexity": 658.6030197158925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195527531.84/warc/CC-MAIN-20190722051628-20190722073628-00162.warc.gz"} |
http://physics.stackexchange.com/questions/62554/is-diracs-description-of-a-photon-in-a-split-beam-still-seen-as-correct-today | # Is Dirac's description of a photon in a split beam still seen as correct today?
This comes from the Interference of Photons section in the book The Principles of Quantum Mechanics by P Dirac:
We shall discuss the description which quantum mechanics provides of the interference. Suppose we have a beam of light which is passed through some kind of interferometer, so that it gets split up into two components and the two components are subsequently made to interfere. We may, as in the preceding section, take an incident beam consisting of only a single photon and inquire what will happen to it as it goes through the apparatus. This will present to us the difficulty of the conflict between the wave and corpuscular theories of light in an acute form.
Corresponding to the description that he had in the case of the polarization, we must now describe the photon as going partly into each of the two components into which the incident beam is split. The photon is then, as we may say, in a translational state given by the superposition of the two translational states associated with the two components. We are thus lead to a generalization of the term "translational state" applied to a photon. For a photon to be in a definite translational state it need not be associated with one single beam of light, but may be associated with two or more beams of light which are the components into which one original beam has been split. In the accurate mathematical theory each translational state is associated with one of the wave functions of ordinary wave optics, which wave function may describe either a single beam or two or more beams into which one original beam has been split. Translational states are thus superposable in a similar way to wave functions.
1. Is this the view held by leading physicists today?
2. Can we really talk of the incident and exiting photon in the beam splitter being the same photon?
3. Is Dirac saying that the photon is partly in the two split beams, but no where else in space?
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## 1 Answer
Let me begin by saying that I hope there are multiple answers to this because it's a question that forces one to make some judgement calls, and I think other peoples' opinions would be valuable. Here's my take:
Whenever reading an old text like this, especially on quantum mechanics, I think it's important to take the language of the text with a grain of semantic salt, and this affects the answers to your questions:
1. I think I understand quantum mechanics (with some conceptual exceptions), and when I read this description, I thought it was pretty decent. I speculate that most physicists who understand quantum mechanics would probably read this description and agree. In my view, one issue with Dirac's description is that that someone just learning the subject might read the following statement and get the wrong idea:
we must now describe the photon as going partly into each of the two components into which the incident beam is split
because it, at least in my opinion, might lead such a person to believe that the photon is somehow in two places at once. I would instead probably describe the situation using probabilities to emphasize that stating where the photon is at any given time is not meaningful in quantum, but stating what the probability is that one will find it in a particular location after measurement is meaningful.
2. I would characterize such questions as being of a philosophical as opposed to physical nature in the realm of quantum mechanics. In classical mechanics, particles are modeled to move along well-defined trajectories in space, so we can keep track of where particles are and whether one particle is the same or different from another. But since this is not possible in the context of quantum mechanics, I am not aware of a reasonable operational definition of "sameness" that would allow one to claim that the outgoing measured photon is the same as the ingoing measured photon.
3. No. It's unclear to me what it would even mean in this context to say that a single photon is "in a beam." A photon is fired, and there are certain probabilities that if one were to measure the position of the photon, then one would find it at a specified location in space.
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https://www.ideals.illinois.edu/handle/2142/85484 | Files in this item
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Title: The Neuronal Basis of Signal Detection in the Presence of Spatially Separated Noise: A Study in the Frog Inferior Colliculus Author(s): Ratnam, Rama Doctoral Committee Chair(s): Feng, Albert S. Department / Program: Biophysics and Computational Biology Discipline: Biophysics and Computational Biology Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Biophysics, General Abstract: Psychophysical results from studies in humans have shown that when an auditory signal and noise are presented simultaneously, the threshold for signal detection decreases when the angular separation between the two sound sources is increased. Signal detection threshold is highest when the two sound sources are colocalized in space. Although this phenomenon has been known for many years, and has recently been confirmed in other vertebrates, the response of neurons in the auditory system to spatial separation of sound sources is not known. To study the neural basis of this phenomenon, the leopard frog Rana pipiens was chosen as the experimental subject. It is known that females of this species use only auditory cues to select and move towards mates calling in a dense heterospecific chorus. Extracellular recordings were made from single auditory units in the inferior colliculus (IC), while the frog was subject to free-field stimulation with sinusoidally amplitude modulated tone bursts (signal) in the presence of spatially separated broad-band noise (masker). Results of the study showed that while many IC neurons (called A-type) demonstrated the psychophysically observed behavior, many other neurons (called U-type) showed the opposite effect. That is, U-type neurons demonstrated an increase in signal detection threshold with an increase in angular separation of sound sources. Together, both these types of neurons accounted for nearly half the population sampled while the remainder of the population was a mix of the two types. When the type of the unit was correlated with the temporal discharge pattern of the unit it was found that most A-type units had phasic discharge patterns whereas most U-type units had tonic discharge patterns. It was also found that the response to signal plus noise was highly nonlinear and that detection thresholds for a unit could not be estimated by examining the neuron's response to signal alone and noise alone. The study concludes with a discussion of the nonlinear interactions in the system and discusses the possible role of A- and U-type units in signal detection. Issue Date: 1998 Type: Text Language: English Description: 105 p.Thesis (Ph.D.)--University of Illinois at Urbana-Champaign, 1998. URI: http://hdl.handle.net/2142/85484 Other Identifier(s): (MiAaPQ)AAI9904569 Date Available in IDEALS: 2015-09-25 Date Deposited: 1998
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http://math.stackexchange.com/questions/209868/explanation-of-a-cross-product-result | Explanation of a cross product result
In my book the result $$(u\times v)\cdot(x\times y)=\begin{vmatrix} u\cdot x & v\cdot x \\u \cdot y & v \cdot y\end{vmatrix},$$ where u, v, x and y are arbitrary vectors, is stated (here '$\cdot$' means the dot product and '$\times$' is the cross product). The book very briefly says that this can be easily done by observing that both sides are linear in u, v, x and y.
I know that if I expand and simply the LHS using the components of a vector the result will be true. However, I don't really understand what it means when the book says ' both sides are linear in u, v, x and y ' and how by noticing this fact, makes this relation easier to prove.
Any help will be greatly appreciated.
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Do you know what a linear function is ? – Phira Oct 9 '12 at 14:28
Yeah, functions which satisfy f(x+y)=f(x)+f(y) and f(ax)=af(x). – SomethingWitty Oct 9 '12 at 14:31
One might also note that both sides of the equation are equal to $\left| \begin{matrix} x \\ y \end{matrix} \right| \cdot \left | \begin{matrix} u & v \end{matrix} \right|$. – Phira Oct 9 '12 at 15:44
2 Answers
Suppose we have two linear functions, $f$ and $g$, which agree on all the basis vectors of some space. Then they must agree for every vector on that space, because they are both linear, and a linear function is completely determined by its values on the basis.
In gory detail, suppose that we know that $f(\vec{e_i}) = g(\vec{e_i})$ for each basis vector $\vec{e_i}$.
Consider some vector $\vec v$. We can express $\vec v$ a a linear combination of basis vectors, say as $$\vec v = c_1\vec{e_1} + \cdots + c_n\vec{e_n}.$$ Then we know that \begin{align} f(\vec v) & = f(c_1\vec{e_1} + \cdots + c_n\vec{e_n}) \\ & = c_1f(\vec{e_1}) + \cdots + c_nf(\vec{e_n}) & \text{(linearity of f)} \\ & = c_1g(\vec{e_1}) + \cdots + c_ng(\vec{e_n}) & \text{(f=g for basis vectors)} \\ & = g(c_1\vec{e_1} + \cdots + c_n\vec{e_n}) & \text{(linearity of g)}\\ & = g(\vec v) \end{align}
One can similarly show an analogous fact for functions of several variables. For example, if $f(u,v,x,y)$ and $g(u,v,x,y)$ are linear functions of $u, v, x,$ and $y$, and if they agree on all combinations of the basis vectors for some space, then they agree on every vector in that space.
Now take $f(u,v,x,y) = (u\times v)\cdot(x\times y)$ and $g(u,v,x,y) =\begin{vmatrix} u\cdot x & v\cdot x \\u \cdot y & v \cdot y\end{vmatrix}$. These are easily seen to be linear, or easy to show to be linear if you don't see it, using properties of cross and dot products (for $f$) and of determinants and dot products (for $g$). So if you can show that they are equal when $u,v,x,$ and $y$ are basis vectors, you are done. And for most choices of basis vectors as arguments, both sides are equal to zero, so this is quick to verify.
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Thanks for the clear and detailed explanation, it helped a lot. – SomethingWitty Oct 9 '12 at 18:51
It means that you can verify the relation just using the standard basis $\{e_1, e_2, e_3 \}$ of three dimensional space. For example you should check $(e_1 \times e_2)\cdot (e_2 \times e_3) =0$ which is the same as the right hand side.
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I'm gonna mull over this; atm I don't really see why it suffices to just verify the relation using the standard basis. – SomethingWitty Oct 9 '12 at 14:43
@SomethingWitty - because any vector is a linear combination of elements of the standard basis, and the function you are computing is linear in each place. – Mark Bennet Oct 9 '12 at 15:09
Any vector is a linear combination of these three basis elements. If a relation is linear, it is enough to verify it on just one basis of the vector space. Suppose you want to show $u\cdot v=v\cdot u$. If it is correct on the $e_i$ then write $u$ and $v$ as linear combinations of the $e_i$, use linearity and you will see the result. – PAD Oct 9 '12 at 15:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9903317093849182, "perplexity": 184.1931682882959}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119648891.34/warc/CC-MAIN-20141024030048-00049-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://mathcracker.com/prime-factorization | # Prime Factorization
Instructions: Use this prime factorization to compute the factors and prime decomposition of a given number you provide in the box below.
The number you want the factors of (Any positive integer number)
## How to use this prime factorization calculator
This calculator will provide you with the factors and the corresponding prime decomposition of a given number. So, you need to provide a valid integer, a positive integer.
Then, once that is provided, you need to click on "Calculate", in order to get all the steps of the calculation shown.
## How do compute the prime decomposition
All you need to do is to find the factors of the corresponding number. These factors are then grouped there will be an exponent associated to each of them (reflecting the number of times the corresponding prime appears in the factorization).
## What are the steps for a prime factorization
• Step 1: Identify the number you want to factor. It must be a positive integer, otherwise you cannot proceed
• Step 2: Find ALL the factors of the number
• Step 3: Count the number of times each factor appears in the decomposition
## Why would need to deal with prime numbers?
Although they are not really covered in basic Algebra, prime numbers play a crucial role in Mathematics, not only Algebra. It seems that primes hold some sort of magic power and they have some incredible properties.
At the basic level, let's consider the fact that every single positive integer admits one and only one prime decomposition to be a sufficiently important property.
### Example: Calculating a prime decomposition
Calculate prime factorization of 3468.
Solution:
First, we need to find all the possible prime divisors of $$n = 3468$$. In this case, it is found that
$3468 = 2\cdot2\cdot3\cdot17\cdot17$
Now, grouping the divisors found above, the following prime decomposition is obtained, in exponential form:
$3468 = 2^2\cdot3\cdot17^2$
This completes the process of the prime decomposition calculation, because no factors can be further decomposed.
### Example: Another prime number
Find the factors of 16.
Solution:< First, we need to find all the possible prime divisors of $$n = 16$$. In this case, it is found that
$16 = 2\cdot2\cdot2\cdot2$
Now, grouping the divisors found above, the following prime decomposition is obtained, in exponential form:
$16 = 2^4$
This completes the process of the prime decomposition calculation, because no factors can be further decomposed.
### Example: Another prime number
Find the factors of 137.
Solution: We need to find all the possible prime divisors of $$n = 137$$. In this case, it is found that the number $$n = 137$$ does not have any factors, and therefore, it is prime, hence, the prime decomposition of $$n = 137$$ is itself.
This completes the process of the prime decomposition calculation, because no factors can be further decomposed.
### Example: Prime decompositions and primes
Is 341 a prime number?
Solution: First, we need to find all the possible prime divisors of $$n = 341$$. In this case, it is found that
$341 = 11\cdot31$
Since there is a factor (11) that is neither 1 nor 341, we conclude that 341 is prime.
## More algebra calculators
Algebra is a very important branch of mathematics, perhaps the most important since it provides the basic grounds for most of the other math fields.
In terms of algebra calculators, you may be interested in computing general algebraic expressions, or also in terms of computing the least common multiple between two numbers, just to mention a few. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8683143854141235, "perplexity": 336.03373485716406}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00842.warc.gz"} |
https://www.physicsforums.com/threads/deriving-probability-density-functions-from-partial-differential-equations.616585/ | # Deriving Probability Density Functions from Partial Differential Equations?
1. Jun 26, 2012
### Annihilator
Deriving Probability Density Functions from Partial Differential Equations???
Hiyas,
I have been told that it is quite normal to get PDFs (Probability Density Functions) from PDEs (Partial Differential Equations). That in PDEs that the function can be a PDF and you can get this by solving the PDE.
Nobody has actually shown me how this is done. I have my doubts because I have learned that Probability Theory is not mixed with PDEs directly this way. That there are other types of mathematics called Stochastic Partial Differential Equations or Stochastic Differential Equations that do something similar.
I was taught to believe that PDFs are found by "reinterpreting" the PDE by approaching it with Probability Theory and looking at from a new way and creating new Probability Mathematics to deal with the PDE.
However others are insisting that I don't start by approaching it with Probability Theory. That I actually just solve PDEs to get PDFs and that I can also get PDFs from PDEs.
Another thing is they are using the term Probability Distribution Function and not Probability Density Function but I think they are claiming them to be the same thing.
Can anyone give me their thoughts on this.
Thank you.
2. Jun 26, 2012
### chiro
Re: Deriving Probability Density Functions from Partial Differential Equations???
Hey Annihilator and welcome to the forums.
The common area for this in financial calculus. What happens is you reduce a stochastic differential equation to a PDE and then you solve it.
One good thing about the PDE approach is that when you get arbitrary pay-off functions for your option contract, you can get an expression for the expectation of that arbitrary function.
The heat equation and the normal distribution are pretty much the same, so when you have the right transformations, you can use things like the heat equation to get a density or expectation value.
The above though is a specialized application of this.
There is however a common tool in probability that is known as the characteristic function which is used to retrieve PDF's from a MGF (Moment Generating Function) where you do a fourier transform on the MGF.
I'm speculating, but it's quite possible that this transform or others like it are able to be translated to PDE's or other similar representations especially if a representation is found to link the fourier transform with PDE's. I know you can relate these to linear ODE's but not sure about PDE's.
3. Jun 26, 2012
### Mute
Re: Deriving Probability Density Functions from Partial Differential Equations???
A PDE tells you how a given function changes with respect to its variables (time, position, etc). If your PDF changes with time and position (i.e., the values you can measure for your random variable change with time and/or position), then one can in principle write down an equation which relates together how the probability changes with time and space, and then solve this equation for the pdf.
See http://en.wikipedia.org/wiki/Master_equation and http://en.wikipedia.org/wiki/Fokker-Planck_equation .
4. Jun 26, 2012
### bigfooted
Re: Deriving Probability Density Functions from Partial Differential Equations???
Deriving probability density functions from pde's is also quite common in classical physics like turbulence. You can go from the Navier stokes equations (pde) to the transport equation of the probability density function of velocity (a Fokker-Planck equation) using not much more than the ito formula and standard calculus. The solution of the Fokker-Planck equation can be obtained by realizing that a stochastic sample of the PDF (where the PDF = the solution of the FP equation) is actually described by a Langevin equation. A discrete solution of the FP equation can be obtained by solving N Langevin equations in a Monte Carlo method.
Actually, every PDE can be written as a stochastic equation, either as a Fokker-Planck equation or a Langevin equation (because every (?) PDE can be written as a diffusion process).
Check the book of Oksendal for an introduction.
5. Jun 26, 2012
### Mute
Re: Deriving Probability Density Functions from Partial Differential Equations???
Be warned, that book is rather formal and makes extensive use of measure theory notation. If you're comfortable with formal proofs and measure theory, go for it, but coming from a physics background I found it pretty dense and hard to read through.
6. Jun 27, 2012
### Annihilator
Re: Deriving Probability Density Functions from Partial Differential Equations???
Ok, nice thank you.
I take it though that not all PDEs can be written as a type of diffusion process without introducing some form of normalization of the PDE so that you can then begin Stochastic calculations. It seems to me some of these master equations types are pure postulates that can't be derived by any general means of getting PDFs from PDEs. It seems more like you have to figure out if the PDE can be modelled as a SPDE or SDE first and then get the PDF from there. Is this right or wrong?
7. Jun 27, 2012
### Mute
Re: Deriving Probability Density Functions from Partial Differential Equations???
It depends on what you want to do, I guess. If you want analytical results, I think you write down an SDE and then try to convert it into a PDE, because we know how to solve PDEs much better than we know how to solve SDEs.
However, if you want numerical results, it's easer to have an SDE which you can simulate because then you just add some noise terms which are really easy to do numerically; easier than solving PDEs numerically, at least.
See
http://en.wikipedia.org/wiki/Feynman-Kac_formula
That discusses the connection between parabolic PDEs and SDEs.
8. Jun 27, 2012
### Annihilator
Re: Deriving Probability Density Functions from Partial Differential Equations???
Cool. I know we are thinking the same because I am looking at the very same thing.
By the way have you heard of cases where people working on PDEs didn't know they where working Stochastic Partial Differential Equations or Stochastic Differential Equations and didn't get the differences at post-graduate level?
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https://blog.stata.com/2016/page/2/ | Archive for 2016
## An ordered-probit inverse probability weighted (IPW) estimator
teffects ipw uses multinomial logit to estimate the weights needed to estimate the potential-outcome means (POMs) from a multivalued treatment. I show how to estimate the POMs when the weights come from an ordered probit model. Moment conditions define the ordered probit estimator and the subsequent weighted average used to estimate the POMs. I use gmm to obtain consistent standard errors by stacking the ordered-probit moment conditions and the weighted mean moment conditions. Read more…
Categories: Statistics Tags:
## Cointegration or spurious regression?
$$\newcommand{\betab}{\boldsymbol{\beta}}$$Time-series data often appear nonstationary and also tend to comove. A set of nonstationary series that are cointegrated implies existence of a long-run equilibrium relation. If such an equlibrium does not exist, then the apparent comovement is spurious and no meaningful interpretation ensues.
Analyzing multiple nonstationary time series that are cointegrated provides useful insights about their long-run behavior. Consider long- and short-term interest rates such as the yield on a 30-year and a 3-month U.S. Treasury bond. According to the expectations hypothesis, long-term interest rates are determined by the average of expected future short-term rates. This implies that the yields on the two bonds cannot deviate from one another over time. Thus, if the two yields are cointegrated, any influence to the short-term rate leads to adjustments in the long-term interest rate. This has important implications in making various policy or investment decisions.
In a cointegration analysis, we begin by regressing a nonstationary variable on a set of other nonstationary variables. Suprisingly, in finite samples, regressing a nonstationary series with another arbitrary nonstationary series usually results in significant coefficients with a high $$R^2$$. This gives a false impression that the series may be cointegrated, a phenomenon commonly known as spurious regression.
In this post, I use simulated data to show the asymptotic properties of an ordinary least-squares (OLS) estimator under cointegration and spurious regression. I then perform a test for cointegration using the Engle and Granger (1987) method. These exercises provide a good first step toward understanding cointegrated processes. Read more…
Categories: Statistics Tags:
## Two faces of misspecification in maximum likelihood: Heteroskedasticity and robust standard errors
For a nonlinear model with heteroskedasticity, a maximum likelihood estimator gives misleading inference and inconsistent marginal effect estimates unless I model the variance. Using a robust estimate of the variance–covariance matrix will not help me obtain correct inference.
This differs from the intuition we gain from linear regression. The estimates of the marginal effects in linear regression are consistent under heteroskedasticity and using robust standard errors yields correct inference.
If robust standard errors do not solve the problems associated with heteroskedasticity for a nonlinear model estimated using maximum likelihood, what does it mean to use robust standard errors in this context? I answer this question using simulations and illustrate the effect of heteroskedasticity in nonlinear models estimated using maximum likelihood. Read more…
Categories: Statistics Tags:
## Group comparisons in structural equation models: Testing measurement invariance
When fitting almost any model, we may be interested in investigating whether parameters differ across groups such as time periods, age groups, gender, or school attended. In other words, we may wish to perform tests of moderation when the moderator variable is categorical. For regression models, this can be as simple as including group indicators in the model and interacting them with other predictors.
We naturally have hypotheses regarding differences in parameters across groups when fitting structural equation models as well. When these models involve latent variables and the corresponding observed measurements, we can test whether those measurements are invariant across groups. Evaluation of measurement invariance typically involves a series of tests for equality of measurement coefficients (factor loadings), equality of intercepts, and equality of error variances across groups.
In this post, I demonstrate how to use the sem command’s group() and ginvariant() options as well as the postestimation command estat ginvariant to easily perform tests of measurement invariance. Read more…
Categories: Statistics Tags:
## Exact matching on discrete covariates is the same as regression adjustment
I illustrate that exact matching on discrete covariates and regression adjustment (RA) with fully interacted discrete covariates perform the same nonparametric estimation. Read more…
Categories: Statistics Tags:
## Vector autoregressions in Stata
Introduction
In a univariate autoregression, a stationary time-series variable $$y_t$$ can often be modeled as depending on its own lagged values:
\begin{align}
y_t = \alpha_0 + \alpha_1 y_{t-1} + \alpha_2 y_{t-2} + \dots
+ \alpha_k y_{t-k} + \varepsilon_t
\end{align}
When one analyzes multiple time series, the natural extension to the autoregressive model is the vector autoregression, or VAR, in which a vector of variables is modeled as depending on their own lags and on the lags of every other variable in the vector. A two-variable VAR with one lag looks like
\begin{align}
y_t &= \alpha_{0} + \alpha_{1} y_{t-1} + \alpha_{2} x_{t-1}
+ \varepsilon_{1t} \\
x_t &= \beta_0 + \beta_{1} y_{t-1} + \beta_{2} x_{t-1}
+ \varepsilon_{2t}
\end{align}
Applied macroeconomists use models of this form to both describe macroeconomic data and to perform causal inference and provide policy advice.
In this post, I will estimate a three-variable VAR using the U.S. unemployment rate, the inflation rate, and the nominal interest rate. This VAR is similar to those used in macroeconomics for monetary policy analysis. I focus on basic issues in estimation and postestimation. Data and do-files are provided at the end. Additional background and theoretical details can be found in Ashish Rajbhandari’s [earlier post], which explored VAR estimation using simulated data. Read more…
Categories: Statistics Tags:
## Multiple-equation models: Estimation and marginal effects using gmm
We estimate the average treatment effect (ATE) for an exponential mean model with an endogenous treatment. We have a two-step estimation problem where the first step corresponds to the treatment model and the second to the outcome model. As shown in Using gmm to solve two-step estimation problems, this can be solved with the generalized method of moments using gmm.
This continues the series of posts where we illustrate how to obtain correct standard errors and marginal effects for models with multiple steps. In the previous posts, we used gsem and mlexp to estimate the parameters of models with separable likelihoods. In the current model, because the treatment is endogenous, the likelihood for the model is no longer separable. We demonstrate how we can use gmm to estimate the parameters in these situations. Read more…
Categories: Statistics Tags:
## Probability differences and odds ratios measure conditional-on-covariate effects and population-parameter effects
$$\newcommand{\Eb}{{\bf E}} \newcommand{\xb}{{\bf x}} \newcommand{\betab}{\boldsymbol{\beta}}$$Differences in conditional probabilities and ratios of odds are two common measures of the effect of a covariate in binary-outcome models. I show how these measures differ in terms of conditional-on-covariate effects versus population-parameter effects. Read more…
Categories: Statistics Tags:
## Doctors versus policy analysts: Estimating the effect of interest
$$\newcommand{\Eb}{{\bf E}}$$The change in a regression function that results from an everything-else-held-equal change in a covariate defines an effect of a covariate. I am interested in estimating and interpreting effects that are conditional on the covariates and averages of effects that vary over the individuals. I illustrate that these two types of effects answer different questions. Doctors, parents, and consultants frequently ask individuals for their covariate values to make individual-specific recommendations. Policy analysts use a population-averaged effect that accounts for the variation of the effects over the individuals. Read more…
Categories: Statistics Tags:
## Effects of nonlinear models with interactions of discrete and continuous variables: Estimating, graphing, and interpreting
I want to estimate, graph, and interpret the effects of nonlinear models with interactions of continuous and discrete variables. The results I am after are not trivial, but obtaining what I want using margins, marginsplot, and factor-variable notation is straightforward. Read more…
Categories: Statistics Tags: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8360849022865295, "perplexity": 1918.3338594204727}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401614309.85/warc/CC-MAIN-20200928202758-20200928232758-00092.warc.gz"} |
https://open.library.ubc.ca/cIRcle/collections/ubctheses/831/items/1.0089794 | # Open Collections
## UBC Theses and Dissertations
### Case-control studies with misclassified exposure : a Bayesian approach Saskin, Refik 2000-12-31
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Case-control Studies with Misclassified Exposure: A Bayesian Approach by Refik Saskin B.Sc, Brock University, 1998 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF Master of Science in THE FACULTY OF GRADUATE STUDIES (Department of Statistics) we accept this thesis as conforming to the required standard The University of British Columbia August 2000 © Refik Saskin, 2000 In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the head of my department or by his or her representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my written permission. Department of The University of British Columbia Vancouver, Canada Date AjJC Of. £°<~ . DE-6 (2/88) Abstract When dealing with the case-control data, it is often the case that the ex posure to a risk factor of interest is subject to miclassification. Methods for correcting the odds-ratio are available when the misclassification probabilities are known. In practice, however, good guesses rather than the exact values are available for these probabilities. We show that when these guesses are treated as exact even the smallest differencies between the true and guessed values can lead to very erroneous odds-ratio estimates. This problem is alle viated by a Bayesian analysis which incorporates the uncertainty about the misclassification probabilities as prior information. In practice, data on the exposure variable are quite often available from more than one source. We review three methods for improving the odds-ratio estimates that combine information from two sources. We then develop a Bayesian approach which is based on latent class analysis, and apply it to the sudden infant death syndrome data. The inference required the use of the Metropolis-Hastings algorithm and/or the Gibbs sampler. ii Contents Abstract ii Contents iiList of Tables v List of Figures vii Acknowledgements xiDedication xiii 1 Introduction 1 2 One Test Case 5 2.1 Definitions and Terminology . 5 2.1.1 Data and Setup 7 2.2 Known Sensitivity and Specificity 10 2.3 The Gibbs Sampler 17 2.3.1 Examples 21 iii 2.4 The Metropolis-Hastings Algorithm 24 2.4.1 Large Sample Case 31 2.4.2 Finite Sample Case 5 2.4.3 Examples . . 36 2.5 Discussion 40 3 Two Test Case 9 3.1 Data and Setup 43.2 Correction Methods 51 3.2.1 The Marshall and Graham Method 51 3.2.2 The Drews, Flanders and Kosinski Method 52 3.2.3 The Kaldor and Clayton Method 54 3.3 Our Method 56 3.3.1 Examples 60 3.4 An Application to Real Data 68 3.5 Discussion 72 4 Conclusion 4 Bibliography 78 IV List of Tables 2.1 Distribution of subjects in a case-control study by disease sta tus and an imperfect measurement of a dichotomous exposure. Symbol + denotes apparently exposed, and — denotes appar ently unexposed 9 2.2 (a) True distribution of exposure among cases and controls, (b) Observed distribution of exposure, given 10% misclassifcation among cases and controls, (c) Estimated distribution of expo sure assuming 14% misclassifcation among cases and controls. 14 2.3 Distribution of observed and latent data when one imperfect classification procedure is used. Symbol + denotes exposed, and — denotes unexposed 18 3.1 Distribution of subjects in a case-control study when two tests are used to assess the exposure status. Symbol + denotes ex posed and — denotes unexposed. 50 v 3.2 Distribution of observed and latent data when two imperfect tests are used, together with the contribution to the likelihood each combination of the observed and latent data makes. Sym bol + denotes exposed, and — denotes unexposed 57 3.3 True parameter values and the empirical coverage of the 80% HPD credible intervals for logc/> for the sample sizes iVj = 200, N{ = 800, and Ni = 3200, i = 0,1. Numbers in brackets repre sent mean interval length. For the empirical coverage and the mean length, 1000 data sets were simulated 67 3.4 A pseudorandom sample of 226 SIDS cases and 226 controls from the NICHD study. Data was classified using medical record (MR) and interview (Int) data 69 3.5 Median estimates of the sensitivities, specificities and the loga rithm of the odds-ratio, and 95% HPD credible intervals for the logarithm of the odds-ratio, using the Gibbs sampler 73 3.6 95% HPD credible intervals for the sensitivities and specificities of the interview data and the medical records 73 3.7 Estimates of the sensitivities, specificities and the logarithm of the odds-ratio, and 95% confidence intervals for the logarithm of the odds-ratio, using the EM algorithm (Drews et al. (1993)). 73 vi List of Figures 2.1 Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in con trols and prevalence of exposure in cases in Example 1. Data was simulated for N0 = Nx = 200 25 2.2 Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in con trols and prevalence of exposure in cases in Example 1. Data was simulated for N0 = Nx = 800 26 2.3 Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in con trols and prevalence of exposure in cases in Example 1. Data was simulated for N0 = Nx = 3200 27 2.4 Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in con trols and prevalence of exposure in cases in Example 2. Data was simulated for N0 = Nx = 200 28 vn 2.5 Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in con trols and prevalence of exposure in cases in Example 2. Data was simulated for N0 = Ni = 800 29 2.6 Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in con trols and prevalence of exposure in cases in Example 2. Data was simulated for N0 = Ni = 3200 . 30 2.7 The support of p(a, P\90, 6\). The shaded rectangles comprise A{e0,el) 32 2.8 Posterior distribution of logc/> in Example 1. The first column gives the M-COR posterior, the second column gives the U-COR posterior, and the third column gives the E-COR posterior. The rows correspond to sample sizes N0 = Nx = 200, N0 = Ni — 800, N0 = N1 = 3200 and 7V0 = Ni = oo 42 2.9 80% highest posterior density credible intervals for logc/> in Ex ample 1. The solid vertical lines represent credible intervals for forty data sets with sample sizes N0 = Nx = 3200, the solid horizontal line indicates the true value of log cp and the dashed horizontal line indicates log </>'. The first panel gives the M-COR intervals, the second panel gives the U-COR intervals, and the third panel gives the E-COR intervals 43 viii 2.10 Prior and posterior samples of the sensitivity and specificity for the datasets in Example 1 44 2.11 Posterior distribution of log</> in Example 2. The first column gives the M-COR posterior, the second column gives the U-COR posterior, and the third column gives the E-COR posterior. The rows correspond to sample sizes N0 = Nx = 200, N0 = Nx = 800, JV0 = Ni= 3200 and NQ = JVI = OO 45 2.12 80% highest posterior density credible intervals for log</> in Ex ample 2. The solid vertical lines represent credible intervals for forty data sets with sample sizes JV0 = Nx = 3200, the solid horizontal line indicates the true value of log <p and the dashed horizontal line indicates log <f>'. The first panel gives the M-COR " intervals, the second panel gives the U-COR intervals, and the third panel gives the E-COR intervals. . 46 2.13 Posterior distribution of log0 in Example 3. The first column gives the M-COR posterior, the second column gives the U-COR posterior, and the third column gives the E-COR posterior. The rows correspond to sample sizes N0 = Nx = 200, N0 = Ni = 800, N0 = N1 = 3200 and N0 = Nx = oo 47 ix 2.14 80% highest posterior density credible intervals for log</> in Ex ample 3. The solid vertical lines represent credible intervals for forty data sets with sample sizes N0 = Nx = 3200, the solid horizontal line indicates the true value of log <f> and the dashed i horizontal line indicates log <$>'. The first panel gives the M-COR intervals, the second panel gives the U-COR intervals, and the third panel gives the E-COR intervals 3.1 Post burn-in output of the five independent chains of the Gibbs sampler for sensitivities, specificities, prevalence of exposure in controls and prevalence of exposure in cases in Example 1. Data was simulated for JV0 = Nx = 200 3.2 Posterior samples of the logarithm of the odds-ratio for the datasets in Example 1 ' 3.3 80% highest posterior density credible intervals for log <j> in Ex ample 1. The solid vertical lines represent credible intervals for forty data sets and the solid horizontal line indicates the true value of log 4>. The panel (a) gives the intervals for the sample size N0 = Ni = 200, the panel (b) gives the intervals for the sample size N0 — Ni — 800, and the panel (c) gives the intervals for the sample size N0 = Nx = 3200 x Posterior samples of the logarithm of the odds-ratio for the sam ple size N0 = JVi = 800 in Example 1. (a) Beta(l, 1) prior used for all six parameters, (b) Beta(7.55, 2.64) prior used for a, and = 1,2, while Beta(l,l) prior used for TIQ and 7Ti (the his togram is identical to that shown in Figure 3.2 (b)). (c) Beta prior for each parameter was chosen so that the mode 7 is cen tred at the true value and 0.95 probability is assigned to the interval 7 ± 0.05 xi Acknowledgements I would like.to thank Dr. Paul Gustafson for his consistent guidance, help and support. His patience and expert advice made the completion of my research possible. I would also like to thank Dr. Nhu Lee for his generous input and willingness to share his expertise with me. Many thanks to the entire UBC Statistics Department for making these last two years most pleasurable. Also, many thanks to my family for all their love, support and encour agement. Finally, I would like to thank my girlfriend Marnie for her continued support and patience. REFIK SASKIN The University of British Columbia August 2000 xii To my mother, God rest her soul. xiii Chapter 1 Introduction Misclassification of exposure is one of the most serious problems in epidemi ology. Even on the smallest scale, exposure misclassification can substantially bias estimates of the relative risk. In particular, nondifferential misclassifica tion of a dichotomous exposure variable always tends to attenuate observed exposure-disease relationships. This attenuation can be particularly large for rare exposures and imperfect specificity or highly prevalent exposure and im perfect sensitivity of the test used to assess the exposure status. The corrected estimates can be obtained easily if the sensitivity and specificity of the clas sification procedure are known, which was illustrated by Barron (1977) and Greenland and Kleinbaum (1983). In most cases, however, these quantities are very difficult to estimate, because no measure of true exposure is available, and even if available, such "gold standard" measures are often very expensive and/or invasive. A more common scenario is when good guesses rather than exact values are available for the sensitivity and specificity. If one treats these 1 guesses as exact, Marshall (1989) has shown that even small discrepancies be tween the true and guessed values of the sensitivity and specificity can lead to very erroneous odds-ratio estimates. The effects of ignoring misclassification and the methods which correct for it have received a considerable attention in the literature. Thomas, Stam and Dwyer (1993) and Walter and Irwig (1988) provide reviews. Although one can rarely obtain completely accurate exposure data for epidemiologic studies, data on the exposure variable are often available from more than one source. For instance, exposure data'are often available from both medical records and interviews, or from two different diagnostic tests performed on a study subject. A variety of methods have been suggested for improving the odds-ratio estimates by combining data from two information sources. One simple approach, as suggested by Marshall and Graham (1984), is to restrict the analysis to subjects for whom the two sources agree about the exposure. As pointed out by Walter (1984), however, this method may result in a substantial loss of precision due to the exclusion of a potentially large proportion of study subjects. Hui and Walter (1980) developed a method for estimating the sensitivity and specificity of two classification schemes in two groups (such as controls and cases) when, given the true exposure status, misclassification is nondifferential and the two schemes classify study subjects independently. More recent work by Drews, Flanders and Kosinski (1993) has extended their method to more complicated settings. They perform numeric maximization rather than providing a closed-form solution. The approach of 2 Drews et al. (1993) requires that the analyst specify the degree of dependency between the two classification schemes. Since, in practice, a gold standard may be unavailable, impractical, or itself measured with error, this method could have limited applicability. In their discussion of latent class analysis, Kaldor and Clayton (1985) give an example of data where replicate measurements are available for some or all of the cases and/or controls. They demonstrate that obtaining a replicate measurements on even a modest proportion of subjects leads to substantially improved estimation of odds-ratio. In the following chapter we will consider a case-control setting where one imperfect classification scheme is used to assess the exposure status. In this context, we will introduce a simple approach to Bayesian inference about the odds-ratio when the misclassification probabilities are known. Then, we will outline a Bayesian approach to odds-ratio inference with only partial knowledge of classification probabilities. Two computational methods will be presented that enable the inference on odds-ratio. Finally, examples will be presented illustrating the use of these methods. In chapter 3 we will look at a case-control setting where the exposure status is measured by two imperfect classification schemes. We will briefly review three existing methods to correct the odds-ratio estimates. Then, we will introduce a Bayesian approach. It is an extension of the Joseph, Gyorkos and Coupal (1995) method for inference on the sensitivities and specificities of two classification procedures applied to one population. We will examine the validity of our method by applying it to nine hypothetical case-control studies, 3 where the true odds-ratio are known. Finally, we will apply our method to a case-control study of sudden infant death syndrome, and compare the results with those obtained using the method of Drews et al. (1993). 4 Chapter 2 One Test Case 2.1 Definitions and Terminology In this section we present some key definitions and terminology, making the subsequent sections easier to follow. In a "case-control study" investigator selects "cases" of a disease, and a comparison group, called "controls". The cases and controls are then compared with respect to a specific risk factor, often referred to as "exposure". The samples of the cases and controls are usually regarded as independent. "Misclassification" or "measurement error" refers to any diference be tween the true value of a variable and its measured value. The term mis classification is almost always used in the context of categorical variables and measurement error when one refers to continuous variables. Errors can be "random" or "systematic". If the error is not randomly distributed around its true value, we say that it is systematic. Overestimating everyone's exposure 5 level by a factor of two would be an example of a systematic error. Systematic and random errors can be either "differential" or "nondif ferential". If the misclassification in the exposure depends upon the disease status, than the misclassification is differential. If on the other hand, the mis classification is the exposure does not depend on the disease status, we say that the misclassification is nondifferential. A more rigorous definition of non-differential misclassification is that the risk of disease depends only on the true exposure, and given this true exposure, the measured exposure does not add any additional information. In clinical medicine and epidemiology, tests are often used to determine the presence or absence of a disease, or whether or not someone was exposed to a risk factor of interest. Ideally, those who are exposed (or with disease) should be classified as being exposed (or with disease), and those who are not exposed (or without disease) should be classified as unexposed (or without disease). The "sensitivity" and "specificity" of a test consider how often such correct classification occurs. The sensitivity of a test is the percentage of exposed study subjects (or with disease) classified as exposed (or with disease). The specificity of a test is the percentage of unexposed study subjects (or without disease) classified as unexposed (or without disease).. Moreover, the "false postive rate" (FPR) of a test is the percentage of unexposed study subjects (or without disease) who are classified as exposed (or with disease). The "false negative rate" (FNR) of a test is the percentage of exposed study subjects (or with disease) who are classified as unexposed (or without disease). Therefore, 6 FPR = 1 - specificity and FNR = 1 - sensitivity. As the methodology we develop in subsequent chapters is of Bayesian nature, we will very briefly introduce some of its key features. In a nutshell, the Bayesian approach uses probability to describe both model parameters and random variables. In the context of the usual statistical model with a random variable X having possible distributions indexed by parameter 9, the data x becomes known to the statistician and the object is to make inference about the unknown parameter. The information about 6 that is available to the statistician prior to observing the data and his/her belief about the parameter are reflected in a "prior distribution". In the Bayesian approach, the statistician will wish to calculate the probability distribution of 9 given X = x, called the "posterior distribution". Once this is accomplished, point and interval estimates of 9 can be calculated, significance tests can be performed, etc. With the recent development of computer techniques has come an increase in the popularity of Bayesian inference. 2.1.1 Data and Setup Suppose we have a case-control study attempting to assess the relationship between disease status and a dichotomous exposure (usually labeled as 0 for unexposed and 1 for exposed). Suppose further that the measurement proce dure used to assess the exposure status is not perfect, i.e., the sensitivity and specificity of that procedure are less than one (so that there is a non-negligible number of false positives and false negatives). The study, therefore, consists of 7 measuring the apparent exposure status for random sample of N0 controls and Ni cases. We will assume that the random samples are independent and that the disease status is known exactly, without error. These assumptions are not particularly unreasonable or limiting. First, the independence between cases and controls is a usual assumption one makes in a case-control study. Second, in order to determine the potential treatment, clinicians concentrate on correctly identifying the disease. For a disease that is difficult to diagnose, multiple tests, sometimes even invasive procedures, are applied. Moreover, the diseased subjects chosen to participate in a case-control study have usually had the condition for quite some time before the start of a study, increasing the chance of correct diagnosis. As a result, the disease status is (usually) very well determined. Throughout this thesis we will assume that the misclassification is non-differential, i.e., independent of disease status. In the case-control setting, the data comes in a two-by-two table, as shown in Table 2.1. Here, Xi and yi are the observed number of positive and negative test results, respectively, in the sample of Xi + yi = study subjects. The subscript i = 0 denotes controls, i = 1 cases. Furthermore, let D, E and E' represent disease status, actual expo sure status and apparent exposure status, respectively. In each case, suppose that one represents presence, and zero represents absence. As nondifferential misclassification is assumed, the relevant parameters are TTO = Pr{E = 1\D = 0), 8 Test + — Controls x0 yo Cases • Xi y\ iVi N Table 2.1: Distribution of subjects in a case-control study by disease status and an imperfect measurement of a dichotomous exposure. Symbol + denotes apparently exposed, and — denotes apparently unexposed. 7Ti = Pr(E = 1\D = 1), a = Pr{E' = l\E = 1), 8 = Pr(E' = 0\E = 0). Here, a is the sensitivity of the classification procedure, (3 is the specificity of the classification procedure, and 7r0 and TT\ are the exposure prevalences among the controls and cases, respectively. The false positive and false negative rates are \ — B and 1 — a, respectively. The odds-ratio, denoted as t/>, is defined in terms of the exposure prevalences as Tri/O-jri) 7r0/(l-7r0)' In general, suppose that R tests are applied to S populations and that misclassification is nondifferential. As R tests are used to determine the expo sure status, there are 2R possible classification outcomes. Furthermore, since 9 the number of subjects is fixed for each population, there are 2R — 1 degrees of freedom per population, giving a total of (2R — 1)5 degrees of freedom avail able. On the other hand, if the misclassification is nondifferential, there are 2R + S parameters to estimate, the sensitivity and specificity of each test and S exposure prevalences. Therefore, in the case of one test and two population (R = 1,5 = 2), the likelihood with no constraint placed on the parameters is overparameterized. Hence, if (a,7r0,7Ti) are unknown, the resulting like lihood function from a case-control study where one classification scheme is used to assess the exposure is nonidentifiable (Hui and Walter (1980)). 2.2 Known Sensitivity and Specificity If the misclassification probabilities a and /? are known, Barron (1977) and Greenland and Kleinbaum (1983) present a relatively straightforward proce dure for the adjustment of odds-ratio estimates. To illustrate their methodol ogy, suppose that a and j3 are the sensitivity and specificity of a test used to assess the exposure, while a* and j3* are the sensitivity and specificity of a test used to assess the disease in a case-control study. Note that their method does not require the assumption that the disease status is known exactly. Suppose further that (xTi,yTi) and {xMi,yMi) are, respectively, true and misclassified cell counts in a 2 x 2 table (see Table 2.1). When there is no misclassification, we estimate the odds-ratio by <h=——> (2.1) XTOVTI 10 and if the exposure is subject to misclassification, we estimate the odds-ratio by (2.2) XMIVMO The misclassified cell counts are related to the true cell counts by the set of four equations m = Ct, where m %M0 %T0 VMO VTO ,t = VM\ VTI and a/3* (1 — (3)8* a(l-a*) {I - (3){l - a*) (l-a)fi* /38* (l-a)(l-a*) 8(1 - a*) a(l-8*) (1 — ^)(1 — aa* (1 - 8)a* (l-a)(l-8*) (3(1-/3*) (l-a)a* pa* Therefore, if (a, (3, a*, 8*) are known and C is invertible, a correction formula is easily derived, since t — C_1m. This, in turn, enables us to compute the estimate of the true odds-ratio </>T-We now present a simple approach to Bayesian inference about odds-ratio when the sensitivity and specificity are known. Let XQ be the number of apparently exposed controls and Xx be the number of apparently exposed cases. A control can be truly exposed and cor rectly classified as exposed with probability (TXQO) or unexposed and incorrectly classified as unexposed with probability (1 — 7r0)(l — B). Similar is true for the 11 cases. Therefore, X0 and X\ are distributed as independent Binomial(Ni,9i) random variables, where .0i = 7TiQ!+(l-7ri)(l- ^,1 = 0,1. (2.3) Here, 90 and$i are apparent exposure prevalences among controls and cases, respectively. Suppose now that a and B are known exactly, without error. Further, note that an estimate of (j> can be obtained from estimates of 9Q and 9X, as 7r0/(l-7r0) (gi+)9-l)/(a-gi) ( . (e0 + p-i)/(a-e0y l'j Thus, inference about the odds-ratio (f> is simple. To illustrate this, suppose that independent Uniform[0,1] priors are chosen for 7r0 and 7Ti, i.e., no prior knowledge is assumed about the prevalence of exposure in controls and cases. This choice of prior implies Uniform [min(l — B,a), max(l — B, a)] priors for 90 and 9i, since (2.3) implies 9i G [min(l - f3, a), max(l — B, a)], i = 0,1. Consequently, 9i\xo,Xx follows a Beta(xi + 1, Ni — Xi + 1) distribution truncated to the interval [min(l — 8, a), max(l — B, a)], yielding the posterior distribution of (j)\x0,xi. Therefore, sampling from the posterior distribution of odds-ratio is straightforward. Examples illustrating this approach will be shown in section 2.4.3. 12 We now examine a more realistic scenario, when good guesses rather than exact values are available for the sensitivity and specificity. These guesses could be available, for instance, via previous studies. If these guesses are treated as exact, Marshall (1989) has demonstrated that even small differ ences between the guessed and true values of misclassification probabilities can lead to very erroneous odds-ratio estimates. More specifically, consider the following example. Table 2(a) below shows the true distribution of ex posure among 300 cases and 300 controls. In this case, logc/> = 1.35, with a 95% confidence interval of (0.90, 1.80). This confidence interval was computed assuming the asymptotic normality of logc/>. Table 2(b) presents the effect the misclassification of 10% has on the odds-ratio. Here, logc/> = 0.85, with a 95% confidence interval of (0.47, 1.23), thus lessening the apparent effect of expo sure. Of the 30 exposed controls, only 90% are correctly identified as exposed and 10% are mistakenly identified as not exposed. Ignoring the sampling error, of the 270 unexposed controls, 90% are correctly classified with 10% classified as exposed. Hence, the 54 controls classified as exposed in Table 2(b) include 90% of 30, or 27, who are exposed and 10% of 270, or 27, who are unexposed. The same process applies to cases. The investigator who knows that the misclassification is 10% can easily adjust the data in Table 2(b) and correctly estimate the odds-ratio. A slight miscorrection can yield an erroneus estimate of the odds-ratio. To illustrate this, suppose that the investigator may guess, on the basis of previous studies, that the misclassification is 14%. Table 2(c) displays the result of adjusting 13 (a) Test + Controls 30 270 300 Cases 90 210 300 600 (b) Test (c) • Test Controls + + — 54 2 If) 300 Controls 17 283 300 102 198 300 Cases 83 217 300 600 600 Table 2.2: (a) True distribution of exposure among cases and controls, (b) Observed distribution of exposure, given 10% misclassifcation among cases and controls, (c) Estimated distribution of exposure assuming 14% misclassifcation among cases and controls. the data in Table 2(b) using the guessed value. Here, log</> = 1.85, with a 95% confidence interval of (1.30, 2.40), thus producing an exaggerated estimate of the effect of exposure. As shown in the above example, odds-ratio estimates can be very sensi tive to small discrepancies between the actual and assumed values of a and 8. We investigate this further by looking at the asymptotic bias of the odds-ratio that arises when incorrect values of the sensitivity and specificity are assumed. 14 To that end, let a' and be the assumed values of a and ft used to correct the odds-ratio estimate. We say that "miscorectioh" occurs when (a1, /?') 7^ (a, 8). Moreover, we say that miscorrection is "asymptotically unde tectable" if both #o and Q\ lie in the interval [min(l — B', a'), max(l — B', a')}. On the other hand, we say that miscorrection is "asymptotically detectable" if a' and 8' are such that one or both of 60 and 0\ lie outside the inter val [min(l — 8', a'), max(l — 8', a')}. This distinction comes from the fact that, in the large sample case, the values of 90 and 0\ are efectively known exactly. So, no amount of data could detect that miscorrection occurs if 6i e [min(l - 8', a'), max(l - B',a')],i = 0,1. Therefore, as N0 and Nx in crease, the exposure prevalences converge to 7r- = (0j + ft' — l)/(a + 8 — 1). Consequently, the posterior distribution of cf> concentrates at * ~ (d0 + 8'-l)/(a'-60y {2-b) The difference $$j>' — 4>\ is the asymptotic bias. The following theorem illustrates potentially very dangerous consequences of a very small miscorrection. Theorem: Suppose a and ft are fixed, with a + 8 > 1. Let ea > 0, e/j > 0 and R > 0 be arbitrary. Then, there exist a', 8', TV0 and TT\ such that (i) there is asymptotically undetectable miscorrection, (ii) \a' — a\ < ea, \fi' — fi\ < ep and (iii) l<\> = R. Proof: Note first that </>' = Rcj), where = [l + cM/[l 4-^/(1-^)] [l + c/7r0]/[l + d/(l-7r0)]' 15 c = (8' - 8)/(a + 8 + 1) and d = (a' - a)/(a + 8 + 1), Note also that the requirement for asymptotic undetectability is equivalent to — c < ^ < 1 + d, i = 0,1. Now, choose a' so that \a' — a\ < ea and d > 1 and choose 8' such that \B' — 8\ < ep and c G (—1,0). Therefore, asymptotic undetectability holds for |c| < 7Tj < 1, z = 0,1. So now equation (2.6) is given by _ [1-lcl/TTx] / [l + d/(l-7n)] . • [1 - |C| /TTo] / [1 + - TTo)] - ^ ' Hence, R increases to infinity as 7To decreases to |c| (or TT\ increases to one) and decreases to zero as 7Ti decreases to |c| (or 7r0 increases to one), as required. Here we considered the case where a' > a and 8' > 8. It is easy to establish the claim for other three cases. It is worth noting that R can go to either zero or infinity without ir0 or 71"! going to either zero or one, which are rather unrealistic cases. Secondly, the assumption that a + 8 > 1 is not unreasonable, given that no case-control study would be carried out if the false positive and false negative rates of the classification procedure are very high (higher than 0.5, say). The previous work, coupled with the result of the above theorem, sug gests that it seems reasonable to include the uncertainty in the available guesses into the analysis. Bayesian methods make this feasible, as the lack of exact knowledge of a and 8 is easily incorporated via the appropriate choice of priors for a and 8. This methodology was introduced by Joseph, Gyorkos and Coupal (1995) in the context of applying R different diagnostic tests to S populations. Specifically, they focused their attention on one population and 16 one test (R = 1,5 = 1) and one population and two tests (R = 2,5 = 1). Consequently, their inference was on ,the sensitivity, specificity and disease prevalence of the population of interest. We will attempt to extend their ap proach to the case-control setting. The Gibbs sampler will be used to draw the posterior samples of a, 8, 7TQ and TX\ and consequently of 0. Let Ai and B~i be the information that is missing when the test used to assess the exposure is imperfect, that is the number of true positive test results out of ai and 6j, respectively. Thus, Ai is the number of true positives and Bi is the number of false negatives, i = 0,1. This missing information is called "latent data" and analysis of such data, called "latent class analysis", has been done by Kaldor and Clayton (1985) and Walter and Irwig (1988). Incorporating this latent information into Table 2.1, we have (Table 2.3). Using the independence between cases and controls, the likelihood func tion of the data in Table 2.3 is given by 2.3 The Gibbs Sampler 1(AQ, B0, Ai, Bi,x0,y0,xuyi\n0,7Ti, a, 8) x [^{\-a)]Bi [(1-^)8] x [7rlC^[(l-^)(l-/?)] \Vi-Bi 17 + Apparent exposure Controls True Exposure Cases True Exposure + — + — x0 - A0 x0 + xx - Ax Xi BQ yo - Bo yo — Bx yi-Bx y\ N0 Table 2.3: Distribution of observed and latent data when one imperfect clas sification procedure is used. Symbol + denotes exposed, and — denotes unex posed. By gathering the like terms, the likelihood becomes Z^o.BcAi.Bi.xo.yo.^i.yikcTi.a,/?) = II [Ail ...(yi- Bi)\ x irfi+Bi(l-TTT)NI-A>-B> x a Ao+Ai Bo+Bi x pyo+yi-Bo-Bi ^ _ p^xo+xi-Ao-Ai (2.8) The likelihood (2.8) is nonidentifiable if (a,/3,7r0,7Ti) are unknown. We now state a more formal definition of nonidentifiability, introduced by Dawid (1979), Suppose that the Bayesian model is denoted by likelihood l(x\A) and prior p(A), where A = (Ai,A2). We say that A2 is nonidentifiable if p(A2|Ax, x) = p(A2|Ai). In other words, A2 is not identified by the data if observing data x 18 does not increase our knowledge about A2 given Ai. However, nonidentifiabil-ity does not imply that there is no Bayesian updating, i.e., it does not imply that p(X2\x) = p(A2). Furthermore, since p(A2|Ai,x) oc l(x\A)p(X2\Xi)p(Xi), A2 is nonidentifiable if and only if l(x\A) is free of A2. Hence, the definition of nonidentifiability, as introduced by Dawid (1979), is equivalent to nonidentifi-ability in the likelihood. We now turn our attention to the choice of prior for a, ft, ir0 and iii. We will choose beta densities to represent the prior information available for a, 8, 7T0 and 7Ti. The reason for this choice of prior is three-fold. First, the beta density is positive on the interval [0,1], which coincides with the range of all parameters of interest. Secondly, the family of beta densities is flexible, in the sense that a variety of shapes can be chosen by selecting different values of the hyperparameters. Finally, it is the conjugate prior distribution for the binomial likelihood, significantly easing the derivation of the posterior distribution. Therefore, suppose that the four parameters are independent a priori, with a ~ Beta(aa, fj,a), 8 ~ Beta(ai8,^), 7T0 ~ Beta(a7ro,^7ro), 7T! ~ Beta(a^1,^7ri), giving a joint prior density p(a,/3,7r0,7ri) = pa(a)pp{P) J[ p^fa). . (2.9) i=0,l 19 Since a posterior density is proportional to a likelihood multiplied by a prior, we have that the posterior density p(ir0, TT\, a, B\Ai, Bi, x0, y0, xx, yi) is propor tional to n i=0,l x a Al\...{yl-Bi)\ A0+Ai+aa-l^ _ ^Bo+Bx+Hc-1 (2.10) Note here that the latent data Ai and Bi, i = 0,1 are not observed, hindering the use of (2.10) in calculating the marginal posterior densities of a, B, ir0 and 7Ti. However, the inference is made possible by using a Gibbs sampler. This is a very useful technique for sampling from a p-dimensional distribution. Here is a brief review. Suppose gx is the joint density distribution function of a p-dimensional random variable X = (Xx,X2,..., Xp), with the univariate conditional densi ties qx1\X2,x3,...,xp,Qx2\xux3,...,Xp, and so on. To implement the Gibbs sampler, we start with initial guesses of the Xi, say .x[°\ X^,..., X^ and simulate xfWUf,...,^0) from qXl\x2,x3,..,xP, X^\X[l\xi°\...,X^ from qx2lxux3,..,xP, X^X11\X£\...,XJ»1 from qx^x,,...^-This is repeated k times, generating the sample X(fc) = (XJfe\ , ..., X^). 20 At each stage the conditional distribution uses the most recent values of all the other components of X. It can be shown that, as k —> oo, the density of the samples approaches qx- In practice, the convergence is usually quite rapid. Once the convergence has occurred, subsequent samples can be gen erated either by restarting the algorithm with the new guesses, or continuing the algorithm at the current value X(fc). We can now use the Gibbs sampler to sample from the posterior distri bution (2.10). We have the following conditional densities: a\A0, B0, Ai, Bx, aa, \ia ~ Beta(A) + Ax + aa, B0 + Bx + fj,a), B\x0,y0, xuyl,AQ, B0, Ax,Bi, o&, up ~ Beta(y0 + Hi - B0 - Bx + ap,x0 + xi - Ao- Ai+ fj.p), ni\Ai, Bi, Xi, yh aWi, /j,^ ~ Beta(Ai + B{ + o^Xi + y, - A{- Bi - nni),i = 0,1, Ai\iri, a, B,Xi ~ Binomial^, v.a+{l7!^%){l_p)), i = 0,1, Bi\irl,a,B,yl ~ Binomial^, v.{1l£)+_1<.)p),i = 0,1. Hence, conditional on knowing the exact values of a, B, 7r0 and nx, we can easily sample from the posterior distributions of the latent variables Ai and Bi. Conversely, conditional on A{ and Bt, sampling from the posterior distributions of a, 8, 7r0 and 7Ti, and consequently from </>, is straightforward. 2.3.1 Examples Example 1 To illustrate the use of the Gibbs sampler, consider a scenario 21 in which the true value of sensitivity is a — 0.85, specificity 8 = 0.90 and the prevalences of exposure in controls and cases is, respectively, ir0 = 0.08 and 7Ti = 0.12. These values imply that 4> = 1-57 or logc/> = 0.45. Data are simulated using the true values for three different sample sizes, namely for JV0 = Nx = 200, N0 = Nx = 800 and N0 = Nx = 3200, with re spective simulated values of (x0,Xi) = (37,41), (x0,xi) = (128,147) and (x0,xi) = (495,603). The prior for the prevalence of exposure in controls and the prevalence of exposure in cases was chosen to be Beta(l, 1). No prior knowledge was assumed for 7r0 and -K\, since the motivation for doing the case-control study is to make inference about odds-ratio. Further, suppose that the researcher's guesses at the sensitivity and specificity are 0.83 and 0.91, respectively, and that these guesses are accurate to within ±0.05. The infor mation of the form 7 ± S could be translated into a Beta prior with mode at 7 and 0.95 probability on the interval 7 ± 6. This implies 5^(2(183.50, 38.38) prior for sensitivity and £eta(128.50,13.61) prior for specificity. Five indepen dent chains of the Gibbs sampler were run with different starting values. The output is shown in Figure 2.1, Figure 2.2 and Figure 2.3. The plots shown in these figures clearly indicate drifiting behaviour by the Gibbs sampler and a lack of adequte mixing, even after the first 10000 observations were discarded. For example, as the plots in Figure 2.1 indicate, convergence was not achieved for none of the parameters. This may not be surprising because, as Gelfand and Sahu (1999) note, drifting behaviour may arise when the Gibbs sampler is applied to nonidentifiable models. The drifting 22 behaviour also seems to be the case for intermediate and large sample sizes, however, it is not as pronounced. Even though more formal diagnostic methods were not used in assessing the convergence of the Gibbs sampler, we believe that the plots in Figure 2.1 through Figure 2.3 present strong enough evidence of the inadequacy of the Gibbs sampler in this case. Example 2 Consider now a different scenario where the true value of sensitivity and specificity is a = (5 = 0.95 and prevalence of exposure in controls and cases is 7r0 = 0.06 and TTX — 0.15, respectively. These values imply 4> = 2.76 or logc/> = 1.02. Using these true values, we simulated data (x0,xi) = (13,44), (x0,xi) = (68,141) and (x0,xi) = (340,595) for N0 = Nx = 200, N0 = Nx = 800 and N0 = Ni = 3200 respectively. As before, Beta(l, 1) was used as prior for both the prevalence of exposure in controls and the prevalence of exposure in cases. As for the sensitivity and specificity, we will assume that our guess at both the sensitivity and specificity is 0.95, and that it is accurate to within ±0.05. This translates into Beta(99.70, 6.19) prior for both the sensitivity and specificity. Five independent chains were run with different starting values. The output is shown in Figure 2.4, Figure 2.5 and Figure 2.6. We observe a similar behaviour of the Gibbs sampler as in Example 1, even though our guesses at the true values of the sensitivity and specificity were exact. For instance, note the behaviour of the fourth chain in Figure 2.4. It exibits a completely different behaviour than the other four chains in the sense that it drifts towards a completely different value in the parameter space. 23 In particular, the mean value of the sensitivity obtained from the fourth chain is 0.03, whereas the mean value obtained from the other four chains is 0.57, a very big discrepancy. Moreover, no point estimates were close to the true value of 0.95 for the sensitivity. The 95% HPD credible intervals of the sensitivity and the prevalence of exposure in cases for every sample size considered did not contain the true values. 2.4 The Metropolis-Hastings Algorithm The behaviour observed above may not be surprising. Gelfand and Sahu (1999) have noted that drifting behaviour in the Gibbs sampler can arise if it is applied to nonidentifiable models. Therefore, the extension of the methodology of Joseph, Gyorkos and Coupal (1995) to the case-control setting does not seem to be appropriate. Instead, we use the the Metropolis-Hastings algorithm in the reparameterization (2.3). This reparameterization separates the identifiable parameters (90,9i) from the nonidentifiable parameters (a, 8). In this new parameterization, (2.9) becomes p(a, 8,90,9,) 1 Pa(a)pp(B) (a + 8-iy x (2-11) since the Jacobian is given by .7(0o,0i) duo d80 dir\ ae0 1 d6i d-Ky 001 (a + B- l)2' 24 J|C4m/ ^iJW/ 3000 INDEX Wf^pwi^ "*^y^V yv^^V ?w****fw 3000 INDEX 3000 INDEX 3000 INDEX Figure 2.1: Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in controls and prevalence of exposure in cases in Example 1. Data was simulated for JV0 = Nx = 200. 25 ^fVAA> wa hjV A,..,. 3000 INDEX 3000 INDEX ^..^.A 3000 INDEX >^ AVXJ 3000 INDEX Figure 2.2: Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in controls and prevalence of exposure in cases in Example 1. Data was simulated for N0 = N1 = 800. 26 3000 INDEX 3000 INDEX 3000 INDEX 3000 INDEX Figure 2.3: Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in controls and prevalence of exposure in cases in Example 1. Data was simulated for iV0 = jVi = 3200. 27 v/w«m wAmi* XjAJ 0 1000 2000 3000 INDEX 4000 5000 600 mV^WW^^ V^^VWW^W^^rf^ 'WV^VfTYWV^^VVVr^ 0 1000 2000 3000 INDEX 4000 5000 60O 4JUU^AJ*« UWUJAWUJW \i/MfMj/^ 0 1000 2000 3000 INDEX 4000 SOOO 600 a/VIW vv\ WW 0 1000 3000 3000 4000 5000 6000 INDEX Figure 2.4: Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in controls and prevalence of exposure in cases in Example 2. Data was simulated for N0 = Ni = 200. 28 ijl v^, w v^v a^t 0 1000 2000 3000 4000 5000 6000 INDEX 0 1000 2000 3000 4000 5000 6000 INDEX °- 0 1000 2000 3000 4000 5000 6000 INDEX ,AAAA^ g 6 i . . . , . . r 0 1000 2000 3000 4000 5000 6000 INDEX Figure 2.5: Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in controls and prevalence of exposure in cases in Example 2. Data was simulated for jV0 = Nx = 800. 29 0 1000 2000 3000 . 4000 5000 6000 INDEX 0 1000 2000 3000 4000 5000 6000 INDEX Figure 2.6: Post burn-in output of the five independent chains of the Gibbs sampler for sensitivity, specificity, prevalence of exposure in controls and prevalence of exposure in cases in Example 2. Data was simulated for N0 = Ni = 3200. 30 Therefore, the joint posterior distribution of (a, B,90,9x) given the data is Here, the prior and posterior conditional distribution of a, 8\9o, 9\ are identical, and hence, according to Dawid (1979), (a, 8) are nonidentifiable parameters. However, as a result of learning about (90,9X), the marginal prior and posterior distributions of the sensitivity and specificity are not equal. Thus, one could learn about (a,B) indirectly, through the updating of {60,9x)-2.4.1 Large Sample Case Let us examine (2.12) further. Since p(90, 9x\xo, xx) and p(90,9x) constitute a regular model and prior, the posterior distribution of 90,9i\x0, x\ will converge to a point mass at the true value of (90,9x), as No —» 00 and iV\ —> 00. The uncertainty in p(a, 8, \90, 9{) remains unchanged by any amount of data. Note that we can express it as p(a,/3|0o,0i) cx p{90,9l\a,B)p{9Q,9l) oc p(90\91,a,8)p(9l\a,p)pa(a)pi}(p) oc p(9o\a, B)p(0i\a, B)pa(a)pp(B) 31 p(a, 8,90,9^X0, xx) p(xQ, XX\90, 9X, a, 8)p(a, 8,90,9±) p(x0,xi)'. p(xo,xl\9o,9l)p{a, B,9Q,9I) p(x0,xi) p(90,91\xo,xi)p(a,B, \90,9x). (2.12) * (a + B- i)2/^o,gi)(Q;> 8)pa(<x)Pf>(P), (2-13) where A(90,9l) = {{a,B) :9< a < 1,1-0 <B< l} U {(a, B) : 0 < a < 0,0 < 8 < 1 - 0} , and 0 = max{0o, #i} and 9 = min{#0, ^j. This region is shown in Figure 2.7. p (1.1) 1-9 1-6 \a+p=l (0. (!) 6 6 a Figure 2.7: The support of p(a, B\QQ, 9X). The shaded rectangles comprise A(90,9l). From (2.13) we see that p(a, B) differs fromp(o;, B\90, 9$$ in two respects. First, the support of p(ot, B) is the whole unit square, whereas the support of 32 p(a, B\90,9i) is A(90,9i). Blettner and Wahrendorf (1984) observed this in the context of finite samples. They describe how to calculate the range of the possible underlying true effects, measured by odds-ratio, and of the possible misclassification probabilities, given the observed data. Second, the density p(a, B\90, 9\) has (a + B — 1)~2 as a factor. It is, of course, infinite along the line a = 1 — B. The rectangles that make up the support A(90,9i) are on either side of this line, implying that (2.13) is bounded. In particular, (Q + p _ 1p J^(0o,fli)0*. P)P*(a)Pp(P) < ^ ]_ e^2lA{B0A)(ai B)pa(a)pp(/3). (2.14) Therefore, we can sample from (2.13) using the "acceptance/rejection algo rithm". To sample from a random variable X, this algorithm makes use of samples of another random variable, say Y,- whose probability density func tion gy is similar to the probability density function of X, pX- The random variable Y is chosen so that we can easily generate samples of it and so that its density gy can be scaled to majorize px, using some constant c; that is, so that cgy(x) > px(x), for all x. The density gY is called the "majorizing density" and cgY is called the "majorizing function". The closer cgy(x) is to px(%), the faster the acceptance/rejection algorithm will be. In our case, the majorizing density is pa(a)pp(B) truncated to the interval A(9o, 9i). For a more detailed description, see Gentle (1998). However, we have not found the acceptance/rejection algorithm to be very efficient, and instead used the "Metropolis-Hastings algorithm". The Metropolis-Hastings algorithm helps when it is difficult to sample from the 33 target distribution, say TT(X), but there is an easy way to sample from an other, say q(x,y), called the "candidate-generating density". In our case, the candidate-generating density is pa{o)pp{B) truncated to the interval A(90,9i), same as the majorizing density above. The Metropolis-Hastings algorithm proceeds by constructing a Markov Chain which has ir(x) as its stationary distribution. We present a brief illustration of how the Metropolis-Hastings algorithm works. For a complete description, see Chib and Greenberg (1995) or Gentle (1998). Let X be an irreducible, recurrent Markov Chain, with the stationary distribution n(x), and let x be its current state. The Metropolis-Hastings algorithm proceeds by sampling a candidate value y from q(x, y), and choosing the next state of the Markov Chain to be either the candidate value y or the current state x, depending on the following acceptance probability a(x,y) = < (2.15) 1 otherwise The process is repeated k times. Of course, the draws are regarded as a sample from the target density ir(x) only as k gets large. The above algorithm has a drawback in comparison with the Gibbs sampler: the degree of serial correlation is increased due to the possibility of rejection of a move. If q(x, y) is such that the probability of a move is very small, the chain will be characterized by long sequences of repeated outcomes. However, a very convenient feature of the Metropolis-Hastings algorithm, apart from the free selection of the candidate-generating density q(x,y), is that we only need to know ir(x) up to a normalizing constant (as it cancels out in the 34 ratio n(y)/7r(x)). Different choices of q(x, y) give rise to different variants of the Metropolis-Hastings algorithm. If q(x,y) is symmetric, i.e., if q(x,y) = q(y,x), the prob ability of a move reduces to min7r(y)/7r(x). This is known as the "Metropolis algorithm". Another family of the candidate-generating densities is given by q(x,y) — q(y — x), characterizing the "random walk" chain. The candidate value y is given by y = x + z, where Z is a random variable drawn from q(z). Our choice of q(x, y) yields what is known as the "independence" chain. It arises when the candidate-generating density is not a function of the current state, i.e, when q(x,y) = q(y), for some density q(-). Other choices of the candidate-generating density are possible. 2.4.2 Finite Sample Case In a finite sample case, we have demonstrated that the Gibbs sampler does not work very well. Instead, we try to sample from (2.12) using the Metropolis-Hastings algorithm. Note that (2.12) can be expressed'as p(a,/5,0o,0i|zo,zi) oc ex00(l-e0)No-xo9^(1-6^-^ x (a + p _ 1)2/-A(go.gi)(Q!» P)pa(a)pp{B). (2.16) The candidate generating density that approximates (2.16) from which we can sample to drive the Metropolis-Hastings algorithm is /(a,)Mo,0i) = ^^/O(0O)/I(0I)/A(0o^O(«^K(«)^(/5)- (2.17). 35 Here, fc(0o,0i) = Pr{(a,£) G A(9O,01)}, so that /(a, /3|0O,#1) is the prior /?) truncated to A(0O, #i)- Furthermore, /, is the Beta(xi + 1, iVj — xi + 1) density, so that p(a, B\9o, 9\) in (2.12) is approximated by f(d0,9i) by replacing the intractable prior for (0O, 9\) by a uniform prior on the unit square. There fore, we can sample from (2.17) by first sampling from f(90, 6\) and then from f{a,B\90,9x). We now present three examples illustrating the use of Metropolis-Hastings algorithm. 2.4.3 Examples Example 1 Consider a scenario in which the true value of sensitivity and specificity is a = B = 0.84, and the prevalences of exposure in controls and cases is, respectively, 7r0 = 0.061 and -KX = 0.15, implying that 90 = 0.20148, 9\ = 0.262 and log0 = 1.00. Furthermore, suppose that our best guesses at the sensitivity and specificity are (a',B') = (0.81,0.81) and that these guesses are accurate to within ±0.05. This leads to a Beta(19A.OO, 46.27) prior for both a and B. Under the guessed values, log^' = 1.94. Data are simulated un der this scenario for three different sample sizes, namely for N0 = Nx = 200, iV0 = Nx — 800 and NQ = Nx = 3200, with respective simulated values of {x0,Xl) = (41,54), (x0,xx) = (162,214) and (x0,xx) = (661,861). Three dif ferent posterior distributions were used to generate the samples of log </> for each of the data sets. The first, named "miscorrected" (M-COR) posterior, originates from assuming (a,B) = (a',B'). The second, named "uncertainty-36 corrected" (U-COR) posterior originates from assuming a prior distribution on a and 8. The third, named "exactly-corrected" (E-COR) posterior arises from the knowledge of the exact values of the sensitivity and specificity. The sam pling from M-COR and E-COR proceeds as described in section 2.2, whereas from U-COR as per section 2.4.2. Furthermore, we consider the large sample case where the M-COR and E-COR concentrate at 4>' and </>, respectively, and the sampling from U-COR proceeds as per section 2.4.1. All posterior distributions are shown in Figure 2.8. There is very little difference between the three posteriors for N0 = Ni = 200. At the intermediate sample size, M-COR and U-COR posteriors are similar, with E-COR posterior being more peaked. A greater difference between M-COR and U-COR poste riors appears for the large sample size. The U-COR posterior covers both the true value of log<?!> and the wrong value log</>', whereas the M-COR posterior misses the true value and is drawn towards log0'. Even in the large-n limit, the U-COR posterior is quite wide, covering both values. However, this should still be preferable to the M-COR posterior which, asymptotically, concentrates at the wrong answer. Convergence is very slow in the present scenario. To assess the coverage of the credible intervals under each posterior, we simulate forty data sets under the same scenario for the large sample size N0 = Ni = 3200 and compute 80% HPD credible intervals for \og(j) under the three posteriors. These intervals appear in Figure 2.9. The empirical coverage rates for the intervals are 22%, 90%, 90% and the average interval widths are 1.76, 1.64, 0.56 for the M-COR, U-COR and E-COR intervals 37 respectively. Note the very poor empirical coverage exhibited by the M-COR credible intervals as many of the intervals are drawn toward log</>'. On the other hand, the U-COR intervals show overcoverage. However, the cost of admitting uncertainty seems to be high, as the average length of the U-COR intervals is 1.64 on the log scale. As mentioned in section 2.4.1, the prior and the posterior distribution for (a, 8) differ in two respects, one of which is the support. Figure 2.10 shows the effects of truncation to A(9o,9i). We see that this effect is most evident for the case NQ = Nx = oo. Furthermore, note that the truncation affects 8 but not a. This will usually be true for small exposure prevalences. When 7r0 and ITi are small, 1 — 90 and 1 — 9± will lie in the region where draws from pp(8) are likely, whereas 90 and 9\ will be in the region where draws from pa (a) are not likely. Example 2 Consider now a scenario in which the true values of the sensitivity an specificity are a = 0.84 and 8 = 0.78, and all other quantities are the same as in Example 1. This implies 90 = 0.25 7 82 , 9\ = 0.313 and log</>' = 0.70. Under these conditions, we simulate data (x0,xi) — (55,67), (z0,zi) = (190,256) and {x0,xx) = (830,966) for N0 = Nx = 200, N0 = Nx = 800 and N0 = Ni = 3200 respectively. Posterior distributions for these data are shown in Figure 2.11. For the small sample, size, there is very little difference between the M-COR and U-COR posteriors. For the intermediate and large sample sizes, the U-COR posterior performs better in the sense that it is closer to the E-COR posterior. The M-COR posterior again misses the 38 true value of the log odds-ratio. Forty data sets are again considered to assess the coverage of the cred ible intervals for the large sample size under each posterior. The 80% HPD credible intervals are shown in Figure 2.12. The empirical coverage rates for the intervals are 37%, 87%, 82% and the average interval widths are 0.40, 0.77, 0.67 for the M-COR, U-COR and E-COR intervals respectively. Again, we see the undercoverage of the M-COR intervals, and the slight overcoverage of the U-COR intervals. In terms of the average widths of the intervals, the cost of admitting uncertainty in this example does not seem to be as high as in Example 1. Example 3 Finally, suppose that our best guesses at the sensitivity and specificity are (a1, 6') = (1,1), analogous to no misclassification. Fur ther, suppose that we feel that there could be small misclassification, so that Beta(58Al, 1) was assigned as a prior to both a and /5. This prior density is a strictly increasing function with finite maximum at 1, and 95% probability assigned to the interval [0.95,1]. Suppose that the prevalences are the same as in Example 1, but the sensitivity and specificity are a = 6 = 0.97. These val ues imply 60 = 0.08734, 9i = 0.171 and log</>' = 0.77. Again, we simulate data under this scenario for NQ = Nx = 200, N0 = N± = 800 and N0 = Nx = 3200, with respective simulated values being (x0,xi) = (23,31), (x0,xi) = (63,140) and (x0,xi) = (285,575). All posteriors appear in Figure 2.13. Again, for the small sample size, there is very little difference between the three poste riors. However, in the case of the intermediate and the large sample size, the 39 U-COR posterior is closer to the E-COR posterior than is M-COR posterior, showing once more the value of admitting uncertainty about the sensitivity and specificity. Again, we simulate 40 data sets under this scenario for No = Nx = 3200. The resulting 80% HPD credible intervals are shown in Figure 2.14. Similar to Examples 1 and 2, the empirical coverage of the M-CQR intervals is very low (3%), while 85% of the U-COR intervals cover the true value. The average interval widths are 0.20, 0.41, 0.28 for the M-COR, U-COR and E-COR intervals respectively. Hence, the cost of admitting uncertainty is not nearly as severe as in Example 1. 2.5 Discussion So far, we have focused our attention to a case-control setting where one imper fect classification scheme is used to assess the exposure status. The odds-ratio estimates obtained from the observed data, if not corrected for misclassifica tion, can be very imprecise. Relatively straightforward correction methods are available if the classification probabilities are known. However, the odds-ratio estimates which are corrected using slightly, inaccurate classification probabil ities may still be quite erroneous. In particular, we have demonstrated that even arbitrarily small differences between the true and assumed classification probabilities can, asymptotically, lead to arbitrarily large difference between the actual odds-ratio and the odds-ratio obtained using the assumed classifica tion probabilities. By admitting uncertainty about the classification prpbabil-40 ities, this problem seems to be alleviated. Two computational approaches to this analysis were suggested: the Gibbs sampler and the Metropolis-Hastings algorithm. The Gibbs sampler did not seem to work particularly well. This was not surprising since, as Gelfand and Sahu (1999) noted, drifting behaviour of the Gibbs sampler may occur if it is applied to a nonidentifiable model. On the other hand, the Metropolis-Hastings algorithm applied to the model that separates the identifiable and the nonidentifiable part seemed to perform well. Despite nonidentifiability, there was considerable learning about the odds-ratio from the data. The U-COR posterior was quite concentrated rela tive to the prior. However, the cost of admitting uncertainty is evident in the increase in variability of the U-COR posterior relative to the E-COR posterior. As a result, the U-COR credible intervals were twice as wide as the E-COR credible intervals. On the other hand, there seemed to be an improvement of the U-COR posterior over the M-COR posterior. This improvement ap peared to arise from marginal learning about the specificity but not about the sensitivity. 41 LOG.-ODDS LOG-ODDS LOG-ODDS LOG-ODDS LOG-ODDS LOG-OODS Figure 2.8: Posterior distribution of log <fi in Example 1. The first column gives the M-COR posterior, the second column gives the U-COR posterior, and the third column gives the E-COR posterior. The rows correspond to sample sizes N0 = Ni = 200, N0 = N1 = 800, iV0 = Ni = 3200 and N0 = Nx = oo. 42 M-COR 1 , 1 | t -J-!— 1 J 1 M"l | 1 -L-L -1 OATASET U-COR Figure 2.9: 80% highest posterior density credible intervals for log</> in Exam ple 1. The solid vertical lines represent credible intervals for forty data sets with sample sizes N0 = Nx = 3200, the solid horizontal line indicates the true value of log(j) and the dashed horizontal line indicates log</>'. The first panel gives the M-COR intervals, the second panel gives the U-COR intervals, and the third panel gives the E-COR intervals. 43 SENSITIVITY (a) Prior (b) Sample size: N0 = Nx = 200 (c) Sample size: N0 = Ni = 3200 (d) Sample size: iVo = Ni = oo Figure 2.10: Prior and posterior samples of the sensitivity and specificity the datasets in Example 1. 44 LOG-ODDS ' LOG-ODDS LOG-ODDS ........... « Figure 2.11: Posterior distribution of log(/> in Example 2. The first column gives the M-COR posterior, the second column gives the U-COR posterior, and the third column gives the E-COR posterior. The rows correspond to sample sizes. N0 = Ni = 200, N0 = Nx = 800, N0 = Nx = 3200 and N0 = Nx = oo. 45 Figure 2.12: 80% highest posterior density credible intervals for log0 in Ex ample 2. The solid vertical lines represent credible intervals for forty data sets with sample sizes JV0 = Nx = 3200, the solid horizontal line indicates the true value of \og(f) and the dashed horizontal line indicates logfi'. The first panel gives the M-COR intervals, the second panel gives the U-COR intervals, and the third panel gives the E-COR intervals. 46 LOG-ODDS LOG-ODDS LOG-ODDS -0.6 0.0 0.5 1.0 1.5 2.0 -0.5 0.0 0.5 1.0 1,5 2.0 -0.5 0.0 0.5 1.0 1.5 2.0 LOG-ODDS LOG-ODDS LOG-ODDS -0.5 • 0.0 0.5 1.0 1.5 2.0 -0.5 0.0 0.5 1.0 1.5 2.0 -0.5 0.0 0.5 1.0 1,5 2.0 LOG-ODDS LOG-ODDS * LOG-ODDS Figure 2.13: Posterior distribution of log</> in Example 3. The first column gives the M-COR posterior, the second column gives the U-COR posterior, and the third column gives the E-COR posterior. The rows correspond to sample sizes N0 = NX = 200, N0 = Nx = 800, N0 =N1 = 3200 and N0 = Nx = oo. 47 M-COR Figure 2.14: 80% highest posterior density credible intervals for log</> in Ex ample 3. The solid vertical lines represent credible intervals for forty data sets with sample sizes N0 = Nx = 3200, the solid horizontal line indicates the true value of log</> and the dashed horizontal line indicates logc//. The first panel gives the M-COR intervals, the second panel gives the U-COR intervals, and the third panel gives the E-COR intervals. 48 Chapter 3 Two Test Case In the previous chapter we examined a case-control setting where exposure is measured with one imperfect test. We now turn our attention to the situa tion where the exposure status is determined by two imperfect classification schemes. We will illustrate the existing methods to correct the odds-ratio es timates, and examine our method via a simulation study and an analysis of data on the sudden infant death syndrome. 3.1 Data and Setup As in chapter 2 we define relevant parameters and introduce the general setup. Consider the data from a case-control study in which each subject has an underlying true, but unobserved exposure (E), coded as 1 for exposed and 0 for unexposed. This exposure is assessed by applying two measures or tests (say, Ti and T2) to each subject. These tests are coded as 1 for positive and 49 0 for negative outcome. We will assume that the classification procedures misclassify the subjects nondifferentially and that the disease status is known exactly, without error. When two tests are used to determine the exposure status, the data usually come in the following form (Table 3.1). Controls Cases Test 2 Test 2 + + a0 b0 a0 + b0 + Test 1 ai bi ax + bi Co d0 c0 + d0 — Ci di ci + di N0 Ni' Table 3.1: Distribution of subjects in a case-control study when two tests are used to assess the exposure status. Symbol + denotes exposed and — denotes unexposed. We define the following parameters to model the probabilities of the different possible outcomes in controls and cases: Oil = Pr(Tx = l\E = 1), a2 = Pr(T2 = 1\E = 1), = Pr(Tx = 0\E = 0), = Pr(T2 = 0\E = 0), with the exposure prevalences 7r0 and 7Ti and the odds-ratio 0 defined as before. 50 Here, ax and a2 are the sensitivities of test 1 and test 2, respectively, and B\ and B2 are the specificities of test 1 and test 2, respectively. 3.2 Correction Methods In this section we illustrate three different methods to correct the odds-ratio estimates using dual measurements. The first method uses only the concor dant data (i.e., the data for which two measurements are in agreement). The other two methods use the EM algorithm to obtain the estimates. This algo rithm is a useful way of obtaining maximum likelihood estimates when some data are missing. Each iteration of the EM algorithm involves two steps: the expectation step (E-step) and the maximization step (M-step). For precise definitions of these steps and the full description of the EM algorithm, see Dempster, Laird and Rubin (1977). 3.2.1 The Marshall and Graham Method Marshall and Graham (1984) consider the problem of exposure to a risk fac tor, where the "true" exposure is unknown. They proposed a simple way to decrease the bias in the odds-ratio estimates caused by misclassification, us ing two independent imperfect tests to gather information on exposure status. The method is based on restricting the analysis to data for which two inde pendent assessments of exposure are concordant, either positive or negative. Subjects for which both tests are positive are considered as exposed, and those for which both tests are negative are considered as unexposed. 51 They have demonstrated that the use of two tests in this fashion can provide a less biased estimate of the odds-ratio. Even if the second test is less accurate than the first, the contrast between subjects about whom there is test agreement yielded a better approximation of the odds-ratio than if only the more accurate report were used. There are, of course, obvious drawbacks of this method. The Marshall-Graham procedure uses only a subset of the data, the concordant observations. Clearly, by using only a part of the data, the statistical efficiency of the esti mate is decreased. Furthermore, this procedure does not completely remove the bias caused by misclassification. It is necessary that both the sensitiv ity and specificity of both assessment procedures be relatively high, in order to closely approximate the relative risk. For example, if the sensitivity and specificity are about 0.9, the prevalence of exposure in cases is 0.3 and the prevalence of exposure in controls is about 0.097, the true odds-ratio of 4.0 will be approximated by about 3.7. 3.2.2 The Drews, Flanders and Kosinski Method Drews, Flanders and Kosinski (1993) examine methods to use two classification schemes to improve the odds ratio estimates in case-control studies. Their assumptions are differ slightly from ours in that they do not assume that errors in one test are independent of errors in the second test. This is reflected in the following parameters used by these authors, in addition to the sensitivity and specificity of test 1, to model the probabilities of the different possible 52 outcomes in cases and controls: = Pr(T2 = 1\E = = 1), = Pr(T2 = 1\E = = 0), 02 = Pr(T2 = 0\E = 0,Ti = 1), &A*0 = Pr(T2 = 0\E = 0,71 = 0). Under this model, [i\ and represent deviations from a model in which, given true exposure status, errors in one test are independent of errors in a second test. Note also that if the misclassification probabilities of T2 are independent of the misclassification probabilities of 7\ (that is, //i = — 1),. a2 and B2 are not the sensitivity and specificity of T2. Since there are two tests applied to two populations, there are six degrees of freedom to estimate six parameters. Hence, the likelihood is identifiable, provided that we treat [ii and fi0 as known, which is precisely what the authors do. This might be a drawback because, in practice, the values of fix and fj,Q are not easily available. To derive the likelihood, consider an individual who was positive on both tests. This person could have been truly exposed and correctly tested positive on both tests (P = 7Tiotxa2), or the person could have been truly unexposed and falsely tested positive on both tests [P = (1 — 7Ti)(l — /?i)(l — B2Ho)]- The net contribution to the likelihood is then \rKiaxa2 + {l — 7r1)(l — B\)(l — (32fj,0)]ai, where ax is the number of individuals who tested positive on test 1 and test 2 and are cases. Similar contributions from other cells in Table 3.1 gives the 53 following likelihood: L = n [KkUi^+(i - 7rfc)(i - - /Mr fc=0,l X [7Tfcai(l - a2) + (1 - 7Tfc)(l - BJB^x x [Tr/bCl-aOCl-^/xO+^l-TrOA^r x [^(l-aOaaMi + Cl-Tr^ACl-^)]"*. (3.1) To estimate the parameters that maximize this likelihood, the authors use the EM algorithm. In this case, the missing data are the true number of exposed and the true number of unexposed subjects in each cell of Table 3.1. A very similar methodology was developed by Hui and Walter (1980). It is a careful application of the work done on the problem of evaluating the accuracy of a new diagnostic test against a standard test with unknown error rates. This method yields a similar likelihood to (3.1). Unlike Drews et al., however, they do not introduce /^ into their analysis and the maximization of the likelihood is done analytically. A drawback of this approach is that these estimates can sometimes lie outside the parameter space (i.e., outside the interval [0,1]). 3.2.3 The Kaldor and Clayton Method The methodology developed by Kaldor and Clayton (1985) uses latent class analysis to correct the odds-ratio estimate. To understand how their approach extends to a two-test case, we first introduce the general setup. 54 Suppose that all measurements made are categorical, and that the dis ease status D is measured without error. Suppose also that some variables Vi,...,Vr are measured without error, but each of the variables W\,..., Ws are measurements of one of t latent classes C\,..., Ct (unobservable true expo sure), where t < s, and are subject to misclassification error. Conditional on Cfc, the variables Wj are mutually independent, and independent of D and V*. Each possible outcome in the (1 + r + s + £)-dimensional table resulting from a cross^classification by all these variables can be represented by the vector x=(d, v,w, c), where d is disease status, and v,w and c are respectively r-, s- and ^-dimensional vectors representing categories of Vi, Wj and Ck- Fur ther, suppose that mx represent the number of individuals in category x. Of course, this variable is unobservable, since the true values of the latent classes are unknown. One can only observe m<ivw^ where the dot indicates summa tion over all categories of the t latent classes. The authors assume, however, that mx can be generated by log-linear model of the form log /J,X = XU^a > where [ix = E(mx). In terms of definitions above, for a model with two measurements per subject, we have r = 0 since there are no variables measured without error, and t — 1 because there is only one latent variable underlying the repeat measurements. Each of the variables D, C and Wj is dichotomous, taking the values 0 and 1. The latent class/logistic model is then of the form log fjiX = 6 + 9f + Bcc + 6™ + ELi(C + CH> where s is the number of repeat measurements made of the risk factor C. The authors utilize the EM algorithm 55 to estimate the relevant parameters. Here, the missing data are unobservable true exposure (C\,..., Ct). 3.3 Our Method Our method is an extension of the Joseph, Gyorkos and Coupal (1995) method for inference on the sensitivities and specificities of two classification proce dures applied to one population. We modify their method to apply to the two test, two population scenario. Let the unobserved latent data Ai} Bt, d, and Di, i = 0,1 represent the number of true positive subjects out of the observed.cell counts ai, bi} Ci, and di, i = 0,1, respectively, in the 2x2x2 table (see Table 3.1). In addition to the assumptions and definitions made in section 3.1, we will assume that the two tests determine the exposure status independently. We believe that this assumption is as reasonable as the assumption made by Drews, Flanders and Kosinski (1993), whereby they assume the degree of dependence is known. Let us now examine the possible outcomes involving the observed and the latent data. Since any.subject, whether truly exposed or not, can test positively or negatively on each test, there are eight possible combinations for both cases and controls. An individual can truly be exposed and correctly classified as exposed by both tests pTiaia^], or can truly be exposed and cor rectly classified by one of the tests and misclassified by the other [7TJQ!I(1 — ct2) or 7Tj(l — ai)o!2], and so on. Similarly, an individual can truly be unexposed and misclassified by both tests [(1 — 7Tj)(l — /?i)(l — 62)}, or can truly be un-56 exposed and correctly classified by both tests [(1 — n^Bifo], and so forth. All eight combinations are summarized in Table 3.2. Number of True Test 1 Test 2 Contribution to subjects exposure (Ti) likelihood Al + + + IXiO.iO.2 Bi + + —. 7TjQ!i(l - a2) Ci + — + 7Ti(l - a1)a2 A + — — 7Tj(l - «i)(l - a2) cii — Ai — + + (\ - - - h) h-B, — + — {l-n)(\-Bx)B2 ci ~ Ci — — + {\-*i)Bx{\-B2) di - Di — — — (1 -Table 3.2: Distribution of observed and latent data when two imperfect tests are used, together with the contribution to the likelihood each combination of the observed and latent data makes. Symbol + denotes exposed, and — denotes unexposed. Under the assumption of independence between controls and cases, the derivation of the likelihood function of the latent and observed data in Ta ble 3.2 is straightforward. Ni l(Ai,Bi,Ci,Di,ai,bi,Ci,di\iTi,ahBi) = JJ i0L>1Ai\...(di-Di)] x [TTiaia2]Ai [7TjQ!i(l - a2)}Bi x [7ri(l - a1)a2fi - ai)(l - a2)f> o-i-Ai [(1 B2 [(1 -7r0(l- B,)B2]BL -Bi [(1 -7T,)A(1 - B2)T -Ci [(1 - ^i)B^B2 \di-Di 57 By gathering the proper exponents and collecting the like terms, the likelihood can be rewritten as Ni n i=0,l ^Ai+Bi+d+Di ^ _ ^Ni-(Ai+Bi+Ci+Di) Ai\ ...(di - Di)\ X aAo+Bo+Ai+Bi ^ _ ^CO+DO+CI+DIQAQ+CO+AI+C!^ _ a^B0+D0+Bi+DX x pC0+d,o+ci+di-(Co+D0+Ci+Di) ^ _ ^^ao+60+ai+61_(A0+Bo+^i+JBi) X pbo+do+bi+di-tBo+Do+Bt+Di)^ _ ^ao+co+ai+ci-(A0+Co+Ai+Ci) (3.2) As in chapter 2, we will choose beta densities to represent the prior information available for at, Bi, and n^i = 0,1. Moreover, we will assume that the six parameters are independent a priori, with ai ~ Beta(crQi, fj,ai), i - 1, 2, Bi ~ Beta(a/3.,/i/8.),i = 1,2, 7r0 ~ Beta(a7ro,^0), 7Ti ~ Beta(crWl,^7ri), giving a joint prior density p(al,/31,a2,p2,ir0,'K1) = II Pf/fa) II P<*j(aj)PPj(Pj)- (3-3) i=0,l j=l,2 Since the posterior density is proportional to the likelihood function (3.2) and the prior distribution (3.3), it is easily seen that the posterior density is 58 X X proportional to TT Ni Aj+Bj+Ci+Di+v^-l, _ ^Ni-iAi+Bt+d+DO+^-l l{l[Al\...{dl-Di)^ V ^ • X af0+B°+Al+Bl+aal~l(l - a^Co+Do + Cy+Dy+lla.-l X aAo+Co+Ai+Ci+<ra2-l ^ _ a^Bo+Do+Bi+Di+na2-l ^co+d0+ci+d1-(Co+r»o+C,i+Ui)+o-^1-l^ _ p^ao+bo+ai+bl_(Ao+Bo+A1+B1)+iJ,i31-l ^6o+do+6i+di-(Bo+£'o+Bi+r'i)+<T/31-l^1 _ ^^ao+Co+ai+Cl_(J40+c0+A1-f-Ci)+^1-1 (3.4) We plan to use the Gibbs sampler to sample from (3.4). The implemen tation of this algorithm should be relatively straightforward as we have the following univariate conditional densities: o.\\Ai, Bi, Ci, Di, aai, jiai ~ Beta(^0 + BQ + Ax + Bx + aai ,C0 + D0 + Cx + DX +fiai), Bx\a,i, bi, dh Ai, Bi, d, Di, opx, (Apt ~ Beta(c0 + d0 + cx + dx - (C0 + D0 + Ci + Dx) + aPl,a0 + b0 + ax + 61 - (A0 + B0 + AX + Bx) +/ip,), ®21 A-i, Bi, Ci, Di, cQ2, fj,a2 ~ Beta(A) + Co + Ax + d + aa2,B0 + D0 + £l+Z>l+Ma2)> AI ^, 6i, q , ^, Ai, Bi, Ci, Di, a02, pp2 ~ Beta(60 + d0 + bx + di - (B0 + D0 + Bx + Dx) + o-^.oo + c0 + ax+cx- (A0 + C0 + Ax + Cx) + iip2), TVi\Ai,Bi,Cl,Di,di,bi,ci,di,ani,^i ~ Beta(Ai + Bt + Ci + D{ + an., Nt -(Ai + Bt + d + Di) + fiv.),i = 0,1, 59 Ai\m, on, Bu at ~ Binomial^, naia2+{1??\$-pl){1-p2)),i = 0,1, B^, ah Bu h ~ Binomial^, v.ai{x^%Zl%x-p^ i = °> L d\-Ki, at, Bi, Cl ~ Binomial^, ni{1_aJ^!}v%i{l_02)), i = 0,1, Aki, ai, A, di ~ Binomial^, J;^^^^ ), i = 0,1, We see that the Gibbs sampler enables us to draw inference not only on the prevalences of exposure (and consequently on the odds-ratio), but also on the the sensitivities and specificities of the two classification schemes. 3.3.1 Examples Here, we plan to investigate the validity of our method by applying it to eight hypothetical case-control studies where the true values of sensitivities, specificities, prevalence of exposure in controls and prevalence of exposure in cases are known. Data will be simulated using these true values for three different sample sizes, 7V0 = Nx = 200, N0 = Nx = 800 and N0 = Nx = 3200. One example will be discussed in more detail and the the results of the remaining eight will be summarized in a table. Example 1 Consider a scenario where the true values of the six param eters are (ax, Bx, a2, fa, TT0, TTI> = (0.85, 0.90, 0.90, 0.88, 0.07, 0.18). The value of 7T0 and iri imply <\> = 2.92 or log0 = 1.07. Data was simulated under this sce nario for three different sample sizes. For the sample size N0 = Nx = 200, simulated values were (a0,b0,c0,d0) = (11,9,21,159) and (ax, bi,cx,di) = (33,8,13,146). For the sample size N0 = Nx = 800, simulated values were 60 (a0, 60, Co, do) = (47, 83, 90, 580) and (auh, Ci, dx) = (113, 78, 90, 519). Finally, for the sample size N0 = Nx = 3200, simulated values were (a0,b0,c0,do) = (181,295,348,2376) and (ai,&i,Ci,di) = (453,297,338,2112). As in chapter 2, the prior for the prevalence of exposure in controls and the prevalence of exposure in cases was chosen to be Beta(l, 1). Furthermore, suppose that the investigator does not have a very good prior knowledge about the sensitivities and specificities of the two test, but is only willing to assume that the prob abilities of correct classification are greater than chance (i.e., they are in the interval [0.5,1]). We will translate this information into a Beta prior which assigns a 0.95 probability to this interval and with its mode at a value in this interval. In this example, we will set the mode at 0.8 for all four classification probabilities. This implies a Beta(7.55, 2.64) prior for the sensitivities and specificities. Five independent chains of the Gibbs sampler were run using different starting values. The sequential output is shown in Figure 3.1. Before attempting any inference, we will first examine the adequacy of the Gibbs sampler. The sequential plots of the post burn-in period (the first 500 observations from each chain were discarded) in Figure 3.1 indicate that convergence has occurred. The five independent chains appear to consistently converge to the same region in the parameter space, with no instances of slow mixing or drifting behaviour. Hence, we will regard Figure 3.1 as informal evidence that our posterior sampling is adequate. The statistical inference will be based on 5 x (2500 — 500) = 10000 sampled pairs of the prevalence of exposure in controls and prevalence of exposure in cases. 61 2000 4000 6000 S0OO Figure 3.1: Post burn-in output of the five independent chains of the Gibbs sampler for sensitivities, specificities, prevalence of exposure in controls and prevalence of exposure in cases in Example 1. Data was simulated for JV0 = Ni = 200. Figure 3.2 shows the histograms of the posterior log odds-ratio for the three sample sizes considered, together with the true log odds-ratio, which is obtained when JVj —> oo, i = 0,1. We see^ that the convergence toward the true value of log odds-ratio is much more rapid than in the case when one test is used. Also, as there is no nonidentifiability in the likelihood, the estimation error is 0(n-1/2), which is witnessed in the widths of the posterior distributions. As the sample size increases from 200 to 800, and from 800 to 62 3200, the width of the posterior distribution decreases roughly by the factor of 2. The posteriors in Figure 3.2 are, of course, based on single datasets. To assess the generality of the findings and the coverage of the credible intervals, we simulate 1000 datasets under the same scenario for the three sample sizes. For each dataset and each sample size we compute 80% HPD credible intervals for log0. The first forty intervals appear in Figure 3.3. The average interval widths are 1.01, 0.51, and 0.28 and the empirical coverage rates for the intervals are 85.7%, 83.8%, and 84.1% for the sample sizes N0 = Nx =.200,'N0 = Ni = 800, and N0- = Ni = 3200, respectively. As the standard error of the empirical coverage rates is approximately 1.26, it appears that 80% HPD credible intervals exhibit slight overcoverage for all three sample sizes. When doing Bayesian inference, it is important to assess the sensitivity of estimates. As we have already seen in Figure 3.1, the Gibbs sampler is not at all sensitive to the choice of starting values; all five chains converged to the same region in the parameter space. We now investigate whether the estimates of log0 are overly sensitive to different choices of the hyperparameters. We re-run the Gibbs sampler algorithm for the intermediate sample size, = Ni = 800, with different values of the hyperparameters. We consider two diametrically opposed scenarios. First, we examine the case where Beta(l, 1) prior is used for all six parameters. This scenario could potentially arise when an investigator has no prior knowledge of any of the six parameters, or is not certain of the validity of the available information. Second, we consider the 63 (a) Sample size: N0 = Nx = 200 (b) Sample size: N0 = Nx = 800 25 30 2.0 2.5 3.0 (c) Sample size: N0 = Nv= 3200 (d) Sample size: No = Ni = oo Figure 3.2: Posterior samples of the logarithm of the odds-ratio for the datasets in Example 1. 64 (a) n nn Ml"1 M 1 1 1 MM1 M -Ml 0 10 20 30 DATASET 40 (b) ! 1 I i 1 1 1 1 1 M i M i i i i iii ,1,1, , 1 i | I 1 | 1 1 11 M i l l 1 i i i | | | i | 1 1 1 i • i | i 0 10 20 30 DATASET 40 (o) i 1 i i 1 1 1 1 1 1 1 1 1 . 1 i . 1 1 1 1 . i 1 1 1 1 1 1 | 1 1 1 1 1 1 1 1 | | 1 1 | 1 1 1 1 1 1 1 1 1 | | 1 1 1 1 | 1 1 0 10 20 30 40 Figure 3.3: 80% highest posterior density credible intervals for \ogcj) in Exam ple 1. The solid vertical lines represent credible intervals for forty data sets and the solid horizontal line indicates the true value of log <j>. The panel (a) gives the intervals for the sample size N0 = Nx = 200, the panel (b) gives the intervals for the sample size N0 = Nx = 800, and the panel (c) gives the intervals for the sample size N0 = Nx = 3200. 65 (a) (b) (c) Figure 3.4: Posterior samples of the logarithm of the odds-ratio for the sample size N0 = Ni = 800 in Example 1. (a) Beta(l, 1) prior used for all six param eters, (b) Beta(7.55, 2.64) prior used for ctj and Bi,i = 1,2, while Beta(l, 1) prior used for ir0 and ITI (the histogram is identical to that shown in Fig ure 3.2 (b)). (c) Beta prior for each parameter was chosen so that the mode 7 is centred at the true value and 0.95 probability is assigned to the interval 7 ±0.05. case where very good prior knowledge is available for all six parameters. The hyperparameters a and /i of Beta(a, fi) priors were chosen so that the mode 7 is centred at the true value of each parameter and 0.95 probability is assigned to the interval 7 ± 0.05. The histograms of log(/> are shown in Figure 3.4, together with the histogram identical to that shown in Figure 3.2 (b). As evidenced in Figure 3.4, the posterior distribution of log 0 is very insensitive to the choice of hyperparameter values in this example, suggesting that the precise knowledge of hyperparameter values is not very important to the analysis. The results for the remaining eight examples are shown in Table 3.3. Under each scenario, 1000 data sets were simulated to assess the coverage of 66 the 80% HPD credible intervals and the mean interval length. It appears that for each scenario and each sample size, 80% HPD credible intervals exhibit slight overcoverage. Parameters Empirical coverage of 80% HPD CI a\ Pi a2 ft \0g(f) Ni = 200 Nt = 800 Ni = 3200 0.96 0.96 0.96 0.96 0.75 83.3% (0.94) 82.8% (0.48) 81.9% (0.25) 0.95 0.90 0.90 0.95 0.75 83.2% (0.96) 84.3% (0.47) 83.1% (0.27) 0.90 0.90 0.90 0.87 0.86 84.4% (0.98) 84.7% (0.50) 84.2% (0.28) 0.85 0.85 0.87 0.87 0.86 83.8% (1.03) 83.6% (0.51) 82.1% (0.24) 0.85 0.83 0.87 0.85 1.54 81.9% (1.02) 84.9% (0.52) 83.0% (0.22) 0.82 0.82 0.82 0.82 1.54 81.8% (1.03) 82.6% (0.52) 82.5% (0.29) 0.85 0.90 0.90 0.88 1.07 83.2% (1.01) 82.5% (0.47) 82.6% (0.28) 0.65 0.65 0.65 0.65 1.11 81.2% (1.04) 82.8% (0.49) 81.6% (0.28) Table 3.3: True parameter values and the empirical coverage of the 80% HPD credible intervals for log</> for the sample sizes = 200, Ni = 800, and = 3200,2 = 0,1. Numbers in brackets represent mean interval length. For the empirical coverage and the mean length, 1000 data sets were simulated. From the above table, we see that the empirical coverage rates do not appear to depend on the values of the six parameters. In particular, the empirical coverage rates for the scenario where relatively high values were chosen for the sensitivities and specificities (the first row in Table 3.3) are not significantly different than the empirical coverage rates for the scenario where relatively low values were chosen for the sensitivities and specificities (the last row in Table 3.3). 67 3.4 An Application to Real Data We will now compare our method to the method of Drews et al. (1993) by analyzing data from a case-control study of sudden infant death syndrome (SIDS). The exposure data was obtained from maternal interviews and medical records. The data appear in Drews et al.(1993), as taken from Hoffman et al. (1988). These data come from the National Institute of Child Health and Human Development's (NICHD) case-control study of SIDS. A total of 844 SIDS victims and two age-matched living infants were included in the case-control study. A first control was individually age-matched to the case in such a way that, at the time of the interview, the control would be the same age as the case had been when he/she died. A second control was matched to the case on birth-weight and race. Drews et al. (1993) were given access to a data on a pseudorandom sample (a systematic sample with a random start) of 226 of the 844 SIDS cases and one of the two age-matched controls for each of the 226 cases. See Hoffman et al. (1988) or Drews, Kraus and Greenland (1990) for a full description. Mothers of cases and controls were interviewed regarding events which had occurred during their pregnancies, their labour and deliveries, and their infants lives within five weeks of the death of the cases. We looked at six dichotomous exposure variables: maternal anemia dur ing pregnancy (ANEM), maternal urinary tract infection during pregnancy (UTI), previous spontaneous abortion (PSA), low pregnancy weight gain, which was defined as weight gain of less than 15 pounds (LPWG), mater-68 Controls Cases MR MR + • — + — ANEM Int + Int -20 15 43 147 24 15 49 125 UTI Int + Int -14 4 10 190 13 14 14 174 PSA Int + Int -21 11 13 175 23 15 12 169 PV Int + Int -91 6 12 66 69 9 9 78 LPWG Int + Int -6 4 3 86 11 5 13 73 PAU Int + Int -21 16 12 168 29 17 22 143 Table 3.4: A pseudorandom sample of 226 SIDS cases and 226 controls from the NICHD study. Data was classified using medical record (MR) and interview (Int) data. 69 nal antibiotic use during pregnancy (PAU), and polio vaccination (PV) before death or interview. Data for these variables were available from both med ical records and maternal interviews. The subjects for which information is missing from either source were not included in the analysis.Further, since the matching of cases and controls was fairly weak, it was ignored. Table 3.4 shows the distribution of controls and cases according to both interview and medical record data for the six variables. Interview data were arbitrarily taken to rep resent test 1 and medical records to represent test 2. This distinction has no ill effect, since we assumed that the errors between two tests are independent. As no reliable prior information about the sensitivity and specificity of the interview data and medical records was available, we assigned a Beta(l, 1) prior to all six parameters. The results of our analysis are shown in Table 3.4 and Table 3.5. Table 3.4 shows median estimates of the sensitivities, specifici ties and log0, together with 95% HPD credible intervals for log</>. Table 3.5 shows 95% HPD credible intervals for the sensitivities and specificities of the interview data and medical records. Five independent chains of the Gibbs sampler were run with different starting values. The inference was based on 5 x (5500 - 500) = 25000 draws. As mentioned before, Drews et al. (1993) used the EM algorithm to obtain estimates of the sensitivity and specificity of the interview data and medical records and the odds-ratio. The results of their analysis appear in Table 3.6. Note that for data set, at least one of the parameters estimated using Drews et al. (1993) method lies on the boundary of the parameter space (i.e., 70 1.00). This is, of course not true when the Gibbs sampler is used. Except for this distinction, the EM algorithm and the Gibbs sampler produce similar estimates of the sensitivities and specificities for the ANEM, PV, LPWG, and PAU data sets. Greater discrepancies occur for the UTI and PSA datasets. For instance, in the UTI dataset, the estimate of the sensitivity of medical records obtained by the Gibbs sampler is 0.67, whereas the estimate obtained using the EM algorithm is 0.56, a difference of 0.11. Also, the estimate of the sensitivity of interview data using the Gibbs sampler is 0.70, while the one obtained via the EM algorithm is 0.60. A similar finding applies to the PSA dataset. The point estimates of log0 are similar in magnitude, though the ones obtained vie the Gibbs sampler are smaller than the EM estimates for all but the ANEM data set. However, the credible intervals are narrower than the confidence intervals for all six data sets. In particular, in five out of six data sets, the credible intervals for log <fi are completely contained in the confidence intervals. This finding is somewhat surprising, since our credible intervals take into account the variability associated with the sensitivities and specificities of the two tests, which is not true for the confidence intervals. The variance used to calculate the confidence interval for log <f> is estimated using the delta method, as follows: Var(log0)= VAR(?RO) 2+ VAR(7FL) 2 ZCOVKTTX) Ml - TTo)]2 Ml - TTx)]2 Ml - TO)] Ml - Tl)] ' The values of Var(-7r0), Var(7Ti), and Cov(7r0,7Ti) can be obtained by substi tuting the parameter estimates into the expected information matrix and in-71 verting this matrix to obtain the estimated covariance matrix. Therefore, Var(log (j)) only depends on the point estimates of the sensitivities and speci ficities, not the estimates of their standard errors. The explicit formulae for the information matrix when the errors of the two classification schemes are independent (/ii — /i0 = 1) are given in Hui and Walter (1980). Since we have established the coverage of the credible intervals through the simulation studies, and the length via the analysis of real data set, we can conclude that they perform better than the confidence intervals. 3.5 Discussion In this chapter we have examined the situation where the exposure status is determined by two imperfect classification schemes. We have seen that the Gibbs sampler works very well in this case, presumably because the likelihood is identifiable. Our method performed well both in the simulation studies and in the analysis of the real data, where it appeared to outperform the EM algorithm approach of Drews et al. (1993). 72 Interview Medical 95% HPD Data Record Cr. Int Variable a P a P for log <f> ANEM 0.75 0.91 0.42 0.94 0.52 (-0.07,1.15) UTI 0.70 0.97 0.67 0.98 0.45 (-0.39,1.07) PSA 0.75 0.96 0.77 0.96 0.15 (-0.42,0.74) PV 0.96 0.94 0.92 0.94 -0.43 • (-0.89,0.01) LPWG 0.78 0.98 0.56 0.97 1.08 (0.15,2.01) PAU 0.77 0.96 0.67 0.94 0.58 (0.02,1.17) Table 3.5: Median estimates of the sensitivities, specificities and the logarithm of the odds-ratio, and 95% HPD credible intervals for the logarithm of the odds-ratio, using the Gibbs sampler. Interview Medical Data Record Variable a P a P ANEM (0.61, 0.90) (0.83, 0.96) (0.30, 0.54) (0.88, 0.97) UTI (0.56, 0.88) (0.91, 1.00) (0.49, 0.87) (0.91, 1.00) PSA (0.63, 0.89) (0.89, 0.99) (0.64, 0.92) (0.90, 0.99) PV (0.89, 0.99) (0.87, 0.98) ' (0.84, 0.97) (0.88, 0.98) LPWG (0.60, 0.91) (0.93, 1.00) (0.41, 0^74) (0.93, 1.00) PAU (0.61, 0.91) (0.89, 1.00) (0.49, 0.87) (0.89, 0.98) Table 3.6: 95% HPD credible intervals for the sensitivities and specificities of the interview data and the medical records. Interview Medical 95% Data Record Conf. Int Variable a P a P log</> for log (j) ANEM 0.78 1.00 0.35 0.93 0.51 (-0.23,1.24) UTI 0.60 0.99 0.56 1.00 0.51 (-0.27,1.30) PSA 0.63 0.93 1.00 1.00 0.21 (-0.44,0.86) PV 1.00 1.00 0.88 0.91 -0.46 (-0.94,0.02) LPWG 0.84 1.00 0.52 0.97 1.17 (0.2,2.31) PAU 0.79 1.00 0.60 0.94 0.62 . (-0.01,1.24) Table 3.7: Estimates of the sensitivities, specificities and the logarithm of the odds-ratio, and 95% confidence intervals for the logarithm of the odds-ratio, using the EM algorithm (Drews et al. (1993)). 73 Chapter 4 Conclusion As we have seen, misclassification of exposure poses serious problems in statis tical analysis. The effects of ignoring misclassification and the methods which correct for it have received a considerable attention in literature. More often than not, exposure misclassification substantially biases estimates of the rel ative risk, even when misclassification rates are very small. Nondifferential misclassification, for example, tends to attenuate observed exposure-disease relationships (odds-ratio, for instance). Corrected estimates of the odds-ratio (p can be easily obtained if the misclassification probabilities (the sensitivity and specificity) are known. Bar ron (1977) and Greenland and Kleinbaum (1983) provide classical methods for correcting the odds-ratio, which rely on the invertibility of matrices. We introduced a Bayesian approach to correcting the odds-ratio when the sensi tivity and specificity are known. The sampling from the posterior distribution of prevalence of exposure among cases and control (and hence, the odds-ratio) 74 proved to be quite simple. However, having the exact values of the sensitivity and specificity is not very common in practice. Instead, good guesses of these values might be available (through previous studies or by comparing the diagnostic test.to a gold standard). It might be very tempting to carry out the above analysis by pretending that these guesses are the true values of the sensitivity and specificity. We showed, however, that even the smallest difference between the true and guessed values can result in very large difference between the miscorrected and true odds-ratio. This result, together with the previous work of Marshall (1989), sug gested that it is reasonable to incorporate the uncertainty into the analysis. Bayesian methods make this feasible, as partial knowledge about the sensitiv ity and specificity is easily represented with the appropriate prior distribution. We suggested two algorithms to sample from the posterior distribution: the Gibbs sampler and the Metropolis-Hastings. The Gibbs sampler did not work particularly well. The possible explanation is that drifting behaviour of the Gibbs sampler may occur when it is applied to the nonidentifiable model. On the other hand, the Metropolis-Hastings algorithm applied to a parameteriza tion that separates the identifiable and nonidentifiable part seemed to perform well. We were able to demonstrate that the approach which admits the un certainty in the guesses of the sensitivity and specificity performs better than the method that ignores it. We also showed that, despite nonidentifiability, the marginal prior and posterior distributions of (a, B) were not equal. As a 75 result of learning about the apparent exposure prevalences (#i,#2), one could learn about (a,B). In chapter 3 we examined the scenario when the exposure data is avail able from two imperfect sources. Various methods for improving the odds-ratio estimates by combining data from two imperfect classification schemes have been suggested, of which we outlined three. The method of Marshall and Graham (1984) uses only the concordant observations to corect the odds-ratio estimates. The methods of Drews et al. (1993) and Kaldor and Clayton (1985) are both classical approaches that incorporate the underlying true exposure into their analysis. We developed a Bayesian latent class aproach, where the latent class was the true exposure. We used the Gibbs sampler to sample from the posterior distributions of the six parameters. Both the adequacy of the Gibbs sampler and the validity of our method were established through a simulation study. The analysis of case-control data on the sudden infant death syndrome showed that our method could easily be applied in practice. The analysis also showed that our method appears to outperform the method of Drews et al. (1993). The following suggestions could be considered for further development of the methodology presented here. One would be to extend the method to the situation where three or more imperfect tests are used to assess the exposure. This should be a very simple extension of our methodology. Other would be to modify the methods to apply to matched-pair case-control studies. And finally, 76 it seems useful to explore the situation where, in addition to the misclassified dichotomous exposure, other covariates are measured, possibly without error. This does not appear to be a matter of simple extension of our method, because it would require the use of a link function between the probability of disease and covariates in question. 77 Bibliography [1] Barron, B.A. "The effects of misclassification on the estimation of relative risk." Biometrics, 33, 414-418 (1977). [2] Blettner, M., Wahrendorf, J. "What does an observed relative risk convey about possible misclassification?." Methods of Information in Medicine, 41, 923-937 (1984). [3] Chib, S., Greenberg, E. "Understanding the Metropolis-Hastings algo rithm." The American Statistician, 49, 327-335 (1995). [4] Dawid, A.P. "Conditional independence in statistical theory (with discus sion)." Journal of the Royal Statistical Society B, 41, 1-31 (1979). [5] Dempster, A.P., Laird, N.M., Rubin, D.B. "Maximum likelihood from incomplete data via the EM algorithm (with discussion)." Journal of the Royal Statistical Society B, 39, 1-38 (1977). [6] Drews, CD., Flanders, W.D., Kosinski, A.S. "Use of two data sources to estimate odds ratios in case-control studies." Epidemiology, 4, 327-335 (1993). 78 [7] Drews, CD., Kraus, J.R., Greenland, S. "Recall bias in a case-control study of sudden infant death syndrome." International Journal of Epi demiology, 19, 405-411 (1990). [8] Gelfand, A.E., Sahu, S.K. "Identifiability, improper priors, and Gibbs sampling for generalized linear models." Journal of the American Statis tical Association, 94, 247-253 (1999). [9] Gentle, J.E. Random Number Generation and Monte Carlo Methods. New York, Springer-Verlag, 1998. [10] Greenland, S., Kleinbaum, D.G. "Correcting for misclassification in two-way tables and matched-pair studies." International Journal of Epidemi ology, 12, 93-97 (1983). [11] Hoffman, H.J., Hunter, J.C., Ellish, N.J., Janerich, D.T., Goldberg,. J. "Adverse reproductive factors and the sudden infant death syndrome." In: Harper, C.M.R., Hoffman, H.J., eds. Sudden Infant Death Syndrome: Risk Factors and Basic Mechanisms. New York, PMA Publishing, 1988. [12] Hui, S.L., Walter, S.D. "Estimating the error rates of diagnostic tests." Biometrics, 36, 167-171 (1980). [13] Joseph, L., Gyorkos, T.W., Coupal, L. "Bayesian estimation of disease prevalence and the parameters of diagnostic tests in the absence of a gold standard." American Journal of Epidemiology, 141, 263-272 (1995). 79 [14] Kaldor, J., Clayton, D. "Latent class analysis in chronic disease epidemi ology." Statistics in Medicine, 4, 327-335 (1985). [15] Marshall, J.R. "The use of dual or multiple reports in epidemiological studies." Statistics in Medicine, 8, 1041-1049 (1989). [16] Marshall, J.R., Graham, S. "Use of dual responses to increase validity of case-control studies." Journal of Chronic Diseases, 37, 125-136 (1984). [17] Thomas, D., Stram, D., Dwyer, J. "Exposure measurement error: influ ence on exposure-disease relationships and methods of correction." An nual Review of Public Health, 14, 69-93 (1993). studies." [18] Walter, S.D. "Commentary on 'Use of dual responses to increase validity of case-control studies' ." Journal of Chronic Diseases, 37, 137-139 (1984). [19] Walter, S.D., Irwig, L.M. "Estimation of test error rates, disease preva lence and relative risk from misclassified data: a review." Journal of Clin ical Epidemiology, 9, 923-937 (1988). 80
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http://iiif.library.ubc.ca/presentation/dsp.831.1-0089794/manifest | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8623082637786865, "perplexity": 1598.4307757111744}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720154.20/warc/CC-MAIN-20161020183840-00529-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/why-did-the-dinosaurs-die-out.184526/ | # Why Did The Dinosaurs Die Out?
1. Sep 14, 2007
### Luke*
Right, I dont know if I can put this here, but I will try.
I know a asteroid, or something, wiped out the dinosaurs. But how exactly were they wiped out, did it just completely destroy the area they live in or what?
Luke.
2. Sep 14, 2007 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8439361453056335, "perplexity": 2342.8715553852994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171900.13/warc/CC-MAIN-20170219104611-00600-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2227819/approximating-specific-integrals-via-sums-tight-error-predictions | Approximating specific integrals via sums: Tight error predictions
Dear math enthusiasts,
I am facing a particular problem where I am looking at integrals of the form $$I = \int_{-\infty}^\infty p(t) {\rm e}^{-t^2} {\rm d}t,$$ where $p(t)$ are certain polynomials. These are either even or odd symmetric, the even symmetric ones being the more interesting case since for odd ones, the integral is zero (in fact, they are auto- and cross-products of Hermite polynomials but I think this detail is not relevant). For now, we can just consider the simplest example $p(t)=t^2$.
The reason why I am looking at the integrals is that I actually have sums of the form $$S(t) = \sum_{n=-\infty}^\infty t_0 p(t-nt_0) {\rm e}^{-(t-nt_0)^2}$$ which I want to quantify. For $t_0$ small enough, these sums are very close to $I$ for any $t$. I need to quantify how close, i.e., I am interested in $$\max_t |S(t)-I|.$$
So, what I did was to interpret $S$ as a quadrature of the integral $I$ and use the residual formulas for quadratures. Since it is a linear quadrature, the standard results predict a residual of the order $|I-S|< {\rm const} \cdot t_0^2 \cdot \max|f''(t)|$, where $f(t) = p(t){\rm e}^{-t^2}$ (which gives $\max|f''(t)| = 2$ for $p(t)=t^2$). In other words, the error should decay quadratically with $t_0$. All this is not surprising and well within what I expected.
Until I tried it and realized that empirically, the error decays much much faster with $t_0$ than this pessimistic bound predicts. Here is an example:
In this example, I computed $S$ for $p(t)=t^2$, varying the grid spacing $t_0$. I plot $\max_t|I-S(t)|/I$ where the exact value $I$ is equal to $\sqrt{\pi}/2$.
Obviously, $t_0=1$ is too coarse but as I make it finer, the error goes to zero very rapidly (Note that the plot is doubly logarithmic!). In fact, around $t=0.35$ it reaches the numerical accuracy of my double floating point but I would expect the exponentially decaying trend to continue. The predicted upper bound is shown in the dashed line (it is a line with slope 2 due to the double logarithmic plot).
So here is my question: Can I make tighter predictions of $|S(t)-I|$ as $t_0\rightarrow 0$? I know it is easy to construct examples where the residual formula is basically tight, so it must have to do with the particular function I am integrating, especially the ${\rm e}^{-t^2}$ term. I keep getting reminded of Gaussian tails (erfc functions of some sort) but I cannot put my hands on how I could get there. Just before posting I stumbled upon Euler-Maclaurin, but it confuses me as well since what I read about it talks about finite sums (and derivatives being evaluated at the borders) while mine seems infinite (and everything becomes zero far enough from $t=0$).
It looks like a standard result and I would not be surprised about a very simple answer that I was just not seeing.
Any hint is appreciated, many thanks in advance!
edit: Thanks to user14717 I got what I needed. For someone stumbling across a similar problem, here is what worked: Theorem 5.1 in [*] says the following: Let $a>0$ such that the function $w(t)$ to be integrated is analytic in the string $|{\rm Im}(t)|<a$ and decays to zero uniformly as $|t|\rightarrow \infty$. Then: $$|I-S| \leq 2\sqrt{\pi} \frac{M}{{\rm e}^{2\pi a/t_0}-1},$$ where $M$ is a constant satisfying $\int |w(t+ib)| {\rm d}t \leq M$ for all $b \in (-a,a)$. If we apply this for $w(t) = {\rm e}^{-t^2}$ (i.e., my $p(t)=1$), we obtain $M={\rm e}^{a^2}$ as the best $M$ for a given $a$. This gives the family of bounds $$|I-S| \leq 2\sqrt{\pi} \frac{{\rm e}^{a^2}}{{\rm e}^{2\pi a/t_0}-1},$$ which is valid for any $a>0$. To obtain the tightest bound, we need to minimize over $a$. This is not possible analytically, however, for $t_0<1$, the value $a=\frac{\pi}{t_0}$ is very close to the optimum. Inserting it gives the bound $$|I-S| \leq \frac{2\sqrt{\pi}}{{\rm e}^{(\pi/t_0)^2}-{\rm e}^{-(\pi/t_0)^2}} \approx 2\sqrt{\pi} {\rm e}^{-(\pi/t_0)^2}.$$
And now, let us plot it:
Adapting this to any $p(t)$ should be a breeze now.
I could barely be happier! :)
Thanks so much!
edit2: For an even symmetric polynomial $p(t)$ it is then very easy to show that $$|S-I|\leq \frac{2{\rm e}^{a^2} h(a)}{{\rm e}^{2\pi a/t_0}-1},$$ where $h(a)$ is a polynomial of same degree as $p(t)$. For $a=\pi/t_0$, this gives $$|S-I|\leq \frac{2{\rm e}^{\pi^2/t_0^2}}{{\rm e}^{2\pi^2/t_0^2}-1}h(\pi/t_0) \approx 2{\rm e}^{-\pi^2/t_0^2}h(\pi/t_0),$$
i.e., still exponential convergence, as expected.
In fact, for a degree $2k$ polynomial $p(t) = \sum_n \alpha_n t^{2(k-n)}$, an explicit form of $h(t)$ is given by $$h(t) = \sqrt{\pi} \sum_n \sum_\ell |\alpha_n|\frac{k! (2\ell)!}{4^\ell (\ell!)^2 (k-\ell)!} b^{2(k-n-\ell)},$$ though it doesn't matter much since the relevant part is the exponential convergence.
It is a standard result. What you are observing is a well-known property of the Euler-Maclaurin expansion \begin{align*} h\frac{f(a) + f(b)}{2} + h\sum_{k = 1}^{n-1} f(a+kh) = \int_{a}^{b} f(x) \, \mathrm{d}x + \sum_{k = 1}^{\infty} \frac{ h^{2k} }{(2k)!} B_{2k}(f^{(2k-1)}(b) - f^{(2k-1)}(a)) \end{align*} Now, you are worried that this doesn't apply since your integrand is infinite. But in finite precision, $\exp(-t^2) = \epsilon$ which means that in practice your integration can only be performed over the range $[-\sqrt{-\log(\epsilon)}, \sqrt{-\log(\epsilon)}]$. In double precision, this is about $[-6, 6]$.
Now, all the derivatives of $f$ are also bell-shaped, so as $a\to -\infty$, $f^{(2k-1)}(a) \to 0$ very fast. Assuming that $B_{2k}f^{(2k-1)}(a)/(2k)!$ is tiny, we can state that your trapezoidal sum converges faster than any power of $h$. Let's examine this claim in more detail: Using the representation of the Bernoulli numbers \begin{align*} B_{2k} = (-1)^{k+1}\frac{2(2k)!}{(2\pi)^{2k}}\zeta(2k) \end{align*} we can write the error as \begin{align*} E(h) := \sum_{k = 1}^{\infty} (-1)^{k+1}h^{2k}\frac{2}{(2\pi)^{2k}}\zeta(2k)(f^{(2k-1)}(b) - f^{(2k-1)}(a)) \end{align*} In your case, it makes sense to make $a = -b$, and since we assume that $f$ is even then all odd-order derivatives of $f$ are odd. Then \begin{align*} E(h) = 4\sum_{k = 1}^{\infty} (-1)^{k+1}\left(\frac{h}{2\pi}\right)^{2k}\zeta(2k)f^{(2k-1)}(b) \end{align*}
Now, $1 < \zeta(2k) < 2 \forall k \in \mathbb{N}$ so \begin{align*} |E(h)| \le 8\sum_{k = 1}^{\infty} \left(\frac{h}{2\pi}\right)^{2k}|f^{(2k-1)}(b)| \end{align*} All that we need is for the error to be smaller than the machine epsilon to get the result, as $f(0) \sim \mathcal{O}(1)$ and hence any corrections smaller than the unit roundoff are not observable.
I have an idea for getting this argument a bit more rigorous, but it needs some more TLC (or a counterexample):
Assume $f(x) = (x^{2n} + \cdots)\exp(-x^2)$ where $\cdots$ are terms with degree lower than $2n$. Then $f^{(2k-1)}(x) = (x^{2n+2k-1} +\cdots)\exp(-x^2)$ and hence for large $x$, $|f^{(2k-1)}(x)| \le C|x^{2k-1}|x^{2n}\exp(-x^2)$, so that \begin{align*} |E(h)| \le 8Cb^{2n-1}\exp(-b^2)\sum_{k = 1}^{\infty} \left(\frac{hb}{2\pi}\right)^{2k} = 8Cb^{2n-1}\exp(-b^2) \frac{(hb/2\pi)^2}{1-(hb/2\pi)^2} \end{align*} assuming $bh/2\pi < 1$. Now just choose $b$ large enough that $|E(h)| < \epsilon$ and the result is proved.
To prove the result without recourse to arguments about finite precision requires complex analysis. This is discussed by Trefenthen here.
• Great, this is helpful! Numerical issues let aside I could simple choose any fixed $[a,b]$ big enough, in a second step even looking at the limit $[a,b]\rightarrow[-\infty,\infty]$. Then I have $f^{(2k-1)}(t) = O(t^{2k+1}e^{-t^2})$ and I can bound $B_{2k}/(2k)!$ via Stirling, right (since it is constant, the $e^{-t^2}$ should take care of it all I guess)? Looks good, tomorrow I'll have a look at what I can do about the error term. I'd really like to get the the correct scaling law (without killing myself on the way). – Florian Apr 10 '17 at 17:03
• Great, the addition helps a lot. Just one more clarification how to get from the last inequality to "faster than any power of $h$": I would assume the argument is that if I were to truncate the sum to any finite number of terms $k=1...K$ then the limit would surely be zero (since $e^{-b^2}h^{2k}$ always goes to zero) - but I cannot say this for the infinite sum (since even if all terms are zero the infinite sum could have any value)? – Florian Apr 10 '17 at 18:54
• Well I guess my $C_k$ are not - if we keep forming derivatives we essentially get Hermite polynomials in front of the exp and their coefficients grow with $k$, if I am not mistaken. But that's fine, I think I can work with this answer. Thank you very much for taking the effort to reply in such detail, greatly appreciated! – Florian Apr 10 '17 at 19:27
• Thanks for the additional edit! The only problem I see is that I cannot choose $b$ "large enough" since we need $bh/2\pi<1$ and $h$ is not necessarily very small. What I'm really after is a point $h$ below which $|R(t)-I|$ is practically zero, so I need a fast decay in $h$ ("a little smaller than one" seems to work, I'd like to quantify that). It seems it's almost there, the only problem being that the $b^{2k}$ is still inside the sum which means I cannot let $b$ grow too much withouth constraining $h$ (and, rigorously speaking I really want to say something about $\lim b\rightarrow \infty$). – Florian Apr 11 '17 at 8:25
• In double precision, $b \sim 6$, so we only need $h < 1$. I don't see why that is a big restriction. As to the limit $b \to \infty$, this mode of analysis isn't going to help much, you'll need another strategy. – user14717 Apr 11 '17 at 13:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9825646281242371, "perplexity": 219.38813165581388}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583513844.48/warc/CC-MAIN-20181021094247-20181021115747-00354.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-7-exponential-and-logarithmic-functions-7-2-graph-exponential-decay-functions-7-2-exercises-problem-solving-page-490/31a | ## Algebra 2 (1st Edition)
$y=200(0.75)^t$
Our initial amount is $200$, and the percent decrease is: $0.25$. Then the model for the value is: $y=200(1-0.25)^t=200(0.75)^t$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9624460339546204, "perplexity": 1190.4517268584116}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711390.55/warc/CC-MAIN-20221209043931-20221209073931-00133.warc.gz"} |
https://www.kseebsolutions.com/2nd-puc-physics-model-question-paper-2/ | # 2nd PUC Physics Model Question Paper 2 with Answers
Students can Download 2nd PUC Physics Model Question Paper 2 with Answers, Karnataka 2nd PUC Physics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka 2nd PUC Physics Model Question Paper 2 with Answers
Time: 3 Hrs 15 Min
Max. Marks: 70
General Instructions:
1. All parts are compulsory.
2. Answers without relevant diagram/figure/circuit wherever necessary will not cany any marks
3. Direct answers to the Numerical problems without detailed solutions will not carry any marks.
Part – A
I. Answer all the following questions ( 10 × 1 = 10 )
Question 1.
Write the SI unit of Electric field.
Newton per coulomb (N/C)
Question 2.
When will the magnetic force on a movi¬ng charge be maximum in a mag notified?
F = qVBsinθ , If θ = 90° (Fm – qυB)
The magnetic force is maximum only when moving charge is ⊥r to field.
Question 3.
Where on the Earth’s surface is the magnetic dip zero?
At equator, dip is zero
Question 4.
State Curie’s law in magnetism.
The magnetic susceptability of a para magnetic substance varies inversely to its absolute temperature(T).
i.e x α $\frac{\mathrm{C}}{\mathrm{T}}$ (x = $\frac{\mathrm{C}}{\mathrm{T}}$)
Question 5.
What is the significance of Lenz’s law?
Law of conservation of energy.
Question 6.
Write the formula for Malus law.
i = i0cos2θ
Where I → is intensity of the light transmitted by the analyser.
I0 → is intensity of the light incident on the analyser.
θ → is angle between the pass axes of the analyser & polariser.
Question 7.
What is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1:3?
Nuclear density is independent of mass number & is approximately constant.
Hence $\frac{P_{1}}{P_{2}}$ = 1
Question 8.
Define current amplification factor in a common – emitter mode of transistor.
It is defined as the ratio of change in collector current ( ΔI)to the change in base current (ΔIb) at constant collector emitter voltage (VCE) when the transistor is in active state
Question 9.
Write the truth table of NAND gate.
Truth table of NAND gate
Question 10.
Why sky wave propagation is not possi¬ble for waves having frequency more than 30 MHz?
The sky wave range is short & it is used in short wave broad cast service.
Part – B
II. Answer anyfive ofthe following questions. ( 5 × 3 = 15 )
Question 11.
Sketch the electric lines of force due to a point charge q. If i) q<0 and ii) q>0
Question 12.
A galvanometer having a coil of resistance 12 Ω gives full scale deflection for a current of 4 mA. How can it be converted into a voltmeter of range 0 to 24V?
Given G = 12 Ω, Ig = 4 x 10-3A. V = 24V
We have,
$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{B}}}-\mathrm{G}$
R = 5988Ω
A resistance of 5988 Ω must be connected in series with galvanometer.
Question 13.
Distinguish between paramagnetic and ferromagnetic substances.
Paramagnetic substances Ferromagnetic Substances 1. Paramagnetics are feebly attracted by magnets. 1. They are strongly attracted by magnets. 2. Magnetic permeability is slightly greater than one i.e. Mr>1 2. Relative permeability is greater than 1000 3. Magnetic susceptability is low & positive. 3. Magnetic susceptability is +ve & large.
Question 14.
What is meant by Self Inductance and Mutual Inductance?
1. The phenomena in which an emf is induced in a coil due to change of current through the same coil is known as self-induction.
2. The phenomena in which an emf is induced in one coil due to change of current is the neighbouring coil is called mutual induction
Question 15.
What are electromagnetic waves? Write the expression for the velocity of electro magnetic waves in terms of permittivity and magnetic permeability of free space.
The waves in which there are a sinusoidal variation of electric & magnetic field vectors at right angles to each other & also right angle to the direction of propagation of the wave is called e.m waves.
∴ The expression for velocity of e.m. wave is
$C=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ Where µ0 → is permeability of free
space, ε0 → is the permittivity of free space.
Question 16.
Write the relation between the path difference and wavelength of light wave used for constructive and destructive interference of light.
a) For constructive interference
∴ Path difference = 2n$\frac{\lambda}{2}$ = nλ
Wheve λ → is wave length of light used
n → 0, 1, 2,
b) For destructive interference :
∴ Path difference = (2n+l)$\frac{\lambda}{2}$
Wheve λ → is wave length of light used n → 0, 1, 2,………
Question 17.
Define :
i) photoelectric work function
ii) electron volt (ev)
i) Photoelectric work fraction : the minimum energy required to remove an electron from the metal surface is called work function, i.e. w = hV0
Where h → is planck’s constant
V0 → is threshold frequency
ii) Electron volt is the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt.
i.e. lev = 1.602 × 10-19J.
Question 18.
Draw the block diagram of a AM receiver.
x → Intermediate frequency stage
y → power amplifier
Part – C
III. Answer any five of the following questions. ( 2 × 5 = 10 )
Question 19.
Derive an expression for potential energy of a system of two charges in the absence of external electric field.
Consider 2 point charges q1 & q2 are separated by a distance ‘r’ are as sho in the fig.
∴ The change q1 is bringing from ∞ to given point A, No work done i.e. w1= 0 |||ly the charge q2 is bringing from ∞ to given point ‘B’ against us field q2
∴work is done it is given by
i.e. w2 = vq2
Question 20.
Arrive at an expression for drift velocity.
Controller a metalic conductor is connected to a battery
Let $\overrightarrow{\mathrm{E}} \rightarrow$ is E.f. setup inside the conductor
m → is mass of electom
e → charge of free elector
Vd → is drift velovity
τ → is ralaxation time
the force experienced by an electron in the field is given by $\overrightarrow{\mathrm{F}}=-\mathrm{e} \overrightarrow{\mathrm{E}}$
-Ve sign shows that divertion of $\overrightarrow{\mathrm{E}}$ & $\overrightarrow{\mathrm{F}}$ are in apposite each other
Question 21.
State and explain Gauss law in magnetism.
The net magnetic flux through any closed surface is always zero
$\sum \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{As}}=0$
consider a closed surface
S in a uniform
$\mathrm{M} . \mathrm{F} \cdot \overrightarrow{\mathrm{B}}$ Let $\overrightarrow{\Delta \mathrm{S}}$
be a small area element of this surface with $\hat{n}$
along its normal
Question 22.
Derive the expression for motional emf induced in a conductor moving in a uniform magnetic field.
Consider a straight metallic rod PQ of length ‘t’ placed in a uniform M.F. $\overrightarrow{\mathrm{B}}$. The rod is moved with a velocity $\overrightarrow{\mathrm{v}}$ is a direction ⊥r to
$\overrightarrow{\mathrm{B}}$ Let the rod moved through a distance ‘x’ in time‘t’ them the area covered by the rod is A= 1 xx The magnetic flux linked with the rod is
Φ = B.A Φ = B1x
∴ The included emf in the rod is
Motional emf (e – Blv)
$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}$
Question 23.
With a diagram, explain the working of a transformer.
AC voltage is applied to the primary, it creates a varying M.F. & hence a changing magnetic flux in the core. Since the secondary coil is magnetically coupled, due to the mutual induction the changing flux causes induced emf in it. Thus, the power is transferred from primary to secondary.
Question 24.
What is total internal reflection? Mention two applications of optical fibres.
The phenomena of complete reflection of light at the interface of two optical media when a ray of light travelling in denser medium is called total internal reflection.
• Optical fibre are used in telecommunication
• It is used to measure rate of flow of blood.
Question 25.
What are the matter waves? Write the expression for De – Broglie wavelength of a particle and explain the terms.
The waves associated with material particles in motion is called matter waves. The expression for de Broglie wavelength is $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
where
h → is Planck’s constant
m → is mass of moving particle
v → is the velocity of moving particle.
Question 26.
Write three differences between n-type and p-type semiconductors.
P – type S.C.
1. It is a semiconductor doped with trivalent impurities.
2. Majority charge carriers are holes
3. The impurit atom is called acceptor impurity.
n-type S.c
1. It is a semiconductor doped with pentavalent impurities.
2. Majority charge carriers are electron
3. The impurity atom is called Sonar impurity
Part – D
IV. Answer any two of the following questions ( 2 × 5 = 10 )
Question 27.
Derive an expression for the electric field at a point due to an infinitely long thin charged straight wire using Gauss law.
Consider an infinitely long thin straight wive with uniform linear charge q density λ. Let P be a point at ⊥r distance r from the wire.
To calculate the E.F $\overrightarrow{\mathrm{E}}$ at P, imagine a cylindrical Gaussian surface.
∴ The surface area of the] curved part S = 2πrl
Total charge enclosed by the Gaussian surface q = λl
Electric fliix through the end Surfaces of the cylinder is Φ = 0
Electric flux through the curved Surfaces of the cylinder is Φ2 = Ecosθ.s
Φ2 = EX1X2πrl
The total electric flux Φ = Φ1 + Φ2
Φ = 0 + E2πrl, Φ2= 2πrlE …………… 1
A/C to Gauss law,
from (a) and (2)
Question 28.
Obtain the bridge balanced condition of Wheatstone’s bridge network by applying Kirchhoff’s rules.
Wheatstone’s bridge is a device used to determine unknown resistance.
It resistances R1, R2, R3 & R4 a galvanometer of resistance G & a battery with key. Let I is the math current & splits into branch current I1 & I2 respectively
i.e. I = I1 +I2
At B, I1 = I3+ Ig………….(1)
A + D, I2 = 4 + Ig ………. (2)
Apply KVL to the mesh ABDA
I1R1 + igG – I2R2 = o ………….. (3)
to the mesh BCDB
I3R3 – I4R4– IgG = 0 ……………… (4)
The wheatstones bridge is said to be balanced
when no current flowing throught the galvano
meter i.e. Ig 0 ∴ eqn (1) is I1 = I3
eqn(2)is I2 = I4
eqn (3) is I1R1 – I2R2 = 0
∴ I1R1 = I2R2 …………. (5)
eqn (4) is I3R3 – I2R2
∴I3R3 = I4R4 …………………. (6)
Question 29.
Two straight parallel conductors are placed at certain distance in free space. The direction of current in both the conductors is same. Find the magnitude and direction of the force between them. Hence define ampere.
Consider two infinitely long straight parallel condu ctors x & y carrying a current I1 & I2 respectively. Let d is the ⊥r distance between them.
Let I1 is the current flowing through x conductor, it produces a M.F. (B1)
Now the conductor Y carrying current I2 in the M.F.B, it experiences a mehanical force of length 1 is
The direction of force can be obtained using ampere’s left-hand rule.
∴ The force bn. two ||le conductors carrying currents in the same direction is attractive. ||ly the force bn two ||le conductors carrying current in opposite direction is repulsive.
Definition of ampere If I1 = I2 = IA, d=lm them F = 2 × 10-7N.
Ampere is the steady current which when flowing through each of 2 infinitely long straight ||le conductors placed in the air at a distance of lm produces a force of 2 × 10-7 N/m.
V. Answer any two of the following question ( 2 × 5 = 10 )
Question 30.
Derive Lens Maker’s formula for a con vex lens.
Consider a thin convex lens of focal length f & R.I (n) placed in air as shown in fig let R1 & R2 → are the of curvatur of the surfaces ABC & ADC of the respectively.
o → Luminous point object on the principle axis.
A ray op invident at p, after refraction, emerges along QI & meet at I on the principal axis.
Image formution takes place in two stages. (1) Refraction at the surface ABC In the observe of ADC, the refracted ray is meet at I1, then
(2) Refraction at the surface ADC the image I1 acts as a virtual object to form a real image at a distance V, then
Question 31.
Assuming the expression for radius of the orbit, derive an expression for total energy of an electron in hydrogen atom.
Consider an electron of mass m, charge, e revoking around the nucleus of radius of r. The charge on the nucleus + Te.
∴ T.E = KE + P.E
T. E = K + U ………… (1)
For circulation
Centripetal force = Electrostate force bn nucleus & electron.
The Potential energy of the electron is the field o nucleus is
Question 32.
With the help of circuit diagram, explain the working of NPN transistor as a common emitter amplifier.
The circuit diagram of a CE amplifier using NPN transistor is as shown m fit. The input circuit is forward biased & the output circuit is reverse biased when the ac input signal to be amplified is fed to the base-emitter circuit. The output voltage V0 varies in accordance with the relation, V0=VCE =Vcc – IcRc, These variations is the collector voltage VCE uppers as amplified output.
During the -t-ve half cycle of ac input signal the forward base of emitter as junction increase Due to this base current IB increase & hence collector current IC increases. As a result of this ICRC increases output voltage VO this indicates that the +ve half cycle of input ac signal voltage is amplified through -ve half cycle.
During the -ve half cycle of ac input signal, the forward bias of emitter-base junction decreases. Due to this base current IB decreases & hence collector current IC decreases. As a result of this ICRC decreases the. output voltage V0 is +ve. This indicates that the -ve half cycle of input ac signal voltage is amplified through +ve half cycle.
Thus, the weak input signal is amplified & output signal is out of phase with the input signal by 180°
VI. Answer any three of the following questions ( 3 × 5 = 15 )
Question 33.
Charges 2 µ C, 4 µ C and 6 µ C are placed at the three corners A, B and C respectively of a square ABCD of side x metre. Find, what charge must be placed at the fourth corner so that the total potential at the center of the square is 0
In the Δle ABC
AC2 = AB2 + BC2
AC2 = x2 + x2
AC2 = 2x2
∴ AC = √2 x,.m
Question 34.
A wire having length 2.0m, diameter 1.0 mm and resistivity 1.963 × 10-8 Ω m is connected in series with a battery of emf 3 V and internal resistance 1 Calculate the resistance of the wire and current in the circuit.
Given,
length (l) = 2m
Diameter (D) = 1mm = 1 × 10-3 m
∴ radius (r) = $\frac{\mathrm{D}}{2}$ = 0.5 × 10-3 m
resistivity (f) = 1.963 × 10-8Ωm
E = 3V Internal senstarce (r) = 1 Ω
R=? I = ?
We have,
Question 35.
An inductor and a bulb are connected in series to an AC source of 220 V, 50 Hz. A current of 11A flows in the circuit and phase angle between voltage and current is $\frac{\pi}{4}$ radians. Calculate the impedance and inductance of the circuit.
Given,
Vrms = 220v, f = 50 Hz
Irms = 11A
Φ = $\frac{\pi}{4}$ radian, Z=?, L =?
Question 36.
In Young’s double slit experiment while using a source of light of wavelength
4500 Å, the fringe width is 5mm. 1f the distance between the screen and the plane of the slits ¡s reduced to half, what should be the wavelength of light to get fringe 4 mm?
Given,
wavelength (λ) 45OOA°= 45OO × 10-10 m
fringe width (w) = 5mm = 5 × 10-3 m
D = $\frac { D }{ 2 }$
wavelength (λ1) = ?
fringe width (w1) = 4mm = 4 × 10-3 m
we have, fringe width (w) = $\frac{\lambda \mathrm{D}}{\mathrm{d}}$
5 × 10-3 =4500 × 1010 × $\frac { D }{ d }$
Question 37.
The activity of a radioactive substance is 4700 per minute. Five minutes later the activity is 2700 per minute. Find
a) decay constant and
b) half – life of the radioactive substance. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 27, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8905866146087646, "perplexity": 2390.6928308741567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657154789.95/warc/CC-MAIN-20200715003838-20200715033838-00538.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=117&t=49610&p=179063 | ## shrodinger equation
$H_{\psi }=E_{\psi }$
1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$
halle young 4A
Posts: 68
Joined: Thu Jul 11, 2019 12:16 am
### shrodinger equation
I know we won't have to do calculations with this equation, but conceptually can someone go over what the equation does? I am still a bit confused on what it means.
thanks
A Raab 1K
Posts: 56
Joined: Sat Aug 24, 2019 12:16 am
### Re: shrodinger equation
I think it just describes the wave function or a path that a given system will take. It has to do with potential energy and kinetic energy to find the total energy, and using it to find the behavior of an electron bound to a nucleus.
Camellia Liu 1J
Posts: 51
Joined: Sat Aug 24, 2019 12:15 am
### Re: shrodinger equation
Shrodinger's equation basically defines a wave function of a particle that has a certain value at any given time for certain points in space. It also helps specify how the waves are altered by external forces.
The equation is HΨ=EΨ,
where Ψ represents the height of a wave at position x, y, z and
Ψ^2 represents the probability of finding an electron
so change in Ψ(x, y, z) = energy (x, y, z)
ValerieChavarin 4F
Posts: 99
Joined: Wed Sep 18, 2019 12:18 am
### Re: shrodinger equation
Emma Popescu 1L
Posts: 105
Joined: Wed Sep 11, 2019 12:16 am
### Re: shrodinger equation
The Schrodinger equation is used to calculate both the wavefunction and the corresponding energy. There is a lot of information about it in Topic 1C in the textbook but we do not need to read that section. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8748894333839417, "perplexity": 1272.73013766335}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703538741.56/warc/CC-MAIN-20210123222657-20210124012657-00453.warc.gz"} |
http://mathhelpforum.com/calculus/66850-concavity-print.html | # Concavity
• January 4th 2009, 01:58 PM
abclarinetuvwxyz
Concavity
The graph of f(x)= integral from 0 to x (15(t^2) - 2(t^3) + 24)dt is concave up on (a,b). Find b-a.
Okay so I know from the curve sketching lesson in AB calculus that to find concavity you take the second derivative and then set that to 0. Do I take the integral and then take the first derivative, and then the second?
• January 4th 2009, 02:17 PM
skeeter
• January 4th 2009, 03:39 PM
Soroban
Hello, abclarinetuvwxyz!
Quote:
The graph of: . $f(x)\;= \;\int^x_0(15t^2 - 2t^3 + 24)\,dt$ . is concave up on $(a,b).$]
Find $b-a$
The first derivative is: . $f'(x) \:=\:15x^2 - 2x^3 + 24$
The second derivative is: . $f''(x) \:=\:30x - 6x^2$
Now where is the graph concave up?
. . Where $f''(x)$ is positive.
So we have: . $30x - 6x^2 \:>\:0\quad\Rightarrow\quad 6x(5 - x) \:>\:0$
This is true when: . $0 \,<\,x\,<\,5$
Therefore, the interval is: . $(0,5) \quad\Rightarrow\quad b - a \:=\:5-0 \:=\:\boxed{5}$
• January 4th 2009, 03:49 PM
abclarinetuvwxyz
Thanks very much! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9844720363616943, "perplexity": 2308.076876516469}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131304444.86/warc/CC-MAIN-20150323172144-00283-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/determine-currents-in-resistors-help.328087/ | # Determine currents in resistors help
1. Jul 31, 2009
### araujo3rd
Iv started physics electronics literacy at university as a requirement for my degree, i didnt do science at school and im very confused with whats going on..How to a determine a current in the resistors, please can someone give me a step by step on how they got to the answer...i have uploaded a picture of the circuit i was given
all help is very much appreciaited, im writing a test on this stuff soon.. i know how to get Req of parrallel resister, but how do i get the currents, specifically of the 3.3k and 560 resistors?
#### Attached Files:
• ###### circuit.GIF
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2.6 KB
Views:
75
2. Jul 31, 2009
### jmb
If you already know how to calculate the effective resistance of parallel and series circuits, then you should be able to use Ohm's Law to calculate the total current through the circuit.
In addition if you use the following two pieces of information: (1) the current along a series circuit is the same everywhere and (2) the total voltage drop across two pieces of circuit that are joined in parallel is the same for each piece, then you should be able to work out the things you need (btw these two rules come from the more general set of rules known as Kirchoff's laws).
Hint: Start out by treating the entire circuit as one single effective resistor and successively break it into smaller pieces (some of which will be series circuits and some will be parallel circuits).
Have a go at it and post your attempt back here and we will give you feedback/corrections...
3. Jul 31, 2009
### araujo3rd
Thanks for your reply, ok so i worked out the total resistance to be 2909.5, then worked out the current passing through I2 (5/2909.5) and got 1.72mA which gave me a voltage of 3.44V, is that correct?
Now what do i do from here, how would i work out the current passing through through the 1k resistor? please show me how and ill then use that knowledge to try calculate the current in the 3 on my own, i just not sure what to do now
4. Jul 31, 2009
### jmb
That's correct.
This is the total current in the whole circuit. If by I2 you meant the 2k resistor then, yes, this is also the current going through the 2k resistor.
Again, assuming I2 means the 2k resistor, then yes that is the correct voltage drop across that resistor.
Since that is the voltage drop across the resistor, the voltage across the rest of the circuit (i.e. if you were to connect a voltmeter between the point just after the 2k resistor and the ground terminal of the battery) is 5V - 3.44V = 1.56V.
Using the second rule I gave you (the voltage drop across two parts of a parallel circuit is the same for each part) this tells you that the voltage across the 3300 Ohm resistor is 1.56V (from which you should be able to work out the current in it directly from Ohm's law) and the voltage drop across the remaining subcircuit (consisting of the 1000, 560 and 470 Ohm resistors) is also 1.56V.
Since you know the voltage drop across the subcircuit you should now be able to treat it on its own as an isolated system. To find the currents in each of its resistors just go through exactly the same process as you did for the original circuit (since it exactly resembles the original circuit but with the subcircuit replaced by a single resistor).
Have a go at finishing it off and post your results here and then we'll check it for you...
5. Aug 1, 2009
### araujo3rd
I have tried am im just not sure how to work them out :( im really confused, because im not sure which values to use for the 1k, 560 and 470 ohm resistors, i know that 560 and 470 must have the same current, but how do i work out the 1k one (because its in parrallel with the 3.3k. Really dono how to get them, should the voltage be of the resistors be calculated first or the current first? thank you for all the feedback
6. Aug 3, 2009
### jmb
No they don't! Remember: only elements in serial will necessarily have the same current going through them.
You've already worked out the voltage drop across what I've called the "subcircuit" (consisting of the 1k, 560 and 470 Ohm resistors) so you can then ignore everything else outside that subcircuit (e.g. the 3.3k resistor) provided you use this calculated voltage drop.
It depends, sometimes it will be easier to get the voltage first, and sometimes the current. The point is to be systematic. Let's recap on what has already been done:
• We started by treating the entire circuit as a single component. Using the laws for adding parallel and series resistors you worked out the total effective resistance for the circuit.
• Since you already knew the total voltage across the circuit (given by the battery) you could combine this with the calculated total resistance to find the total current in the circuit (Ohm's Law).
• Since the current won't change until there's a junction (because of the first rule I gave you: the current is the same across series parts of the circuit) we knew that the current going through the 2k resistor was equal to this total current. So this time we could use Ohm's law to find the voltage drop across the 2k resistor. This in turn told us how much voltage was 'left over' to drive those parts of the circuit that come after the 2k resistor.
• I told you (the 2nd rule I gave you) that the voltage drop is the same across parts of a parallel circuit, so that we knew the voltage across the 3.3k resistor was the 'left over' calculated above. Since we know resistance and voltage, this should be enough to let you calculate the current going through it using Ohm's Law again.
• This 'left over' voltage is also the voltage drop across the 'subcircuit' (consisting of the 1k 560 and 470 Ohm resistors).
As I said, the remaining subcircuit can be solved by repeating the same process again. Try and think what this is for yourself and then check it against the below:
• Calculate the total effective resistance for the subcircuit using the parallel and series resistor rules.
• Use this, the already calculated voltage drop across the subcircuit and Ohm's Law to calculate the total current in the subcircuit.
• This total subcircuit current must also be the current going through the 1k resistor (first rule I gave you), thus you can use Ohm's Law to cacluate the voltage drop across the 1k resistor.
• You then know the 'left over' voltage is the voltage drop across the parallel 560 and 470 Ohm resistors. Since the voltage drop across both will be the same (2nd rule I gave you) you can again just use Ohm's Law to calculate the current in each.
Have a go at doing this and see what numbers you get. You should see that current is conserved at junctions (i.e. the outgoing current always splits into amounts that sum up to the original input current at that junction).
If you still have difficulty then there are some additional rules I can give you. However, these rules are the result of applying the rules I've already given you to the general case. So if you can do things without them the first few times you'll end up with a much better understanding of how circuits work...
Let us know how you get on!
7. Aug 4, 2009
### araujo3rd
ok thank you so mch i think i got it...the current in the 1k is 1.24V, leaving 0.32V in the 560 and 470 ohm resistors..therefore current in the 570 ohm is 571 micro amps and the 470 is 680 micro amps
8. Aug 5, 2009
### jmb
Good that is correct, but be careful with your rounding. If you want to quote the current in the 560 Ohm resistor to 3 figures then you need to keep your intermediate calculations more accurately (the answer should be 568 microAmps but you have a little rounding error in your own calculations, similarly the 470 Ohm resistor has 677 microAmps).
As I mentioned in my last post you should notice that the current is conserved. In fact you could have used this fact to save a little time in your calculations.
However I would suggest you keep using the simple approach we went through here until you feel happy with how things work --- this should give you a good feeling for circuits.
Once you are comfortable with the methods I would advise you to do this:
Consider a simple series circuit consisting of just two resistors of resistance A and B under a total voltage V (this is often called a potential divider). Use the rules we've been using to calculate an expression for the voltage drop across the first resistor (A) and hence the remaining voltage that is applied to resistor B. The resulting formula should allow you tackle these kind of problems much faster since you should then be able to immediately calculate the voltage across each resistor and so then get the current straight from Ohm's Law. The advantage (in my opinion) is that you will also by this point understand the formula rather than just be applying it blindly.
Similar Discussions: Determine currents in resistors help | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8242232203483582, "perplexity": 396.29902908927187}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948580416.55/warc/CC-MAIN-20171215231248-20171216013248-00626.warc.gz"} |
https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2427270 | Public User
# On an Optimal Stopping Problem of an Insider
6 Pages Posted: 23 Apr 2014
## Erhan Bayraktar
University of Michigan at Ann Arbor - Department of Mathematics
## Zhou Zhou
University of Minnesota - Twin Cities
Date Written: April 21, 2014
### Abstract
We consider the optimal problem $\sup_{\tau\in\mathcal{T}_{\eps,T}}\mathbb{E}\left[\sum_{i=1}^n \phi_{(\tau-\eps^i)^ }^i\right]$, where $T>0$ is a fixed time horizon, $(\phi_t^i)_{0\leq t\leq T}$ is progressively measurable with respect to the Brownian filtration, $\eps^i\in[0,T]$ is a constant, $i=1,\dotso,n$, and $\mathcal{T}_{\eps,T}$ is the set of stopping times that lie between a constant $\eps\in[0,T]$ and $T$. We solve this problem by conditioning and then using the theory of reflected backward stochastic differential equations (RBSDEs). As a corollary, we provide a solution to the optimal stopping problem $\sup_{\tau\in\mathcal{T}_{0,T}}\mathbb{E}B_{(\tau-\eps)^ }$ recently posed by Shiryaev at the International Conference on Advanced Stochastic Optimization Problems organized by the Steklov Institute of Mathematics in September 2012. We also provide its asymptotic order as $\eps\searrow 0$.
Keywords: optimal stopping problem of an insider, Reflected Backward Stochastic Differential Equations (RBSDEs), Levy's modulus for Brownian motion
Suggested Citation
Bayraktar, Erhan and Zhou, Zhou, On an Optimal Stopping Problem of an Insider (April 21, 2014). Available at SSRN: https://ssrn.com/abstract=2427270 or http://dx.doi.org/10.2139/ssrn.2427270 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9100989103317261, "perplexity": 829.9289029893802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886107744.5/warc/CC-MAIN-20170821080132-20170821100132-00246.warc.gz"} |
https://byjus.com/maths/square-root-questions/ | # Square Root Questions
It has been often said that “practice makes perfect.” This phrase is actually very relatable when it comes to math as students can be good at the subject only by practicing and solving problems more and more. While it may not be easy but developing math skills and fluency is important. Mathematical skills of investigation, problem solving, and logical thinking can further help students progress in other subjects as well.
That being said, here in this page we will be discussing about the topic of square roots. Well, generally when we define the term, the square root of a number is a value that, when multiplied by itself, gives the number. For example, let’s say when you multiply 4 × 4 you get 16. So the square root of 16 is 4. This is the basic concept.
It is denoted by the symbol √ and it means that is a positive or perfect square root. For example, √36 = 6 (6 x 6 = 36). There are also square negative numbers. For example, (-5) X (-5) = 25. When we square a negative number we get a positive result.
Moving on, if you want to know how to find the square root of a number, then there are a lot of methods. However, the most basic method that can be used is the prime factorization method or the popular square root long division method. You can use this method to find the square root of a number that is satisfactory or accurate enough for you. An example of division method showing how to take the square root is given below;
Additionally, you can also check out BYJU’S YouTube channel to learn how to find the square of any number using different methods:
1.Which of the following numbers is a perfect square?
(a) 141
(b) 196
(c) 124
(d) 222
2.A perfect square number can never have the digit ….. at the units place.
(a) 1
(b) 4
(c) 8
(d) 9
3.Evaluate √6084
(a) 75
(b) 77
(c) 78
(d) 68
4.Find the square root of 5929.
(a) 49
(b) 33
(c) 77
(d) 73
5.Evaluate √1471369.
(a) 1213
(b) 1223
(c) 1233
(d) 1243 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8221369981765747, "perplexity": 309.1363779098857}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249490870.89/warc/CC-MAIN-20190223061816-20190223083816-00604.warc.gz"} |
http://math.stackexchange.com/questions/387611/calculating-the-galois-group-of-the-splitting-field-of-f-x3-3 | # Calculating The Galois Group of the Splitting Field of $f=x^3-3$
If we let $f=x^3-3$ the let L be the splitting field for this polynomial I am trying to find $\Gamma(L:\mathbb{Q})$ and all intermediate field extensions.
Now as this is a splitting field and finite it is a finite normal extension and so we can use the fundemental theorem of Galois Theory so we know that:
$|\Gamma(L:\mathbb{Q})|=[L:\mathbb{Q}]$
Now we can see that if we define $\zeta=\exp{\frac{2\pi i}{3}}$ then we have that $L=\mathbb{Q}(\zeta,(3)^{\frac{1}{3}})$ and the minimal polynomial of $(3)^{\frac{1}{3}}$ over $\mathbb{Q}$ has degree 3 and the minimal polynomial of $\zeta$ has degree 2 over $\mathbb{Q}((3)^{\frac{1}{3}})$ so from the tower law we can see that:
$[L:\mathbb{Q}]=6$
Now we can see that a basis for $L$ is given by $\displaystyle{\{1,\zeta,(3)^{\frac{1}{3}},(3)^{\frac{2}{3}},(3)^{\frac{1}{3}}\zeta,(3)^{\frac{2}{3}}\zeta\}}$
So if we now consider $\mathbb{Q}$ automorphisms of $L$ then we need only consider it's action on the basis elements, so we can see that if $f$ is a $\mathbb{Q}$ automorphism then we must have that:
$f(\zeta)=\zeta,(3)^{\frac{1}{3}}\zeta$ and
$f((3)^{\frac{1}{3}})=(3)^{\frac{1}{3}},(3)^{\frac{2}{3}},-\zeta$
which will give us the 6 elements of $\Gamma(L:\mathbb{Q})$
Is what I have done so far correct? and if so how do I now find the intermediate fields? I know that they are related to the subgroups but I am unsure how?
Thanks for any help
-
In the very first line I think that + must be an = ... – DonAntonio May 10 '13 at 14:21
If you write $\,(-1)^{1/3}\,$ chances are people will think this is $\,-1\,$ – DonAntonio May 10 '13 at 14:22
@DonAntonio yeah thats what it was supposed to be, what should I write instead of $(-1)^{\frac{1}{3}}$ just $\alpha$ where $\alpha$ is a non-real third root of unity? – hmmmm May 10 '13 at 14:40
it is customary to write $\,\zeta_3=e^{2\pi i/3}\,$ to avoid confusions, or simply $\,\zeta\,$ is there's only one such root of unity. – DonAntonio May 10 '13 at 14:42
@DonAntonio cool thanks I did not know that, I will edit it now and just put $\exp{\frac{2\pi i}{3}}$ in instead I can see that this is much more clear. – hmmmm May 10 '13 at 14:44
I'll add a few things to this. You know that the Galois group has order 6, so there are only two choices: $S_3$ and $Z_3\times Z_2$. The latter group is abelian, so this means that every subgroup is normal. Thus, let $L/\mathbb{Q}$ be the splitting field and denote the Galois group by $G$.
The fundamental theorem also says that there's a bijective correspondence between subextensions $L/K/\mathbb{Q}$ such that $K/\mathbb{Q}$ is Galois and normal subgroups of $G$. Take the subextension $\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q}$. This extension is not normal, so the subgroup $\textrm{Aut}(L/\mathbb{Q}(\sqrt[3]{3}))\leq G$ is not a normal subgroup. This implies that $G$ can't be abelian, so it must be $S_3$.
Now $S_3$ has the following subgroups:
• A unique subgroup of order $3$.
• $3$ subgroups of order $2$ that are all conjugate. These are the ones generated by $(1\, 2)$, $(1\, 3)$ and $(2\, 3)$.
The unique order $3$ subgroup is normal, so it corresponds to a unique quadratic Galois extension $K/\mathbb{Q}$. This must be the field extension $\mathbb{Q}(\zeta_3)/\mathbb{Q}$, which we know is Galois.
Next, we find the three non-Galois cubic subextensions of $\mathbb{Q}$. These are the field extensions corresponding to each root of $X^3-3$, so we get the extensions:
$$\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q},\;\;\mathbb{Q}(\zeta_3\sqrt[3]{3})/\mathbb{Q},\;\;\mathbb{Q}(\zeta_3^2\sqrt[3]{3})/\mathbb{Q}$$
To show that these extensions are all distinct, note that if e.g. $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\zeta_3\sqrt[3]{3})$, then $\zeta_3 = \zeta_3\sqrt[3]{3}/\sqrt[3]{3}\in \mathbb{Q}(\sqrt[3]{3})$. This is impossible, since $L\neq \mathbb{Q}(\sqrt[3]{3})$. The same argument applies to the other two pairs.
-
An easier way to see why the group is $S_3$ is that the Galois group of (the splitting field of) a polynomial of degree $n$ acts by permuting the roots of the polynomial, and thus is always a subgroup of $S_n$. – Hurkyl May 13 '13 at 0:25
Yes, there are much slicker ways of doing this depending on how much you know. I chose this method, since for trickier cases you have to look at the subgroup lattice. – Edvard Fagerholm May 16 '13 at 4:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9746586680412292, "perplexity": 97.13139127037213}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049281876.4/warc/CC-MAIN-20160524002121-00019-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/19611-functions-models.html | 1. functions and models
The following problems are the only problems from my homework that i couldn't not solve or rather figure out. If i could by any chance receive any help to that would be great. -please if you help me solve these show me the steps so i at least know how to do it... thank you.
1)Find an expression for the function whose graph consists of the line segment from the point (-2,2) to the point (-1,0) together with the top half of the circle with center the origin and radius 1.
F(x)= -2x-2 if -2 < x < -1
root(1-x^2) if -1< x <1
why?
2) if f(x) = ln x and g(x) = x^2-9 find the functions of a, b ,c ,d and their domains
a)(f o g)
ln(x^2-9)
b)(g o f) (ln x)^2 -9
c) (f o f) ln(ln(x))
d) (g o g) (x^2-9)^2 -9 = x^4-18x^2 +72
3) if f(x) = 2x+ ln(x) find inverse of f(2)
the answer is 1 but how?
4) Solve each equation for x.
a) e^x = 5
x = 5th root of e? am i right?
b) ln(x) = 2
c)e^e^x = 2
x = ln(ln(2))
5) sketch the graph of the function g(x) = |x^2-1| - |x^2=4|
Any help with any of the above problems would be greatly appreciated it. Those are some of the few problems that i seem to be having a hard time. I also showed some of the work that i did in regular font(no bold).
2. Originally Posted by nuttynutz
The following problems are the only problems from my homework that i couldn't not solve or rather figure out. If i could by any chance receive any help to that would be great. -please if you help me solve these show me the steps so i at least know how to do it... thank you.
1)Find an expression for the function whose graph consists of the line segment from the point (-2,2) to the point (-1,0) together with the top half of the circle with center the origin and radius 1.
F(x)= -2x-2 if -2 < x < -1
root(1-x^2) if -1< x <1
why?
The first part of the function is because $f(x) = -2x - 2$ is the line containing the points (-2, 2) and (-1, 0).
For the second part note that the circle of radius 1 and center at the origin is
$x^2 + y^2 = 1$
So solving for $y = f(x) = \sqrt{1 - x^2}$.
-Dan
3. Originally Posted by nuttynutz
2) if f(x) = ln x and g(x) = x^2-9 find the functions of a, b ,c ,d and their domains
a)(f o g)
ln(x^2-9)
b)(g o f) (ln x)^2 -9
c) (f o f) ln(ln(x))
d) (g o g) (x^2-9)^2 -9 = x^4-18x^2 +72
So far so good. Now what the their domains?
-Dan
4. i can't find their domain... i don't know how to solve for ln at least not yet. so, i can't find the domain
5. Originally Posted by nuttynutz
3) if f(x) = 2x+ ln(x) find inverse of f(2)
the answer is 1 but how?
Don't you mean the inverse function evaluated at 2? ie. $f^{-1}(2)$?
Unless you get lucky and see the answer probably the best thing to do would be to graph the function and look for the x value such that f(x) = 2. See the graph below.
-Dan
Attached Thumbnails
6. Originally Posted by nuttynutz
4) Solve each equation for x.
a) e^x = 5
x = 5th root of e? am i right?
b) ln(x) = 2
c)e^e^x = 2
x = ln(ln(2))
a)
$e^x = 5$
$ln(e^x) = ln(5)$
$x = ln(5)$
b)
$ln(x) = 2$
$e^{ln(x)} = e^2$
$x = e^2$
c)
$e^{e^x} = 2$
$ln \left ( e^{e^x} \right ) = ln(2)$
$e^x = ln(2)$
$ln(e^x) = ln(ln(2))$
$x = ln(ln(2))$
-Dan
7. Originally Posted by nuttynutz
5) sketch the graph of the function g(x) = |x^2-1| - |x^2=4|
Worst comes to worst, pick a bunch of x values and find the g(x)'s.
$g(x) = |x^2 - 1| - |x^2 - 4|$
$g(x) = |(x + 1)(x - 1)| - |(x + 2)(x - 2)|$
Critical points are where the arguments of the absolute value bars is zero, so I have critical points at $x = -2, -1, 1, 2$.
For $(-\infty, -2)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$.
So on $(-\infty, -2)$ $g(x) = (x + 1)(x - 1) - (x + 2)(x - 2) = 3$
For $(-2, -1)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) < 0 \implies |(x + 2)(x - 2)| = -(x + 2)(x - 2)$.
So on $(-2, -1)$ $g(x) = (x + 1)(x - 1) + (x + 2)(x - 2) = 2x^2 - 5$
So far we have:
$g(x) = \begin{cases} 3, \text{ for }-\infty \leq x \\ 2x^2 - 5, \text{ for } -2 < x \leq -1 \end{cases}$
I'll let you figure the other two intervals.
-Dan
8. Originally Posted by nuttynutz
2) if f(x) = ln x and g(x) = x^2-9 find the functions of a, b ,c ,d and their domains
a)(f o g)
ln(x^2-9)
b)(g o f) (ln x)^2 -9
c) (f o f) ln(ln(x))
d) (g o g) (x^2-9)^2 -9 = x^4-18x^2 +72
If you haven't worked with the ln function then why are you even trying to do these??
You should still be able to do d). This is simply a polynomial. What is the domain of a polynomial function?
-Dan
9. Originally Posted by topsquark
If you haven't worked with the ln function then why are you even trying to do these??
You should still be able to do d). This is simply a polynomial. What is the domain of a polynomial function?
-Dan
all real numbers =) lol sometimes i don't think...
10. Originally Posted by topsquark
Don't you mean the inverse function evaluated at 2? ie. $f^{-1}(2)$?
Unless you get lucky and see the answer probably the best thing to do would be to graph the function and look for the x value such that f(x) = 2. See the graph below.
-Dan
could this be done without the use of a calculator? if so then how?
11. Originally Posted by topsquark
Worst comes to worst, pick a bunch of x values and find the g(x)'s.
$g(x) = |x^2 - 1| - |x^2 - 4|$
$g(x) = |(x + 1)(x - 1)| - |(x + 2)(x - 2)|$
Critical points are where the arguments of the absolute value bars is zero, so I have critical points at $x = -2, -1, 1, 2$.
For $(-\infty, -2)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$.
So on $(-\infty, -2)$ $g(x) = (x + 1)(x - 1) - (x + 2)(x - 2) = 3$
For $(-2, -1)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) < 0 \implies |(x + 2)(x - 2)| = -(x + 2)(x - 2)$.
So on $(-2, -1)$ $g(x) = (x + 1)(x - 1) + (x + 2)(x - 2) = 2x^2 - 5$
So far we have:
$g(x) = \begin{cases} 3, \text{ for }-\infty \leq x \\ 2x^2 - 5, \text{ for } -2 < x \leq -1 \end{cases}$
I'll let you figure the other two intervals.
-Dan
alright i did a lot of research and sort of reviewing to try and figure everything out and i think i have it all done correctly my only problem is the question above.. i really am clueless when it comes to it. Any help would be appreciated it
12. Originally Posted by topsquark
Don't you mean the inverse function evaluated at 2? ie. $f^{-1}(2)$?
Unless you get lucky and see the answer probably the best thing to do would be to graph the function and look for the x value such that f(x) = 2. See the graph below.
-Dan
Originally Posted by nuttynutz
could this be done without the use of a calculator? if so then how?
The only way to do it without a calculator is to look at the equation and see what your intuition says. I can't guide you with that any further, except to say that I would likely have (if I hadn't seen the solution right below it) tried a few integer x to see what came out of it. If you did that you would have noted that $f(1) = 2$ and thus you could easily get $f^{-1}(2) = 1$.
-Dan
13. Originally Posted by topsquark
Worst comes to worst, pick a bunch of x values and find the g(x)'s.
$g(x) = |x^2 - 1| - |x^2 - 4|$
$g(x) = |(x + 1)(x - 1)| - |(x + 2)(x - 2)|$
Critical points are where the arguments of the absolute value bars is zero, so I have critical points at $x = -2, -1, 1, 2$.
For $(-\infty, -2)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$.
So on $(-\infty, -2)$ $g(x) = (x + 1)(x - 1) - (x + 2)(x - 2) = 3$
For $(-2, -1)$ $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) < 0 \implies |(x + 2)(x - 2)| = -(x + 2)(x - 2)$.
So on $(-2, -1)$ $g(x) = (x + 1)(x - 1) + (x + 2)(x - 2) = 2x^2 - 5$
So far we have:
$g(x) = \begin{cases} 3, \text{ for }-\infty \leq x \\ 2x^2 - 5, \text{ for } -2 < x \leq -1 \end{cases}$
I'll let you figure the other two intervals.
-Dan
Originally Posted by nuttynutz
alright i did a lot of research and sort of reviewing to try and figure everything out and i think i have it all done correctly my only problem is the question above.. i really am clueless when it comes to it. Any help would be appreciated it
The solution method I gave you is pretty systematic and exploits the fact that
$|x| = \begin{cases} -x; x < 0 \\ x; x \geq 0 \end{cases}$
You look at your function and see where the argument inside the absolute value bars is negative and then equate that to the negative of the argument. Case in point:
For $(-\infty, -2)$, $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$
from above. (Sorry, I guess I could've used some separators in there for clarity.)
I've given you examples of how to do it with two of the intervals, and I'm asking you to do it yourself for the other three intervals: $(-1, 1) \text{ and } (1, 2) \text{ and } (2, \infty )$.
Just find where each expression inside the absolute value bars is negative, then replace the absolute value with the negative of that expression. Then simplify.
Edit: If the problem is notation, let me break down the above quote a bit:
For $(-\infty, -2)$, $(x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $(x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$
This means that I am considering the case where x is between $-\infty$ and -2. So we know that $x + 1$ is negative and $x - 1$ is negative so $(x + 1)(x - 1)$ is positive, etc.
-Dan | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 73, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9022791385650635, "perplexity": 411.3531189846847}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661768.23/warc/CC-MAIN-20160924173741-00294-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/203601-related-rates-problem.html | # Thread: Related rates problem.. :)
1. ## Related rates problem.. :)
Hello So first this is the problem:
A Swimming pool is 20 ft. wide, 40 ft. long, 3 ft. deep at the shallow end and 9 ft. deep at it's deepest point. A cross section is shown in the figure. If the pool is being filled at a rate of 0.8 cu ft. per min., how fast is the water level rising when the depth at the deepest end is 5ft.
Thank you.. mind if you please show me the step by step procedure thank you...
2. ## Re: Related rates problem.. :)
Was that picture part of the given problem or did you draw it yourself? It is not what I would get from your description- nor is it my experience with swimming pools. You have a "lowered" section on the left with slanted sides and then up to a height of 3 feet beow the top, then flat for the rest of the forty foot length. Where did you get that "6- 12- 6- 16" along the bottom (which I assume are distances). I would, rather, have drawn the side view of the swimming pool as a single trapezoid with the two parallel ends of lengths 3 and 9 feet and the "height" 40 feet. I don't know if you were given that rather strange picture or drew it yourself.
3. ## Re: Related rates problem.. :)
No sir, I didn't... It was given to our midterms exam just a while ago, me and my other colleague discussed the solutions and answered all of our questions.. excepts one... that One exactly... yep, i tried to use the area, which then substituted it to the volume then differentiated it.. but I'm pretty sure I'm still lost..
4. ## Re: Related rates problem.. :)
Hello, kspkido!
A swimming pool is 20 ft. wide, 40 ft. long, 3 ft. deep at the shallow end and 9 ft. deep at it's deepest point.
A cross-section is shown in the figure.
If the pool is being filled at a rate of 0.8 cu ft. per min.,
. . how fast is the water level rising when the depth at the deepest end is 5 ft?
Code:
: - - - - - - - - - - 40 - - - - - - - - - :
- *-------------------------------------------*
3 | |
- * - - - - - - - *-------------------*
:*~~~~~~~~~~~~~~~~~~~~~*: - - - 16 - - - - :
6 : *:::::::::|:::::::::* :
: -*::::::::|h:::::::* :
- + - *-------------- * - :
: 6 : - - 12 - - - : 6 :
Since $h \le 6$, we are concerned with the trapezoid only.
The cross-section looks like this:
Code:
h : 12 : h
*-------+-------*
*::::::|::::::*
*:::::|h::::*
*::::|::::*
*---*---*
12
The area is: . $A \:=\:\tfrac{h}{2}(12 + 12 + 2h) \:=\:12h+h^2$
The volume is: . $V \:=\:20A \:=\:20(12h+h^2)$
Then: . $\frac{dV}{dt} \:=\:20(12 + 2h)\frac{dh}{dt}$
When $\frac{dV}{dt} = 0.8,\;h - 5$, we have: . $0.8 \:=\:20(22)\frac{dh}{dt}$
. . $\frac{dh}{dt} \:=\:\frac{0.8}{440} \:=\:\frac{1}{55}$
The water is rising at the rate of $\tfrac{1}{55}$ ft/min.
5. ## Re: Related rates problem.. :)
Wow... I never thought you could do that... h=6, but you used it as an integer just so that you could derive ayt?... but one question sir... h-5? is equals to 5? how is that? but thank you sir...
6. ## Re: Related rates problem.. :)
you've posted this problem before ... and it was answered.
Rate Problem Involving Changing Height.. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8655416369438171, "perplexity": 1146.0022773493145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818693363.77/warc/CC-MAIN-20170925201601-20170925221601-00417.warc.gz"} |
https://pegaswitch.com/trending/what-is-the-principle-of-an-analog-multiplier/ | Trending
# What is the principle of an analog multiplier?
## What is the principle of an analog multiplier?
ANALOG MULTIPLIER BASICS The signal at the output is the product of the two input signals. If both input and output signals are voltages, the transfer characteristic is the product of the two voltages divided by a scaling factor, K, which has the dimension of voltage as shown in Figure 1.
How does frequency doubler work?
Frequency multipliers consist of a nonlinear circuit that distorts the input signal and consequently generates harmonics of the input signal. A subsequent bandpass filter selects the desired harmonic frequency and removes the unwanted fundamental and other harmonics from the output.
### Where is analog multiplier IC used?
Analog multiplier is a circuit whose output voltage at any instant is proportional to the product of instantaneous value of two individual input voltages. Important applications of these multipliers are multiplication, division, squaring and square – rooting of signals, modulation and demodulation.
Is an example for a frequency multiplier?
__________ is an example for a frequency multiplier. Explanation: A non linear device has the ability to generate the harmonics of the input sinusoidal signal. Transistor and diodes are non linear devices and hence can be used as a frequency multiplier.
#### What are the applications of analog multiplier?
In electronics, an analog multiplier is a device which takes two analog signals and produces an output which is their product. Such circuits can be used to implement related functions such as squares (apply same signal to both inputs), and square roots.
How does BJT act as multiplier?
The capacitance multiplier circuit operation is quite straightforward. It acts as a simple emitter follower. The resistor R1 provides bias for the base emitter junction, and the capacitor provides smoothing. This considerably reduces the levels on noise on the output, i.e. Vout.
## What device improves output at low frequency?
A frequency changer or frequency converter is an electronic or electromechanical device that converts alternating current (AC) of one frequency to alternating current of another frequency.
How do you increase the frequency of a signal?
If we want to increase the frequency, we need to do is:
1. take the fourier transform.
2. leave the DC offset unchanged.
3. shift the positive part of the spectrum to the right.
4. shift the negative part of the spectrum to the left.
### Which diode is used in frequency multiplier circuit?
Planar Schottky varactor diodes
Planar Schottky varactor diodes are commonly used in frequency multipliers, taking the advantage of GaAs substrateless technology to reduce substrate loss. The drive sources can be BWOs or solid-state sources such as Gunn and IMPATT oscillators, with relatively high output power in the range of 50 GHz to 150 GHz.
What analog means?
1 : something that is similar or comparable to something else either in general or in some specific detail : something that is analogous to something else historical analogues to the current situation an aspirin analogue.
#### What is the application of analog multiplier?
What is multiplier circuit?
A multiplier is a combinational logic circuit that we use to multiply binary digits. Just like the adder and the subtractor, a multiplier is an arithmetic combinational logic circuit. It is also known as a binary multiplier or a digital multiplier.
## Which is an example of an analog multiplier?
INTRODUCTION 2. Analog multiplier are basic circuit building blocks for analog signal processing in instrumentation and communication systems such as a variable gain amplifier, automatic gain control amplifier, frequency doubler, phase locked loop, amplitude locked loop, small signal rectifier, etc.
How is the multiplier of an operational amplifier achieved?
The opamp supply current sensing technique [3]-[4] is employed to obtain the square of the sum and difference of two input signals. The multiplication is achieved by the quarter-square algebraic identity. The purpose of this article is to propose analog multiplier using operational amplifier.
### How are multipliers used in the feedback loop?
USING MULTIPLIERS WITH OP AMPS TO PERFORM ARITHMETIC FUNCTIONS . Multipliers can be placed in the feedback loop of op amps to form several useful functions. Figure 9 illustrates the basic principle of analog computation that a function generator in a negative feedback loop computes the inverse function (provided, of course, that the function is
What do you need to know about multipliers?
PREFACE CONTENTS MULTIPLIERS – SOME BASICS MULTIPLICATION DIVISION AND ROOTING POLARITY SPECIAL MULTIPLICATION-DIVISION FUNCTIONS SCALING BRIEF DEFINITIONS APPLICATIONS ii 2 2 3 3 3 4 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8868049383163452, "perplexity": 1213.8489960547226}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103850139.45/warc/CC-MAIN-20220630153307-20220630183307-00195.warc.gz"} |
http://www.physicsforums.com/printthread.php?t=21017 | Physics Forums (http://www.physicsforums.com/index.php)
- General Astronomy (http://www.physicsforums.com/forumdisplay.php?f=71)
- - Finite hyperbolic universe and large scale structure patterns (http://www.physicsforums.com/showthread.php?t=21017)
hellfire Apr16-04 04:13 PM
Finite hyperbolic universe and large scale structure patterns
This paper :
Hyperbolic Universes with a Horned Topology and the CMB Anisotropy
http://arxiv.org/astro-ph/0403597
...press release:
http://www.newscientist.com/news/news.jsp?id=ns99994879
proposes a universe with the shape of a horn. This is a hyperbolic space with negative curvature.
The paper mentions an interesting issue: the relation between finite hyperbolic spaces and chaos. Finite hyperbolic spaces generate chaotic mixing of trayectories, leading to fractal structure formation. See e.g.:
Chaos and order in a finite universe
http://arxiv.org/abs/astro-ph/9907288
A fractal nature of large scale structures was already suggested due to the self-similarity of the distribution of galaxies and clusters (similar correlation functions AFAIK).
My knowledge of chaotic systems is almost non existent, thus I would like to know qualitatively why finite hyperbolic spaces do have such properties in relation to chaos and infinite flat spaces do not (although you can find an interesting remark in the previous cited paper about the cosmological constant in infinite flat spaces).
But there is another thing that bothers me. In the paper it is claimed that the CMB data would not reflect the negative curvature. But why? Usually it is assumed that the angular scale of the first peak of the CMB anisotropies gives a measure of the curvature.
Regards.
meteor Apr16-04 06:16 PM
An universe infinitely long but with finite volume: it remembers me a surface called Gabriel's Horn
And this thing called Picard topology must be an invention of F.Steiner. i did a google on "Picard topology", and only appeared 5 entries, and the 5 related to this horn-shaped-universe theory
Mike2 Apr16-04 08:11 PM
Is this like a universe that grow from a singularity infinitely in the past in an accelerated manner?
All times are GMT -5. The time now is 11:14 PM. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.972237229347229, "perplexity": 1494.3585803738006}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394021306108/warc/CC-MAIN-20140305120826-00073-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://julia.quantecon.org/tools_and_techniques/finite_markov.html | • Lectures
• Code
• Notebooks
• Community
# Finite Markov Chains¶
## Overview¶
Markov chains are one of the most useful classes of stochastic processes, being
• simple, flexible and supported by many elegant theoretical results
• valuable for building intuition about random dynamic models
• central to quantitative modeling in their own right
You will find them in many of the workhorse models of economics and finance.
In this lecture we review some of the theory of Markov chains.
We will also introduce some of the high quality routines for working with Markov chains available in QuantEcon.jl.
Prerequisite knowledge is basic probability and linear algebra.
### Setup¶
In [1]:
using InstantiateFromURL
# optionally add arguments to force installation: instantiate = true, precompile = true
github_project("QuantEcon/quantecon-notebooks-julia", version = "0.8.0")
In [2]:
using LinearAlgebra, Statistics
using Distributions, Plots, Printf, QuantEcon, Random
gr(fmt = :png);
## Definitions¶
The following concepts are fundamental.
### Stochastic Matrices¶
A stochastic matrix (or Markov matrix) is an $n \times n$ square matrix $P$ such that
1. each element of $P$ is nonnegative, and
2. each row of $P$ sums to one
Each row of $P$ can be regarded as a probability mass function over $n$ possible outcomes.
It is too not difficult to check [1] that if $P$ is a stochastic matrix, then so is the $k$-th power $P^k$ for all $k \in \mathbb N$.
### Markov Chains¶
There is a close connection between stochastic matrices and Markov chains.
To begin, let $S$ be a finite set with $n$ elements $\{x_1, \ldots, x_n\}$.
The set $S$ is called the state space and $x_1, \ldots, x_n$ are the state values.
A Markov chain $\{X_t\}$ on $S$ is a sequence of random variables on $S$ that have the Markov property.
This means that, for any date $t$ and any state $y \in S$,
$$\mathbb P \{ X_{t+1} = y \,|\, X_t \} = \mathbb P \{ X_{t+1} = y \,|\, X_t, X_{t-1}, \ldots \} \tag{1}$$
In other words, knowing the current state is enough to know probabilities for future states.
In particular, the dynamics of a Markov chain are fully determined by the set of values
$$P(x, y) := \mathbb P \{ X_{t+1} = y \,|\, X_t = x \} \qquad (x, y \in S) \tag{2}$$
By construction,
• $P(x, y)$ is the probability of going from $x$ to $y$ in one unit of time (one step)
• $P(x, \cdot)$ is the conditional distribution of $X_{t+1}$ given $X_t = x$
We can view $P$ as a stochastic matrix where
$$P_{ij} = P(x_i, x_j) \qquad 1 \leq i, j \leq n$$
Going the other way, if we take a stochastic matrix $P$, we can generate a Markov chain $\{X_t\}$ as follows:
• draw $X_0$ from some specified distribution
• for each $t = 0, 1, \ldots$, draw $X_{t+1}$ from $P(X_t,\cdot)$
By construction, the resulting process satisfies (2).
### Example 1¶
Consider a worker who, at any given time $t$, is either unemployed (state 1) or employed (state 2).
Suppose that, over a one month period,
1. An unemployed worker finds a job with probability $\alpha \in (0, 1)$.
2. An employed worker loses her job and becomes unemployed with probability $\beta \in (0, 1)$.
In terms of a Markov model, we have
• $S = \{ 1, 2\}$
• $P(1, 2) = \alpha$ and $P(2, 1) = \beta$
We can write out the transition probabilities in matrix form as
$$P = \left( \begin{array}{cc} 1 - \alpha & \alpha \\ \beta & 1 - \beta \end{array} \right)$$
Once we have the values $\alpha$ and $\beta$, we can address a range of questions, such as
• What is the average duration of unemployment?
• Over the long-run, what fraction of time does a worker find herself unemployed?
• Conditional on employment, what is the probability of becoming unemployed at least once over the next 12 months?
We’ll cover such applications below.
### Example 2¶
Using US unemployment data, Hamilton [Ham05] estimated the stochastic matrix
$$P = \left( \begin{array}{ccc} 0.971 & 0.029 & 0 \\ 0.145 & 0.778 & 0.077 \\ 0 & 0.508 & 0.492 \end{array} \right)$$
where
• the frequency is monthly
• the first state represents “normal growth”
• the second state represents “mild recession”
• the third state represents “severe recession”
For example, the matrix tells us that when the state is normal growth, the state will again be normal growth next month with probability 0.97.
In general, large values on the main diagonal indicate persistence in the process $\{ X_t \}$.
This Markov process can also be represented as a directed graph, with edges labeled by transition probabilities
Here “ng” is normal growth, “mr” is mild recession, etc.
## Simulation¶
One natural way to answer questions about Markov chains is to simulate them.
(To approximate the probability of event $E$, we can simulate many times and count the fraction of times that $E$ occurs)
Nice functionality for simulating Markov chains exists in QuantEcon.jl.
• Efficient, bundled with lots of other useful routines for handling Markov chains.
However, it’s also a good exercise to roll our own routines — let’s do that first and then come back to the methods in QuantEcon.jl.
In these exercises we’ll take the state space to be $S = 1,\ldots, n$.
### Rolling our own¶
To simulate a Markov chain, we need its stochastic matrix $P$ and either an initial state or a probability distribution $\psi$ for initial state to be drawn from.
The Markov chain is then constructed as discussed above. To repeat:
1. At time $t=0$, the $X_0$ is set to some fixed state or chosen from $\psi$.
2. At each subsequent time $t$, the new state $X_{t+1}$ is drawn from $P(X_t, \cdot)$.
In order to implement this simulation procedure, we need a method for generating draws from a discrete distributions.
For this task we’ll use a Categorical random variable (i.e. a discrete random variable with assigned probabilities)
In [3]:
d = Categorical([0.5, 0.3, 0.2]) # 3 discrete states
@show rand(d, 5)
@show supertype(typeof(d))
@show pdf(d, 1) # the probability to be in state 1
@show support(d)
@show pdf.(d, support(d)); # broadcast the pdf over the whole support
rand(d, 5) = [3, 2, 2, 2, 1]
supertype(typeof(d)) = Distribution{Univariate,Discrete}
pdf(d, 1) = 0.5
support(d) = Base.OneTo(3)
pdf.(d, support(d)) = [0.5, 0.3, 0.2]
We’ll write our code as a function that takes the following three arguments
• A stochastic matrix P
• An initial state init
• A positive integer sample_size representing the length of the time series the function should return
In [4]:
function mc_sample_path(P; init = 1, sample_size = 1000)
@assert size(P)[1] == size(P)[2] # square required
N = size(P)[1] # should be square
# create vector of discrete RVs for each row
dists = [Categorical(P[i, :]) for i in 1:N]
# setup the simulation
X = fill(0, sample_size) # allocate memory, or zeros(Int64, sample_size)
X[1] = init # set the initial state
for t in 2:sample_size
dist = dists[X[t-1]] # get discrete RV from last state's transition distribution
X[t] = rand(dist) # draw new value
end
return X
end
Out[4]:
mc_sample_path (generic function with 1 method)
Let’s see how it works using the small matrix
$$P := \left( \begin{array}{cc} 0.4 & 0.6 \\ 0.2 & 0.8 \end{array} \right) \tag{3}$$
As we’ll see later, for a long series drawn from P, the fraction of the sample that takes value 1 will be about 0.25.
If you run the following code you should get roughly that answer
In [5]:
P = [0.4 0.6; 0.2 0.8]
X = mc_sample_path(P, sample_size = 100_000); # note 100_000 = 100000
μ_1 = count(X .== 1)/length(X) # .== broadcasts test for equality. Could use mean(X .== 1)
Out[5]:
0.24773
### Using QuantEcon’s Routines¶
As discussed above, QuantEcon.jl has routines for handling Markov chains, including simulation.
Here’s an illustration using the same P as the preceding example
In [6]:
P = [0.4 0.6; 0.2 0.8];
mc = MarkovChain(P)
X = simulate(mc, 100_000);
μ_2 = count(X .== 1)/length(X) # or mean(x -> x == 1, X)
Out[6]:
0.25181
#### Adding state values and initial conditions¶
If we wish to, we can provide a specification of state values to MarkovChain.
These state values can be integers, floats, or even strings.
The following code illustrates
In [7]:
mc = MarkovChain(P, ["unemployed", "employed"])
simulate(mc, 4, init = 1) # start at state 1
Out[7]:
4-element Array{String,1}:
"unemployed"
"employed"
"employed"
"employed"
In [8]:
simulate(mc, 4, init = 2) # start at state 2
Out[8]:
4-element Array{String,1}:
"employed"
"employed"
"unemployed"
"employed"
In [9]:
simulate(mc, 4) # start with randomly chosen initial condition
Out[9]:
4-element Array{String,1}:
"employed"
"employed"
"employed"
"employed"
In [10]:
simulate_indices(mc, 4)
Out[10]:
4-element Array{Int64,1}:
2
2
2
2
## Marginal Distributions¶
Suppose that
1. $\{X_t\}$ is a Markov chain with stochastic matrix $P$
2. the distribution of $X_t$ is known to be $\psi_t$
What then is the distribution of $X_{t+1}$, or, more generally, of $X_{t+m}$?
### Solution¶
Let $\psi_t$ be the distribution of $X_t$ for $t = 0, 1, 2, \ldots$.
Our first aim is to find $\psi_{t + 1}$ given $\psi_t$ and $P$.
To begin, pick any $y \in S$.
Using the law of total probability, we can decompose the probability that $X_{t+1} = y$ as follows:
$$\mathbb P \{X_{t+1} = y \} = \sum_{x \in S} \mathbb P \{ X_{t+1} = y \, | \, X_t = x \} \cdot \mathbb P \{ X_t = x \}$$
In words, to get the probability of being at $y$ tomorrow, we account for all ways this can happen and sum their probabilities.
Rewriting this statement in terms of marginal and conditional probabilities gives.
$$\psi_{t+1}(y) = \sum_{x \in S} P(x,y) \psi_t(x)$$
There are $n$ such equations, one for each $y \in S$.
If we think of $\psi_{t+1}$ and $\psi_t$ as row vectors (as is traditional in this literature), these $n$ equations are summarized by the matrix expression.
$$\psi_{t+1} = \psi_t P \tag{4}$$
In other words, to move the distribution forward one unit of time, we postmultiply by $P$.
By repeating this $m$ times we move forward $m$ steps into the future.
Hence, iterating on (4), the expression $\psi_{t+m} = \psi_t P^m$ is also valid — here $P^m$ is the $m$-th power of $P$.
As a special case, we see that if $\psi_0$ is the initial distribution from which $X_0$ is drawn, then $\psi_0 P^m$ is the distribution of $X_m$.
This is very important, so let’s repeat it
$$X_0 \sim \psi_0 \quad \implies \quad X_m \sim \psi_0 P^m \tag{5}$$
and, more generally,
$$X_t \sim \psi_t \quad \implies \quad X_{t+m} \sim \psi_t P^m \tag{6}$$
### Multiple Step Transition Probabilities¶
We know that the probability of transitioning from $x$ to $y$ in one step is $P(x,y)$.
It turns out that the probability of transitioning from $x$ to $y$ in $m$ steps is $P^m(x,y)$, the $(x,y)$-th element of the $m$-th power of $P$.
To see why, consider again (6), but now with $\psi_t$ putting all probability on state $x$.
• 1 in the $x$-th position and zero elsewhere.
Inserting this into (6), we see that, conditional on $X_t = x$, the distribution of $X_{t+m}$ is the $x$-th row of $P^m$.
In particular
$$\mathbb P \{X_{t+m} = y \} = P^m(x, y) = (x, y) \text{-th element of } P^m$$
### Example: Probability of Recession¶
Recall the stochastic matrix $P$ for recession and growth considered above.
Suppose that the current state is unknown — perhaps statistics are available only at the end of the current month.
We estimate the probability that the economy is in state $x$ to be $\psi(x)$.
The probability of being in recession (either mild or severe) in 6 months time is given by the inner product
$$\psi P^6 \cdot \left( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right)$$
### Example 2: Cross-Sectional Distributions¶
The marginal distributions we have been studying can be viewed either as probabilities or as cross-sectional frequencies in large samples.
To illustrate, recall our model of employment / unemployment dynamics for a given worker discussed above.
Consider a large (i.e., tending to infinite) population of workers, each of whose lifetime experiences are described by the specified dynamics, independently of one another.
Let $\psi$ be the current cross-sectional distribution over $\{ 1, 2 \}$.
• For example, $\psi(1)$ is the unemployment rate.
The cross-sectional distribution records the fractions of workers employed and unemployed at a given moment.
The same distribution also describes the fractions of a particular worker’s career spent being employed and unemployed, respectively.
## Irreducibility and Aperiodicity¶
Irreducibility and aperiodicity are central concepts of modern Markov chain theory.
### Irreducibility¶
Let $P$ be a fixed stochastic matrix.
Two states $x$ and $y$ are said to communicate with each other if there exist positive integers $j$ and $k$ such that
$$P^j(x, y) > 0 \quad \text{and} \quad P^k(y, x) > 0$$
In view of our discussion above, this means precisely that
• state $x$ can be reached eventually from state $y$, and
• state $y$ can be reached eventually from state $x$
The stochastic matrix $P$ is called irreducible if all states communicate; that is, if $x$ and $y$ communicate for all $(x, y)$ in $S \times S$.
For example, consider the following transition probabilities for wealth of a fictitious set of households
We can translate this into a stochastic matrix, putting zeros where there’s no edge between nodes
$$P := \left( \begin{array}{ccc} 0.9 & 0.1 & 0 \\ 0.4 & 0.4 & 0.2 \\ 0.1 & 0.1 & 0.8 \end{array} \right)$$
It’s clear from the graph that this stochastic matrix is irreducible: we can reach any state from any other state eventually.
We can also test this using QuantEcon.jl’s MarkovChain class
In [11]:
P = [0.9 0.1 0.0; 0.4 0.4 0.2; 0.1 0.1 0.8];
mc = MarkovChain(P)
is_irreducible(mc)
Out[11]:
true
Here’s a more pessimistic scenario, where the poor are poor forever
This stochastic matrix is not irreducible, since, for example, rich is not accessible from poor.
Let’s confirm this
In [12]:
P = [1.0 0.0 0.0; 0.1 0.8 0.1; 0.0 0.2 0.8];
mc = MarkovChain(P);
is_irreducible(mc)
Out[12]:
false
We can also determine the “communication classes,” or the sets of communicating states (where communication refers to a nonzero probability of moving in each direction).
In [13]:
communication_classes(mc)
Out[13]:
2-element Array{Array{Int64,1},1}:
[1]
[2, 3]
It might be clear to you already that irreducibility is going to be important in terms of long run outcomes.
For example, poverty is a life sentence in the second graph but not the first.
We’ll come back to this a bit later.
### Aperiodicity¶
Loosely speaking, a Markov chain is called periodic if it cycles in a predictible way, and aperiodic otherwise.
Here’s a trivial example with three states
The chain cycles with period 3:
In [14]:
P = [0 1 0; 0 0 1; 1 0 0];
mc = MarkovChain(P);
period(mc)
Out[14]:
3
More formally, the period of a state $x$ is the greatest common divisor of the set of integers
$$D(x) := \{j \geq 1 : P^j(x, x) > 0\}$$
In the last example, $D(x) = \{3, 6, 9, \ldots\}$ for every state $x$, so the period is 3.
A stochastic matrix is called aperiodic if the period of every state is 1, and periodic otherwise.
For example, the stochastic matrix associated with the transition probabilities below is periodic because, for example, state $a$ has period 2
We can confirm that the stochastic matrix is periodic as follows
In [15]:
P = zeros(4, 4);
P[1, 2] = 1;
P[2, 1] = P[2, 3] = 0.5;
P[3, 2] = P[3, 4] = 0.5;
P[4, 3] = 1;
mc = MarkovChain(P);
period(mc)
Out[15]:
2
In [16]:
is_aperiodic(mc)
Out[16]:
false
## Stationary Distributions¶
As seen in (4), we can shift probabilities forward one unit of time via postmultiplication by $P$.
Some distributions are invariant under this updating process — for example,
In [17]:
P = [.4 .6; .2 .8];
ψ = [0.25, 0.75];
ψ' * P
Out[17]:
1×2 Adjoint{Float64,Array{Float64,1}}:
0.25 0.75
Such distributions are called stationary, or invariant.
Formally, a distribution $\psi^*$ on $S$ is called stationary for $P$ if $\psi^* = \psi^* P$.
From this equality we immediately get $\psi^* = \psi^* P^t$ for all $t$.
This tells us an important fact: If the distribution of $X_0$ is a stationary distribution, then $X_t$ will have this same distribution for all $t$.
Hence stationary distributions have a natural interpretation as stochastic steady states — we’ll discuss this more in just a moment.
Mathematically, a stationary distribution is a fixed point of $P$ when $P$ is thought of as the map $\psi \mapsto \psi P$ from (row) vectors to (row) vectors.
Theorem. Every stochastic matrix $P$ has at least one stationary distribution.
(We are assuming here that the state space $S$ is finite; if not more assumptions are required)
For a proof of this result you can apply Brouwer’s fixed point theorem, or see EDTC, theorem 4.3.5.
There may in fact be many stationary distributions corresponding to a given stochastic matrix $P$.
• For example, if $P$ is the identity matrix, then all distributions are stationary.
Since stationary distributions are long run equilibria, to get uniqueness we require that initial conditions are not infinitely persistent.
Infinite persistence of initial conditions occurs if certain regions of the state space cannot be accessed from other regions, which is the opposite of irreducibility.
This gives some intuition for the following fundamental theorem.
Theorem. If $P$ is both aperiodic and irreducible, then
1. $P$ has exactly one stationary distribution $\psi^*$.
2. For any initial distribution $\psi_0$, we have $\| \psi_0 P^t - \psi^* \| \to 0$ as $t \to \infty$.
For a proof, see, for example, theorem 5.2 of [Haggstrom02].
(Note that part 1 of the theorem requires only irreducibility, whereas part 2 requires both irreducibility and aperiodicity)
A stochastic matrix satisfying the conditions of the theorem is sometimes called uniformly ergodic.
One easy sufficient condition for aperiodicity and irreducibility is that every element of $P$ is strictly positive
• Try to convince yourself of this
### Example¶
Recall our model of employment / unemployment dynamics for a given worker discussed above.
Assuming $\alpha \in (0,1)$ and $\beta \in (0,1)$, the uniform ergodicity condition is satisfied.
Let $\psi^* = (p, 1-p)$ be the stationary distribution, so that $p$ corresponds to unemployment (state 1).
Using $\psi^* = \psi^* P$ and a bit of algebra yields
$$p = \frac{\beta}{\alpha + \beta}$$
This is, in some sense, a steady state probability of unemployment — more on interpretation below.
Not surprisingly it tends to zero as $\beta \to 0$, and to one as $\alpha \to 0$.
### Calculating Stationary Distributions¶
As discussed above, a given Markov matrix $P$ can have many stationary distributions.
That is, there can be many row vectors $\psi$ such that $\psi = \psi P$.
In fact if $P$ has two distinct stationary distributions $\psi_1, \psi_2$ then it has infinitely many, since in this case, as you can verify,
$$\psi_3 := \lambda \psi_1 + (1 - \lambda) \psi_2$$
is a stationary distribution for $P$ for any $\lambda \in [0, 1]$.
If we restrict attention to the case where only one stationary distribution exists, one option for finding it is to try to solve the linear system $\psi (I_n - P) = 0$ for $\psi$, where $I_n$ is the $n \times n$ identity.
But the zero vector solves this equation.
Hence we need to impose the restriction that the solution must be a probability distribution.
A suitable algorithm is implemented in QuantEcon.jl — the next code block illustrates
In [18]:
P = [.4 .6; .2 .8];
mc = MarkovChain(P);
stationary_distributions(mc)
Out[18]:
1-element Array{Array{Float64,1},1}:
[0.25, 0.7499999999999999]
The stationary distribution is unique.
### Convergence to Stationarity¶
Part 2 of the Markov chain convergence theorem stated above tells us that the distribution of $X_t$ converges to the stationary distribution regardless of where we start off.
This adds considerable weight to our interpretation of $\psi^*$ as a stochastic steady state.
The convergence in the theorem is illustrated in the next figure
In [19]:
P = [0.971 0.029 0.000
0.145 0.778 0.077
0.000 0.508 0.492] # stochastic matrix
ψ = [0.0 0.2 0.8] # initial distribution
t = 20 # path length
x_vals = zeros(t)
y_vals = similar(x_vals)
z_vals = similar(x_vals)
colors = [repeat([:red], 20); :black] # for plotting
for i in 1:t
x_vals[i] = ψ[1]
y_vals[i] = ψ[2]
z_vals[i] = ψ[3]
ψ = ψ * P # update distribution
end
mc = MarkovChain(P)
ψ_star = stationary_distributions(mc)[1]
x_star, y_star, z_star = ψ_star # unpack the stationary dist
plt = scatter([x_vals; x_star], [y_vals; y_star], [z_vals; z_star], color = colors,
gridalpha = 0.5, legend = :none)
plot!(plt, camera = (45,45))
Out[19]:
Here
• $P$ is the stochastic matrix for recession and growth considered above
• The highest red dot is an arbitrarily chosen initial probability distribution $\psi$, represented as a vector in $\mathbb R^3$
• The other red dots are the distributions $\psi P^t$ for $t = 1, 2, \ldots$
• The black dot is $\psi^*$
The code for the figure can be found here — you might like to try experimenting with different initial conditions.
## Ergodicity¶
Under irreducibility, yet another important result obtains: For all $x \in S$,
$$\frac{1}{m} \sum_{t = 1}^m \mathbf{1}\{X_t = x\} \to \psi^*(x) \quad \text{as } m \to \infty \tag{7}$$
Here
• $\mathbf{1}\{X_t = x\} = 1$ if $X_t = x$ and zero otherwise
• convergence is with probability one
• the result does not depend on the distribution (or value) of $X_0$
The result tells us that the fraction of time the chain spends at state $x$ converges to $\psi^*(x)$ as time goes to infinity.
This gives us another way to interpret the stationary distribution — provided that the convergence result in (7) is valid.
The convergence in (7) is a special case of a law of large numbers result for Markov chains — see EDTC, section 4.3.4 for some additional information.
### Example¶
Recall our cross-sectional interpretation of the employment / unemployment model discussed above.
Assume that $\alpha \in (0,1)$ and $\beta \in (0,1)$, so that irreducibility and aperiodicity both hold.
We saw that the stationary distribution is $(p, 1-p)$, where
$$p = \frac{\beta}{\alpha + \beta}$$
In the cross-sectional interpretation, this is the fraction of people unemployed.
In view of our latest (ergodicity) result, it is also the fraction of time that a worker can expect to spend unemployed.
Thus, in the long-run, cross-sectional averages for a population and time-series averages for a given person coincide.
This is one interpretation of the notion of ergodicity.
## Computing Expectations¶
We are interested in computing expectations of the form
$$\mathbb E [ h(X_t) ] \tag{8}$$
and conditional expectations such as
$$\mathbb E [ h(X_{t + k}) \mid X_t = x] \tag{9}$$
where
• $\{X_t\}$ is a Markov chain generated by $n \times n$ stochastic matrix $P$
• $h$ is a given function, which, in expressions involving matrix algebra, we’ll think of as the column vector
$$h = \left( \begin{array}{c} h(x_1) \\ \vdots \\ h(x_n) \\ \end{array} \right)$$
The unconditional expectation (8) is easy: We just sum over the distribution of $X_t$ to get
$$\mathbb E [ h(X_t) ] = \sum_{x \in S} (\psi P^t)(x) h(x)$$
Here $\psi$ is the distribution of $X_0$.
Since $\psi$ and hence $\psi P^t$ are row vectors, we can also write this as
$$\mathbb E [ h(X_t) ] = \psi P^t h$$
For the conditional expectation (9), we need to sum over the conditional distribution of $X_{t + k}$ given $X_t = x$.
We already know that this is $P^k(x, \cdot)$, so
$$\mathbb E [ h(X_{t + k}) \mid X_t = x] = (P^k h)(x) \tag{10}$$
The vector $P^k h$ stores the conditional expectation $\mathbb E [ h(X_{t + k}) \mid X_t = x]$ over all $x$.
### Expectations of Geometric Sums¶
Sometimes we also want to compute expectations of a geometric sum, such as $\sum_t \beta^t h(X_t)$.
In view of the preceding discussion, this is
$$\mathbb{E} \left[ \sum_{j=0}^\infty \beta^j h(X_{t+j}) \mid X_t = x \right] = [(I - \beta P)^{-1} h](x)$$
where
$$(I - \beta P)^{-1} = I + \beta P + \beta^2 P^2 + \cdots$$
Premultiplication by $(I - \beta P)^{-1}$ amounts to “applying the resolvent operator”.
## Exercises¶
### Exercise 1¶
According to the discussion above, if a worker’s employment dynamics obey the stochastic matrix
$$P = \left( \begin{array}{cc} 1 - \alpha & \alpha \\ \beta & 1 - \beta \end{array} \right)$$
with $\alpha \in (0,1)$ and $\beta \in (0,1)$, then, in the long-run, the fraction of time spent unemployed will be
$$p := \frac{\beta}{\alpha + \beta}$$
In other words, if $\{X_t\}$ represents the Markov chain for employment, then $\bar X_m \to p$ as $m \to \infty$, where
$$\bar X_m := \frac{1}{m} \sum_{t = 1}^m \mathbf{1}\{X_t = 1\}$$
Your exercise is to illustrate this convergence.
First,
• generate one simulated time series $\{X_t\}$ of length 10,000, starting at $X_0 = 1$
• plot $\bar X_m - p$ against $m$, where $p$ is as defined above
Second, repeat the first step, but this time taking $X_0 = 2$.
In both cases, set $\alpha = \beta = 0.1$.
The result should look something like the following — modulo randomness, of course
(You don’t need to add the fancy touches to the graph — see the solution if you’re interested)
### Exercise 2¶
A topic of interest for economics and many other disciplines is ranking.
Let’s now consider one of the most practical and important ranking problems — the rank assigned to web pages by search engines.
(Although the problem is motivated from outside of economics, there is in fact a deep connection between search ranking systems and prices in certain competitive equilibria — see [DLP13])
To understand the issue, consider the set of results returned by a query to a web search engine.
For the user, it is desirable to
1. receive a large set of accurate matches
2. have the matches returned in order, where the order corresponds to some measure of “importance”
Ranking according to a measure of importance is the problem we now consider.
The methodology developed to solve this problem by Google founders Larry Page and Sergey Brin is known as PageRank.
To illustrate the idea, consider the following diagram
Imagine that this is a miniature version of the WWW, with
• each node representing a web page
• each arrow representing the existence of a link from one page to another
Now let’s think about which pages are likely to be important, in the sense of being valuable to a search engine user.
One possible criterion for importance of a page is the number of inbound links — an indication of popularity.
By this measure, m and j are the most important pages, with 5 inbound links each.
However, what if the pages linking to m, say, are not themselves important?
Thinking this way, it seems appropriate to weight the inbound nodes by relative importance.
The PageRank algorithm does precisely this.
A slightly simplified presentation that captures the basic idea is as follows.
Letting $j$ be (the integer index of) a typical page and $r_j$ be its ranking, we set
$$r_j = \sum_{i \in L_j} \frac{r_i}{\ell_i}$$
where
• $\ell_i$ is the total number of outbound links from $i$
• $L_j$ is the set of all pages $i$ such that $i$ has a link to $j$
This is a measure of the number of inbound links, weighted by their own ranking (and normalized by $1 / \ell_i$).
There is, however, another interpretation, and it brings us back to Markov chains.
Let $P$ be the matrix given by $P(i, j) = \mathbf 1\{i \to j\} / \ell_i$ where $\mathbf 1\{i \to j\} = 1$ if $i$ has a link to $j$ and zero otherwise.
The matrix $P$ is a stochastic matrix provided that each page has at least one link.
With this definition of $P$ we have
$$r_j = \sum_{i \in L_j} \frac{r_i}{\ell_i} = \sum_{\text{all } i} \mathbf 1\{i \to j\} \frac{r_i}{\ell_i} = \sum_{\text{all } i} P(i, j) r_i$$
Writing $r$ for the row vector of rankings, this becomes $r = r P$.
Hence $r$ is the stationary distribution of the stochastic matrix $P$.
Let’s think of $P(i, j)$ as the probability of “moving” from page $i$ to page $j$.
The value $P(i, j)$ has the interpretation
• $P(i, j) = 1/k$ if $i$ has $k$ outbound links, and $j$ is one of them
• $P(i, j) = 0$ if $i$ has no direct link to $j$
Thus, motion from page to page is that of a web surfer who moves from one page to another by randomly clicking on one of the links on that page.
Here “random” means that each link is selected with equal probability.
Since $r$ is the stationary distribution of $P$, assuming that the uniform ergodicity condition is valid, we can interpret $r_j$ as the fraction of time that a (very persistent) random surfer spends at page $j$.
Your exercise is to apply this ranking algorithm to the graph pictured above, and return the list of pages ordered by rank.
When you solve for the ranking, you will find that the highest ranked node is in fact g, while the lowest is a.
### Exercise 3¶
In numerical work it is sometimes convenient to replace a continuous model with a discrete one.
In particular, Markov chains are routinely generated as discrete approximations to AR(1) processes of the form
$$y_{t+1} = \rho y_t + u_{t+1}$$
Here ${u_t}$ is assumed to be i.i.d. and $N(0, \sigma_u^2)$.
The variance of the stationary probability distribution of $\{ y_t \}$ is
$$\sigma_y^2 := \frac{\sigma_u^2}{1-\rho^2}$$
Tauchen’s method [Tau86] is the most common method for approximating this continuous state process with a finite state Markov chain.
A routine for this already exists in QuantEcon.jl but let’s write our own version as an exercise.
As a first step we choose
• $n$, the number of states for the discrete approximation
• $m$, an integer that parameterizes the width of the state space
Next we create a state space $\{x_0, \ldots, x_{n-1}\} \subset \mathbb R$ and a stochastic $n \times n$ matrix $P$ such that
• $x_0 = - m \, \sigma_y$
• $x_{n-1} = m \, \sigma_y$
• $x_{i+1} = x_i + s$ where $s = (x_{n-1} - x_0) / (n - 1)$
Let $F$ be the cumulative distribution function of the normal distribution $N(0, \sigma_u^2)$.
The values $P(x_i, x_j)$ are computed to approximate the AR(1) process — omitting the derivation, the rules are as follows:
1. If $j = 0$, then set
$$P(x_i, x_j) = P(x_i, x_0) = F(x_0-\rho x_i + s/2)$$
1. If $j = n-1$, then set
$$P(x_i, x_j) = P(x_i, x_{n-1}) = 1 - F(x_{n-1} - \rho x_i - s/2)$$
1. Otherwise, set
$$P(x_i, x_j) = F(x_j - \rho x_i + s/2) - F(x_j - \rho x_i - s/2)$$
The exercise is to write a function approx_markov(rho, sigma_u, m = 3, n = 7) that returns $\{x_0, \ldots, x_{n-1}\} \subset \mathbb R$ and $n \times n$ matrix $P$ as described above.
• Even better, write a function that returns an instance of QuantEcon.jl’s MarkovChain type.
## Solutions¶
### Exercise 1¶
Compute the fraction of time that the worker spends unemployed, and compare it to the stationary probability.
In [20]:
α = 0.1 # probability of getting hired
β = 0.1 # probability of getting fired
N = 10_000
p̄ = β / (α + β) # steady-state probabilities
P = [1 - α α
β 1 - β] # stochastic matrix
mc = MarkovChain(P)
labels = ["start unemployed", "start employed"]
y_vals = Array{Vector}(undef, 2) # sample paths holder
for x0 in 1:2
X = simulate_indices(mc, N; init = x0) # generate the sample path
X̄ = cumsum(X .== 1) ./ (1:N) # compute state fraction. ./ required for precedence
y_vals[x0] = X̄ .- p̄ # plot divergence from steady state
end
plot(y_vals, color = [:blue :green], fillrange = 0, fillalpha = 0.1,
ylims = (-0.25, 0.25), label = reshape(labels, 1, length(labels)))
Out[20]:
### Exercise 2¶
In [21]:
web_graph_data = sort(Dict('a' => ['d', 'f'],
'b' => ['j', 'k', 'm'],
'c' => ['c', 'g', 'j', 'm'],
'd' => ['f', 'h', 'k'],
'e' => ['d', 'h', 'l'],
'f' => ['a', 'b', 'j', 'l'],
'g' => ['b', 'j'],
'h' => ['d', 'g', 'l', 'm'],
'i' => ['g', 'h', 'n'],
'j' => ['e', 'i', 'k'],
'k' => ['n'],
'l' => ['m'],
'm' => ['g'],
'n' => ['c', 'j', 'm']))
┌ Warning: sort(d::Dict; args...) is deprecated, use sort!(OrderedDict(d); args...) instead.
│ caller = top-level scope at In[21]:1
└ @ Core In[21]:1
Out[21]:
OrderedCollections.OrderedDict{Char,Array{Char,1}} with 14 entries:
'a' => ['d', 'f']
'b' => ['j', 'k', 'm']
'c' => ['c', 'g', 'j', 'm']
'd' => ['f', 'h', 'k']
'e' => ['d', 'h', 'l']
'f' => ['a', 'b', 'j', 'l']
'g' => ['b', 'j']
'h' => ['d', 'g', 'l', 'm']
'i' => ['g', 'h', 'n']
'j' => ['e', 'i', 'k']
'k' => ['n']
'l' => ['m']
'm' => ['g']
'n' => ['c', 'j', 'm']
In [22]:
nodes = keys(web_graph_data)
n = length(nodes)
Q = fill(false, n, n)
for (node, edges) in enumerate(values(web_graph_data))
Q[node, nodes .∈ Ref(edges)] .= true
end
# create the corresponding stochastic matrix
P = Q ./ sum(Q, dims = 2)
mc = MarkovChain(P)
r = stationary_distributions(mc)[1] # stationary distribution
ranked_pages = Dict(zip(keys(web_graph_data), r)) # results holder
# print solution
println("Rankings\n ***")
sort(collect(ranked_pages), by = x -> x[2], rev = true) # print sorted
Rankings
***
Out[22]:
14-element Array{Pair{Char,Float64},1}:
'g' => 0.16070778858515053
'j' => 0.15936158342833578
'm' => 0.119515123584059
'n' => 0.10876973827831275
'k' => 0.0910628956751643
'b' => 0.0832646081451476
'e' => 0.05312052780944526
'i' => 0.05312052780944526
'c' => 0.04834210590147233
'h' => 0.04560118369030004
'l' => 0.032017852378295776
'd' => 0.030562495452009602
'f' => 0.011642855410289372
'a' => 0.002910713852572343
### Exercise 3¶
A solution from QuantEcon.jl can be found here.
Footnotes
[1] Hint: First show that if $P$ and $Q$ are stochastic matrices then so is their product — to check the row sums, try postmultiplying by a column vector of ones. Finally, argue that $P^n$ is a stochastic matrix using induction.
• Share page | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9186151027679443, "perplexity": 1113.33083290041}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704824728.92/warc/CC-MAIN-20210127121330-20210127151330-00119.warc.gz"} |
http://clay6.com/qa/27524/refractive-index-of-glass-with-respect-to-water-is-1-125-if-the-absolute-re | Browse Questions
# Refractive index of glass with respect to water is $1.125$ . If the absolute refractive index of glass is $1.5$ find the absolute refractive index of water.
$(a)\;1.23 \\ (b)\;1.5 \\ (c)\;1.33 \\ (d)\;1.77$
Here refractive index of glass with respect to water
$\qquad= 1.125$ and absolute refractive index of glass $\mu_g=1.5$
We know that $a \mu_w=\large\frac{a \mu _g}{w _\mu g}=\frac {1.5}{1.125}$
$\qquad=1.33$
Hence c is the correct answer. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9760460257530212, "perplexity": 290.6679177021857}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171171.24/warc/CC-MAIN-20170219104611-00243-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://pureportal.spbu.ru/ru/publications/embedded-spaces-of-hermite-splines | # Embedded spaces of hermite splines
Результат исследований: Научные публикации в периодических изданияхстатья
## Аннотация
This paper is devoted to the processing of large numerical signals which arise in different technical problems (for example, in positioning systems, satellite maneuvers, in the prediction a lot of phenomenon, and so on). The main tool of the processing is polynomial and nonpolynomial splines of the Hermite type, which are obtained by the approximation relations. These relations allow us to construct splines with approximate properties, which are asymptotically optimal as to N-width of the standard compact sets. The interpolation properties of the mentioned splines are investigated. Such properties give opportunity to obtain the solution of the interpolation Hermite problems without solution of equation systems. The calibration relations on embedded grids are established in the case of deleting the grid knots and in the case of the addition of the last one. A consequence of the obtained results is the embedding of the Hermite spline spaces on the embedded grids. The mentioned embedding allows us to obtain wavelet decomposition of the Hermite spline spaces.
Язык оригинала английский 222-234 13 WSEAS Transactions on Applied and Theoretical Mechanics 14 Опубликовано - 1 янв 2019
## Предметные области Scopus
• Вычислительная механика
• Городское и структурное проектирование
• Сопротивление материалов
• Общее машиностроение
• Гидродинамика и трансферные процессы | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.889480710029602, "perplexity": 2411.106478507554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107911792.65/warc/CC-MAIN-20201030212708-20201031002708-00322.warc.gz"} |
http://ijgt.ui.ac.ir/article_23524.html | # The character table of a sharply 5-transitive subgroup of the alternating group of degree 12
Document Type: Research Paper
Authors
1 University of South Wales
2 University of Southampton
Abstract
In this paper we calculate the character table of a sharply $5$-transitive subgroup of ${rm Alt}(12)$, and of a sharply $4$-transitive subgroup of ${rm Alt}(11)$. Our presentation of these calculations is new because we make no reference to the sporadic simple Mathieu groups, and instead deduce the desired character tables using only the existence of the stated multiply transitive permutation representations.
Keywords
Main Subjects | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8680877685546875, "perplexity": 420.3158105448096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540517557.43/warc/CC-MAIN-20191209041847-20191209065847-00385.warc.gz"} |
http://tronprog.blogspot.com/2007/08/eclipse-fonts-in-linux.html?showComment=1227090120000 | ## Saturday, August 25, 2007
### Eclipse Fonts in Linux
As you might have guessed I'm using Linux most of the times (Windows only seldom), but I noticed that the Eclipse font layout looked better in Windows: actually, the Eclipse font size in Windows permits fitting much more information in a view. This is even more crucial when I'm using my laptop with a resolution of 1280x800.
Of course, I had tried to adjust the font size using the Eclipse Preferences, and actually managed to reduce the size of the editors and view titles (also for dialogs), but not other fonts, such as menu fonts, and, more importantly, the size of tables and trees (e.g., the Java Package Explorer). That's actually due to the fact that under Linux Eclipse relies on Gtk, and thus inherits the preferences of the Gnome Desktop but I'm using KDE :-)
Fortunately, it's quite easy to adjust Gnome font settings even from KDE, by running gnome-control-center from the command line and then set the font size from there. Now, my eclipse shows much more information due to the reduced font size.
And by the way, this reduced the fonts also for other Gtk based applications I'm using, e.g., Thunderbird and Firefox!
Anonymous said...
Very nice betto ! You're the first one, who gave me a solution for that problem and the reason, that i can now use eclipse under Linux ! Greetings from (a Croat in) Germany.
betto said...
Hi there! Glad you liked the solution :-)
However, this seems to be a temporary one, since each time you restart KDE you should also call the GNOME configurator :-(
This other solution seems better: http://tronprog.blogspot.com/2007/08/eclipse-fonts-in-linux-part-2.html
Anonymous said...
In Ubuntu you can use the System menu / Preferences / Appearance / Fonts tab to set application and window title fonts size.
I have been looking to solve this Eclipse package explorer issue for a long time. Very much appreciated. Thanks
Anonymous said...
thanks a lot I was being driven crazy on this...
Anonymous said...
thank you very much!
Anonymous said...
Another solution:
http://techtavern.wordpress.com/2008/09/24/smaller-font-sizes-for-eclipse-on-linux/ | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8341561555862427, "perplexity": 3272.3236911466383}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368698090094/warc/CC-MAIN-20130516095450-00043-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://docmadhattan.fieldofscience.com/2015/06/the-chirality-at-beginning-of-universe.html | The chirality at the beginning of the universe
A new clue about the #quarkgluonplasma from @RHIC_STAR at @BrookhavenLab
Within the particles that constitute atomic nuclei, protons and neutrons, there are the quarks, the elementary particles with fractional charges, linked to each other thanks to the gluons, bosons that carry the nuclear interaction. Thanks to the gluons it is impossible to observe, at present, free quarks, but it is expected that in the very first stage of the universe, matter was in a state called quark-gluon plasma. Thanks to the observation of so-called quark jets we know, indirectly, that the interior of the accelerators RHIC and LHC, in particular in heavy ion collisions, this kind of plasmas were created and, according to the theory, in the presence of axial anomalies, dued by the presence of strong electromagnetic fields, we can create two special effects: the Chiral Magnetic Effect (CME) and the Chiral Separation Effect (CSE).
The CME is the phenomenon of electric charge separation along the axis of the applied magnetic field in the presence of fluctuating topological charge.
The Chiral Separation Effect (CSE) refers to the separation of chiral charge along the axis of external magnetic field at finite density of vector charge (e.g. at finite baryon number density)(5)
These two effects are generated by the topology of the system: indeed, within the theory(1, 2, 3) is contemplated the existence of some numbers (called topological invariants, or winding number(4)) that, while not associated with an observable, still generate effects physically relevant because of their link with the fundamental symmetries of the system.
In particular the two excitations CME and CSE are generated, and when they happen at the same time, they interact generating a new excited state inside the quark-gluon plasma. This new state of matter is called Chiral Magnetic Wave (CMW).
This excitation stems from the coupling between the density waves of electric and chiral charge. Let us illustrate this statement by a qualitative argument (...): consider a local fluctuation of electric charge density; (...) it will induce a local fluctuation of axial current. This fluctuation of axial current would in turn induce a local fluctuation of the axial chemical potential, and thus (...) a fluctuation of electric current. The resulting fluctuation of electric charge density completes the cycle leading to the excitation that combines the density waves of electric and chiral charges(5)
It is clear, therefore, that the CMW, although in an indirect manner, is linked to the quark-gluon plasma. The exciting news is that this particular excitement was observed within the RHIC(6) in the collision of gold ions, a step that brings us very closer to the big bang.
(1) Belavina A.A., Polyakova A.M., Schwartza A.S. & Tyupkina Y.S. (1975). Pseudoparticle solutions of the Yang-Mills equations, Physics Letters B, 59 (1) 85-87. DOI:
(2) Hooft G.'t. (1976). Symmetry Breaking through Bell-Jackiw Anomalies, Physical Review Letters, 37 (8) DOI:
(3) Hooft G.'t. (1976). Computation of the quantum effects due to a four-dimensional pseudoparticle, Physical Review D, 14 DOI:
(4) Fukushima K., Kharzeev D.E. & Warringa H.J. (2008). The Chiral Magnetic Effect, Physics Review D, 78 (7) DOI: (arXiv)
(5) Kharzeev D.E. & Yee H.U. (2011). Chiral Magnetic Wave, Physical Review D, 83 DOI: (arXiv)
(6) STAR Collaboration (2015). Observation of charge asymmetry dependence of pion elliptic flow and the possible chiral magnetic wave in heavy-ion collisions, arXiv: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.940592348575592, "perplexity": 1510.7745725284349}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986697760.44/warc/CC-MAIN-20191019191828-20191019215328-00190.warc.gz"} |
http://www.nag.com/numeric/cl/nagdoc_cl23/html/M01/m01dsc.html | nag_rank_sort (m01dsc) (PDF version)
m01 Chapter Contents
m01 Chapter Introduction
NAG C Library Manual
# NAG Library Function Documentnag_rank_sort (m01dsc)
## 1 Purpose
nag_rank_sort (m01dsc) ranks a vector of arbitrary data type objects in ascending or descending order.
## 2 Specification
#include #include
void nag_rank_sort (const Pointer vec, size_t n, ptrdiff_t stride,
Integer (*compare)(const Nag_Pointer a, const Nag_Pointer b),
Nag_SortOrder order, size_t ranks[], NagError *fail)
## 3 Description
nag_rank_sort (m01dsc) ranks a set of $n$ data objects of arbitrary type, which are stored in the elements of an array at intervals of length stride. The ranks are in the range 0 to $n-1$.
Either ascending or descending ranking order may be specified.
nag_rank_sort (m01dsc) uses a variant of list merging as described by Knuth (1973).
## 4 References
Knuth D E (1973) The Art of Computer Programming (Volume 3) (2nd Edition) Addison–Wesley
## 5 Arguments
1: vec[${\mathbf{n}}$]const Pointer Input
On entry: the array of objects to be ranked.
2: nsize_tInput
On entry: the number $n$ of objects.
Constraint: ${\mathbf{n}}\ge 0$.
3: strideptrdiff_tInput
On entry: the increment between data items in vec to be ranked.
Note: if stride is positive, vec should point at the first data object; otherwise vec should point at the last data object. It should be noted that $\left|{\mathbf{stride}}\right|$ must be greater than or equal to size_of (data objects), for correct ranks to be produced. However, the code performs no check for violation of this constraint.
Constraint: $\left|{\mathbf{stride}}\right|>0$.
4: comparefunction, supplied by the userExternal Function
nag_rank_sort (m01dsc) compares two data objects. If its arguments are pointers to a structure, this function must allow for the offset of the data field in the structure (if it is not the first).
The function must return:
$-1$ if the first data field is less than the second, $\phantom{-}0$ if the first data field is equal to the second, $\phantom{-}1$ if the first data field is greater than the second.
The specification of compare is:
Integer compare (const Nag_Pointer a, const Nag_Pointer b)
1: aconst Nag_Pointer Input
On entry: the first data field.
2: bconst Nag_Pointer Input
On entry: the second data field.
5: orderNag_SortOrderInput
On entry: specifies whether the array is to be ranked into ascending or descending order.
Constraint: ${\mathbf{order}}=\mathrm{Nag_Ascending}$ or $\mathrm{Nag_Descending}$.
6: ranks[n]size_tOutput
On exit: the ranks of the corresponding data elements in vec.
7: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_BAD_PARAM
On entry, argument order had an illegal value.
NE_INT_ARG_EQ
On entry, ${\mathbf{stride}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{stride}}=0$.
NE_INT_ARG_GT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\le 〈\mathit{\text{value}}〉$.
On entry, ${\mathbf{stride}}=〈\mathit{\text{value}}〉$.
Constraint: $\left|{\mathbf{stride}}\right|\le 〈\mathit{\text{value}}〉$.
These arguments are limited to an implementation-dependent size which is printed in the error message.
NE_INT_ARG_LT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 0$.
Not applicable.
## 8 Further Comments
The time taken by nag_rank_sort (m01dsc) is approximately proportional to $n\mathrm{log}n$.
## 9 Example
The example program reads a list of real numbers and ranks them into ascending order.
### 9.1 Program Text
Program Text (m01dsce.c)
### 9.2 Program Data
Program Data (m01dsce.d)
### 9.3 Program Results
Program Results (m01dsce.r)
nag_rank_sort (m01dsc) (PDF version)
m01 Chapter Contents
m01 Chapter Introduction
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https://dsp.stackexchange.com/questions/21964/oversampling-and-aliasing | # Oversampling and Aliasing
I am really confused about oversampling and upsampling. I know that upsampling means stretching our signal in time domain by a factor of K, then interpolating some values or using just zeros. As I found out stretching in time domain is shrinking in frequency domain by the same rate. For example, stretch by factor of 2 in time domain mean shrinking by factor of 2 in frequency domain.
Let talk about oversampling, we just increase the sampling frequency more than Nyquist rate. Then we have too much information. Is there any chance to have aliasing? I mean in time domain, lets say I am sampling my signal (frequency is 70 Hz) by sampling frequency equals to 280 Hz (Nyquist*2) so, is there aliasing in frequency domain? Sorry if my explanation is not clear. It is because I am bewildered with these concepts; oversampling, upsampling, downsampling, undersampling, critical sampling and aliasing.
theoretically, I know them but when it comes to practical view, I have serious problem.
When upsampling, you don't really stretch the signal in time. You insert new samples between the existing ones, without modifying the times at which those samples were taken. One property of upsampling is that the waveform remains exactly the same before and after the process; you just increase the number of samples. Note that the main difference between oversampling and upsampling is that the former occurs at the time of sampling, and the latter occurs after sampling has already been done. If there is no aliasing, in theory both produce the same result.
Aliasing is only present when there is a signal at the input of your sampler whose frequency is higher than the Nyquist frequency. In your example, if you sample a 70 Hz signal at 280 samples per second, you will not have aliasing. However, in practice you will not always have precise knowledge or control over the signal you're sampling. For example, you may want to sample a signal coming from an antenna. You don't know in advance what signals are going to be picked up by the antenna. In a case like this, one approach is to low-pass filter the signal before sampling, to ensure no aliasing will happen.
• But what about down sampling and under sampling. According to your description, they act the same way. Down sampling after sampling occurred and under sampling when the signal is going to be sampled. Right? Mar 11 '15 at 0:24
• On the other hand, you say that we have aliasing just when we have under sampling? Mar 11 '15 at 0:25
• @David, you're right, downsampling happens after sampling. Say you sample a narrowband, high frequency signal, and then you shift it (in DSP) to low frequency. After shifting, you can safely reduce your sampling rate and decrease the running time of your code. On the other hand, undersampling is almost always undesirable -- it means you didn't sample at the rate you needed to satisfy Nyquist. Both can introduce aliasing. However, most DSP programs low-pass filter the signal before actually downsampling it, avoiding any aliasing.
– MBaz
Mar 11 '15 at 1:13
• "When upsampling, you don't really stretch the signal in time. You insert new samples between the existing ones, without modifying the times at which those samples were taken." well, M, you can play the new buffer (with inserted samples) back at the same sample rate and a slower tempo (and lower in pitch). also, upsampling need not be just inserting samples between existing samples (which would be equivalent to Lagrange or a windowed sinc() function). it could be some optimal interpolation kernel that is different from a windowed sinc(). Mar 11 '15 at 3:56
• @robertbristow-johnson: of course; the main point is that upsampling doesn't stretch or compress the signal in time, which is what the OP was confused about.
– MBaz
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https://xianblog.wordpress.com/tag/hamiltonian-monte-carlo/ | accelerating HMC by learning the leapfrog scale
Posted in Books, Statistics with tags , , , , , , , , on October 12, 2018 by xi'an
In this new arXiv submission that was part of Changye Wu’s thesis [defended last week], we try to reduce the high sensitivity of the HMC algorithm to its hand-tuned parameters, namely the step size ε of the discretisation scheme, the number of steps L of the integrator, and the covariance matrix of the auxiliary variables. By calibrating the number of steps of the Leapfrog integrator towards avoiding both slow mixing chains and wasteful computation costs. We do so by learning from the No-U-Turn Sampler (NUTS) of Hoffman and Gelman (2014) which already automatically tunes both the step size and the number of leapfrogs.
The core idea behind NUTS is to pick the step size via primal-dual averaging in a burn-in (warmup, Andrew would say) phase and to build at each iteration a proposal based on following a locally longest path on a level set of the Hamiltonian. This is achieved by a recursive algorithm that, at each call to the leapfrog integrator, requires to evaluate both the gradient of the target distribution and the Hamiltonianitself. Roughly speaking an iteration of NUTS costs twice as much as regular HMC with the same number of calls to the integrator. Our approach is to learn from NUTS the scale of the leapfrog length and use the resulting empirical distribution of the longest leapfrog path to randomly pick the value of L at each iteration of an HMC scheme. This obviously preserves the validity of the HMC algorithm.
While a theoretical comparison of the convergence performances of NUTS and this eHMC proposal seem beyond our reach, we ran a series of experiments to evaluate these performances, using as a criterion an ESS value that is calibrated by the evaluation cost of the logarithm of target density function and of its gradient, as this is usually the most costly part of the algorithms. As well as a similarly calibrated expected square jumping distance. Above is one such illustration for a stochastic volatility model, the first axis representing the targeted acceptance probability in the Metropolis step. Some of the gains in either ESS or ESJD are by a factor of ten, which relates to our argument that NUTS somewhat wastes computation effort using a uniformly distributed proposal over the candidate set, instead of being close to its end-points, which automatically reduces the distance between the current position and the proposal.
Hamiltonian tails
Posted in Books, Kids, R, Statistics, University life with tags , , , , , , on July 17, 2018 by xi'an
“We demonstrate HMC’s sensitivity to these parameters by sampling from a bivariate Gaussian with correlation coefficient 0.99. We consider three settings (ε,L) = {(0.16; 40); (0.16; 50); (0.15; 50)}” Ziyu Wang, Shakir Mohamed, and Nando De Freitas. 2013
In an experiment with my PhD student Changye Wu (who wrote all R codes used below), we looked back at a strange feature in an 2013 ICML paper by Wang, Mohamed, and De Freitas. Namely, a rather poor performance of an Hamiltonian Monte Carlo (leapfrog) algorithm on a two-dimensional strongly correlated Gaussian target, for very specific values of the parameters (ε,L) of the algorithm.
The Gaussian target associated with this sample stands right in the middle of the two clouds, as identified by Wang et al. And the leapfrog integration path for (ε,L)=(0.15,50)
keeps jumping between the two ridges (or tails) , with no stop in the middle. Changing ever so slightly (ε,L) to (ε,L)=(0.16,40) does not modify the path very much
but the HMC output is quite different since the cloud then sits right on top of the target
with no clear explanation except for a sort of periodicity in the leapfrog sequence associated with the velocity generated at the start of the code. Looking at the Hamiltonian values for (ε,L)=(0.15,50)
and for (ε,L)=(0.16,40)
does not help, except to point at a sequence located far in the tails of this Hamiltonian, surprisingly varying when supposed to be constant. At first, we thought the large value of ε was to blame but much smaller values still return poor convergence performances. As below for (ε,L)=(0.01,450)
generalizing Hamiltonian Monte Carlo with neural networks
Posted in Statistics with tags , , , on April 25, 2018 by xi'an
Daniel Levy, Matthew Hoffman, and Jascha Sohl-Dickstein pointed out to me a recent paper of theirs submitted to and accepted by ICLR 2018, with the above title. This allowed me to discover the open source handling of paper reviews at ICLR, which I find quite convincing, except for not using MathJax or another medium for LaTeX formulas. And which provides a collection of comments besides mine’s. (Disclaimer: I was not involved in the processing of this paper for ICLR!)
“Ultimately our goal (and that of HMC) is to produce a proposal that mixes efficiently, not to simulate Hamiltonian dynamics accurately.”
The starting concept is the same as GANs (generative adversarial networks) discussed here a few weeks ago. Complemented by a new HMC that also uses deep neural networks to represent the HMC trajectory. (Also seen in earlier papers by e.g. Strathman.) The novelty in the HMC seems to be a binary direction indicator on top of the velocity. The leapfrog integrator is also modified, with a location scale generalisation for the velocity and a half-half location scale move for the original target x. The functions appearing in the location scale aspects are learned by neural nets. Towards minimising lag-one auto-correlation. Plus an extra penalty for not moving enough. Reflecting on the recent MCMC literature and in particular on the presentations at BayesComp last month, judging from comments of participants, this inclusion of neural tools in the tuning of MCMC algorithms sounds like a steady trend in the community. I am slightly at a loss about the adaptive aspects of the trend with regards to the Markovianity of the outcome.
“To compute the Metropolis-Hastings acceptance probability for a deterministic transition, the operator
must be invertible and have a tractable Jacobian.”
A remark (above) that seems to date back at least to Peter Green’s reversible jump. Duly mentioned in the paper. When reading about the performances of this new learning HMC, I could not see where the learning steps for the parameters of the leapfrog operators were accounted for, although the authors mention an identical number of gradient computations (which I take to mean the same thing). One evaluation of this method against earlier ones (Fig.2) checks successive values of the likelihood, which may be intuitive enough but does not necessarily qualify convergence to the right region since the posterior may concentrate away from the maximal likelihood.
unbiased HMC
Posted in Books, pictures, Statistics with tags , , , , , , , on September 25, 2017 by xi'an
Jeremy Heng and Pierre Jacob arXived last week a paper on unbiased Hamiltonian Monte Carlo by coupling, following the earlier paper of Pierre and co-authors on debiasing by coupling a few weeks ago. The coupling within the HMC amounts to running two HMC chains with common random numbers, plus subtleties!
“As with any other MCMC method, HMC estimators are justified in the limit of the number of iterations. Algorithms which rely on such asymptotics face the risk of becoming obsolete if computational power keeps increasing through the number of available processors and not through clock speed.”
The main difficulty here is to have both chains meet (exactly) with large probability, since coupled HMC can only bring these chain close to one another. The trick stands in using both coupled HMC and coupled Hastings-Metropolis kernels, since the coupled MH kernel allows for exact meetings when the chains are already close, after which they remain happily and forever together! The algorithm is implemented by choosing between the kernels at random at each iteration. (Unbiasedness follows by the Glynn-Rhee trick, which is eminently well-suited for coupling!) As pointed out from the start of the paper, the appeal of this unbiased version is that the algorithm can be (embarrassingly) parallelised since all processors in use return estimators that are iid copies of one another, hence easily merged into a better estimator.
a conceptual introduction to HMC [reply from the author]
Posted in Statistics with tags , , , , , , , , on September 8, 2017 by xi'an
[Here is the reply on my post from Michael Bétancourt, detailed enough to be promoted from comment to post!]
As Dan notes this is meant as an introduction for those without a strong mathematical background, hence the focus on concepts rather than theorems! There’s plenty of maths deeper in the references. ;-)
I am not sure I get this sentence. Either it means that an expectation remains invariant under reparameterisation. Or something else and more profound that eludes me. In particular because Michael repeats later (p.25) that the canonical density does not depend on the parameterisation.
What I was trying to get at is that expectations and really all of measure theory are reparameteriztion invariant, but implementations of statistical algorithms that depend on parameterization-dependent representations, namely densities, are not. If your algorithm is sensitive to these parameterization dependencies then you end up with a tuning problem — which parameterization is best? — which makes it harder to utilize the algorithm in practice.
Exact implementations of HMC (i.e. without an integrator) are fully geometric and do not depend on any chosen parameterization, hence the canonical density and more importantly the Hamiltonian being an invariant objects. That said, there are some choices to be made in that construction, and those choices often look like parameter dependencies. See below!
“Every choice of kinetic energy and integration time yields a new Hamiltonian transition that will interact differently with a given target distribution (…) when poorly-chosen, however, the performance can suffer dramatically.”
This is exactly where it’s easy to get confused with what’s invariant and what’s not!
The target density gives rise to a potential energy, and the chosen density over momenta gives rise to a kinetic energy. The two energies transform in opposite ways under a reparameterization so their sum, the Hamiltonian, is invariant.
Really there’s a fully invariant, measure-theoretic construction where you use the target measure directly and add a “cotangent disintegration”.
In practice, however, we often choose a default kinetic energy, i.e. a log density, based on the parameterization of the target parameter space, for example an “identify mass matrix” kinetic energy. In other words, the algorithm itself is invariant but by selecting the algorithmic degrees of freedom based on the parameterization of the target parameter space we induce an implicit parameter dependence.
This all gets more complicated when we introducing the adaptation we use in Stan, which sets the elements of the mass matrix to marginal variances which means that the adapted algorithm is invariant to marginal transformations but not joint ones…
The explanation of the HMC move as a combination of uniform moves along isoclines of fixed energy level and of jumps between energy levels does not seem to translate into practical implementations, at least not as explained in the paper. Simulating directly the energy distribution for a complex target distribution does not seem more feasible than moving up likelihood levels in nested sampling.
Indeed, being able to simulate exactly from the energy distribution, which is equivalent to being able to quantify the density of states in statistical mechanics, is intractable for the same reason that marginal likelihoods are intractable. Which is a shame, because conditioned on those samples HMC could be made embarrassingly parallel!
Instead we draw correlated samples using momenta resamplings between each trajectory. As Dan noted this provides some intuition about Stan (it reduced random walk behavior to one dimension) but also motivates some powerful energy-based diagnostics that immediately indicate when the momentum resampling is limiting performance and we need to improve it by, say, changing the kinetic energy. Or per my previous comment, by keeping the kinetic energy the same but changing the parameterization of the target parameter space. :-)
In the end I cannot but agree with the concluding statement that the geometry of the target distribution holds the key to devising more efficient Monte Carlo methods.
Yes! That’s all I really want statisticians to take away from the paper. :-)
a conceptual introduction to HMC
Posted in Books, Statistics with tags , , , , , , , on September 5, 2017 by xi'an
“…it has proven a empirical success on an incredibly diverse set of target distributions encountered in applied problems.”
In January this year (!), Michael Betancourt posted on arXiv a detailed introduction to Hamiltonian Monte Carlo that recouped some talks of his I attended. Like the one in Boston two years ago. I have (re)read through this introduction to include an HMC section in my accelerating MCMC review for WIREs (which writing does not accelerate very much…)
“…this formal construction is often out of reach of theoretical and applied statisticians alike.”
With the relevant provision of Michael being a friend and former colleague at Warwick, I appreciate the paper at least as much as I appreciated the highly intuitive approach to HMC in his talks. It is not very mathematical and does not provide theoretical arguments for the defence of one solution versus another, but it (still) provides engaging reasons for using HMC.
“One way to ensure computational inefficiency is to waste computational resources evaluating the target density and relevant functions in regions of parameter space that have negligible contribution to the desired expectation.”
The paper starts by insisting on the probabilistic importance of the typical set, which amounts to a ring for Gaussian-like distributions. Meaning that in high dimensions the mode of the target is not a point that is particularly frequently visited. I find this notion quite compelling and am at the same time [almost] flabbergasted that I have never heard of it before.
“we will consider only a single parameterization for computing expectations, but we must be careful to ensure that any such computation does not depend on the irrelevant details of that parameterization, such as the particular shape of the probability density function.”
I am not sure I get this sentence. Either it means that an expectation remains invariant under reparameterisation. Or something else and more profound that eludes me. In particular because Michael repeats later (p.25) that the canonical density does not depend on the parameterisation.
“Every choice of kinetic energy and integration time yields a new Hamiltonian transition that will interact differently with a given target distribution (…) when poorly-chosen, however, the performance can suffer dramatically.”
When discussing HMC, Michael tends to get a wee bit overboard with superlatives!, although he eventually points out the need for calibration as in the above quote. The explanation of the HMC move as a combination of uniform moves along isoclines of fixed energy level and of jumps between energy levels does not seem to translate into practical implementations, at least not as explained in the paper. Simulating directly the energy distribution for a complex target distribution does not seem more feasible than moving up likelihood levels in nested sampling. (Unless I have forgotten something essential about HMC!) Similarly, when discussing symplectic integrators, the paper intuitively conveys the reason these integrators avoid Euler’s difficulties, even though one has to seek elsewhere for rigorous explanations. In the end I cannot but agree with the concluding statement that the geometry of the target distribution holds the key to devising more efficient Monte Carlo methods.
Bouncing bouncy particle papers
Posted in Books, pictures, Statistics, University life with tags , , , , on July 27, 2017 by xi'an
Yesterday, two papers on bouncy particle samplers simultaneously appeared on arXiv, arxiv:1707.05200 by Chris Sherlock and Alex Thiery, and arxiv:1707.05296 by Paul Vanetti, Alexandre Bouchard-Côté, George Deligiannidis, and Arnaud Doucet. As a coordinated move by both groups of authors who had met the weeks before at the Isaac Newton Institute in Cambridge.
The paper by Sherlock and Thiery, entitled a discrete bouncy particle sampler, considers a delayed rejection approach that only requires point-wise evaluations of the target density. The delay being into making a speed flip move after a proposal involving a flip in the speed and a drift in the variable of interest is rejected. To achieve guaranteed ergodicity, they add a random perturbation as in our recent paper, plus another perturbation based on a Brownian argument. Given that this is a discretised version of the continuous-time bouncy particle sampler, the discretisation step δ need be calibrated. The authors follow a rather circumvoluted argument to argue in favour of seeking a maximum number of reflections (for which I have obviously no intuition). Overall, I find it hard to assess how much of an advance this is, even when simulations support the notion of a geometric convergence.
“Our results provide a cautionary example that in certain high-dimensional scenarios, it is still preferable to perform refreshment even when randomized bounces are used.” Vanetti et al.
The paper by Paul Vanetti and co-authors has a much more ambitious scale in that it unifies most of the work done so far in this area and relates piecewise deterministic processes, Hamiltonian Monte Carlo, and discrete versions, containing on top fine convergence results. The main idea is to improve upon the existing deterministic methods by taking (more) into account the target density. Hence the use of a bouncy particle sampler associated with the Hamiltonian (as in HMC). This borrows from an earlier slice sampler idea of Iain Murray, Ryan Adams, and David McKay (AISTATS 2010), exploiting an exact Hamiltonian dynamics for an approximation to the true target to explore its support. Except that bouncing somewhat avoids the slice step. The [eight] discrete bouncy particle particle samplers derived from this framework are both correct against the targeted distribution and do not require the simulation of event times. The paper distinguishes between global and local versions, the later exploiting conditional independence properties in the (augmented) target. Which sounds like a version of multiple slice sampling. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8266028761863708, "perplexity": 938.7415735395477}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743011.30/warc/CC-MAIN-20181116111645-20181116133645-00337.warc.gz"} |
https://www.physicsforums.com/threads/heat-of-fusion.112313/ | # Heat of fusion?
1. Feb 26, 2006
### metalmagik
I have a graph here separated into 5 different parts (A-B, B-C, C-D, D-E, E-F) It is a curve going upwards...I just need to know how to calculate the heat of fusion of the substance using the curve in the graph...I can use the formula and calculate the heat of fusion for each little piece but...do I add them after that? I'm just not really sure...any help is once again appreciated, if you need me to clarify or draw the graph I will, gladly. Thank you.
2. Feb 26, 2006
### Staff: Mentor
What is the curve?
http://en.wikipedia.org/wiki/Heat_of_fusion
and perhaps better - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase2.html#c1
3. Feb 26, 2006
### metalmagik
It's a positive curve going upwards but its not really a curve it has constants in temperature every now and then but its going positive. I still dont understand how to interpret graphs from those links haha im sorry please help me
4. Feb 26, 2006
### Staff: Mentor
During a phase change, temperature is essentially constant. The energy goes into transforming from solid to liquid.
What are the abscissa (x or horizontal scale) and ordinate (y or vertical scale) of the graph?
If the graphs show rise - constant - rise - constant - rise, then it may correspond to solid - melting - liquid - vaporizing - vapor (gas). The first constant temperature interval may coincide with the heat of fusion.
5. Feb 26, 2006
### metalmagik
ah yes thats how it is exactly...thank you...how do i calcualte the heat of fusion of the unknown substance by simply using that? There is also a chart which asks me for the change on kinetic energy and the change in potential energy how do i do that!
6. Feb 26, 2006
### metalmagik
By the heat of fusion i mean for the ENTIRE graph (the whole substance and not just for the different segments
7. Feb 26, 2006
### Staff: Mentor
Answer this question - What are the abscissa (x or horizontal scale) and ordinate (y or vertical scale) of the graph? - and I will try to explain it. If the constant parts are horizontal, then the ordinate is temperature. If the constant parts are vertical then the abscissa is temperature.
Kinetic or potential of what?
Increasing a temperature of a substance increases the kinetic energy of the molecules, more so for vapor than liquid, and more so for liquid than solid. Increasing temperature also increases the potential energy or potential to do work.
8. Feb 26, 2006
### metalmagik
ordinate is temperature yes and abscissa is time in minutes. it is asking for the delta KE and delta PE for each line segment. How would I calculate this?
9. Feb 26, 2006
### Staff: Mentor
Heat of fusion applies only to the energy absorbed when a substance melts, i.e. changes from solid to liquid or liquid to solid.
Heat in to a material would imply solid to liquid transformation. Heat out (removal) would imply 'freezing' or transformation from liquid to solid.
Unless, one has two different substances which melt at two different temperatures.
10. Feb 26, 2006
### metalmagik
So does that mean that I can only calculate heat of fusion for those parts?
11. Feb 26, 2006
### Staff: Mentor
What other information is given.
If one is given a heat rate (or power, which = energy/time) then simply integrate the area of power * time to get energy, and the energy in the constant period would be heat of fusion for the solid to liquid transformation.
Is there a discussion in your text on the change in kinetic energy or potential energy as a function of temperature in a liquid or vapor?
Are you looking at the kinetic or potential energy from a molecular perspective?
See this - http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c2
12. Feb 26, 2006
### metalmagik
I do not know how I am looking at the KE or PE...i have only looked through packets and sheets to understand this topic becuse I Was given no instruction. but so heat rate = energy/time...and if I find heat of fusion for each segment is THAT the energy? if so then I will be able to do this
13. Feb 26, 2006
### Staff: Mentor
If there are two constant temperature parts, then I expect one is for melting (heat of fusion) and the other is for boiling or vaporization (heat of vaporization).
Yes, power (heat rate) = energy/time, so power * time = energy. Time is the time interval (e.g. t2-t1) over which heat is added (or subtracted if heat is removed).
A temperature increase would result in a change in potential energy or possibly kinetic energy, but I would need further information.
Last edited: Feb 26, 2006
14. Feb 26, 2006
### metalmagik
Oh ok thank you, yes, I understand the heat of fusion and vaporization now.
15. Feb 26, 2006
### metalmagik
oh wait im sorry it tells me the energy here, 200 J/minute
16. Feb 26, 2006
### metalmagik
I can figure out the PE and KE thank you very much...there is another question concerning the Average KE of molecules in a larger solid compared to average KE of molecules in a smaller solid. It also says to compare their internal energies. Block A is 1 kg, Block B is 1 Gram, both temperatures are at 300 K
17. Feb 26, 2006
### Staff: Mentor
18. Feb 26, 2006
### Staff: Mentor
200 Joules/minute is the heat rate or power going into the substance. Make sure the time is compatible, i.e. in minutes.
The energy is just the area under the curve.
If the temperature is constant for 10 minutes, then at 200 J/min, 2000 J would be put into the substance.
Do you have a mass or number of moles into which the heat is input? Usually one works with energy/unit mass.
19. Feb 26, 2006
### metalmagik
the mass is 10 kg, i dont have a number of moles...im still trying to find the KE of this substance, what exactly was the formula again? I remember you told me power = energy/time and i found all of that but how do I get the KE and PE?
20. Feb 26, 2006
### Staff: Mentor
Just use the mass. Heat of fusion is often given by energy/unit mass.
See examples - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase2.html#c1
How are potential and kinetic energy defined? Certainly increasing the temperature of a substance increases the molecular kinetic energy, as well as the potential energy. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9016436338424683, "perplexity": 1496.7447340387926}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128319575.19/warc/CC-MAIN-20170622135404-20170622155404-00075.warc.gz"} |
https://community.jmp.com/t5/Discussions/cubic-spline-regression/m-p/12835 | Choose Language Hide Translation Bar
Highlighted
Community Trekker
## cubic spline regression
Hello everybody,
I'm looking for a solution to regress a bivariate data set with a cubic spline regression which has exactly 15 knots. It would be great if someone has a solution for me!
Greetings
1 ACCEPTED SOLUTION
Accepted Solutions
Community Trekker
## Re: cubic spline regression
Hello and thanks for your answer! Exactly something like that, but the question is, does lambda represent the amount of knots for my cubic spline?
Edit// Ok I have read now, that your suggestion is a smoothing spline. Smoothing splines are not calculating with knots instead you have penalty function with the factor lambda. So the question is, is there another function in JMP that I can calculate the splines with knots?
By the way is there a nice feature how I can extract the formula of the spline regression and calculate "new" height values by just input the weight values?
Edit// I think I found a solution. The knotted spline effect should be fine for me, it is hidden under Analyze->FitModel->Attributes
To extract formulas and calculate new values, just click on the red red triangle->save columns->prediction formula
In the next step just write down the new x-values in the same column were the old x-values are already. The predicted values are displayed in the new column!
3 REPLIES 3
Staff
## Re: cubic spline regression
Something like this
Bivariate(
Y( :height ),
X( :weight ),
Fit Spline( 15, {Line Color( {208, 64, 86} )} )
);
Community Trekker
## Re: cubic spline regression
Hello and thanks for your answer! Exactly something like that, but the question is, does lambda represent the amount of knots for my cubic spline?
Edit// Ok I have read now, that your suggestion is a smoothing spline. Smoothing splines are not calculating with knots instead you have penalty function with the factor lambda. So the question is, is there another function in JMP that I can calculate the splines with knots?
By the way is there a nice feature how I can extract the formula of the spline regression and calculate "new" height values by just input the weight values?
Edit// I think I found a solution. The knotted spline effect should be fine for me, it is hidden under Analyze->FitModel->Attributes
To extract formulas and calculate new values, just click on the red red triangle->save columns->prediction formula
In the next step just write down the new x-values in the same column were the old x-values are already. The predicted values are displayed in the new column!
Community Trekker
## Re: cubic spline regression
I see the option of cubic splines is available in JMP at graph builder when you test the moderating effect of a categorical variable on the effect of a categorical variable to the outcome which happens to be... you guessed it...also categorical variable. Does this make sense? Is that possible? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8475621342658997, "perplexity": 1294.2359260323183}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256494.24/warc/CC-MAIN-20190521162634-20190521184634-00183.warc.gz"} |
http://mathhelpforum.com/calculus/20963-anti-derivatives-print.html | # ...Anti-Derivatives?
• October 20th 2007, 06:55 PM
Super Mallow
...Anti-Derivatives?
I guess this are called Anti-Derivatives, which is strange since we never discussed it in class.
Anyways, I'm given a function's derivative and they want me to find the actual function. It's not too bad, but then they ask us to find the function through a certain point, and that's when I get lost
Here's the questions I'm stuck on...
Find the function with the given derivative whose graph passes through the given point.
A) 1/xSquared + 2x, Point = (-1,1)
B) 8-CscSquared, Point = (Pi/4,0)
C) SecXTanX - 1, Point (0,0)
Any help is appreciated!
• October 20th 2007, 07:08 PM
Jhevon
Quote:
Originally Posted by Super Mallow
I guess this are called Anti-Derivatives, which is strange since we never discussed it in class.
Anyways, I'm given a function's derivative and they want me to find the actual function. It's not too bad, but then they ask us to find the function through a certain point, and that's when I get lost
Here's the questions I'm stuck on...
Find the function with the given derivative whose graph passes through the given point.
A) 1/xSquared + 2x, Point = (-1,1)
B) 8-CscSquared, Point = (Pi/4,0)
C) SecXTanX - 1, Point (0,0)
Any help is appreciated!
how far did you get with anti-derivatives? they weren't discussed at all? nothing?
if so, i guess you have no idea what the power rule is. here's what you do. look up all your derivative formulas. the things you have here will be the right side of the equation. the anti-derivative will be on the left. for example,
you should find that: $\frac d{dx}\sec x = \sec x \tan x$
so the anti-derivative of $\sec x \tan x$ is $\sec x + C$ since sec(x) is what you derived to get the sec(x)tan(x) so taking the antiderivative of that brings you back to sec(x). the + C is a constant. remember that when we take the derivative, if there is a constant, it's derivative is zero, so we put + C just in case there was a constant that got wiped out by the differentiation.
continuing with this example:
we have $\sec x \tan x - 1$
the anti-derivative of this is: $\sec x - x + C$ (when we differentiate x we get 1).
we want this to pass through (0,0), which means we want when x = 0, y = 0
so, plug that in: $\sec 0 - 0 + C = 0$
now solve for C
we get: $1 + C = 0 \implies C = -1$
so the antiderivative is: $\sec x - x - 1$
try the others
• October 20th 2007, 07:14 PM
Super Mallow
Yeah, we had no talks about Anti-Derivatives whatsoever. My roomate told me that they are preparing us for it in the upcoming weeks.
This may be a dumb question and may be why I don't understand it...what is C? Why did we put it in there?
• October 20th 2007, 07:23 PM
Jhevon
Quote:
Originally Posted by Super Mallow
Yeah, we had no talks about Anti-Derivatives whatsoever. My roomate told me that they are preparing us for it in the upcoming weeks.
This may be a dumb question and may be why I don't understand it...what is C? Why did we put it in there?
did you read my post, or did you not understand it?
Quote:
Originally Posted by Jhevon
...the + C is a constant. remember that when we take the derivative, if there is a constant, it's derivative is zero [it gets wiped out], so we put + C just in case there was a constant that got wiped out by the differentiation....
• October 20th 2007, 07:34 PM
Super Mallow
My bad: I didn't really understand it
• October 20th 2007, 07:39 PM
Jhevon
Quote:
Originally Posted by Super Mallow
My bad: I didn't really understand it
ok, let's add a little example shall we.
let's say we had a function, a simple one, like y = x + 1
now, take it's derivative, we get:
y' = 1
now, if we take the anti-derivative, we should get back the original function.
but let's say we said the anti-derivative was x. that would not be true, there was a 1 there, so we would have something missing from our original function. now if we said, the anti-derivative is x + C, that is fine, there is nothing missing, but the specific value of C, which we can find if we are given the appropriate clues. so that's why when finding anti-derivatives, we add an arbitrary constant, just in case there was a constant there that got wiped out by differentiation | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.858788788318634, "perplexity": 733.9975954532255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454702032759.79/warc/CC-MAIN-20160205195352-00305-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/124249-solve-following-trigonometric-integral-3-a.html | # Thread: Solve The Following Trigonometric Integral. (3)
1. ## Solve The Following Trigonometric Integral. (3)
Hello
here is it:
$\int ( cos(x) + 1 )^{\frac{3}{2}} dx$
I used $u=cos(x)+1$
Then, I got :
$- \int \frac{u^{\frac{3}{2}} }{ \sqrt{ 1 - (u-1)^2 }} du$
Using the trigonometric substitution $u-1=sin(\theta)$, I got:
$- \int (sin(\theta) + 1)^{\frac{3}{2}} du$
I think I will do the same work again and I will get the original integral in another variable.
then I will moved it the left hand side.
Since $\int f(x) dx = \int f(u) du = \int f(\theta) d\theta ....$
And I will devide by the resulting coefficient.
I think you understood what I mean.
2. Hint: $\cos x=2\cos ^{2}\frac{x}{2}-1.$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9982388615608215, "perplexity": 1093.7031147474893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00579.warc.gz"} |
https://www.andlearning.org/percentage-formula/ | Home » Math Formulas » Percentage Formula
# Percentage Change Formula | Decrease, Yield, Increase
## Percentage Formula
In simple terms, percent means per hundred. Whenever you have to express the number between one and zero then percentage formula is needed. Usually, the percentage is defined as the fraction of 100. The symbol for percentage is “%” and the major application of percentage is to compare or find the ratios. Here, we have given percentage formula in mathematics –
#### The Percentage Formula is given as,
$\ Percentage = \frac{Value}{Total\:Value}\times 100$
### Percentage Change Formula
Whenever we find the difference between two numbers then it would be either greater or less than earlier. To compare any two values or to find the differences, we need a percentage change formula. They are named as the percentage increase or percentage decrease in mathematics. Let us discuss on each of the concepts in the next sections deeply.
#### Percentage change formula is
$\ Percentage\;Change=\frac{Old\;Value-New\;Value}{Old\;Value}\times 100%$
### Percentage Decrease Formula
To find the percentage decrease, you first need to note down the numbers you are comparing. First, find the difference i.e. Decrease Value = Original number – New number. Now you have to divide the decreased value with original value and multiply the same with 100. If your final output is a negative number then this would be percent increase otherwise percent decrease. With the same technique, this is possible to calculate the percentage differences of multiple numbers.
The Percentage Decrease Formula is given as,
$\ Percentage\;Decrease = \frac{Decrease\;in\;value}{Original\; value} \times 100$
### Percentage Increase Formula
To find the percentage decrease, you first need to note down the numbers you are comparing. First, find the difference i.e. Increase Value = New number Original number. Now you have to divide the Increase value with original value and multiply the same with 100. If your final output is a negative number then this would be a percent decrease otherwise percent increase. The Percentage Increase Formula in mathematics is given is given as,
The Percentage Increase Formula is given as,
$\large Percentage\;Increase=\frac{Increased\;Value}{Original\;Value}\times 100$
Note down, the final value could be positive, negative or zero. For zero value, it means there is no change in percent at all.
### Percentage Yield Formula
Every time when we are experimenting, the final value would be a little different from your predictions. In case of chemistry, this difference is popular with the name percentage yield. Take an example of the online recipe where actual ingredients with proper quantity are given. Also, it will give you a perfect idea of servings. But, the quantity may be off or extra in most of the cases. There are different reasons for the same like wrong cup measures, food spoilage, or different stove etc.
$\ Percentage\; Yield = \frac{actual\;yield}{theoretical\;yield} \times 100\;%$
The same concept is applicable when we do perform experiments in Chemistry lab to make a Compound. However, the experiment should be performed in ideal conditions only. Still, most of the times, there are experimental errors most of the times and your predictions would be slightly different from the actual one.
Here, is given a quick formula to calculate the percentage yield in mathematics and chemistry. When you apply this formula, make sure that actual yield value and theoretical yield values are in the same units otherwise there would be a definite error in the final value. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9499363899230957, "perplexity": 641.2990459501905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578527566.44/warc/CC-MAIN-20190419101239-20190419123239-00353.warc.gz"} |
https://math.stackexchange.com/questions/3011647/given-technology-find-number-of-distinguishable-ways-the-letters-can-be-arrang | # Given TECHNOLOGY , find number of distinguishable ways the letters can be arranged in which letters T,E and N are together [closed]
Given TECHNOLOGY , find number of distinguishable ways the letters can be arranged in which letters T,E and N are together
This is my working-
$$3! \cdot \frac{7!}{2!}$$
is this correct ?
## closed as off-topic by Don Thousand, MisterRiemann, NCh, Shailesh, John BDec 1 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Don Thousand, MisterRiemann, NCh, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
• Solutions to such exercises should usually be presented in such a way that one can understand how you reached the answer, i.e. you should explain how you got that particular number, instead of just presenting the final answer. – MisterRiemann Nov 24 '18 at 15:00
• Be careful. TECHNOLOGY has ten letters, so you have a block of three letters and seven other letters to arrange. – N. F. Taussig Nov 24 '18 at 15:04
Assume $$T,E, N$$ as single letter therefore, total number of letters in the word technology is 8 this can be arranged in $$8!$$ ways and number of ways in which $$T,E, N$$ can be arranged $$3!$$ ways and since $$O$$ is repeating two times hence answer is $$\frac{8!×3!}{2!}$$.
Consider $$\text{TEN}$$ together as a block and all other letters as single block. Then you have $$8$$ blocks. So there are $$8!$$ permutations possible and $$3!$$ permutations of $$TEN$$. Also the letter $$\text{O}$$ is not distinguishable.
So total number of ways is $$\dfrac{8!\cdot 3!}{2!}$$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8130512237548828, "perplexity": 545.7132811196825}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00032.warc.gz"} |
https://math.stackexchange.com/questions/773742/express-skew-commutative-product-of-whitney-sum-of-vector-bundles-in-tensor-prod | # express skew-commutative product of Whitney sum of vector bundles in tensor products
Let $\xi$ and $\eta$ be vector bundles over a paracompact space $B$ and $\xi\oplus\eta$ be their Whitney sum. Can we write $\Lambda(\xi\oplus\eta)\cong \Lambda(\xi)\otimes \Lambda(\eta)$ as (graded) vector bundles?
For vector spaces $V,W$, we have $\Lambda(V\oplus W)\cong \Lambda(V)\otimes \Lambda(W)$ are isomorphic as graded vector spaces. But I am confused for vector bundle case...
• You just transplant what you know about vector spaces to the trivializations of your bundles, so you're done. :D – Ted Shifrin Apr 29 '14 at 3:56
• In Milnor's book (whose notation you seem to be using), on chap. 3, p. 31, he answers your question: how to transfer linear algebra constructions to vector bundles. It is a bit abstract, but very clear and complete, as is usual with Milnor. – Gil Bor May 1 '14 at 4:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9598910808563232, "perplexity": 440.55133254571365}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573080.8/warc/CC-MAIN-20190917141045-20190917163045-00079.warc.gz"} |
https://www.arxiv-vanity.com/papers/1305.5939/ | arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.
The common ancestor process revisited
Sandra Kluth Thiemo Hustedt Ellen Baake
Technische Fakultät, Universität Bielefeld, Box 100131, 33501 Bielefeld, Germany
E-mail: {skluth, thustedt, ebaake}@techfak.uni-bielefeld.de
Abstract
Abstract. We consider the Moran model in continuous time with two types, mutation, and selection. We concentrate on the ancestral line and its stationary type distribution. Building on work by Fearnhead (J. Appl. Prob. 39 (2002), 38-54) and Taylor (Electron. J. Probab. 12 (2007), 808-847), we characterise this distribution via the fixation probability of the offspring of all individuals of favourable type (regardless of the offspring’s types). We concentrate on a finite population and stay with the resulting discrete setting all the way through. This way, we extend previous results and gain new insight into the underlying particle picture.
2000 Mathematics Subject Classification: Primary 92D15; Secondary 60J28.
Key words: Moran model, ancestral process with selection, ancestral line, common ancestor process, fixation probabilities.
1 Introduction
Understanding the interplay of random reproduction, mutation, and selection is a major topic of population genetics research. In line with the historical perspective of evolutionary research, modern approaches aim at tracing back the ancestry of a sample of individuals taken from a present population. Generically, in populations that evolve at constant size over a long time span without recombination, the ancestral lines will eventually coalesce backwards in time into a single line of descent. This ancestral line is of special interest. In particular, its type composition may differ substantially from the distribution at present time. This mirrors the fact that the ancestral line consists of those individuals that are successful in the long run; thus, its type distribution is expected to be biased towards the favourable types.
This article is devoted to the ancestral line in a classical model of population genetics, namely, the Moran model in continuous time with two types, mutation, and selection (i.e., one type is ‘fitter’ than the other). We are particularly interested in the stationary distribution of the types along the ancestral line, to be called the ancestral type distribution. We build on previous work by Fearnhead [9] and Taylor [20]. Fearnhead’s approach is based on the ancestral selection graph, or ASG for short [14, 16]. The ASG is an extension of Kingman’s coalescent [12, 13], which is the central tool to describe the genealogy of a finite sample in the absence of selection. The ASG copes with selection by including so-called virtual branches in addition to the real branches that define the true genealogy. Fearnhead calculates the ancestral type distribution in terms of the coefficients of a series expansion that is related to the number of (‘unfit’) virtual branches.
Taylor uses diffusion theory and a backward-forward construction that relies on a description of the full population. He characterises the ancestral type distribution in terms of the fixation probability of the offspring of all ‘fit’ individuals (regardless of the offspring’s types). This fixation probability is calculated via a boundary value problem.
Both approaches rely strongly on analytical tools; in particular, they employ the diffusion limit (which assumes infinite population size, weak selection and mutation) from the very beginning. The results only have partial interpretations in terms of the graphical representation of the model (i.e., the representation that makes individual lineages and their interactions explicit). The aim of this article is to complement these approaches by starting from the graphical representation for a population of finite size and staying with the resulting discrete setting all the way through, performing the diffusion limit only at the very end. This will result in an extension of the results to arbitrary selection strength, as well as a number of new insights, such as an intuitive explanation of Taylor’s boundary value problem in terms of the particle picture, and an alternative derivation of the ancestral type distribution.
The paper is organised as follows. We start with a short outline of the Moran model (Section 2). In Section 3, we introduce the common ancestor type process and briefly recapitulate Taylor’s and Fearnhead’s approaches. We concentrate on a Moran model of finite size and trace the descendants of the initially ‘fit’ individuals forward in time. Decomposition according to what can happen after the first step gives a difference equation, which turns into Taylor’s diffusion equation in the limit. We solve this difference equation and obtain the fixation probability in the finite-size model in closed form. In Section 5, we derive the coefficients of the ancestral type distribution within the discrete setting. Section 6 summarises and discusses the results.
2 The Moran model with mutation and selection
We consider a haploid population of fixed size in which each individual is characterised by a type . If an individual reproduces, its single offspring inherits the parent’s type and replaces a randomly chosen individual, maybe its own parent. This way the replaced individual dies and the population size remains constant.
Individuals of type reproduce at rate , whereas individuals of type reproduce at rate , . Accordingly, type- individuals are termed ‘fit’, type- individuals are ‘unfit’. In line with a central idea of the ASG, we will decompose reproduction events into neutral and selective ones. Neutral ones occur at rate and happen to all individuals, whereas selective events occur at rate and are reserved for type- individuals.
Mutation occurs independently of reproduction. An individual of type mutates to type at rate , , , . This is to be understood in the sense that every individual, regardless of its type, mutates at rate and the new type is with probability . Note that this includes the possibility of ‘silent’ mutations, i.e., mutations from type to type .
The Moran model has a well-known graphical representation as an interacting particle system (cf. Fig. 1). The vertical lines represent the individuals, with time running from top to bottom in the figure. Reproduction events are represented by arrows with the reproducing individual at the base and the offspring at the tip. Mutation events are marked by bullets.
We are now interested in the process , where is the number of individuals of type at time . When the number of type- individuals is , it increases by one at rate and decreases by one at rate , where
λNk=k(N−k)N(1+sN)+(N−k)uNν0andμNk=k(N−k)N+kuNν1. (1)
Thus, is a birth-death process with birth rates and death rates . For and its stationary distribution is with
πNZ(k)=CNk∏i=1λNi−1μNi,0⩽k⩽N, (2)
where is a normalising constant (cf. [4, p. 19]). (As usual, an empty product is understood as .)
To arrive at a diffusion, we consider the usual rescaling
(XNt)t⩾0:=1N(ZNNt)t⩾0,
and assume that , , and , . As , we obtain the well-known diffusion limit
(Xt)t⩾0:=limN→∞(XNt)t⩾0.
Given , a sequence with and , is characterised by the drift coefficient
a(x)=limN→∞(λNkN−μNkN)=(1−x)θν0−xθν1+(1−x)xσ (3)
and the diffusion coefficient
b(x)=limN→∞1N(λNkN+μNkN)=2x(1−x). (4)
Hence, the infinitesimal generator of the diffusion is defined by
Af(x)=(1−x)x∂2∂x2f(x)+[(1−x)θν0−xθν1+(1−x)xσ]∂∂xf(x), f∈C2([0,1]).
The stationary density – known as Wright’s formula – is given by
πX(x)=C(1−x)θν1−1xθν0−1exp(σx), (5)
where is a normalising constant. See [5, Ch. 7, 8] or [8, Ch. 4, 5] for reviews of diffusion processes in population genetics and [11, Ch. 15] for a general survey of diffusion theory.
In contrast to our approach starting from the Moran model, [9] and [20] choose the diffusion limit of the Wright-Fisher model as the basis for the common ancestor process. This is, however, of minor importance, since both diffusion limits differ only by a rescaling of time by a factor of (cf. [5, Ch. 7], [8, Ch. 5] or [11, Ch. 15]).
3 The common ancestor type process
Assume that the population is stationary and evolves according to the diffusion process . Then, at any time , there almost surely exists a unique individual that is, at some time , ancestral to the whole population; cf. Fig. 2. (One way to see this is via [14, Thm. 3.2, Corollary 3.4], which shows that the expected time to the ultimate ancestor in the ASG remains bounded if the sample size tends to infinity.) We say that the descendants of this individual become fixed and call it the common ancestor at time . The lineage of these distinguished individuals over time defines the so-called ancestral line. Denoting the type of the common ancestor at time by , , we term the common ancestor type process or CAT process for short. Of particular importance is its stationary type distribution , to which we will refer as the ancestral type distribution. Unfortunately, the CAT process is not Markovian. But two approaches are available that augment by a second component to obtain a Markov process. They go back to Fearnhead [9] and Taylor [20]; we will recapitulate them below.
3.1 Taylor’s approach
For ease of exposition, we start with Taylor’s approach [20]. It relies on a description of the full population forward in time (in the diffusion limit of the Moran model as ) and builds on the so-called structured coalescent [2]. The process is , with states , and . In [20] this process is termed common ancestor process (CAP).
Define as the probability that the common ancestor at a given time is of type , provided that the frequency of type- individuals at this time is . Obviously, , . Since the process is time-homogeneous, is independent of time. Denote the (stationary) distribution of by . Its marginal distributions are (with respect to the first variable) and (with respect to the second variable). may then be written as the product of the marginal density and the conditional probability (cf. [20]):
πT(0,x)dx =h(x)πX(x)dx, πT(1,x)dx =(1−h(x))πX(x)dx.
Since is well known (5), it remains to specify . Taylor uses a backward-forward construction within diffusion theory to derive a boundary value problem for , namely:
12b(x)h′′(x)+a(x)h′(x)−(θν1x1−x+θν01−xx)h(x)+θν1x1−x=0,h(0)=0,h(1)=1. (6)
Taylor shows that (6) has a unique solution. The stationary distribution of is thus determined in a unique way as well. The function is smooth in and its derivative can be continuously extended to (cf. [20, Lemma 2.3, Prop. 2.4]).
In the neutral case (i.e., without selection, ), all individuals reproduce at the same rate, independently of their types. For reasons of symmetry, the common ancestor thus is a uniform random draw from the population; consequently, . In the presence of selection, Taylor determines the solution of the boundary value problem via a series expansion in (cf. [20, Sec. 4] and see below), which yields
h(x)=x+σx−θν0(1−x)−θν1exp(−σx)∫x0(~x−p)pθν0(1−p)θν1exp(σp)dp (7) (8)
The stationary type distribution of the ancestral line now follows via marginalisation:
α0=∫10h(x)πX(x)dx \ and \ α1=∫10(1−h(x))πX(x)dx. (9)
Following [20], we define and write
h(x)=x+ψ(x). (10)
Since is the conditional probability that the common ancestor is fit, is the part of this probability that is due to selective reproduction.
Substituting (10) into (6) leads to a boundary value problem for :
12b(x)ψ′′(x)+a(x)ψ′(x)−(θν1x1−x+θν01−xx)ψ(x)+σx(1−x)=0,ψ(0)=ψ(1)=0. (11)
Here, the smooth inhomogeneous term is more favourable as compared to the divergent inhomogeneous term in (6). Note that Taylor actually derives the boundary value problems (6) and (11) for the more general case of frequency-dependent selection, but restricts himself to frequency-independence to derive solution (7).
We can only give a brief introduction to Fearnhead’s approach [9] here. On the basis of the ASG, the ancestry of a randomly chosen individual from the present (stationary) population is traced backwards in time. More precisely, one considers the process with values in , where is the type of the individual’s ancestor at time before the present (that is, at forward time ). Obviously, there is a minimal time so that, for all , (see also Fig. 2), provided the underlying process is extended to .
To make the process Markov, the true ancestor (known as the real branch) is augmented by a collection of virtual branches (see [1, 14, 16, 19] for the details). Following [9, Thm. 1], certain virtual branches may be removed (without compromising the Markov property) and the remaining set of virtual branches contains only unfit ones. We will refer to the resulting construction as the pruned ASG. It is described by the process , where (with values in ) is the number of virtual branches (of type ). is termed common ancestor process in [9] (but keep in mind that it is that is called CAP in [20]). Reversing the direction of time in the pruned ASG yields an alternative augmentation of the CAT process (for ).
Fearnhead provides a representation of the stationary distribution of the pruned process, which we will denote by . This stationary distribution is expressed in terms of constants defined by the following backward recursion:
ρ(k)k+1=0 \ and \ ρ(k)j−1=σj+σ+θ−(j+θν1)ρ(k)j, k∈N,2⩽j⩽k+1. (12)
The limit exists (cf. [9, Lemma 1]) and the stationary distribution of the pruned ASG is given by (cf. [9, Thm. 3])
πF(i,n)={anEπX(X(1−X)n),if i=0,(an−an+1)EπX((1−X)n+1),if i=1, with \ an:=n∏j=1ρj
for all . Fearnhead proves this result by straightforward verification of the stationarity condition; the calculation is somewhat cumbersome and does not yield insight into the connection with the graphical representation of the pruned ASG. Marginalising over the number of virtual branches results in the stationary type distribution of the ancestral line, namely,
αi=∑n⩾0πF(i,n). (13)
Furthermore, this reasoning points to an alternative representation of respectively (cf. [20]):
h(x)=x+x∑n⩾1an(1−x)n \ respectively \ ψ(x)=x∑n⩾1an(1−x)n. (14)
The , to which we will refer as Fearnhead’s coefficients, can be shown [20] to follow the second-order forward recursion
(2+θν1)a2−(2+σ+θ)a1+σ=0,(n+θν1)an−(n+σ+θ)an−1+σan−2=0,n⩾3. (15)
Indeed, (14) solves the boundary problem (6) and, therefore, equals (7) (cf. [20, Lemma 4.1]).
The forward recursion (15) is greatly preferable to the backward recursion (12), which can only be solved approximately with initial value for some large . What is still missing is the initial value, . To calculate it, Taylor defines (cf. [20, Sec. 4.1])
v(x):=h(x)−xx=ψ(x)x=∑n⩾1an(1−x)n (16)
and uses111Note the missing factor of in his equation (28).
an=(−1)nn!v(n)(1). (17)
This way a straightforward (but lengthy) calculation (that includes a differentiation of expression (7)) yields
a1=−v′(1)=−ψ′(1)=σ1+θν1(1−~x). (18)
4 Discrete approach
Our focus is on the stationary type distribution of the CAT process. We have seen so far that it corresponds to the marginal distribution of both and , with respect to the first variable. Our aim now is to establish a closer connection between the properties of the ancestral type distribution and the graphical representation of the Moran model. In a first step we re-derive the differential equations for and in a direct way, on the basis of the particle picture for a finite population. This derivation will be elementary and, at the same time, it will provide a direct interpretation of the resulting differential equations.
4.1 Difference and differential equations for h and ψ
Equations for . Since it is essential to make the connection with the graphical representation explicit, we start from a population of finite size , rather than from the diffusion limit. Namely, we look at a new Markov process with the natural filtration , where . is the number of fit individuals as before and holds the number of descendants of types and at time of an unordered sample with composition collected at time . More precisely, we start with a -measurable state (this means that must be independent of the future evolution; but note that it need not be a random sample) and observe the population evolve in forward time. At time , count the type- descendants and the type- descendants of our initial sample and summarise the results in the unordered sample . Together with , this gives the current state (cf. Fig. 3).
As soon as the initial sample is ancestral to all individuals, it clearly will be ancestral to all individuals at all later times. Therefore,
AN:={(m,k):k∈{0,…,N},m0⩽k,|m|=N},
where for a sample , is a closed (or invariant) set of the Markov chain. (Given a Markov chain in continuous time on a discrete state space , a non-empty subset is called closed (or invariant) provided that , , (cf. [17, Ch. 3.2]).)
From now on we restrict ourselves to the initial value , i.e. the population consists of fit individuals and the initial sample contains them all. Our aim is to calculate the probability of absorption in , which will also give us the fixation probability of the descendants of the type- individuals. In other words, we are interested in the probability that the common ancestor at time 0 belongs to our fit sample . Let us define as the equivalent of in the case of finite population size , that is, is the probability that one of the fit individuals is the common ancestor given . Equivalently, is the absorption probability of in , conditional on . Obviously, , . It is important to note that, given absorption in , the common ancestor is a random draw from the initial sample. Therefore,
P(a specific type-0 individual will fix∣ZN0=k)=hNkk. (19)
Likewise,
P(a specific type-1 individual will fix∣ZN0=k)=1−hNkN−k. (20)
We will now calculate the absorption probabilities with the help of ‘first-step analysis’ (cf. [17, Thm. 3.3.1], see also [5, Thm. 7.5]). Let us recall the method for convenience.
Lemma 1 (‘first-step analysis’).
Assume that is a Markov chain in continuous time on a discrete state space , is a closed set and , , is the waiting time to leave the state . Then for all ,
P(Y absorbs in A∣Y(0)=y) =∑z∈E:z≠yP(Y(Ty)=z∣Y(0)=y) =∑z∈E:z≠x×P(Y absorbs in A∣Y(0)=z).
So let us decompose the event ‘absorption in ’ according to the first step away from the initial state. Below we analyse all possible transitions (which are illustrated in Fig. 4), state the transition rates and calculate absorption probabilities, based upon the new state. We assume throughout that .
(a)
:
One of the sample individuals of type reproduces and replaces a type- individual. We distinguish according to the kind of the reproduction event.
(a1)
Neutral reproduction rate: .
(a2)
Selective reproduction rate: .
In both cases, the result is a sample containing all fit individuals. Now starts afresh in the new state , with absorption probability .
(b)
:
A type- individual reproduces and replaces a (sample) individual of type . This occurs at rate and leads to a sample that consists of all fit individuals. The absorption probability, if we start in the new state, is .
(c)
:
This transition describes a mutation of a type- individual to type and occurs at rate . The new sample contains all fit individuals, plus a single unfit one. Starting now from , the absorption probability has two contributions: First, by definition, with probability , one of the fit individuals will be the common ancestor. In addition, by (20), the single unfit individual has fixation probability , so the probability to absorb in when starting from the new state is
P(absorption in AN∣(M0,ZN0)=((k−1,1),k−1)) =hNk−1+1−hNk−1N−(k−1).
(d)
:
This is a mutation from type to type , which occurs at rate . We then have fit individuals in the population altogether, but the new sample contains only of them. Arguing as in (c) and this time using (19), we get
P(absorption in AN∣(M0,ZN0)=((k,0),k+1)) =hNk+1−hNk+1k+1.
Note that, in steps (c) and (d) (and already in (19) and (20)), we have used the permutation invariance of the fit (respectively unfit) lines to express the absorption probabilities as a function of (the number of fit individuals in the population) alone. This way, we need not cope with the full state space of . Taking together the first-step principle with the results of (a)–(d), we obtain the linear system of equations for (with the rates and as in (1)):
(λNk+μNk)hNk=λNkhNk+1+μNkhNk−1+kuNν11−hNk−1N−(k−1)−(N−k)uNν0hNk+1k+1, (21)
, which is complemented by the boundary conditions , . Rearranging results in
(22)
Let us now consider a sequence with and . The probabilities converge to as (for the stationary case a proof is given in the Appendix). Equation (22), with replaced by , together with (3) and (4) leads to Taylor’s boundary value problem (6).
Equations for . As before, we consider with start in , and now introduce the new function . is the part of the absorption probability in that goes back to selective reproductions (in comparison to the neutral case). We therefore speak of (as well as of ) as the ‘extra’ absorption probability.
Substituting in (21) yields the following difference equation for :
(λNk+μNk)ψNk=λNkψNk+1+μNkψNk−1+k(N−k)N2sN−kuNν1ψNk−1N−(k−1)−(N−k)uNν0ψNk+1k+1 (23)
, together with the boundary conditions . It has a nice interpretation, which is completely analogous to that of except in case (a2): If one of the fit sample individuals reproduces via a selective reproduction event, the extra absorption probability is (rather than ). Here, is the neutral fixation probability of the individual just created via the selective event; is the extra absorption probability of all type- individuals present after the event. The neutral contribution gives rise to the term on the right-hand side of (23). Performing in the same way as for , we obtain Taylor’s boundary value problem (11) and now have an interpretation in terms of the graphical representation to go with it.
4.2 Solution of the difference equation
In this Section, we derive an explicit expression for the fixation probabilities , that is, a solution of the difference equation (21), or equivalently, (23). Although the calculations only involve standard techniques, we perform them here explicitly since this yields additional insight. Since there is no danger of confusion, we omit the subscript (or superscript) for economy of notation.
The following Lemma specifies the extra absorption probabilities in terms of a recursion.
Lemma 2.
Let . Then
ψN−k=k(N−k)μN−k(μN−1N−1ψN−1+λN−k+1(k−1)(N−k+1)ψN−k+1−s(k−1)N2). (24)
Remark 1.
The quantity is well defined for all , and is well defined even for .
Proof of Lemma 2.
Let . Set in (23) and divide by to obtain
(λii(N−i)+μii(N−i))ψi =(1+sN+uν0i+1)ψi+1+(1N+uν1N−(i−1))ψi−1+sN2 =λi+1(i+1)(N−i−1)ψi+1+μi−1(i−1)(N−i+1)ψi−1+sN2. (25)
Together with
(λ1N−1+μ1N−1)ψ1 =λ22(N−2)ψ2+sN2, (26) (λN−1N−1+μN−1N−1)ψN−1 =μN−22(N−2)ψN−2+sN2, (27)
and the boundary conditions , we obtain a new linear system of equations for the vector Summation over the last equations yields
N−1∑i=N−k+1(λii(N−i)+μii(N−i))ψi= N−2∑i=N−k+1λi+1(i+1)(N−i−1)ψi+1 +N−1∑i=N−k+1μi−1(i−1)(N−i+1)ψi−1+s(k−1)N2,
which proves the assertion.
Lemma 2 allows for an explicit solution for .
Theorem 1.
For , let
χnℓ:=n∏i=ℓλiμi \ and \ K:=N−1∑n=0χn1. (28)
The solution of recursion is then given by
ψN−k=k(N−k)μN−kN−1∑n=N−kχnN−k+1(μN−1N−1ψN−1−s(N−1−n)N2) (29)
with
ψN−1=1KN−1μN−1sN2N−2∑n=0(N−1−n)χn1. (30)
An alternative representation is given by
ψN−k=1Kk(N−k)μN−ksN2N−k−1∑ℓ=0N−1∑n=N−k(n−ℓ)χℓ1χnN−k+1. (31)
Proof.
We first prove (29) by induction over . For , (29) is easily checked to be true. Inserting the induction hypothesis for some into recursion (24) yields
ψN−k= k(N−k)μN−k[μN−1N−1ψN−1 +λN−k+1μN−k+1N−1∑n=N−k+1χnN−k+2(μN−1N−1ψN−1−s(N−1−n)N2)−s(k−1)N2],
which immediately leads to (29). For , (29) gives (30), since and is well defined by Remark 1. We now check (31) by inserting (30) into (29) and then use the expression for as in (28):
ψN−k =1Kk(N−k)μN−ksN2N−1∑n=N−kχnN−k+1[N−1∑ℓ=0(N−1−ℓ)χℓ1−N−1∑ℓ=0(N−1−n)χℓ1] =1Kk(N−k)μN−ksN2N−1∑ℓ=0N−1∑n=N−k(n−ℓ)χℓ1χnN−k+1. Then we split the first sum according to whether ℓ⩽N−k−1 or ℓ⩾N−k, and use χℓ1=χN−k1χℓN−k+1 in the latter case: ψN−k =1Kk(N−k)μN−ksN2[N−k−1∑ℓ=0N−1∑n=N−k(n−ℓ)χℓ1χnN−k+1 =+χN−k1N−1∑ℓ=N−kN−1∑n=N−k(n−ℓ)χℓN−k+1χnN−k+1].
The first sum is the right-hand side of (31) and the second sum disappears due to symmetry.
Let us note that the fixation probabilities thus obtained have been well known for the case with selection in the absence of mutation (see, e.g., [5, Thm. 6.1]), but to the best of our knowledge, have not yet appeared in the literature for the case with mutation.
4.3 The solution of the differential equation
As a little detour, let us revisit the boundary value problem (6). To solve it, Taylor assumes that can be expanded in a power series in . This yields a recursive series of boundary value problems (for the various powers of ), which are solved by elementary methods and combined into a solution of (cf. [20]).
However, the calculations are slightly long-winded. In what follows, we show that the boundary value problem (6) (or equivalently (11)) may be solved in a direct and elementary way, without the need for a series expansion. Defining
c(x):=−θν1x1−x−θν01−xx
and remembering the drift coefficient (cf. (3)) and the diffusion coefficient (cf. (4)), differential equation (11) reads
12b(x)ψ′′(x)+a(x)ψ′(x)+c(x)ψ(x)=−σx(1−x)
or, equivalently,
(32)
Since
c(x)b(x)=ddxa(x)b(x), (33)
(32) is an exact differential equation (for the concept of exactness, see [10, Ch. 3.11] or [3, Ch. 2.6]). Solving it corresponds to solving its primitive
ψ′(x)+2a(x)b(x)ψ(x)=−σ(x−~x). (34)
The constant plays the role of an integration constant and will be determined by the initial conditions later. (Obviously, (32) is recovered by differentiating (34) and observing (33).) As usual, we consider the homogeneous equation
φ′(x)+2a(x)b(x)φ(x)=φ′(x)+(σ−θν11−x+θν0x)φ(x)=0
first. According to [5, Ch. 7.4] and [8, Ch. 4.3], its solution is given by
φ1(x)=exp(∫x−2a(z)b(z)dz)=γ(1−x)−θν1x−θν0exp(−σx)=2Cγb(x)πX(x).
(Note the link to the stationary distribution provided by the last expression (cf. [5, Thm. 7.8] and [8, Ch. 4.5]).) Of course, the same expression is obtained via separation of variables. Again we will deal with the constant later.
Variation of parameters yields the solution of the inhomogeneous equation (34):
φ2(x)=φ1(x)∫xβ−σ(p−~x)φ1(p)dp=σφ1(x)∫xβ~x−pφ1(p)dp. (35)
Finally, it remains to specify the constants of integration , and the constant to comply with . We observe that the factor cancels in (35), thus its choice is arbitrary. diverges for and , so the choice of and has to guarantee , where . Hence, and
~x∫101φ1(p)dp=∫10pφ1(p)dp ⇔ ~x=∫10pφ1(p)dp∫101φ1(p)dp.
For the sake of completeness, l’Hôpital’s rule can be used to check that . The result indeed coincides with Taylor’s (cf. (7)).
We close this Section with a brief consideration of the initial value of the recursions (15). Since, by (18), , it may be obtained by analysing the limit of (34). In the quotient , numerator and denominator disappear as . According to l’Hôpital’s rule, we get
limx→1a(x)ψ(x)b(x)=limx→1(−θν0−θν1+σ(1−2x))ψ(x)+a(x)ψ′(x)2(1−2x)=12θν1ψ′(1),
therefore, the limit of (34) yields
−ψ′(1)(1+θν1)=σ(1−~x).
Thus, we obtain without the need to differentiate expression (7).
5 Derivation of Fearnhead’s coefficients in the discrete setting
Let us now turn to the ancestral type distribution and Fearnhead’s coefficients that characterise it. To this end, we start from the linear system of equations for in (25)-(27). Let
˜ψNk:=ψNkk(N−k), (36)
for . In terms of these new variables, (27) reads
−μNN−1˜ψNN−1+μNN−2˜ψNN | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9475066065788269, "perplexity": 617.4526878164663}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655881763.20/warc/CC-MAIN-20200706160424-20200706190424-00389.warc.gz"} |
http://mathhelpforum.com/algebra/218867-cannot-life-me-follow-simplification-print.html | # Cannot for the life of me follow this simplification
• May 13th 2013, 12:33 AM
lukasaurus
Cannot for the life of me follow this simplification
http://i.imgur.com/6HvVncv.jpg
I understand how the sqrt comes down, from the right half of the top, but how does the top left part lose the square root?
After simplifying, I can get
sqrt(x^2+1) - x^2 / (x^2+1)^(3/2)
However, the answer that all online calcs give and my tutorial notes give is
1/ (x^2+1)^(3/2)
And I can see where that is going from the working up the top, but can't figure out how the sqrt(x^2+1) turns into (x^2+1)
• May 13th 2013, 01:16 AM
agentmulder
Re: Cannot for the life of me follow this simplification
One way to see this is to factor, first cancel the 2 with 1/2 at top right then factor
$\frac{(x^2 + 1)^{-\frac{1}{2}}[(x^2 + 1) - x^2]}{x^2 + 1}$
Notice if you distribute, the bases are the same so you add exponents and keep the base , -1/2 + 1 = 1/2 so we're good
Next simplify within brackets = 1 then bring what's left to the denominator so it's exponent is positive, finally keep the same base and add the exponents. Let me know if you need more details.
:)
• May 13th 2013, 01:36 AM
lukasaurus
Re: Cannot for the life of me follow this simplification
Yep,I definitely need more details with that first factoring. I have been sitting, trying to work this out for over two hours
• May 13th 2013, 01:40 AM
ibdutt
1 Attachment(s)
Re: Cannot for the life of me follow this simplification
• May 13th 2013, 02:26 AM
agentmulder
Re: Cannot for the life of me follow this simplification
Quote:
Originally Posted by lukasaurus
Yep,I definitely need more details with that first factoring. I have been sitting, trying to work this out for over two hours
Let's just look at only the numerator for a moment , after canceling 1/2 with 2 and let's represent the radical with an equivalent exponent.
$(x^2 + 1)^{\frac{1}{2}} - x^2(x^2 + 1)^{-\frac{1}{2}}$
is it easier to see the same base now?
We must factor from both terms because we have subtraction between the 2 terms.
Factoring from the term on the right is easy, just pull it out, factoring from the term on the left is a bit harder but if you can understand it you save time and space. Essentialy , to figure out what exponent to put you subtract the exponents
We identified the common base
$(x^2 + 1)$
We decide to factor out
$(x^2 + 1)^{-\frac{1}{2}}$
So far we have
$(x^2 + 1)^{-\frac{1}{2}}[ \ \ \ \ \ \ - x^2]$
what do we put in the space? Well, it's going to be the same base with an exponent of...
$\frac{1}{2} - (- \frac{1}{2} ) = 1$
the positive 1/2 is the exponent on the base that is getting factored the -1/2 is the exponent on the base that is doing the factoring.
$(x^2 + 1)^{-\frac{1}{2}}[(x^2 + 1)^1 - x^2]$
or simply
$(x^2 + 1)^{-\frac{1}{2}}[x^2 + 1 - x^2]$
The rest of it is obvious, i hope to have helped you understand the factoring. Let me know if anything is not clear.
:) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8528382778167725, "perplexity": 738.1014638906853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678687395/warc/CC-MAIN-20140313024447-00071-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/hall-effect-data.383060/ | # Hall Effect data
1. Mar 2, 2010
### Kazkek
Hi, I am doing research in Hall Mobility and Half-heusler compounds.
I've taken some data on several samples now and I am kind of perplexed. We use a very stable magnetic field between -.6 Tesla and .6 Tesla to do the measurements.
Currently we take data points at intervals of .1 tesla to acquire a line of Magnetic Field vs. Resistance (or voltage since Current is held constant). And for the most part we take the slope of that line to get a "B / R" ratio and then calculate the carrier concentration.
Most of the samples have run smoothly and we've gotten linear data, but a few of the samples are returning Parabolic data.
This meaning that we start at .6 Tesla and go to -.6 Tesla in .1 Tesla increments. Taking data at each point.
http://img682.imageshack.us/img682/2230/graph3t.png" [Broken]
We are also using a Van der pauw geometry instead of bar geometry (due to the shape of the samples that are made).
my question is, What does a parabolic curve mean for Hall effect data. I thought it should be mostly linear.
If you need any more info that I've left out, just ask I should be able to tell you. Thanks.
Last edited by a moderator: May 4, 2017
Can you offer guidance or do you also need help?
Draft saved Draft deleted
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https://arxiv.org/abs/1504.05661 | math.OC
(what is this?)
# Title: Control of Generalized Energy Storage Networks
Abstract: The integration of intermittent and volatile renewable energy resources requires increased flexibility in the operation of the electric grid. Storage, broadly speaking, provides the flexibility of shifting energy over time; network, on the other hand, provides the flexibility of shifting energy over geographical locations. The optimal control of general storage networks in uncertain environments is an important open problem. The key challenge is that, even in small networks, the corresponding constrained stochastic control problems with continuous spaces suffer from curses of dimensionality, and are intractable in general settings. For large networks, no efficient algorithm is known to give optimal or near-optimal performance. This paper provides an efficient and provably near-optimal algorithm to solve this problem in a very general setting. We study the optimal control of generalized storage networks, i.e., electric networks connected to distributed generalized storages. Here generalized storage is a unifying dynamic model for many components of the grid that provide the functionality of shifting energy over time, ranging from standard energy storage devices to deferrable or thermostatically controlled loads. An online algorithm is devised for the corresponding constrained stochastic control problem based on the theory of Lyapunov optimization. We prove that the algorithm is near-optimal, and construct a semidefinite program to min- imize the sub-optimality bound. The resulting bound is a constant that depends only on the parameters of the storage network and cost functions, and is independent of uncertainty realizations. Numerical examples are given to demonstrate the effectiveness of the algorithm.
Comments: This report, written in January 2014, is a longer version of the conference paper [1] (See references in the report). This version contains a somewhat more general treatment for the cases with sub-differentiable objective functions and Markov disturbance. arXiv admin note: substantial text overlap with arXiv:1405.7789 Subjects: Optimization and Control (math.OC) Cite as: arXiv:1504.05661 [math.OC] (or arXiv:1504.05661v1 [math.OC] for this version)
## Submission history
From: Junjie Qin [view email]
[v1] Wed, 22 Apr 2015 05:53:32 GMT (1527kb) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8975402116775513, "perplexity": 915.6503428281238}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158766.65/warc/CC-MAIN-20180923000827-20180923021227-00408.warc.gz"} |
https://www.deepdyve.com/lp/oxford-university-press/global-structural-properties-of-random-graphs-UIkbSR5ban | # Global Structural Properties of Random Graphs
Global Structural Properties of Random Graphs Abstract We study two global structural properties of a graph $$\Gamma$$, denoted $$\mathcal{AS}$$ and $$\mathcal{CFS}$$, which arise in a natural way from geometric group theory. We study these properties in the Erdős–Rényi random graph model $${\mathcal G}(n,p)$$, proving the existence of a sharp threshold for a random graph to have the $$\mathcal{AS}$$ property asymptotically almost surely, and giving fairly tight bounds for the corresponding threshold for the $$\mathcal{CFS}$$ property. As an application of our results, we show that for any constant $$p$$ and any $$\Gamma\in{\mathcal G}(n,p)$$, the right-angled Coxeter group $$W_\Gamma$$ asymptotically almost surely has quadratic divergence and thickness of order $$1$$, generalizing and strengthening a result of Behrstock–Hagen–Sisto [8]. Indeed, we show that at a large range of densities a random right-angled Coxeter group has quadratic divergence. 1 Introduction In this article, we consider two properties of graphs motivated by geometric group theory. We show that these properties are typically present in random graphs. We repay the debt to geometric group theory by applying our (purely graph-theoretic) results to the large-scale geometry of Coxeter groups. Random graphs Let $${\mathcal G}(n,p)$$ be the random graph model on $$n$$ vertices obtained by including each edge independently at random with probability $$p=p(n)$$. The parameter $$p$$ is often referred to as the density of $${\mathcal G}(n,p)$$. The model $${\mathcal G}(n,p)$$ was introduced by Gilbert [23], and the resulting random graphs are usually referred to as the “Erdős–Rényi random graphs” in honor of Erdős and Rényi’s seminal contributions to the field, and we follow this convention. We say that a property $$\mathcal{P}$$ holds asymptotically almost surely (a.a.s.) in $${\mathcal G}(n,p)$$ if for $$\Gamma\in{\mathcal G}(n,p),$$ we have $$\mathbb{P}(\Gamma \in \mathcal{P})\rightarrow 1$$ as $$n\rightarrow \infty$$. In this article, we will be interested in proving that certain global properties hold a.a.s. in $${\mathcal G}(n,p)$$ both for a wide range of probabilities $$p=p(n)$$. A graph property is (monotone) increasing if it is closed under the addition of edges. A paradigm in the theory of random graphs is that global increasing graph properties exhibit sharp thresholds in $${\mathcal G}(n,p)$$: for many global increasing properties $$\mathcal{P}$$, there is a critical density$$p_c=p_c(n)$$ such that for any fixed $$\epsilon>0$$ if $$p<(1-\epsilon)p_c$$ then a.a.s. $$\mathcal{P}$$ does not hold in $${\mathcal G}(n,p)$$, while if $$p>(1+\epsilon)p_c$$ then a.a.s. $$\mathcal{P}$$ holds in $${\mathcal G}(n,p)$$. A quintessential example is the following classical result of Erdős and Rényi which provides a sharp threshold for connectedness: Theorem (Erdős–Rényi; [21]). There is a sharp threshold for connectivity of a random graph with critical density $$p_c(n)=\frac{\log(n)}{n}$$. □ The local structure of the Erdős–Rényi random graph is well understood, largely due to the assumption of independence between the edges. For example, Erdős–Rényi and others have obtained threshold densities for the existence of certain subgraphs in a random graph (see e.g., [21, Theorem 1, Corollaries 1–5]). In earlier applications of random graphs to geometric group theory, this feature of the model was successfully exploited in order to analyze the geometry of right-angled Artin and Coxeter groups presented by random graphs; this is notable, for example, in the work of Charney and Farber [14]. In particular, the presence of an induced square implies non-hyperbolicity of the associated right-angled Coxeter group [14, 21, 31]. In this article, we take a more global approach. Earlier work established a correspondence between some fundamental geometric properties of right-angled Coxeter groups and large-scale structural properties of the presentation graph, rather than local properties such as the presence or absence of certain specified subgraphs. The simplest of these properties is the property of being the join of two subgraphs that are not cliques. One large scale graph property relevant in the present context is a property studied in [8] which, roughly, says that the graph is constructed in a particular organized, inductive way from joins. In this article, we discuss a refined version of this property, $$\mathcal{CFS}$$, which is a slightly-modified version of a property introduced by Dani–Thomas [17]. We also study a stronger property, $$\mathcal{AS}$$, and show it is generic in random graphs for a large range of $$p(n)$$, up $$1-\omega(n^{-2})$$. $$\boldsymbol{\mathcal{AS}}$$ graphs The first class of graphs we study is the class of augmented suspensions, which we denote $$\mathcal{AS}$$. A graph is an augmented suspension if it contains an induced subgraph which is a suspension (see Section 2 for a precise definition of this term), and any vertex which is not in that suspension is connected by edges to at least two nonadjacent vertices of the suspension. Theorem 4.4 and 4.5 (Sharp Threshold for $$\boldsymbol{\mathcal{AS}}$$).Let $$\epsilon>0$$ be fixed. If $$p=p(n)$$ satisfies $$p\ge(1+\epsilon)\left(\displaystyle\frac{\log n}{n}\right)^{\frac{1}{3}}$$ and $$(1-p)n^2\to\infty$$, then $$\Gamma\in{\mathcal G}(n,p(n))$$ is a.a.s. in $$\mathcal{AS}$$. On the other hand, if $$p\le(1-\epsilon)\left(\displaystyle\frac{\log{n}}{n}\right)^{\frac{1}{3}}$$, then $$\Gamma\in{\mathcal G}(n,p)$$ a.a.s. does not lie in $$\mathcal{AS}$$. Intriguingly, Kahle proved that a function similar to the critical density in Theorem 4.4 is the threshold for a random simplicial complex to have vanishing second rational cohomology [28]. Remark (Behavior near $$p=1$$). Note that property $$\mathcal{AS}$$ is not monotone increasing, since it requires the presence of a number of non-edges. In particular, complete graphs are not in $$\mathcal{AS}$$. Thus unlike the global properties typically studied in the theory of random graphs, $$\mathcal{AS}$$ will cease to hold a.a.s. when the density $$p$$ is very close to $$1$$. In fact, [8, Theorem 3.9] shows that if $$p(n)=1-\Omega\left(\frac{1}{n^2}\right)$$, then a.a.s. $$\Gamma$$ is either a clique or a clique minus a fixed number of edges whose endpoints are all disjoint. Thus, with positive probability, $$\Gamma\in\mathcal{AS}$$. However, [14, Theorem 1] shows that if $$(1-p)n^2\to 0$$ then $$\Gamma$$ is asymptotically almost surely a clique, and hence not in $$\mathcal{AS}$$. □ $$\boldsymbol{\mathcal{CFS}}$$ graphs The second family of graphs, which we call $$\mathcal{CFS}$$ graphs (“Constructed From Squares”), arise naturally in geometric group theory in the context of the large–scale geometry of right–angled Coxeter groups, as we explain below and in Section 3. A special case of these graphs was introduced by Dani–Thomas to study divergence in triangle-free right-angled Coxeter groups [17]. The graphs we study are intimately related to a property called thickness, a feature of many key examples in geometric group theory and low dimensional topology that is, closely related to divergence, relative hyperbolicity, and a number of other topics. This property is, in essence, a connectivity property because it relies on a space being “connected” through sequences of “large” subspaces. Roughly speaking, a graph is $$\mathcal{CFS}$$ if it can be built inductively by chaining (induced) squares together in such a way that each square overlaps with one of the previous squares along a diagonal (see Section 2 for a precise definition). We explain in the next section how this class of graphs generalizes $$\mathcal{AS}$$. Our next result about genericity of $$\mathcal{CFS}$$ combines with Proposition 3.1 below to significantly strengthen [8, Theorem VI]. This result is an immediate consequence of Theorems 5.1 and 5.7, which, in fact, establish slightly more precise, but less concise, bounds. Theorem 5.1 and 5.7 Suppose $$(1-p)n^2\to\infty$$ and let $$\epsilon>0$$. Then $$\Gamma\in{\mathcal G}(n,p)$$ is a.a.s. in $$\mathcal{CFS}$$ whenever $$p(n)>n^{-\frac{1}{2}+\epsilon}$$. Conversely, $$\Gamma\in{\mathcal G}(n,p)$$ is a.a.s. not in $$\mathcal{CFS}$$, whenever $$p(n)<n^{-\frac{1}{2} -\epsilon}.$$ We actually show, in Theorem 5.1, that at densities above $$5\sqrt{\frac{\log n}{n}}$$, with $$(1-p)n^2\to\infty$$, the random graph is a.a.s. in $$\mathcal{CFS}$$, while in Theorem 5.7 we show a random graph a.a.s. not in $$\mathcal{CFS}$$ at densities below $$\frac{1}{\sqrt{n}\log{n}}$$. Theorem 5.1 applies to graphs in a range strictly larger than that in which Theorem 4.4 holds (though our proof of Theorem 5.1 relies on Theorem 4.4 to deal with the large $$p$$ case). Theorem 5.1 combines with Theorem 4.5 to show that, for densities between $$\left(\log n/n\right)^{\frac{1}{2}}$$ and $$\left(\log n/n\right)^{\frac{1}{3}}$$, a random graph is asymptotically almost surely in $$\mathcal{CFS}$$ but not in $$\mathcal{AS}$$. We also note that Babson–Hoffman–Kahle [3] proved that a function of order $$n^{-\frac{1}{2}}$$ appears as the threshold for simple-connectivity in the Linial–Meshulam model for random 2–complexes [29]. It would be interesting to understand whether there is a connection between genericity of the $$\mathcal{CFS}$$ property and the topology of random 2-complexes. Unlike our results for the $$\mathcal{AS}$$ property, we do not establish a sharp threshold for the $$\mathcal{CFS}$$ property. In fact, we believe that neither the upper nor lower bounds, given in Theorem 5.1 and Theorem 5.7, for the critical density around which $$\mathcal{CFS}$$ goes from a.a.s. not holding to a.a.s. holding are sharp. Indeed, we believe that there is a sharp threshold for the $$\mathcal{CFS}$$ property located at $$p_c(n)=\theta(n^{-\frac{1}{2}})$$. This conjecture is linked to the emergence of a giant component in the “square graph” of $$\Gamma$$ (see the next section for a definition of the square graph and the heuristic discussion after the proof of Theorem 5.7). Applications to geometric group theory Our interest in the structure of random graphs was sparked largely by questions about the large-scale geometry of right-angled Coxeter groups. Coxeter groups were first introduced in [15] as a generalization of reflection groups, that is, discrete groups generated by a specified set of reflections in Euclidean space. A reflection group is right-angled if the reflection loci intersect at right angles. An abstract right-angled Coxeter group generalizes this situation: it is defined by a group presentation in which the generators are involutions and the relations are obtained by declaring some pairs of generators to commute. Right-angled Coxeter groups (and more general Coxeter groups) play an important role in geometric group theory and are closely-related to some of that field’s most fundamental objects, for example, CAT(0) cube complexes [18, 27, 33] and (right-angled) Bruhat-Tits buildings (see e.g., [18]). A right-angled Coxeter group is determined by a unique finite simplicial presentation graph: the vertices correspond to the involutions generating the group, and the edges encode the pairs of generators that commute. In fact, the presentation graph uniquely determines the right-angled Coxeter group [32]. In this article, as an application of our results on random graphs, we continue the project of understanding large-scale geometric features of right-angled Coxeter groups in terms of the combinatorics of the presentation graph, begun in [8, 14, 17]. Specifically, we study right-angled Coxeter groups defined by random presentation graphs, focusing on the prevalence of two important geometric properties: relative hyperbolicity and thickness. Relative hyperbolicity, in the sense introduced by Gromov and equivalently formulated by many others [10, 22, 24, 35], when it holds, is a powerful tool for studying groups. On the other hand, thickness of a finitely-generated group (more generally, a metric space) is a property introduced by Behrstock–Druţu–Mosher in [6] as a geometric obstruction to relative hyperbolicity and has a number of powerful geometric applications. For example, thickness gives bounds on divergence (an important quasi-isometry invariant of a metric space) in many different groups and spaces [5, 7, 11, 17, 37]. Thickness is an inductive property: in the present context, a finitely generated group $$G$$ is thick of order $$0$$ if and only if it decomposes as the direct product of two infinite subgroups. The group $$G$$ is thick of order $$n$$ if there exists a finite collection $$\mathcal H$$ of undistorted subgroups of $$G$$, each thick of order $$n-1$$, whose union generates a finite-index subgroup of $$G$$ and which has the following “chaining” property: for each $$g,g'\in G$$, one can construct a sequence $$g\in g_1H_1,g_2H_2,\ldots,g_kH_k\ni g'$$ of cosets, with each $$H_i\in\mathcal H$$, so that consecutive cosets have infinite coarse intersection. Many of the best-known groups studied by geometric group theorists are thick, and indeed thick of order $$1$$: one-ended right-angled Artin groups, mapping class groups of surfaces, outer automorphism groups of free groups, fundamental groups of three-dimensional graph manifolds, etc. [6]. The class of Coxeter groups contains many examples of hyperbolic and relatively hyperbolic groups. There is a criterion for hyperbolicity purely in terms of the presentation graph due to Moussong [31] and an algebraic criterion for relative hyperbolicity due to Caprace [13]. The class of Coxeter groups includes examples which are non-relatively hyperbolic, for instance, those constructed by Davis–Januszkiewicz [19] and, also, ones studied by Dani–Thomas [17]. In fact, in [8], this is taken further: it is shown that every Coxeter group is actually either thick, or hyperbolic relative to a canonical collection of thick Coxeter subgroups. Further, there is a simple, structural condition on the presentation graph, checkable in polynomial time, which characterizes thickness. This result is needed to deduce the applications below from our graph theoretic results. Charney and Farber initiated the study of random graph products (including right-angled Artin and Coxeter groups) using the Erdős-Rényi model of random graphs [14]. The structure of the group cohomology of random graph products was obtained in [20]. In [8], various results are proved about which random graphs have the thickness property discussed above, leading to the conclusion that, at certain low densities, random right-angled Coxeter groups are relatively hyperbolic (and thus not thick), while at higher densities, random right-angled Coxeter groups are thick. In this article, we improve significantly on one of the latter results, and also prove something considerably more refined: we isolate not just thickness of random right-angled Coxeter groups, but thickness of a specified order, namely $$1$$: Corollary 3.2 (Random Coxeter groups are thick of order 1.)There exists a constant $$C>0$$ such that if $$p\colon{\mathbb{N}}\rightarrow(0,1)$$ satisfies $$\left(\displaystyle\frac{C\log{n}}{n}\right)^{\frac{1}{2}}\leq p(n)\leq 1-\displaystyle\frac{(1+\epsilon)\log{n}}{n}$$ for some $$\epsilon>0$$, then the random right-angled Coxeter group $$W_{G_{n,p}}$$ is asymptotically almost surely thick of order exactly $$1$$, and in particular has quadratic divergence. Corollary 3.2 significantly improves on Theorem 3.10 of [8], as discussed in Section 3. This theorem follows from Theorems 5.1 and 4.4, the latter being needed to treat the case of large $$p(n)$$, including the interesting special case in which $$p$$ is constant. Remark 1.1. We note that characterizations of thickness of right-angled Coxeter groups in terms of the structure of the presentation graph appear to generalize readily to graph products of arbitrary finite groups and, probably, via the action on a cube complex constructed by Ruane and Witzel in [36], to arbitrary graph products of finitely generated abelian groups, using appropriate modifications of the results in [8]. □ Organization of the article In Section 2, we give the formal definitions of $$\mathcal{AS}$$ and $$\mathcal{CFS}$$ and introduce various other graph-theoretic notions we will need. In Section 3, we discuss the applications of our random graph results to geometric group theory, in particular, to right-angled Coxeter groups and more general graph products. In Section 4, we obtain a sharp threshold result for $$\mathcal{AS}$$ graphs. Section 5 is devoted to $$\mathcal{CFS}$$ graphs. Finally, Section 6 contains some simulations of random graphs with density near the threshold for $$\mathcal{AS}$$ and $$\mathcal{CFS}$$. 2 Definitions Convention 2.1 A graph is a pair of finite sets $$\Gamma=(V,E)$$, where $$V=V(\Gamma)$$ is a set of vertices, and $$E=E(\Gamma)$$ is a collection of pairs of distinct elements of $$V$$, which constitute the set of edges of $$G$$. A subgraph of $$\Gamma$$ is a graph $$\Gamma'$$ with $$V(\Gamma')\subseteq V(\Gamma)$$ and $$E(\Gamma')\subseteq E(\Gamma)$$; $$\Gamma'$$ is said to be an induced subgraph of $$\Gamma$$ if $$E(\Gamma')$$ consists exactly of those edges from $$E(\Gamma)$$ whose vertices lie in $$V(\Gamma')$$. In this article, we focus on induced subgraphs, and we generally write “subgraph” to mean “induced subgraph”. In particular, we often identify a subgraph with the set of vertices inducing it, and we write $$\vert \Gamma\vert$$ for the order of $$\Gamma$$, that is, the number of vertices it contains. A clique of size $$t$$ is a complete graph on $$t\geq0$$ vertices. This includes the degenerate case of the empty graph on $$t= 0$$ vertices. □ Fig. 1. View largeDownload slide A graph in $$\mathcal{AS}$$. A block exhibiting inclusion in $$\mathcal{AS}$$ is shown in bold; the two (left-centrally located) ends of the bold block are highlighted. (A colour version of this figure is available from the journal’s website.) Fig. 1. View largeDownload slide A graph in $$\mathcal{AS}$$. A block exhibiting inclusion in $$\mathcal{AS}$$ is shown in bold; the two (left-centrally located) ends of the bold block are highlighted. (A colour version of this figure is available from the journal’s website.) Definition 2.2 (Link, join). Given a graph $$\Gamma$$, the link of a vertex $$v\in\Gamma$$, denoted $${\mathrm Lk}_\Gamma(v)$$, is the subgraph spanned by the set of vertices adjacent to $$v$$. Given graphs $$A,B$$, the join$$A\star B$$ is the graph formed from $$A\sqcup B$$ by joining each vertex of $$A$$ to each vertex of $$B$$ by an edge. A suspension is a join where one of the factors $$A,B$$ is the graph consisting of two vertices and no edges. □ We now describe a family of graphs, denoted $$\mathcal{CFS}$$, which satisfy the global structural property that they are “constructed from squares.” Definition 2.3 ($$\boldsymbol{\mathcal{CFS}}$$). Given a graph $$\Gamma$$, let $$\square(\Gamma)$$ be the auxiliary graph whose vertices are the induced $$4$$–cycles from $$\Gamma$$, with two distinct $$4$$–cycles joined by an edge in $$\square(\Gamma)$$ if and only if they intersect in a pair of non-adjacent vertices of $$\Gamma$$ (i.e., in a diagonal). We refer to $$\square(\Gamma)$$ as the square-graph of $$\Gamma$$. A graph $$\Gamma$$ belongs to $$\mathcal{CFS}$$ if $$\Gamma=\Gamma'\star K$$, where $$K$$ is a (possibly empty) clique and $$\Gamma'$$ is a non-empty subgraph such that $$\square(\Gamma')$$ has a connected component $$C$$ such that the union of the $$4$$–cycles from $$C$$ covers all of $$V(\Gamma')$$. Given a vertex $$F\in\square(\Gamma)$$, we refer to the vertices in the $$4$$–cycle in $$\Gamma$$ associated to $$F$$ as the support of $$F$$. □ Remark 2.4. Dani–Thomas introduced component with full support graphs in [17], a subclass of the class of triangle-free graphs. We note that each component with full support graph is constructed from squares, but the converse is not true. Indeed, since we do not require our graphs to be triangle-free, our definition necessarily only counts induced 4–cycles and allows them to intersect in more ways than in [17]. This distinction is relevant to the application to Coxeter groups, which we discuss in Section 3. □ Definition 2.5 (Augmented suspension). The graph $$\Gamma$$ is an augmented suspension if it contains an induced subgraph $$B=\{w,w'\}\star \Gamma'$$, where $$w,w'$$ are nonadjacent and $$\Gamma'$$ is not a clique, satisfying the additional property that if $$v\in\Gamma-B$$, then $$\mathrm{Lk}_\Gamma(v)\cap \Gamma'$$ is not a clique. Let $$\mathcal{AS}$$ denote the class of augmented suspensions. Figure 1 shows a graph in $$\mathcal{AS}$$. □ Remark 2.6. Neither the $$\mathcal{CFS}$$ nor the $$\mathcal{AS}$$ properties introduced above are monotone with respect to the addition of edges. This stands in contrast to the most commonly studied global properties of random graphs. □ Definition 2.7 (Block, core, ends). A block in $$\Gamma$$ is a subgraph of the form $$B(w,w')=\{w,w'\}\star \Gamma'$$ where $$\{w,w'\}$$ is a pair of non-adjacent vertices and $$\Gamma'\subset \Gamma$$ is a subgraph of $$\Gamma$$ induced by a set of vertices adjacent to both $$w$$ and $$w'$$. A block is maximal if $$V(\Gamma')=\mathrm{Lk}_{\Gamma}(w)\cap \mathrm{Lk}_{\Gamma}(w')$$. Given a block $$B=B(w,w')$$, we refer to the non-adjacent vertices $$w,w'$$ as the ends of $$B$$, denoted $$\mathrm{end}(B)$$, and the vertices of $$\Gamma'$$ as the core of $$B$$, denoted $$\mathrm{core}(B)$$. □ Note that $$\mathcal{AS}\subsetneq\mathcal{CFS}$$, indeed Theorems 5.1 and 4.5 show that there must exist graphs in $$\mathcal{CFS}$$ that are not in $$\mathcal{AS}$$. Here we explain how any graph in $$\mathcal{AS}$$ is in $$\mathcal{CFS}$$. Lemma 2.8. Let $$\Gamma$$ be a graph in $$\mathcal{AS}$$. Then $$\Gamma \in \mathcal{CFS}$$. □ Proof Let $$B(w,w')=\{w,w'\}\star \Gamma'$$ be a maximal block in $$\Gamma$$ witnessing $$\Gamma \in \mathcal{AS}$$. Write $$\Gamma'=A\star D$$, where $$D$$ is the collection of all vertices of $$\Gamma'$$ which are adjacent to every other vertex of $$\Gamma'$$. Note that $$D$$ induces a clique in $$\Gamma$$. By definition of the $$\mathcal{AS}$$ property $$\Gamma'$$ is not a clique, whence, $$A$$ contains at least one pair of non-adjacent vertices. Furthermore, by the definition of $$D$$, for every vertex $$a\in A$$ there exists $$a' \in A$$ with $$\{a,a'\}$$ non-adjacent. The $$4$$–cycles induced by $$\{w,w'a,a'\}$$ for non-adjacent pairs $$a, a'$$ from $$A$$ are connected in $$\square(\Gamma)$$. Denote the component of $$\square(\Gamma)$$ containing them by $$C$$. Consider now a vertex $$v\in \Gamma - B(w,w')$$. Since $$B(w,w')$$ is maximal, we have that at least one of $$w,w'$$ is not adjacent to $$v$$ — without loss of generality, let us assume that it is $$w$$. By the $$\mathcal{AS}$$ property, $$v$$ must be adjacent to a pair $$a,a'$$ of non-adjacent vertices from $$A$$. Then $$\{w, a,a', v\}$$ induces a $$4$$–cycle, which is adjacent to $$\{w,w',a,a'\}\in\square(\Gamma)$$ and hence lies in $$C$$. Finally, consider a vertex $$d \in D$$. If $$v$$ is adjacent to all vertices of $$\Gamma$$, then $$\Gamma$$ is the join of a graph with a clique containing $$d$$, and we can ignore $$d$$ with respect to establishing the $$\mathcal{CFS}$$ property. Otherwise $$d$$ is not adjacent to some $$v\in \Gamma-B(w,w')$$. By the $$\mathcal{AS}$$ property, $$v$$ is connected by edges to a pair of non-adjacent vertices $$\{a,a'\}$$ from $$A$$. Thus $$\{d,a,a',v\}$$ induces a $$4$$–cycle. Since (as established above) there is some $$4$$–cycle in $$C$$ containing $$\{a,a',v\}$$, we have that $$\{d,a,a',v\} \in C$$ as well. Thus $$\Gamma= \Gamma'' \star K$$, where $$K$$ is a clique and $$V(\Gamma'')$$ is covered by the union of the $$4$$–cycles in $$C$$, so that $$\Gamma \in \mathcal{CFS}$$ as claimed. ■ 3 Geometry of right-angled Coxeter groups If $$\Gamma$$ is a finite simplicial graph, the right-angled Coxeter group$$W_\Gamma$$presented by$$\Gamma$$ is the group defined by the presentation ⟨Vert(Γ)∣{w2,uvu−1v−1:u,v,w∈Vert(Γ),{u,v}∈Edge(Γ)⟩. A result of Mühlherr [32] shows that the correspondence $$\Gamma\leftrightarrow W_\Gamma$$ is bijective. We can thus speak of “the random right-angled Coxeter group” — it is the right-angled Coxeter group presented by the random graph. (We emphasize that the above presentation provides the definition of a right-angled Coxeter group: this definition abstracts the notion of a reflection group – a subgroup of a linear group generated by reflections — but infinite Coxeter groups need not admit representations as reflection groups.) Recent articles have discussed the geometry of Coxeter groups, especially relative hyperbolicity and closely-related quasi-isometry invariants like divergence and thickness, cf. [8, 13, 17]. In particular, Dani–Thomas introduced a property they call having a component of full support for triangle-free graphs (which is exactly the triangle-free version of $$\mathcal{CFS}$$) and they prove that under the assumption $$\Gamma$$ is triangle-free, $$W_\Gamma$$ is thick of order at most $$1$$ if and only if it has quadratic divergence if and only if $$\Gamma$$ is in $$\mathcal{CFS}$$, see [17, Theorem 1.1 and Remark 4.8]. Since the densities where random graphs are triangle-free are also square-free (and thus not $$\mathcal{CFS}$$ — in fact, they are disconnected!), we need the following slight generalization of the result of Dani–Thomas: Proposition 3.1. Let $$\Gamma$$ be a finite simplicial graph. If $$\Gamma$$ is in $$\mathcal{CFS}$$ and $$\Gamma$$ does not decompose as a nontrivial join, then $$W_\Gamma$$ is thick of order exactly $$1$$. □ Proof Theorem II of [8] shows immediately that, if $$\Gamma\in\mathcal{CFS}$$, then $$W_\Gamma$$ is thick, being formed by a series of thick unions of $$4$$–cycles; since each $$4$$–cycle is a join, it follows that $$\Gamma$$ is thick of order at most $$1$$. On the other hand, [8, Proposition 2.11] shows that $$W_\Gamma$$ is thick of order at least $$1$$ provided $$\Gamma$$ is not a join. ■ Our results about random graphs yield: Corollary 3.2. There exists $$k>0$$ so that if $$p\colon{\mathbb{N}}\rightarrow(0,1)$$ and $$\epsilon>0$$ are such that $$\sqrt{\frac{k\log n}{n}}\leq p(n)\leq 1-\displaystyle\frac{(1-\epsilon)\log{n}}{n}$$ for all sufficiently large $$n$$, then for $$\Gamma\in{\mathcal G}(n,p)$$ the group $$W_\Gamma$$ is asymptotically almost surely thick of order exactly $$1$$ and hence has quadratic divergence. □ Proof Theorem 5.1 shows that any such $$\Gamma$$ is asymptotically almost surely in $$\mathcal{CFS}$$, whence $$W_\Gamma$$ is thick of order at most $$1$$. We emphasize that to apply this result for sufficiently large functions $$p(n)$$ the proof of Theorem 5.1 requires an application of Theorem 4.4 to establish that $$\Gamma$$ is a.a.s. in $$\mathcal{AS}$$ and hence in $$\mathcal{CFS}$$ by Lemma 2.8. By Proposition 3.1, to show that the order of thickness is exactly one, it remains to rule out the possibility that $$\Gamma$$ decomposes as a nontrivial join. However, this occurs if and only if the complement graph is disconnected, which asymptotically almost surely does not occur whenever $$p(n)\le 1-\frac{(1-\epsilon)\log{n}}{n}$$, by the sharp threshold for connectivity of $${\mathcal G}(n,1-p)$$ established by Erdős and Rényi in [21]. Since this holds for $$p(n)$$ by assumption, we conclude that asymptotically almost surely, $$W_\Gamma$$ is thick of order at least $$1$$. Since $$W_\Gamma$$ is CAT(0) and thick of order exactly $$1$$, the consequence about divergence now follows from [5]. ■ This corollary significantly generalizes Theorem 3.10 of [8], which established that, if $$\Gamma\in{\mathcal G}(n,\frac{1}{2})$$, then $$W_\Gamma$$ is asymptotically almost surely thick. Theorem 3.10 of [8] does not provide effective bounds on the order of thickness and its proof is significantly more complicated than the proof of Corollary 3.2 given above — indeed, it required several days of computation (using 2013 hardware) to establish the base case of an inductive argument. Remark 3.3 (Higher-order thickness). A lower bound of $$p(n)=n^{-\frac{5}{6}}$$ for membership in a larger class of graphs whose corresponding Coxeter groups are thick can be found in [8, Theorem 3.4]. In fact, this argument can be adapted to give a simple proof that a.a.s. thickness does not occur at densities below $$n^{-\frac{3}{4}}$$. The correct threshold for a.a.s. thickness is, however, unknown. □ Remark 3.4 (Random graph products versus random presentations). Corollary 3.2 and Remark 3.3 show that the random graph model for producing random right-angled Coxeter groups generates groups with radically different geometric properties. This is in direct contrast to other methods of producing random groups, most notably Gromov’s random presentation model [25, 26] where, depending on the density of relators, groups are either almost surely hyperbolic or finite (with order at most $$2$$). This contrast speaks to the merits of considering a random right-angled Coxeter group as a natural place to study random groups. For instance, Calegari–Wilton recently showed that in the Gromov model a random group contains many subgroups which are isomorphic to the fundamental group of a compact hyperbolic 3–manifold [12]; does the random right-angled Coxeter group also contain such subgroups? Right-angled Coxeter groups, and indeed thick ones, are closely related to Gromov’s random groups in another way. When the parameter for a Gromov random groups is $$<\frac{1}{6}$$ such a group is word-hyperbolic [25] and acts properly and cocompactly on a CAT(0) cube complex [34]. Hence the Gromov random group virtually embeds in a right-angled Artin group [4]. Moreover, at such parameters such a random group is one-ended [16], whence the associated right-angled Artin group is as well. By [4] this right-angled Artin group is thick of order 1. Since any right-angled Artin group is commensurable with a right-angled Coxeter group [19], one obtains a thick of order $$1$$ right-angled Coxeter group containing the randomly presented group. □ 4 Genericity of $$\boldsymbol{\mathcal{AS}}$$ We will use the following standard Chernoff bounds (see e.g., [2, Theorems A.1.11 and A.1.13]): Lemma 4.1 (Chernoff bounds). Let $$X_1,\ldots,X_n$$ be independent identically distributed random variables taking values in $$\{0,1\}$$, let $$X$$ be their sum, and let $$\mu=\mathbb E[X]$$. Then for any $$\delta\in(0,2/3)$$ P(|X−μ|≥δμ)≤2e−δ2μ3. □ Corollary 4.2. Let $$\varepsilon, \delta>0$$ be fixed. (i) If $$p(n)\ge \left(\frac{(6+\varepsilon)\log n}{\delta^2n}\right)^{1/2}$$, then a.a.s. for all pairs of distinct vertices $$\{x,y\}$$ in $$\Gamma \in {\mathcal G}(n,p)$$ we have $$\left\vert \vert \mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)\vert-p^2(n-2)\right\vert < \delta p^2(n-2)$$. (ii) If $$p(n)\ge \left(\frac{(9+\varepsilon)\log n}{\delta^2n}\right)^{1/3}$$, then a.a.s. for all triples of distinct vertices $$\{x,y, z\}$$ in $$\Gamma \in {\mathcal G}(n,p)$$ we have $$\left\vert \vert \mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)\cap \mathrm{Lk}_{\Gamma}(z)\vert-p^3(n-3)\right\vert < \delta p^3(n-3)$$. □ Proof For (i), let $$\{x,y\}$$ be any pair of distinct vertices. For each vertex $$v \in \Gamma-\{x,y\}$$, set $$X_v$$ to be the indicator function of the event that $$v\in \mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)$$, and set $$X=\sum_v X_v$$ to be the size of $$\mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)$$. We have $$\mathbb{E}X=p^2(n-2)$$ and so by the Chernoff bounds above, $$\Pr\left(\vert X-p^2(n-2)\vert\geq \delta p^2(n-2)\right) \leq 2e^{-\frac{\delta^2p^2(n-2)}{3}}$$. Applying Markov’s inequality, the probability that there exists some “bad pair” $$\{x,y\}$$ in $$\Gamma$$ for which $$\vert\mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)\vert$$ deviates from its expected value by more than $$\delta p^2(n-2)$$ is at most (n2)2e−δ2p2(n−2)3=o(1), provided $$\delta^2 p^2n\ge (6+\varepsilon) \log n$$ and $$\varepsilon, \delta>0$$ are fixed. Thus for this range of $$p=p(n)$$, a.a.s. no such bad pair exists. The proof of (ii) is nearly identical. ■ Lemma 4.3. (i) Suppose $$1-p\geq \frac{\log n}{2n}$$. Then asymptotically almost surely, the order of a largest clique in $$\Gamma\in{\mathcal G}(n,p)$$ is $$o(n)$$. (ii) Let $$\eta$$ be fixed with $$0<\eta<1$$. Suppose $$1-p\geq \eta$$. Then asymptotically almost surely, the order of a largest clique in $$\Gamma \in {\mathcal G}(n,p)$$ is $$O(\log n)$$. □ Proof For (i), set $$r=\alpha n$$, for some $$\alpha$$ bounded away from $$0$$. Write $$H(\alpha)=\alpha\log\frac{1}{\alpha}+(1-\alpha)\log\frac{1}{1-\alpha}$$. Using the standard entropy bound $$\binom{n}{\alpha n}\leq e^{H(\alpha)n}$$ and our assumption for $$(1-p)$$, we see that the expected number of $$r$$-cliques in $$\Gamma$$ is (nr)p(r2) ≤eH(α)nelog(1−(1−p))(α2n22+O(n))≤exp(−α22nlogn+O(n))=o(1). Thus by Markov’s inequality, a.a.s. $$\Gamma$$ does not contain a clique of size $$r$$, and the order of a largest clique in $$\Gamma$$ is $$o(n)$$. The proof of (ii) is similar: suppose $$1-p>\eta$$. Then for any $$r\leq n$$, (nr)p(r2) <nr(1−η)r(r−1)/2=exp(r(logn−r−12log11−η)), which for $$\eta>0$$ fixed and $$r-1>\frac{2}{\log (1/(1-\eta))}(1+ \log n )$$ is as most $$n^{-\frac{2}{\log (1/(1-\eta))}}=o(1)$$. We may thus conclude as above that a.a.s. a largest clique in $$\Gamma$$ has order $$O(\log n)$$. ■ Theorem 4.4 (Genericity of $$\mathcal{AS}$$). Suppose $$p(n)\ge(1+\epsilon)\left(\displaystyle\frac{\log{n}}{n}\right)^{\frac{1}{3}}$$ for some $$\epsilon>0$$ and $$(1-p)n^2\to\infty$$. Then, a.a.s. $$\Gamma\in{\mathcal G}(n,p)$$ is in $$\mathcal{AS}$$. □ Proof Let $$\delta>0$$ be a small constant to be specified later (the choice of $$\delta$$ will depend on $$\epsilon$$). By Corollary 4.2 (i) for $$p(n)$$ in the range we are considering, a.a.s. all joint links have size at least $$(1-\delta)p^2(n-2)$$. Denote this event by $$\mathcal{E}_1$$. We henceforth condition on $$\mathcal{E}_1$$ occurring (not this only affects the values of probabilities by an additive factor of $$\mathbb{P}(\mathcal{E}_1^c)=O(n^{-\varepsilon})=o(1)$$). With probability $$1-p^{\binom{n}{2}}=1-o(1)$$, $$\Gamma$$ is not a clique, whence, there there exist non-adjacent vertices in $$\Gamma$$. We henceforth assume $$\Gamma\neq K_n$$, and choose $$v_1, v_2\in \Gamma$$ which are not adjacent. Let $$B$$ be the maximal block associated with the pair $$(v_1,v_2)$$. We separate the range of $$p$$ into three. Case 1: $$\boldsymbol{p}$$ is “far” from both the threshold and $$1$$. Let $$\alpha>0$$ be fixed, and suppose $$\alpha n^{-1/4}\leq p \leq 1- \frac{\log n}{2n}$$. Let $$\mathcal{E}_2$$ be the event that for every vertex $$v\in\Gamma-B$$ the set $$\mathrm{Lk}_{\Gamma}(v)\cap B$$ has size at least $$\frac{1}{2}p^3(n-3)$$. By Corollary 4.2, a.a.s. event $$\mathcal{E}_2$$ occurs, that is, all vertices in $$\Gamma-B$$ have this property. We claim that a.a.s. there is no clique of order at least $$\frac{1}{2}p^3(n-3)$$ in $$\Gamma$$. Indeed, if $$p<1-\eta$$ for some fixed $$\eta>0$$, then by Lemma 4.3 part (ii), a largest clique in $$\Gamma$$ has order $$O(\log n)=o(p^3n)$$. On the other hand, if $$1-\eta <p\leq 1- \frac{\log n}{2n}$$, then by Lemma 4.3 part (i), a largest clique in $$\Gamma$$ has order $$o(n)=o(p^3n)$$. Thus in either case a.a.s. for every$$v\in \Gamma-B$$, $$\mathrm{Lk}_{\Gamma}(v)\cap B$$ is not a clique and hence $$v \in \overline{B}$$, so that a.a.s. $$\overline{B}=\Gamma$$, and $$\Gamma \in \mathcal{AS}$$ as required. Case 2: $$\boldsymbol{p}$$ is “close” to the threshold. Suppose that $$(1+\epsilon)\left(\displaystyle\frac{\log{n}}{n}\right)^{\frac{1}{3}}\le p(n)$$ and $$np^4\to 0$$. Let $$\vert B\vert=m+2$$. By our conditioning, we have $$(1-\delta)(n-2)p^2\leq m \leq (1+\delta)(n-2)p^2$$. The probability that a given vertex $$v\in \Gamma$$ is not in $$\overline{B}$$ is given by: P(v∉B¯|{|B|=m})=(1−p)m+mp(1−p)m+1+∑r=2m(mr)pr(1−p)m−rp(r2). (1) In this equation, the first two terms come from the case where $$v$$ is connected to $$0$$ and $$1$$ vertex in $$B\setminus\{v_1,v_2\}$$ respectively, while the third term comes from the case where the link of $$v$$ in $$B\setminus\{v_1,v_2\}$$ is a clique on $$r\geq 2$$ vertices. As we shall see, in the case $$np^4\to 0$$ which we are considering, the contribution from the first two terms dominates. Let us estimate their order: (1−p)m+mp(1−p)m−1=(1+mp1−p)(1−p)m ≤(1+mp1−p)e−mp ≤(1+(1−δ)(1+ϵ)3logn1−p)n−(1−δ)(1+ϵ)3. Taking $$\delta<1-\frac{1}{(1+\epsilon)^3}$$ this expression is $$o(n^{-1})$$. We now treat the sum making up the remaining terms in Equation 1. To do so, we will analyze the quotient of successive terms in the sum. Fixing $$2\le r\le m-1$$ we see: (mr+1)pr+1(1−p)m−r−1p(r+12)(mr)pr(1−p)m−rp(r2)=m−r−1r+1⋅pr+11−p≤mpr+1≤mp3. Since $$np^4\to 0$$ (by assumption), this also tends to zero as $$n\to\infty$$. The quotients of successive terms in the sum thus tend to zero uniformly as $$n\to \infty$$, and we may bound the sum by a geometric series: ∑r=2m(mr)pr(1−p)m−rp(r2)≤(m2)p3(1−p)m−2∑i=0m−2(mp3)i≤(12+o(1))m2p3(1−p)m−2. Now, $$m^2p^3(1-p)^{m-2}=\frac{mp^2}{1-p}\cdot mp(1-p)^{m-1}$$. The second factor in this expression was already shown to be $$o(n^{-1})$$, while $$mp^2\le(1+\delta)np^4 \to 0$$ by assumption, so the total contribution of the sum is $$o(n^{-1})$$. Thus for any value of $$m$$ between $$(1-\delta)p^2(n-2)$$ and $$(1+\delta)p^2(n-2)$$, the right hand side of Equation (1) is $$o(n^{-1})$$, and we conclude: P(v∉B¯|E1)≤o(n−1). Thus, by Markov’s inequality, P(B¯=Γ)≥P(E1)(1−∑vP(v∉B¯|E1))=1−o(1), establishing that a.a.s. $$\Gamma\in\mathcal{AS}$$, as claimed. Case 3: $$p$$ is “close” to $$1$$. Suppose $$n^{-2}\ll (1-p)\leq \frac{\log n}{2n}$$. Consider the complement of $$\Gamma$$, $$\Gamma^c\in {\mathcal G}(n, 1-p)$$. In the range of the parameter $$\Gamma^c$$ a.a.s. has at least two connected components that contain at least two vertices. In particular, taking complements, we see that $$\Gamma$$ is a.a.s. a join of two subgraphs, neither of which is a clique. It is a simple exercise to see that such as graph is in $$\mathcal{AS}$$, thus a.a.s. $$\Gamma\in\mathcal{AS}$$. ■ As we now show, the bound obtained in the above theorem is actually a sharp threshold. Analogous to the classical proof of the connectivity threshold [21], we consider vertices which are “isolated” from a block to prove that graphs below the threshold strongly fail to be in $$\mathcal{AS}$$. Theorem 4.5. If $$p\le\left(1-\epsilon\right)\left(\displaystyle\frac{\log{n}}{n}\right)^{\frac{1}{3}}$$ for some $$\epsilon>0$$, then $$\Gamma\in{\mathcal G}(n,p)$$ is asymptotically almost surely not in $$\mathcal{AS}$$. □ Proof We will show that, for $$p$$ as hypothesized, every block has a vertex “isolated” from it. Explicitly, let $$\Gamma\in{\mathcal G}(n,p)$$ and consider $$B=B_{v,w}=\mathrm{Lk}(v)\cap \mathrm{Lk}(w)\cup\{v,w\}$$. Let $$X(v,w)$$ be the event that every vertex of $$\Gamma-B$$ is connected by an edge to some vertex of $$B$$. Clearly $$\Gamma\in \mathcal{AS}$$ only if the event $$X(v,w)$$ occurs for some pair of non-adjacent vertices $$\{v,w\}$$. Set $$X=\bigcup_{\{v,w\}} X(v,w)$$. Note that $$X$$ is a monotone event, closed under the addition of edges, so that the probability it occurs in $$\Gamma\in {\mathcal G}(n,p)$$ is a non-decreasing function of $$p$$. We now show that when $$p=\left(1-\epsilon\right)\left(\log{n}/ n\right)^{\frac{1}{3}}$$, a.a.s. $$X$$ does not occur, completing the proof. Consider a pair of vertices $$\{v,w\}$$, and set $$k=\vert B_{v,w}\vert$$. Conditional on $$B_{v,w}$$ having this size and using the standard inequality $$(1-x)\le e^{-x}$$, we have that P(X(v,w))=(1−(1−p)k)n−k≤e−(n−k)(1−p)k. Now, the value of $$k$$ is concentrated around its mean: by Corollary 4.2, for any fixed $$\delta>0$$ and all $$\{v,w\}$$, the order of $$B_{v,w}$$ is a.a.s. at most $$(1+\delta)np^2$$. Conditioning on this event $$\mathcal{E}$$, we have that for any pair of vertices $$v,w$$, P(X(v,w)|E)≤maxk≤(1+δ)np2e−(n−k)(1−p)k=e−(n−(1+δ)np2)(1−p)(1+δ)np2. Now $$(1-p)^{(1+\delta)np^2} = e^{(1+\delta)np^2\log(1-p)}$$ and by Taylor’s theorem $$\log(1-p)=-p+O(p^2)$$, so that: P(X(v,w)|E)≤e−n(1+O(p2))e−(1+δ)np3(1+O(p))=e−n1−(1+δ)(1−ϵ)3+o(1). Choosing $$\delta<\displaystyle\frac{1}{(1-\epsilon)^3}-1$$, the expression above is $$o(n^{-2})$$. Thus P(X) ≤P(Ec)+∑{v,w}P(X(v,w)|E) =o(1)+(n2)o(n−2)=o(1). Thus a.a.s. the monotone event $$X$$ does not occur in $$\Gamma\in{\mathcal G}(n,p)$$ for $$p=\left(1-\epsilon\right)\left(\log{n}/ n\right)^{\frac{1}{3}}$$, and hence a.a.s. the property $$\mathcal{AS}$$ does not hold for $$\Gamma\in{\mathcal G}(n,p)$$ and $$p(n)\leq \left(1-\epsilon\right)\left(\log{n}/ n\right)^{\frac{1}{3}}$$. ■ 5 Genericity of $$\boldsymbol{\mathcal{CFS}}$$ The two main results in this section are upper and lower bounds for inclusion in $$\mathcal{CFS}$$. These results are established in Theorems 5.1 and 5.7. Theorem 5.1. If $$p\colon{\mathbb{N}}\rightarrow(0,1)$$ satisfies $$(1-p)n^2\to\infty$$ and $$p(n)\geq 5\sqrt{\frac{\log n}{n}}$$ for all sufficiently large $$n$$, then a.a.s. $$\Gamma \in {\mathcal G}(n,p)$$ lies in $$\mathcal{CFS}$$. □ The proof of Theorem 5.1 divides naturally into two ranges. First of all for large $$p$$, namely for $$p(n)\geq 2\left(\log{n}/n\right)^{\frac{1}{3}}$$, we appeal to Theorem 4.4 to show that a.a.s. a random graph $$\Gamma\in {\mathcal G}(n,p)$$ is in $$\mathcal{AS}$$ and hence, by Lemma 2.8, in $$\mathcal{CFS}$$. In light of our proof of Theorem 4.4, we may think of this as the case when we can “beam up” every vertex of the graph $$\Gamma$$ to a single block $$B_{x,y}$$ in an appropriate way, and thus obtain a connected component of $$\square(\Gamma)$$ whose support is all of $$V(\Gamma)$$ Secondly we have the case of “small $$p$$” where 5lognn≤p(n)≤2(lognn)13, which is the focus of the remainder of the proof. Here we construct a path of length of order $$n/\log n$$ in $$\square(\Gamma)$$ on to which every vertex $$v\in V(\Gamma)$$ can be “beamed up” by adding a $$4$$–cycle whose support contains $$v$$. This is done in the following manner: we start with an arbitrary pair of non-adjacent vertices contained in a block $$B_{0}$$. We then pick an arbitrary pair of non-adjacent vertices in the block $$B_0$$ and let $$B_1$$ denote the intersection of the block they define with $$V(\Gamma)\setminus B_0$$. We repeat this procedure, to obtain a chain of blocks $$B_0, B_1, B_2, \ldots, B_t$$, with $$t=O(n/\log n)$$, whose union contains a positive proportion of $$V(\Gamma)$$, and which all belong to the same connected component $$C$$ of $$\square(\Gamma)$$. This common component $$C$$ is then large enough that every remaining vertex of $$V(\Gamma)$$ can be attached to it. The main challenge is showing that our process of recording which vertices are included in the support of a component of the square graph does not die out or slow down too much, that is, that the block sizes $$\vert B_i\vert$$ remains relatively large at every stage of the process and that none of the $$B_i$$ form a clique. Having described our strategy, we now fill in the details, beginning with the following upper bound on the probability of $$\Gamma\in {\mathcal G}(n,p)$$ containing a copy of $$K_{10}$$, the complete graph on $$10$$ vertices. The following lemma is a variant of [21, Corollary 4]: Lemma 5.2. Let $$\Gamma\in{\mathcal G}(n,p)$$. If $$p=o(n^{-\frac{1}{4}})$$, then the probability that $$\Gamma\in {\mathcal G}(n,p)$$ contains a clique with at least $$10$$ vertices is at most $$o(n^{-\frac{5}{4}})$$. □ Proof The expected number of copies of $$K_{10}$$ in $$\Gamma$$ is (n10)p(102)≤n10p45=o(n−5/4). The statement of the lemma then follows from Markov’s inequality. ■ Proof of Theorem 5.1. As remarked above, Theorem 4.4 proves Theorem 5.1 for “large” $$p$$, so we only need to deal with the case where 5lognn≤p(n)≤2(lognn)13. We iteratively build a chain of blocks, as follows. Let $$\{x_0,y_0\}$$ be a pair of non-adjacent vertices in $$\Gamma$$, if such a pair exists, and an arbitrary pair of vertices if not. Let $$B_0$$ be the block with ends $$\{x_0, y_0\}$$. Now assume we have already constructed the blocks $$B_0, \ldots, B_i$$, for $$i\geq 0$$. Let $$C_i=\bigcup_i B_i$$ (for convenience we let $$C_{-1}=\emptyset$$). We will terminate the process and set $$t=i$$ if any of the three following conditions occur: $$\vert \mathrm{core}(B_i)\vert\leq 6\log n$$ or $$i\geq n/6\log n$$ or $$\vert V(\Gamma)\setminus C_i\vert \leq n/2$$. Otherwise, we let $$\{x_{i+1}, y_{i+1}\}$$ be a pair of non-adjacent vertices in $$\mathrm{core}(B_i)$$, if such a pair exists, and an arbitrary pair of vertices from $$\mathrm{core}(B_i)$$ otherwise. Let $$B_{i+1}$$ denote the intersection of the block whose ends are $$\{x_{i+1}, y_{i+1}\}$$ and the set $$\left(V(\Gamma)\setminus(C_i)\right)\cup\{x_{i+1}, y_{i+1}\}$$. Repeat. Eventually this process must terminate, resulting in a chain of blocks $$B_0, B_1, \ldots, B_t$$. We claim that a.a.s. both of the following hold for every $$i$$ satisfying $$0 \leq i \leq t$$: (i) $$\vert\mathrm{core}(B_i)\vert > 6\log n$$; and (ii) $$\{x_i,y_i\}$$ is a non-edge in $$\Gamma$$. Part (i) follows from the Chernoff bound given in Lemma 4.1: for each $$i\geq -1$$ the set $$V(\Gamma)\setminus C_i$$ contains at least $$n/2$$ vertices by construction. For each vertex $$v\in V(\Gamma)\setminus C_i$$, let $$X_v$$ be the indicator function of the event that $$v$$ is adjacent to both of $$\{x_{i+1}, y_{i+1}\}$$. The random variables $$(X_v)$$ are independent identically distributed Bernoulli random variables with mean $$p$$. Their sum $$X=\sum_v X_v$$ is exactly the size of the core of $$B_{i+1}$$, and its expectation is at least $$p^2n/2$$. Applying Lemma 4.1, we get that P(X<6logn) ≤P(X≤12EX) ≤2e−(12)225logn6n=2e−2524logn. Thus the probability that $$\vert \mathrm{core}(B_i)\vert <6\log n$$ for some $$i$$ with $$0\leq i \leq t$$ is at most: t2e−2524logn≤4n5logn2e−2524logn=o(1). Part (ii) is a trivial consequence of part (i) and Lemma 5.2: a.a.s. $$\mathrm{core}(B_i)$$ has size at least $$6\log n$$ for every $$i$$ with $$0 \leq i \leq t$$, and a.a.s. $$\Gamma$$ contains no clique on $$10 < \log n$$ vertices, so that a.a.s. at each stage of the process we could choose an non-adjacent pair $$\{x_i,y_i\}$$. From now on we assume that both (i) and (ii) occur, and that $$\Gamma$$ contains no clique of size $$10$$. In addition, we assume that $$\vert \mathrm{core}(B_0)\vert < 8n^{\frac{1}{3}}(\log{n})^{\frac{2}{3}}$$, which occurs a.a.s. by the Chernoff bound. Since $$\mathrm{core}(B_i)\geq 6\log n$$ for every $$i$$, we must have that by time $$0<t\leq n/6\log n$$ the process will have terminated with $$C_t=\bigcup_{i=0}^t B_i$$ supported on at least half of the vertices of $$V(\Gamma)$$. Lemma 5.3. Either one of the assumptions above fails or there exists a connected component $$F$$ of $$\square(\Gamma)$$ such that: (i) for every $$i$$ with $$0 \leq i \leq t$$ and every pair of non-adjacent vertices $$\{v,v'\}\in B_i$$, there is a vertex in $$F$$ whose support in $$\Gamma$$ contains the pair $$\{v,v'\}$$; and (ii) the support in $$\Gamma$$ of the $$4$$–cycles corresponding to vertices of $$F$$ contains all of $$C_t$$ with the exception of at most $$9$$ vertices of $$\mathrm{core}(B_0)$$; moreover, these exceptional vertices are each adjacent to all the vertices of $$\mathrm{core}(B_0)$$. □ Proof By assumption the ends $$\{x_0, y_0\}$$ of $$B_0$$ are non-adjacent. Thus, every pair of non-adjacent vertices $$\{v,v'\}$$ in $$\mathrm{core}(B_0)$$ induces a $$4$$–cycle in $$\Gamma$$ when taken together with $$\{x_0,y_0\}$$, and all of these squares clearly lie in the same component $$F$$ of $$\square(\Gamma)$$. Repeating the argument with the non-adjacent pair $$\{x_1, y_1\}\in \mathrm{core}(B_0)$$ and the block $$B_1$$, and then the non-adjacent pair $$\{x_2, y_2\}\in \mathrm{core}(B_1)$$ and the block $$B_2$$, and so on, we see that there is a connected component $$F$$ in $$\square(\Gamma)$$ such that for every $$0\leq i \leq t$$, every pair of non adjacent vertices $$\{v,v'\} \in B_i$$ lies in a $$4$$–cycle corresponding to a vertex of $$F$$. This establishes (i). We now show that the support of $$F$$ contains all of $$C_t$$ except possibly some vertices in $$B_0$$. We already established that every pair $$\{x_i,y_i\}$$ is in the support of some vertex of $$F$$. Suppose $$v\in \mathrm{core}(B_i)$$ for some $$i>0$$. By construction, $$v$$ is not adjacent to at least one of $$\{x_{i-1}, y_{i-1}\}$$, say $$x_{i-1}$$. Thus, $$\{x_{i-1},x_i, y_i, v\}$$ induces a $$4$$–cycle which contains $$v$$ and is associated to a vertex of $$F$$. Finally, suppose $$v\in \mathrm{core}(B_0)$$. By (i), $$v$$ fails to be in the support of $$F$$ only if $$v$$ is adjacent to all other vertices of $$\mathrm{core}(B_0)$$. Since, by assumption, $$\Gamma$$ does not contain any clique of size $$10$$, there are at most $$9$$ vertices not in the support of $$F$$, proving (ii). ■ Lemma 5.3, shows that a.a.s. we have a “large” component $$F$$ in $$\square(\Gamma)$$ whose support contains “many” pairs of non-adjacent vertices. In the last part of the proof, we use these pairs to prove that the remaining vertices of $$V(\Gamma)$$ are also supported on our connected component. For each $$i$$ satisfying $$0 \leq i \leq t$$, consider a a maximal collection, $$M_{i}$$, of pairwise-disjoint pairs of vertices in $$\mathrm{core}(B_i)\setminus\{x_{i+1},y_{i+1}\}$$. Set $$M=\bigcup_i M_i$$, and let $$M'$$ be the subset of $$M$$ consisting of pairs, $$\{v,v'\}$$, for which $$v$$ and $$v'$$ are not adjacent in $$\Gamma$$. We have |M|=∑i=1t(⌊12|core(Bi)|⌋−1)≥|Ct|2−2t≥n4(1−o(1)). The expected size of $$M'$$ is thus $$(1-p)n(\frac{1}{4}-o(1))=\frac{n}{4}(1-o(1))$$, and by the Chernoff bound from Lemma 4.1 we have P(|M′|≤n5) ≤2e−(15+o(1))2(1−p)n12≤e−(1300+o(1))n, which is $$o(1)$$. Thus a.a.s. $$M'$$ contains at least $$n/5$$ pairs, and by Lemma 5.3 each of these lies in some $$4$$–cycle of $$F$$. We now show that we can “beam up” every vertex not yet supported on $$F$$ by a $$4$$–cycle using a pair from $$M'$$. By construction we have at most $$n/2$$ unsupported vertices from $$V(\Gamma)\setminus C_t$$ and at most $$9$$ unsupported vertices from $$\mathrm{core}(B_0)$$. Assume that $$\vert M' \vert \geq n/5$$. Fix a vertex $$w\in V(\Gamma)\setminus C_t$$. For each pair $$\{v,v'\}\in M'$$, let $$X_{v,v'}$$ be the event that $$w$$ is adjacent to both $$v$$ and $$v'$$. We now observe that if $$X_{v,v'}$$ occurs for some pair $$\{v,v'\}\in M'\cap \mathrm{core}(B_i)$$, then $$w$$ is supported on $$F$$. By construction, $$w$$ is not adjacent to at least one of $$\{x_i, y_i\}$$, let us say without loss of generality $$x_i$$. Hence, $$\{x_i,v,v',w\}$$ is an induced $$4$$–cycle in $$\Gamma$$ which contains $$w$$ and which corresponds to a vertex of $$F$$. The probability that $$X_{v,v'}$$ fails to happen for every pair $$\{v,v'\}\in M'$$ is exactly (1−p2)|M′|≤(1−p2)n/5≤e−p2n5=e−5logn. Thus the expected number of vertices $$w \in V(\Gamma)\setminus C_t$$ which fail to be in the support of $$F$$ is at most $$\frac{n}{2}e^{-5\log n}=o(1)$$, whence by Markov’s inequality a.a.s. no such bad vertex $$w$$ exists. Finally, we deal with the possible $$9$$ left-over vertices $$b_1, b_2, \ldots b_9$$ from $$\mathrm{core}(B_0)$$ we have not yet supported. We observe that since $$\mathrm{core}(B_0)$$ contains at most $$8n^{\frac{1}{3}}(\log{n})^{\frac{2}{3}}$$ vertices (as we are assuming and as occurs a.a.s. , see the discussion before Lemma 5.3 ), we do not stop the process with $$B_0$$, $$\mathrm{core}(B_1)$$ is non-empty and contains at least $$6\log n$$ vertices. As stated in Lemma 5.3, each unsupported vertex $$b_i$$ is adjacent to all other vertices in $$\mathrm{core}(B_0)$$, and in particular to both of $$\{x_1, y_1\}$$. If $$b_i$$ fails to be adjacent to some vertex $$v\in \mathrm{core}(B_1)$$, then the set $$\{b_i, x_1, y_1, v\}$$ induces a $$4$$–cycle corresponding to a vertex of $$F$$ and whose support contains $$b_i$$. The probability that there is some $$b_i$$ not supported in this way is at most 9P(bi adjacent to all of core(B1)) =9p6logn=o(1). Thus a.a.s. we can “beam up” each of the vertices $$b_1, \ldots b_9$$ to $$F$$ using a vertex $$v\in \mathrm{core}(B_1)$$, and the support of the component $$F$$ in the square graph $$\square(\Gamma)$$ contains all vertices of $$\Gamma$$. This shows that a.a.s. $$\Gamma \in \mathcal{CFS}$$, and concludes the proof of the theorem. ■ Remark 5.4. The constant $$5$$ in Theorem 5.1 is not optimal, and indeed it is not hard to improve on it slightly, albeit at the expense of some tedious calculations. We do not try to obtain a better constant, as we believe that the order of the upper bound we have obtained is not sharp. We conjecture that the actual threshold for $$\mathcal{CFS}$$ occurs when $$p(n)$$ is of order $$n^{-1/2}$$ (see the discussion below Theorem 5.7), but a proof of this is likely to require significantly more involved and sophisticated arguments than the present article. □ A simple lower bound for the emergence of the $$\mathcal{CFS}$$ property can be obtained from the fact that if $$\Gamma\in\mathcal{CFS}$$, then $$\Gamma$$ must contain at least $$n-3$$ squares; if $$p(n)\ll n^{-\frac{3}{4}}$$, then by Markov’s inequality a.a.s. a graph in $${\mathcal G}(n,p)$$ contains fewer than $$o(n)$$ squares and thus cannot be in $$\mathcal{CFS}$$. Below, in Theorem 5.7, we prove a better lower bound, showing that the order of the upper bound we proved in Theorem 5.1 is not off by a factor of more than $$(\log n)^{3/2}$$. Lemma 5.5. Let $$\Gamma$$ be a graph and let $$C$$ be the subgraph of $$\Gamma$$ supported on a given connected component of $$\square(\Gamma)$$. Then there exists an ordering $$v_1<v_2< \cdots <v_{\vert C\vert}$$ of the vertices of $$C$$ such that for all $$i\ge3$$, $$v_i$$ is adjacent in $$\Gamma$$ to at least two vertices preceding it in the order. □ Proof As $$C$$ is a component of $$\square(\Gamma)$$, it contains at least one induced $$4$$–cycle. Let $$v_1, v_2$$ be a pair of non-adjacent vertices from such an induced $$4$$–cycle. Then the two other vertices $$\{v_3,v_4\}$$ of the $$4$$–cycle are both adjacent in $$\Gamma$$ to both of $$v_{1}$$ and $$v_{2}$$. If this is all of $$C$$, then we are done. Otherwise, we know that each $$4$$–cycle in $$C$$ is “connected” to the cycle $$F=\{v_{1},v_{2},v_{3},v_{4}\}$$ via a sequence of induced $$4$$–cycles pairwise intersecting in pairwise non-adjacent vertices. In particular, there is some such $$4$$–cycle whose intersection with $$F$$ is either a pair of non-adjacent vertices in $$F$$ or three vertices of $$F$$; either way, we may add the new vertex next in the order. Continuing in this way and using the fact that the number of vertices not yet reached is a monotonically decreasing set of positive integers, the lemma follows. ■ Proposition 5.6. Let $$\delta>0$$. Suppose $$p\leq\frac{1}{\sqrt{n}\log n}$$. Then a.a.s. for $$\Gamma\in{\mathcal G}(n,p)$$, no component of $$\square(\Gamma)$$ has support containing more than $$4\log n$$ vertices of $$\Gamma$$. □ Proof Let $$\delta>0$$. Let $$m=\left\lceil \min\left(4\log n, 4\log \left(\frac{1}{p}\right) \right)\right\rceil$$, with $$p\leq 1/\left(\sqrt{n}\log n\right)$$. We shall show that a.a.s. there is no ordered $$m$$–tuple of vertices $$v_1<v_2<\cdots <v_m$$ from $$\Gamma$$ such that for every $$i\ge 2$$ each vertex $$v_i$$ is adjacent to at least two vertices from $$\{v_j: 1\leq j <i\}$$. By Lemma 5.5, this is enough to establish our claim. Let $$v_1<v_2<\cdots < v_m$$ be an arbitrary ordered $$m$$–tuple of vertices from $$V(\Gamma)$$. For $$i\geq 2$$, let $$A_i$$ be the event that $$v_i$$ is adjacent to at least two vertices in the set $$\{v_j: 1\leq j <i\}$$. We have: Pr(Ai)=∑j=2i−1(i−1j)pj(1−p)i−j−1. (2) As in the proof of Theorem 4.5 we consider the quotients of successive terms in the sum to show that its order is given by the term $$j=2$$. To see this, observe: (i−1j+1)pj+1(1−p)i−j−2(i−1j)pj(1−p)i−j−1≤i−j−1j+1⋅p1−p<mp where the final inequality holds for $$n$$ sufficiently large and $$p=p(n)$$ satisfying our assumption. Since $$m=O\left(\log n\right)$$ and $$p=o(n^{-1/2})$$ we have, again for $$n$$ large enough, that $$mp=o(1)$$, and we may bound the sum in equation (2) by a geometric series to obtain the bound: Pr(Ai) =(i−12)p2(1−p)i−3(1+O(mp)) ≤(i−1)22p2(1+O(mp)). Now let $$A=\bigcap_{i=1}^m A_i$$. Note that the events $$A_i$$ are mutually independent, since they are determined by disjoint edge-sets. Thus we have: Pr(A)=∏i=3mPr(Ai) ≤∏i=3m((i−1)22p2(1+O(mp))) =((m−1)!)2p2m−42m−2(1+O(m2p)), where in the last line we used the equality $$(1+O(mp))^{m-2} =1+O(m^2p)$$ to bound the error term. Thus we have that the expected number $$X$$ of ordered $$m$$–tuples of vertices from $$\Gamma$$ for which $$A$$ holds is at most: E(X)=n!(n−m)!P(A) ≤nm4e2(m2p2−4me22)m(1+O(m2p)) =4e2(nm2p2−4/m2e2)m(1+O(m2p)), where in the first line we used the inequality $$(m-1)!\leq e (m/e)^{m}$$. We now consider the quantity f(n,m,p)=nm2p2−4/m2e2 which is raised to the $$m^{\textrm{th}}$$ power in the inequality above. We claim that $$f(n,m,p)\leq e^{-1+\log 2+o(1)}$$. We have two cases to consider: Case 1: $$m=\lceil 4\log n\rceil$$. Since $$4\log n \leq 4\log(1/p)$$, we deduce that $$p\leq n^{-1}$$. Then f(n,m,p) =n(4logn)2p2−o(1)2e2≤n−1+o(1)≤e−1+log2+o(1). Case 2: $$m=\lceil 4\log(1/p)\rceil$$. First, note that $$p^{-4/m}=\exp\left(\frac{4\log(1/p)}{\lceil 4 \log 1/p\rceil}\right)\leq e$$. Also, for $$p$$ in the range $$[0, n^{-1/2}(\log n)^{-1}]$$ and $$n$$ large enough, $$p^2(\log(1/p))^2$$ is an increasing function of $$p$$ and is thus at most: 1n(logn)2(12logn)2(1+O(loglognlogn))=14n−1(1+o(1)). Plugging this into the expression for $$f(n,m,p)$$, we obtain: f(n,m,p) =(1+o(1))16n(log(1/p))2p2−4/m2e2 ≤(1+o(1))2e=e−1+log2+o(1). Thus, in both cases (1) and (2) we have $$f(n,m,p)\leq e^{-1 +\log 2+o(1)}$$, as claimed, whence E(X) ≤4e2(f(n,m,p))m(1+O(m2p))≤4e2e−(1−log2)m+o(m)(1+O(m2p))=o(1). It follows from Markov’s inequality that the non-negative, integer-valued random variable $$X$$ is a.a.s. equal to $$0$$. In other words, a.a.s. there is no ordered $$m$$–tuple of vertices in $$\Gamma$$ for which the event $$A$$ holds and, hence by Lemma 5.5, no component in $$\square(\Gamma)$$ covering more than $$m\leq 4\log n$$ vertices of $$\Gamma$$. ■ Theorem 5.7. Suppose $$p\leq\frac{1}{\sqrt{n} \log n}$$. Then a.a.s. $$\Gamma\in{\mathcal G}(n,p)$$ is not in $$\mathcal{CFS}$$. □ Proof To show that $$\Gamma\not\in\mathcal{CFS}$$, we first show that, for $$p\leq \frac{1}{\sqrt{n} \log n}$$, a.a.s. there is no non-empty clique $$K$$ such that $$\Gamma=\Gamma' \star K$$. Indeed the standard Chernoff bound guarantees that we have a.a.s. no vertex in $$\Gamma$$ with degree greater than $$\sqrt{n}$$. Thus to prove the theorem, it is enough to show that a.a.s. there is no connected component $$C$$ in $$\square(\Gamma)$$ containing all the vertices in $$\Gamma$$. Theorem 5.6 does this by establishing the stronger bound that a.a.s. there is no connected component $$C$$ covering more than $$4\log n$$ vertices. ■ While Theorem 5.7 improves on the trivial lower bound of $$n^{-3/4}$$, it is still off from the upper bound for the emergence of the $$\mathcal{CFS}$$ property established in Theorem 5.1. It is a natural question to ask where the correct threshold is located. Remark 5.8. We strongly believe that there is a sharp threshold for the $$\mathcal{CFS}$$ property analogous to the one we established for the $$\mathcal{AS}$$ property. What is more, we believe this threshold should essentially coincide with the threshold for the emergence of a giant component in the auxiliary square graph $$\square(\Gamma)$$. Indeed, our arguments in Proposition 5.6 and Theorems 5.1 both focus on bounding the growth of a component in $$\square(\Gamma)$$. Heuristically, we would expect a giant component to emerge in $$\square(\Gamma)$$ to emerge for $$p(n)=cn^{-1/2}$$, for some constant $$c>0$$, when the expected number of common neighbors of a pair of non-adjacent vertices in $$\Gamma$$ is $$c^2$$, and thus the expected number of distinct vertices in $$4$$–cycles which meet a fixed $$4$$–cycle in a non-edge is $$2c^2$$. What the precise value of $$c$$ should be is not entirely clear (a branching process heuristics suggests $$\sqrt{\displaystyle\frac{\sqrt{17}-3}{2}}$$ as a possible value, see Remark 6.2), however, and the dependencies in the square graph make its determination a delicate matter. □ 6 Experiments Theorems 4.4 and 4.5 show that $$\left(\log n/n\right)^{\frac{1}{3}}$$ is a sharp threshold for the family $$\mathcal{AS}$$ and Theorem 5.1 shows that $$n^{-\frac{1}{2}}$$ is the right order of magnitude of the threshold for $$\mathcal{CFS}$$. Below we provide some empirical results on the behavior of random graphs near the threshold for $$\mathcal{AS}$$ and the conjectured threshold for $$\mathcal{CFS}$$. We also compare our experimental data with analogous data at the connectivity threshold. Our experiments are based on various algorithms that we implemented in $$\texttt{C++}$$; the source code is available from the authors. At www.wescac.net/research.html or math.columbia.edu/~jason, all source code and data can be downloaded. We begin with the observation that computer simulations of $$\mathcal{AS}$$ and $$\mathcal{CFS}$$ are tractable. Indeed, it is easily seen that there are polynomial-time algorithms for deciding whether a given graph is in $$\mathcal{AS}$$ and/or $$\mathcal{CFS}$$. Testing for $$\mathcal{AS}$$ by examining each block and determining whether it witnesses $$\mathcal{AS}$$ takes $$O(n^5)$$ steps, where $$n$$ is the number of vertices. The $$\mathcal{CFS}$$ property is harder to detect, but essentially reduces to determining the component structure of the square graph. The square graph can be produced in polynomial time and, in polynomial time, one can find its connected components and check the support of these components in the original graph. Using our software, we tested random graphs in $${\mathcal G}(n,p)$$ for membership in $$\mathcal{AS}$$ for $$n\in\{300+100k\mid0\leq k\leq12\}$$ and $$\{p(n)=\alpha\left(\log n/n\right)^{\frac{1}{3}}\mid \alpha=0.80+0.1k,\,0\leq k\leq 9\}$$. For each pair $$(n,p)$$ of this type, we generated $$400$$ random graphs and tested each for membership in $$\mathcal{AS}$$. (This number of tests ensures that, with probability approximately $$95\%$$, the measured proportion of $$\mathcal{AS}$$ graphs is within at most $$0.05$$ of the actual proportion.) The results are summarized in Figure 2. The data suggest that, fixing $$n$$, the probability that a random graph is in $$\mathcal{AS}$$ increases monotonically in the range of $$p$$ we are considering, rising sharply from almost zero to almost one. Fig. 2. View largeDownload slide Experimental prevalence of $$\mathcal{AS}$$ at density $$\alpha\left(\frac{\log n}{n}\right)^{\frac{1}{3}}$$. (A colour version of this figure is available from the journal’s website.) Fig. 2. View largeDownload slide Experimental prevalence of $$\mathcal{AS}$$ at density $$\alpha\left(\frac{\log n}{n}\right)^{\frac{1}{3}}$$. (A colour version of this figure is available from the journal’s website.) In Figure 3, we display the results of testing random graphs for membership in $$\mathcal{CFS}$$ for $$n\in\{100+100k\mid 0\leq k\leq 15\}$$ and $$\{p(n)=\alpha n^{-\frac{1}{2}} \mid \alpha=0.700+0.025k,\,0\leq k\leq 8\}$$. For each pair $$(n,p)$$ of this type, we generated $$400$$ random graphs and tested each for membership in $$\mathcal{CFS}$$. (This number of tests ensures that, with probability approximately $$95\%$$, the measured proportion of $$\mathcal{AS}$$ graphs is within at most $$0.05$$ of the actual proportion.) The data suggest that, fixing $$n$$, the probability that a random graph is in $$\mathcal{CFS}$$ increases monotonically in considered range of $$p$$: rising sharply from almost zero to almost one inside a narrow window. Fig. 3. View largeDownload slide Experimental prevalence of $$\mathcal{CFS}$$ membership at density $$\alpha n^{-\frac{1}{2}}$$. (A colour version of this figure is available from the journal’s website.) Fig. 3. View largeDownload slide Experimental prevalence of $$\mathcal{CFS}$$ membership at density $$\alpha n^{-\frac{1}{2}}$$. (A colour version of this figure is available from the journal’s website.) Remark 6.1 (Block and core sizes). For each graph $$\Gamma$$ tested, the $$\mathcal{AS}$$ software also keeps track of how many nonadjacent pairs $$\{x,y\}$$ — that is, how many blocks — were tested before finding one sufficient to verify membership in $$\mathcal{AS}$$; if no such block is found, then all non-adjacent pairs have been tested and the graph is not in $$\mathcal{AS}$$. ({A set of such data comes with the source code, and more is available upon request.}) At densities near the threshold, this number of blocks is generally very large compared to the number of blocks tested at densities above the threshold. For example, in one instance with $$(n,\alpha)=(1000,0.89)$$, verifying that the graph was in $$\mathcal{AS}$$ was accomplished after testing just $$86$$ blocks, while at $$(1000,0.80)$$, a typical test checked all $$422961$$ blocks (expected number: $$423397$$) before concluding that the graph is not $$\mathcal{AS}$$. At the same $$(n,\alpha)$$, another test found that the graph was in $$\mathcal{AS}$$, but only after $$281332$$ tests. These data are consonant with the spirit of our proofs of Theorems 4.5 and 4.4: in the former case, we showed that no “good” block exists, while in the latter we show that every block is good. We believe that right at the threshold we should have some intermediate behavior, with the expected number of “good” blocks increasing continuously from $$0$$ to $$(1-p)\binom{n}{2}(1-o(1))$$. What is more, we expect that the more precise threshold for the $$\mathcal{AS}$$ property, coinciding with the appearance of a single “good” block, should be located “closer” to our lower bound than to our upper one, that is, at $$p(n)= \left((1-\epsilon)\log n/n\right)^{1/3}$$, where $$\epsilon(n)$$ is a sequence of strictly positive real numbers tending to $$0$$ as $$n\rightarrow \infty$$ (most likely decaying at a rate just faster than $$(\log n)^{-2}$$, see below). Our experimental data, which exhibit a steep rise in $$\Pr(\mathcal{AS})$$ strictly before the value $$\alpha$$ hits one, gives some support to this guess. Finally, our observations on the number of blocks suggests a natural way to understand the influence of higher-order terms on the emergence of the $$\mathcal{AS}$$ property: at exactly the threshold for $$\mathcal{AS}$$, the event $$E(v,w)$$ that a pair of non-adjacent vertices $$\{v,w\}$$ gives rise to a “good” block is rare and, despite some mild dependencies, the number $$N$$ of pairs $$\{v,w\}$$ for which $$E(v,w)$$ occurs is very likely to be distributed approximatively like a Poisson random variable. The probability $$\Pr(N\geq 1)$$ would then be a very good approximation for $$\Pr(\mathcal{AS})$$. “Good” blocks would thus play a role for the emergence of the $$\mathcal{AS}$$ property in random graphs analogous to that of isolated vertices for connectivity in random graphs. When $$p=\left((1-\epsilon) \log n/n\right)^{\frac{1}{3}}$$, the expectation of $$N$$ is roughly $$ne^{-n^{\epsilon}(1-\epsilon)\log n}$$. This expectation is $$o(1)$$ when $$0<\epsilon(n)=\Omega(1/n)$$ and is $$1/2$$ when $$\epsilon(n)=(1+o(1)) \log 2/(\log n)^2$$. This suggest that the emergence of $$\mathcal{AS}$$ should occur when $$\epsilon(n)$$ decays just a little faster than $$(\log n)^2$$. □ Remark 6.2. Our data suggest that the prevalence of $$\mathcal{CFS}$$ is closely related to the emergence of a giant component in the square graph. Indeed, below the experimentally observed threshold for $$\mathcal{CFS}$$, not only is the support of the largest component in the square graph not all of $$\Gamma \in {\mathcal G}(n,p)$$, but in fact the size of the support of the largest component is an extremely small proportion of the vertices (see Figure 4). In the Erdős–Rényi random graph, a giant component emerges when $$p$$ is around $$1/n$$, that is, when vertices begin to expect at least one neighbor; this corresponds to a paradigmatic condition of expecting at least one child for survival of a Galton–Watson process (see [9] for a modern treatment of the topic). The heuristic observation that when $$p=cn^{-1/2}$$ the vertices of a diagonal $$e$$ in a fixed $$4$$-cycle $$F$$ expect to be adjacent to $$cn^2$$ vertices outside the $$4$$-cycle, giving rise to an expected $$\binom{cn^2+2}{2}-1$$ new $$4$$-cycles connected to $$F$$ through $$e$$ in $$\square(\Gamma)$$ suggests that $$c=\sqrt{\displaystyle\frac{\sqrt{17}-3}{2}}\approx 0.7494$$ could be a reasonable guess for the location of the threshold for the $$\mathcal{CFS}$$ property. 2 Our data, although not definitive, appears somewhat supportive of this guess: see Figure 4 which is based on the same underlying data set as Figure 3. We note that unlike an Erdős–Rényi random graph, the square graph $$\square(\Gamma)$$ exhibits some strong local dependencies, which may make the determination of the exact location of its phase transition a much more delicate affair. □ Fig. 4. View largeDownload slide Fraction of vertices supporting the largest $$\mathcal{CFS}$$–subgraph at density $$\alpha n^{-\frac{1}{2}}$$. (A colour version of this figure is available from the journal’s website.) Fig. 4. View largeDownload slide Fraction of vertices supporting the largest $$\mathcal{CFS}$$–subgraph at density $$\alpha n^{-\frac{1}{2}}$$. (A colour version of this figure is available from the journal’s website.) Remark 6.3. For comparison with the threshold data for $$\mathcal{AS}$$ and $$\mathcal{CFS}$$, we include below a similar figure of experimental data for connectivity for $$\alpha$$ from $$0.8$$ to $$1.4$$, where $$p=\frac{\alpha\log n}{n}$$. Given, what we know about the thresholds for connectivity and the $$\mathcal{AS}$$ property, this last set of data together with Figure 2 should serve as a warning not to draw overly strong conclusions: the graphs we tested are sufficiently large for the broader picture to emerge, but probably not large enough to allow us to pinpoint the exact location of the threshold for $$\mathcal{CFS}$$. □ Fig. 5. View largeDownload slide Experimental prevalence of connectedness at density $$\alpha\frac{\log n}{n}$$. (A colour version of this figure is available from the journal’s website.) Fig. 5. View largeDownload slide Experimental prevalence of connectedness at density $$\alpha\frac{\log n}{n}$$. (A colour version of this figure is available from the journal’s website.) Funding This work was supported by National Science Foundation [DMS-0739392 to J.B, 1045119 to M.F.H.] and also supported by a Simons Fellowship to J.B. Acknowledgments J.B. thanks the Barnard/Columbia Mathematics Department for their hospitality during the writing of this article. J.B. thanks Noah Zaitlen for introducing him to the joy of cluster computing and for his generous time spent answering remedial questions about programming. Also, thanks to Elchanan Mossel for several interesting conversations about random graphs. M.F.H. and T.S. thank the organizers of the 2015 Geometric Groups on the Gulf Coast conference, at which some of this work was completed. This work benefited from several pieces of software, the results of one of which is discussed in Section 6. Some of the software written by the authors incorporates components previously written by J.B. and M.F.H. jointly with Alessandro Sisto. Another related useful program was written by Robbie Lyman under the supervision of J.B. and T.S. during an REU program supported by NSF Grant DMS-0739392, see [30]. 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# Global Structural Properties of Random Graphs
, Volume 2018 (5) – Mar 1, 2018
31 pages
/lp/oxford-university-press/global-structural-properties-of-random-graphs-UIkbSR5ban
Publisher
Oxford University Press
ISSN
1073-7928
eISSN
1687-0247
D.O.I.
10.1093/imrn/rnw287
Publisher site
See Article on Publisher Site
### Abstract
Abstract We study two global structural properties of a graph $$\Gamma$$, denoted $$\mathcal{AS}$$ and $$\mathcal{CFS}$$, which arise in a natural way from geometric group theory. We study these properties in the Erdős–Rényi random graph model $${\mathcal G}(n,p)$$, proving the existence of a sharp threshold for a random graph to have the $$\mathcal{AS}$$ property asymptotically almost surely, and giving fairly tight bounds for the corresponding threshold for the $$\mathcal{CFS}$$ property. As an application of our results, we show that for any constant $$p$$ and any $$\Gamma\in{\mathcal G}(n,p)$$, the right-angled Coxeter group $$W_\Gamma$$ asymptotically almost surely has quadratic divergence and thickness of order $$1$$, generalizing and strengthening a result of Behrstock–Hagen–Sisto [8]. Indeed, we show that at a large range of densities a random right-angled Coxeter group has quadratic divergence. 1 Introduction In this article, we consider two properties of graphs motivated by geometric group theory. We show that these properties are typically present in random graphs. We repay the debt to geometric group theory by applying our (purely graph-theoretic) results to the large-scale geometry of Coxeter groups. Random graphs Let $${\mathcal G}(n,p)$$ be the random graph model on $$n$$ vertices obtained by including each edge independently at random with probability $$p=p(n)$$. The parameter $$p$$ is often referred to as the density of $${\mathcal G}(n,p)$$. The model $${\mathcal G}(n,p)$$ was introduced by Gilbert [23], and the resulting random graphs are usually referred to as the “Erdős–Rényi random graphs” in honor of Erdős and Rényi’s seminal contributions to the field, and we follow this convention. We say that a property $$\mathcal{P}$$ holds asymptotically almost surely (a.a.s.) in $${\mathcal G}(n,p)$$ if for $$\Gamma\in{\mathcal G}(n,p),$$ we have $$\mathbb{P}(\Gamma \in \mathcal{P})\rightarrow 1$$ as $$n\rightarrow \infty$$. In this article, we will be interested in proving that certain global properties hold a.a.s. in $${\mathcal G}(n,p)$$ both for a wide range of probabilities $$p=p(n)$$. A graph property is (monotone) increasing if it is closed under the addition of edges. A paradigm in the theory of random graphs is that global increasing graph properties exhibit sharp thresholds in $${\mathcal G}(n,p)$$: for many global increasing properties $$\mathcal{P}$$, there is a critical density$$p_c=p_c(n)$$ such that for any fixed $$\epsilon>0$$ if $$p<(1-\epsilon)p_c$$ then a.a.s. $$\mathcal{P}$$ does not hold in $${\mathcal G}(n,p)$$, while if $$p>(1+\epsilon)p_c$$ then a.a.s. $$\mathcal{P}$$ holds in $${\mathcal G}(n,p)$$. A quintessential example is the following classical result of Erdős and Rényi which provides a sharp threshold for connectedness: Theorem (Erdős–Rényi; [21]). There is a sharp threshold for connectivity of a random graph with critical density $$p_c(n)=\frac{\log(n)}{n}$$. □ The local structure of the Erdős–Rényi random graph is well understood, largely due to the assumption of independence between the edges. For example, Erdős–Rényi and others have obtained threshold densities for the existence of certain subgraphs in a random graph (see e.g., [21, Theorem 1, Corollaries 1–5]). In earlier applications of random graphs to geometric group theory, this feature of the model was successfully exploited in order to analyze the geometry of right-angled Artin and Coxeter groups presented by random graphs; this is notable, for example, in the work of Charney and Farber [14]. In particular, the presence of an induced square implies non-hyperbolicity of the associated right-angled Coxeter group [14, 21, 31]. In this article, we take a more global approach. Earlier work established a correspondence between some fundamental geometric properties of right-angled Coxeter groups and large-scale structural properties of the presentation graph, rather than local properties such as the presence or absence of certain specified subgraphs. The simplest of these properties is the property of being the join of two subgraphs that are not cliques. One large scale graph property relevant in the present context is a property studied in [8] which, roughly, says that the graph is constructed in a particular organized, inductive way from joins. In this article, we discuss a refined version of this property, $$\mathcal{CFS}$$, which is a slightly-modified version of a property introduced by Dani–Thomas [17]. We also study a stronger property, $$\mathcal{AS}$$, and show it is generic in random graphs for a large range of $$p(n)$$, up $$1-\omega(n^{-2})$$. $$\boldsymbol{\mathcal{AS}}$$ graphs The first class of graphs we study is the class of augmented suspensions, which we denote $$\mathcal{AS}$$. A graph is an augmented suspension if it contains an induced subgraph which is a suspension (see Section 2 for a precise definition of this term), and any vertex which is not in that suspension is connected by edges to at least two nonadjacent vertices of the suspension. Theorem 4.4 and 4.5 (Sharp Threshold for $$\boldsymbol{\mathcal{AS}}$$).Let $$\epsilon>0$$ be fixed. If $$p=p(n)$$ satisfies $$p\ge(1+\epsilon)\left(\displaystyle\frac{\log n}{n}\right)^{\frac{1}{3}}$$ and $$(1-p)n^2\to\infty$$, then $$\Gamma\in{\mathcal G}(n,p(n))$$ is a.a.s. in $$\mathcal{AS}$$. On the other hand, if $$p\le(1-\epsilon)\left(\displaystyle\frac{\log{n}}{n}\right)^{\frac{1}{3}}$$, then $$\Gamma\in{\mathcal G}(n,p)$$ a.a.s. does not lie in $$\mathcal{AS}$$. Intriguingly, Kahle proved that a function similar to the critical density in Theorem 4.4 is the threshold for a random simplicial complex to have vanishing second rational cohomology [28]. Remark (Behavior near $$p=1$$). Note that property $$\mathcal{AS}$$ is not monotone increasing, since it requires the presence of a number of non-edges. In particular, complete graphs are not in $$\mathcal{AS}$$. Thus unlike the global properties typically studied in the theory of random graphs, $$\mathcal{AS}$$ will cease to hold a.a.s. when the density $$p$$ is very close to $$1$$. In fact, [8, Theorem 3.9] shows that if $$p(n)=1-\Omega\left(\frac{1}{n^2}\right)$$, then a.a.s. $$\Gamma$$ is either a clique or a clique minus a fixed number of edges whose endpoints are all disjoint. Thus, with positive probability, $$\Gamma\in\mathcal{AS}$$. However, [14, Theorem 1] shows that if $$(1-p)n^2\to 0$$ then $$\Gamma$$ is asymptotically almost surely a clique, and hence not in $$\mathcal{AS}$$. □ $$\boldsymbol{\mathcal{CFS}}$$ graphs The second family of graphs, which we call $$\mathcal{CFS}$$ graphs (“Constructed From Squares”), arise naturally in geometric group theory in the context of the large–scale geometry of right–angled Coxeter groups, as we explain below and in Section 3. A special case of these graphs was introduced by Dani–Thomas to study divergence in triangle-free right-angled Coxeter groups [17]. The graphs we study are intimately related to a property called thickness, a feature of many key examples in geometric group theory and low dimensional topology that is, closely related to divergence, relative hyperbolicity, and a number of other topics. This property is, in essence, a connectivity property because it relies on a space being “connected” through sequences of “large” subspaces. Roughly speaking, a graph is $$\mathcal{CFS}$$ if it can be built inductively by chaining (induced) squares together in such a way that each square overlaps with one of the previous squares along a diagonal (see Section 2 for a precise definition). We explain in the next section how this class of graphs generalizes $$\mathcal{AS}$$. Our next result about genericity of $$\mathcal{CFS}$$ combines with Proposition 3.1 below to significantly strengthen [8, Theorem VI]. This result is an immediate consequence of Theorems 5.1 and 5.7, which, in fact, establish slightly more precise, but less concise, bounds. Theorem 5.1 and 5.7 Suppose $$(1-p)n^2\to\infty$$ and let $$\epsilon>0$$. Then $$\Gamma\in{\mathcal G}(n,p)$$ is a.a.s. in $$\mathcal{CFS}$$ whenever $$p(n)>n^{-\frac{1}{2}+\epsilon}$$. Conversely, $$\Gamma\in{\mathcal G}(n,p)$$ is a.a.s. not in $$\mathcal{CFS}$$, whenever $$p(n)<n^{-\frac{1}{2} -\epsilon}.$$ We actually show, in Theorem 5.1, that at densities above $$5\sqrt{\frac{\log n}{n}}$$, with $$(1-p)n^2\to\infty$$, the random graph is a.a.s. in $$\mathcal{CFS}$$, while in Theorem 5.7 we show a random graph a.a.s. not in $$\mathcal{CFS}$$ at densities below $$\frac{1}{\sqrt{n}\log{n}}$$. Theorem 5.1 applies to graphs in a range strictly larger than that in which Theorem 4.4 holds (though our proof of Theorem 5.1 relies on Theorem 4.4 to deal with the large $$p$$ case). Theorem 5.1 combines with Theorem 4.5 to show that, for densities between $$\left(\log n/n\right)^{\frac{1}{2}}$$ and $$\left(\log n/n\right)^{\frac{1}{3}}$$, a random graph is asymptotically almost surely in $$\mathcal{CFS}$$ but not in $$\mathcal{AS}$$. We also note that Babson–Hoffman–Kahle [3] proved that a function of order $$n^{-\frac{1}{2}}$$ appears as the threshold for simple-connectivity in the Linial–Meshulam model for random 2–complexes [29]. It would be interesting to understand whether there is a connection between genericity of the $$\mathcal{CFS}$$ property and the topology of random 2-complexes. Unlike our results for the $$\mathcal{AS}$$ property, we do not establish a sharp threshold for the $$\mathcal{CFS}$$ property. In fact, we believe that neither the upper nor lower bounds, given in Theorem 5.1 and Theorem 5.7, for the critical density around which $$\mathcal{CFS}$$ goes from a.a.s. not holding to a.a.s. holding are sharp. Indeed, we believe that there is a sharp threshold for the $$\mathcal{CFS}$$ property located at $$p_c(n)=\theta(n^{-\frac{1}{2}})$$. This conjecture is linked to the emergence of a giant component in the “square graph” of $$\Gamma$$ (see the next section for a definition of the square graph and the heuristic discussion after the proof of Theorem 5.7). Applications to geometric group theory Our interest in the structure of random graphs was sparked largely by questions about the large-scale geometry of right-angled Coxeter groups. Coxeter groups were first introduced in [15] as a generalization of reflection groups, that is, discrete groups generated by a specified set of reflections in Euclidean space. A reflection group is right-angled if the reflection loci intersect at right angles. An abstract right-angled Coxeter group generalizes this situation: it is defined by a group presentation in which the generators are involutions and the relations are obtained by declaring some pairs of generators to commute. Right-angled Coxeter groups (and more general Coxeter groups) play an important role in geometric group theory and are closely-related to some of that field’s most fundamental objects, for example, CAT(0) cube complexes [18, 27, 33] and (right-angled) Bruhat-Tits buildings (see e.g., [18]). A right-angled Coxeter group is determined by a unique finite simplicial presentation graph: the vertices correspond to the involutions generating the group, and the edges encode the pairs of generators that commute. In fact, the presentation graph uniquely determines the right-angled Coxeter group [32]. In this article, as an application of our results on random graphs, we continue the project of understanding large-scale geometric features of right-angled Coxeter groups in terms of the combinatorics of the presentation graph, begun in [8, 14, 17]. Specifically, we study right-angled Coxeter groups defined by random presentation graphs, focusing on the prevalence of two important geometric properties: relative hyperbolicity and thickness. Relative hyperbolicity, in the sense introduced by Gromov and equivalently formulated by many others [10, 22, 24, 35], when it holds, is a powerful tool for studying groups. On the other hand, thickness of a finitely-generated group (more generally, a metric space) is a property introduced by Behrstock–Druţu–Mosher in [6] as a geometric obstruction to relative hyperbolicity and has a number of powerful geometric applications. For example, thickness gives bounds on divergence (an important quasi-isometry invariant of a metric space) in many different groups and spaces [5, 7, 11, 17, 37]. Thickness is an inductive property: in the present context, a finitely generated group $$G$$ is thick of order $$0$$ if and only if it decomposes as the direct product of two infinite subgroups. The group $$G$$ is thick of order $$n$$ if there exists a finite collection $$\mathcal H$$ of undistorted subgroups of $$G$$, each thick of order $$n-1$$, whose union generates a finite-index subgroup of $$G$$ and which has the following “chaining” property: for each $$g,g'\in G$$, one can construct a sequence $$g\in g_1H_1,g_2H_2,\ldots,g_kH_k\ni g'$$ of cosets, with each $$H_i\in\mathcal H$$, so that consecutive cosets have infinite coarse intersection. Many of the best-known groups studied by geometric group theorists are thick, and indeed thick of order $$1$$: one-ended right-angled Artin groups, mapping class groups of surfaces, outer automorphism groups of free groups, fundamental groups of three-dimensional graph manifolds, etc. [6]. The class of Coxeter groups contains many examples of hyperbolic and relatively hyperbolic groups. There is a criterion for hyperbolicity purely in terms of the presentation graph due to Moussong [31] and an algebraic criterion for relative hyperbolicity due to Caprace [13]. The class of Coxeter groups includes examples which are non-relatively hyperbolic, for instance, those constructed by Davis–Januszkiewicz [19] and, also, ones studied by Dani–Thomas [17]. In fact, in [8], this is taken further: it is shown that every Coxeter group is actually either thick, or hyperbolic relative to a canonical collection of thick Coxeter subgroups. Further, there is a simple, structural condition on the presentation graph, checkable in polynomial time, which characterizes thickness. This result is needed to deduce the applications below from our graph theoretic results. Charney and Farber initiated the study of random graph products (including right-angled Artin and Coxeter groups) using the Erdős-Rényi model of random graphs [14]. The structure of the group cohomology of random graph products was obtained in [20]. In [8], various results are proved about which random graphs have the thickness property discussed above, leading to the conclusion that, at certain low densities, random right-angled Coxeter groups are relatively hyperbolic (and thus not thick), while at higher densities, random right-angled Coxeter groups are thick. In this article, we improve significantly on one of the latter results, and also prove something considerably more refined: we isolate not just thickness of random right-angled Coxeter groups, but thickness of a specified order, namely $$1$$: Corollary 3.2 (Random Coxeter groups are thick of order 1.)There exists a constant $$C>0$$ such that if $$p\colon{\mathbb{N}}\rightarrow(0,1)$$ satisfies $$\left(\displaystyle\frac{C\log{n}}{n}\right)^{\frac{1}{2}}\leq p(n)\leq 1-\displaystyle\frac{(1+\epsilon)\log{n}}{n}$$ for some $$\epsilon>0$$, then the random right-angled Coxeter group $$W_{G_{n,p}}$$ is asymptotically almost surely thick of order exactly $$1$$, and in particular has quadratic divergence. Corollary 3.2 significantly improves on Theorem 3.10 of [8], as discussed in Section 3. This theorem follows from Theorems 5.1 and 4.4, the latter being needed to treat the case of large $$p(n)$$, including the interesting special case in which $$p$$ is constant. Remark 1.1. We note that characterizations of thickness of right-angled Coxeter groups in terms of the structure of the presentation graph appear to generalize readily to graph products of arbitrary finite groups and, probably, via the action on a cube complex constructed by Ruane and Witzel in [36], to arbitrary graph products of finitely generated abelian groups, using appropriate modifications of the results in [8]. □ Organization of the article In Section 2, we give the formal definitions of $$\mathcal{AS}$$ and $$\mathcal{CFS}$$ and introduce various other graph-theoretic notions we will need. In Section 3, we discuss the applications of our random graph results to geometric group theory, in particular, to right-angled Coxeter groups and more general graph products. In Section 4, we obtain a sharp threshold result for $$\mathcal{AS}$$ graphs. Section 5 is devoted to $$\mathcal{CFS}$$ graphs. Finally, Section 6 contains some simulations of random graphs with density near the threshold for $$\mathcal{AS}$$ and $$\mathcal{CFS}$$. 2 Definitions Convention 2.1 A graph is a pair of finite sets $$\Gamma=(V,E)$$, where $$V=V(\Gamma)$$ is a set of vertices, and $$E=E(\Gamma)$$ is a collection of pairs of distinct elements of $$V$$, which constitute the set of edges of $$G$$. A subgraph of $$\Gamma$$ is a graph $$\Gamma'$$ with $$V(\Gamma')\subseteq V(\Gamma)$$ and $$E(\Gamma')\subseteq E(\Gamma)$$; $$\Gamma'$$ is said to be an induced subgraph of $$\Gamma$$ if $$E(\Gamma')$$ consists exactly of those edges from $$E(\Gamma)$$ whose vertices lie in $$V(\Gamma')$$. In this article, we focus on induced subgraphs, and we generally write “subgraph” to mean “induced subgraph”. In particular, we often identify a subgraph with the set of vertices inducing it, and we write $$\vert \Gamma\vert$$ for the order of $$\Gamma$$, that is, the number of vertices it contains. A clique of size $$t$$ is a complete graph on $$t\geq0$$ vertices. This includes the degenerate case of the empty graph on $$t= 0$$ vertices. □ Fig. 1. View largeDownload slide A graph in $$\mathcal{AS}$$. A block exhibiting inclusion in $$\mathcal{AS}$$ is shown in bold; the two (left-centrally located) ends of the bold block are highlighted. (A colour version of this figure is available from the journal’s website.) Fig. 1. View largeDownload slide A graph in $$\mathcal{AS}$$. A block exhibiting inclusion in $$\mathcal{AS}$$ is shown in bold; the two (left-centrally located) ends of the bold block are highlighted. (A colour version of this figure is available from the journal’s website.) Definition 2.2 (Link, join). Given a graph $$\Gamma$$, the link of a vertex $$v\in\Gamma$$, denoted $${\mathrm Lk}_\Gamma(v)$$, is the subgraph spanned by the set of vertices adjacent to $$v$$. Given graphs $$A,B$$, the join$$A\star B$$ is the graph formed from $$A\sqcup B$$ by joining each vertex of $$A$$ to each vertex of $$B$$ by an edge. A suspension is a join where one of the factors $$A,B$$ is the graph consisting of two vertices and no edges. □ We now describe a family of graphs, denoted $$\mathcal{CFS}$$, which satisfy the global structural property that they are “constructed from squares.” Definition 2.3 ($$\boldsymbol{\mathcal{CFS}}$$). Given a graph $$\Gamma$$, let $$\square(\Gamma)$$ be the auxiliary graph whose vertices are the induced $$4$$–cycles from $$\Gamma$$, with two distinct $$4$$–cycles joined by an edge in $$\square(\Gamma)$$ if and only if they intersect in a pair of non-adjacent vertices of $$\Gamma$$ (i.e., in a diagonal). We refer to $$\square(\Gamma)$$ as the square-graph of $$\Gamma$$. A graph $$\Gamma$$ belongs to $$\mathcal{CFS}$$ if $$\Gamma=\Gamma'\star K$$, where $$K$$ is a (possibly empty) clique and $$\Gamma'$$ is a non-empty subgraph such that $$\square(\Gamma')$$ has a connected component $$C$$ such that the union of the $$4$$–cycles from $$C$$ covers all of $$V(\Gamma')$$. Given a vertex $$F\in\square(\Gamma)$$, we refer to the vertices in the $$4$$–cycle in $$\Gamma$$ associated to $$F$$ as the support of $$F$$. □ Remark 2.4. Dani–Thomas introduced component with full support graphs in [17], a subclass of the class of triangle-free graphs. We note that each component with full support graph is constructed from squares, but the converse is not true. Indeed, since we do not require our graphs to be triangle-free, our definition necessarily only counts induced 4–cycles and allows them to intersect in more ways than in [17]. This distinction is relevant to the application to Coxeter groups, which we discuss in Section 3. □ Definition 2.5 (Augmented suspension). The graph $$\Gamma$$ is an augmented suspension if it contains an induced subgraph $$B=\{w,w'\}\star \Gamma'$$, where $$w,w'$$ are nonadjacent and $$\Gamma'$$ is not a clique, satisfying the additional property that if $$v\in\Gamma-B$$, then $$\mathrm{Lk}_\Gamma(v)\cap \Gamma'$$ is not a clique. Let $$\mathcal{AS}$$ denote the class of augmented suspensions. Figure 1 shows a graph in $$\mathcal{AS}$$. □ Remark 2.6. Neither the $$\mathcal{CFS}$$ nor the $$\mathcal{AS}$$ properties introduced above are monotone with respect to the addition of edges. This stands in contrast to the most commonly studied global properties of random graphs. □ Definition 2.7 (Block, core, ends). A block in $$\Gamma$$ is a subgraph of the form $$B(w,w')=\{w,w'\}\star \Gamma'$$ where $$\{w,w'\}$$ is a pair of non-adjacent vertices and $$\Gamma'\subset \Gamma$$ is a subgraph of $$\Gamma$$ induced by a set of vertices adjacent to both $$w$$ and $$w'$$. A block is maximal if $$V(\Gamma')=\mathrm{Lk}_{\Gamma}(w)\cap \mathrm{Lk}_{\Gamma}(w')$$. Given a block $$B=B(w,w')$$, we refer to the non-adjacent vertices $$w,w'$$ as the ends of $$B$$, denoted $$\mathrm{end}(B)$$, and the vertices of $$\Gamma'$$ as the core of $$B$$, denoted $$\mathrm{core}(B)$$. □ Note that $$\mathcal{AS}\subsetneq\mathcal{CFS}$$, indeed Theorems 5.1 and 4.5 show that there must exist graphs in $$\mathcal{CFS}$$ that are not in $$\mathcal{AS}$$. Here we explain how any graph in $$\mathcal{AS}$$ is in $$\mathcal{CFS}$$. Lemma 2.8. Let $$\Gamma$$ be a graph in $$\mathcal{AS}$$. Then $$\Gamma \in \mathcal{CFS}$$. □ Proof Let $$B(w,w')=\{w,w'\}\star \Gamma'$$ be a maximal block in $$\Gamma$$ witnessing $$\Gamma \in \mathcal{AS}$$. Write $$\Gamma'=A\star D$$, where $$D$$ is the collection of all vertices of $$\Gamma'$$ which are adjacent to every other vertex of $$\Gamma'$$. Note that $$D$$ induces a clique in $$\Gamma$$. By definition of the $$\mathcal{AS}$$ property $$\Gamma'$$ is not a clique, whence, $$A$$ contains at least one pair of non-adjacent vertices. Furthermore, by the definition of $$D$$, for every vertex $$a\in A$$ there exists $$a' \in A$$ with $$\{a,a'\}$$ non-adjacent. The $$4$$–cycles induced by $$\{w,w'a,a'\}$$ for non-adjacent pairs $$a, a'$$ from $$A$$ are connected in $$\square(\Gamma)$$. Denote the component of $$\square(\Gamma)$$ containing them by $$C$$. Consider now a vertex $$v\in \Gamma - B(w,w')$$. Since $$B(w,w')$$ is maximal, we have that at least one of $$w,w'$$ is not adjacent to $$v$$ — without loss of generality, let us assume that it is $$w$$. By the $$\mathcal{AS}$$ property, $$v$$ must be adjacent to a pair $$a,a'$$ of non-adjacent vertices from $$A$$. Then $$\{w, a,a', v\}$$ induces a $$4$$–cycle, which is adjacent to $$\{w,w',a,a'\}\in\square(\Gamma)$$ and hence lies in $$C$$. Finally, consider a vertex $$d \in D$$. If $$v$$ is adjacent to all vertices of $$\Gamma$$, then $$\Gamma$$ is the join of a graph with a clique containing $$d$$, and we can ignore $$d$$ with respect to establishing the $$\mathcal{CFS}$$ property. Otherwise $$d$$ is not adjacent to some $$v\in \Gamma-B(w,w')$$. By the $$\mathcal{AS}$$ property, $$v$$ is connected by edges to a pair of non-adjacent vertices $$\{a,a'\}$$ from $$A$$. Thus $$\{d,a,a',v\}$$ induces a $$4$$–cycle. Since (as established above) there is some $$4$$–cycle in $$C$$ containing $$\{a,a',v\}$$, we have that $$\{d,a,a',v\} \in C$$ as well. Thus $$\Gamma= \Gamma'' \star K$$, where $$K$$ is a clique and $$V(\Gamma'')$$ is covered by the union of the $$4$$–cycles in $$C$$, so that $$\Gamma \in \mathcal{CFS}$$ as claimed. ■ 3 Geometry of right-angled Coxeter groups If $$\Gamma$$ is a finite simplicial graph, the right-angled Coxeter group$$W_\Gamma$$presented by$$\Gamma$$ is the group defined by the presentation ⟨Vert(Γ)∣{w2,uvu−1v−1:u,v,w∈Vert(Γ),{u,v}∈Edge(Γ)⟩. A result of Mühlherr [32] shows that the correspondence $$\Gamma\leftrightarrow W_\Gamma$$ is bijective. We can thus speak of “the random right-angled Coxeter group” — it is the right-angled Coxeter group presented by the random graph. (We emphasize that the above presentation provides the definition of a right-angled Coxeter group: this definition abstracts the notion of a reflection group – a subgroup of a linear group generated by reflections — but infinite Coxeter groups need not admit representations as reflection groups.) Recent articles have discussed the geometry of Coxeter groups, especially relative hyperbolicity and closely-related quasi-isometry invariants like divergence and thickness, cf. [8, 13, 17]. In particular, Dani–Thomas introduced a property they call having a component of full support for triangle-free graphs (which is exactly the triangle-free version of $$\mathcal{CFS}$$) and they prove that under the assumption $$\Gamma$$ is triangle-free, $$W_\Gamma$$ is thick of order at most $$1$$ if and only if it has quadratic divergence if and only if $$\Gamma$$ is in $$\mathcal{CFS}$$, see [17, Theorem 1.1 and Remark 4.8]. Since the densities where random graphs are triangle-free are also square-free (and thus not $$\mathcal{CFS}$$ — in fact, they are disconnected!), we need the following slight generalization of the result of Dani–Thomas: Proposition 3.1. Let $$\Gamma$$ be a finite simplicial graph. If $$\Gamma$$ is in $$\mathcal{CFS}$$ and $$\Gamma$$ does not decompose as a nontrivial join, then $$W_\Gamma$$ is thick of order exactly $$1$$. □ Proof Theorem II of [8] shows immediately that, if $$\Gamma\in\mathcal{CFS}$$, then $$W_\Gamma$$ is thick, being formed by a series of thick unions of $$4$$–cycles; since each $$4$$–cycle is a join, it follows that $$\Gamma$$ is thick of order at most $$1$$. On the other hand, [8, Proposition 2.11] shows that $$W_\Gamma$$ is thick of order at least $$1$$ provided $$\Gamma$$ is not a join. ■ Our results about random graphs yield: Corollary 3.2. There exists $$k>0$$ so that if $$p\colon{\mathbb{N}}\rightarrow(0,1)$$ and $$\epsilon>0$$ are such that $$\sqrt{\frac{k\log n}{n}}\leq p(n)\leq 1-\displaystyle\frac{(1-\epsilon)\log{n}}{n}$$ for all sufficiently large $$n$$, then for $$\Gamma\in{\mathcal G}(n,p)$$ the group $$W_\Gamma$$ is asymptotically almost surely thick of order exactly $$1$$ and hence has quadratic divergence. □ Proof Theorem 5.1 shows that any such $$\Gamma$$ is asymptotically almost surely in $$\mathcal{CFS}$$, whence $$W_\Gamma$$ is thick of order at most $$1$$. We emphasize that to apply this result for sufficiently large functions $$p(n)$$ the proof of Theorem 5.1 requires an application of Theorem 4.4 to establish that $$\Gamma$$ is a.a.s. in $$\mathcal{AS}$$ and hence in $$\mathcal{CFS}$$ by Lemma 2.8. By Proposition 3.1, to show that the order of thickness is exactly one, it remains to rule out the possibility that $$\Gamma$$ decomposes as a nontrivial join. However, this occurs if and only if the complement graph is disconnected, which asymptotically almost surely does not occur whenever $$p(n)\le 1-\frac{(1-\epsilon)\log{n}}{n}$$, by the sharp threshold for connectivity of $${\mathcal G}(n,1-p)$$ established by Erdős and Rényi in [21]. Since this holds for $$p(n)$$ by assumption, we conclude that asymptotically almost surely, $$W_\Gamma$$ is thick of order at least $$1$$. Since $$W_\Gamma$$ is CAT(0) and thick of order exactly $$1$$, the consequence about divergence now follows from [5]. ■ This corollary significantly generalizes Theorem 3.10 of [8], which established that, if $$\Gamma\in{\mathcal G}(n,\frac{1}{2})$$, then $$W_\Gamma$$ is asymptotically almost surely thick. Theorem 3.10 of [8] does not provide effective bounds on the order of thickness and its proof is significantly more complicated than the proof of Corollary 3.2 given above — indeed, it required several days of computation (using 2013 hardware) to establish the base case of an inductive argument. Remark 3.3 (Higher-order thickness). A lower bound of $$p(n)=n^{-\frac{5}{6}}$$ for membership in a larger class of graphs whose corresponding Coxeter groups are thick can be found in [8, Theorem 3.4]. In fact, this argument can be adapted to give a simple proof that a.a.s. thickness does not occur at densities below $$n^{-\frac{3}{4}}$$. The correct threshold for a.a.s. thickness is, however, unknown. □ Remark 3.4 (Random graph products versus random presentations). Corollary 3.2 and Remark 3.3 show that the random graph model for producing random right-angled Coxeter groups generates groups with radically different geometric properties. This is in direct contrast to other methods of producing random groups, most notably Gromov’s random presentation model [25, 26] where, depending on the density of relators, groups are either almost surely hyperbolic or finite (with order at most $$2$$). This contrast speaks to the merits of considering a random right-angled Coxeter group as a natural place to study random groups. For instance, Calegari–Wilton recently showed that in the Gromov model a random group contains many subgroups which are isomorphic to the fundamental group of a compact hyperbolic 3–manifold [12]; does the random right-angled Coxeter group also contain such subgroups? Right-angled Coxeter groups, and indeed thick ones, are closely related to Gromov’s random groups in another way. When the parameter for a Gromov random groups is $$<\frac{1}{6}$$ such a group is word-hyperbolic [25] and acts properly and cocompactly on a CAT(0) cube complex [34]. Hence the Gromov random group virtually embeds in a right-angled Artin group [4]. Moreover, at such parameters such a random group is one-ended [16], whence the associated right-angled Artin group is as well. By [4] this right-angled Artin group is thick of order 1. Since any right-angled Artin group is commensurable with a right-angled Coxeter group [19], one obtains a thick of order $$1$$ right-angled Coxeter group containing the randomly presented group. □ 4 Genericity of $$\boldsymbol{\mathcal{AS}}$$ We will use the following standard Chernoff bounds (see e.g., [2, Theorems A.1.11 and A.1.13]): Lemma 4.1 (Chernoff bounds). Let $$X_1,\ldots,X_n$$ be independent identically distributed random variables taking values in $$\{0,1\}$$, let $$X$$ be their sum, and let $$\mu=\mathbb E[X]$$. Then for any $$\delta\in(0,2/3)$$ P(|X−μ|≥δμ)≤2e−δ2μ3. □ Corollary 4.2. Let $$\varepsilon, \delta>0$$ be fixed. (i) If $$p(n)\ge \left(\frac{(6+\varepsilon)\log n}{\delta^2n}\right)^{1/2}$$, then a.a.s. for all pairs of distinct vertices $$\{x,y\}$$ in $$\Gamma \in {\mathcal G}(n,p)$$ we have $$\left\vert \vert \mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)\vert-p^2(n-2)\right\vert < \delta p^2(n-2)$$. (ii) If $$p(n)\ge \left(\frac{(9+\varepsilon)\log n}{\delta^2n}\right)^{1/3}$$, then a.a.s. for all triples of distinct vertices $$\{x,y, z\}$$ in $$\Gamma \in {\mathcal G}(n,p)$$ we have $$\left\vert \vert \mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)\cap \mathrm{Lk}_{\Gamma}(z)\vert-p^3(n-3)\right\vert < \delta p^3(n-3)$$. □ Proof For (i), let $$\{x,y\}$$ be any pair of distinct vertices. For each vertex $$v \in \Gamma-\{x,y\}$$, set $$X_v$$ to be the indicator function of the event that $$v\in \mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)$$, and set $$X=\sum_v X_v$$ to be the size of $$\mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)$$. We have $$\mathbb{E}X=p^2(n-2)$$ and so by the Chernoff bounds above, $$\Pr\left(\vert X-p^2(n-2)\vert\geq \delta p^2(n-2)\right) \leq 2e^{-\frac{\delta^2p^2(n-2)}{3}}$$. Applying Markov’s inequality, the probability that there exists some “bad pair” $$\{x,y\}$$ in $$\Gamma$$ for which $$\vert\mathrm{Lk}_{\Gamma}(x)\cap \mathrm{Lk}_{\Gamma}(y)\vert$$ deviates from its expected value by more than $$\delta p^2(n-2)$$ is at most (n2)2e−δ2p2(n−2)3=o(1), provided $$\delta^2 p^2n\ge (6+\varepsilon) \log n$$ and $$\varepsilon, \delta>0$$ are fixed. Thus for this range of $$p=p(n)$$, a.a.s. no such bad pair exists. The proof of (ii) is nearly identical. ■ Lemma 4.3. (i) Suppose $$1-p\geq \frac{\log n}{2n}$$. Then asymptotically almost surely, the order of a largest clique in $$\Gamma\in{\mathcal G}(n,p)$$ is $$o(n)$$. (ii) Let $$\eta$$ be fixed with $$0<\eta<1$$. Suppose $$1-p\geq \eta$$. Then asymptotically almost surely, the order of a largest clique in $$\Gamma \in {\mathcal G}(n,p)$$ is $$O(\log n)$$. □ Proof For (i), set $$r=\alpha n$$, for some $$\alpha$$ bounded away from $$0$$. Write $$H(\alpha)=\alpha\log\frac{1}{\alpha}+(1-\alpha)\log\frac{1}{1-\alpha}$$. Using the standard entropy bound $$\binom{n}{\alpha n}\leq e^{H(\alpha)n}$$ and our assumption for $$(1-p)$$, we see that the expected number of $$r$$-cliques in $$\Gamma$$ is (nr)p(r2) ≤eH(α)nelog(1−(1−p))(α2n22+O(n))≤exp(−α22nlogn+O(n))=o(1). Thus by Markov’s inequality, a.a.s. $$\Gamma$$ does not contain a clique of size $$r$$, and the order of a largest clique in $$\Gamma$$ is $$o(n)$$. The proof of (ii) is similar: suppose $$1-p>\eta$$. Then for any $$r\leq n$$, (nr)p(r2) <nr(1−η)r(r−1)/2=exp(r(logn−r−12log11−η)), which for $$\eta>0$$ fixed and $$r-1>\frac{2}{\log (1/(1-\eta))}(1+ \log n )$$ is as most $$n^{-\frac{2}{\log (1/(1-\eta))}}=o(1)$$. We may thus conclude as above that a.a.s. a largest clique in $$\Gamma$$ has order $$O(\log n)$$. ■ Theorem 4.4 (Genericity of $$\mathcal{AS}$$). Suppose $$p(n)\ge(1+\epsilon)\left(\displaystyle\frac{\log{n}}{n}\right)^{\frac{1}{3}}$$ for some $$\epsilon>0$$ and $$(1-p)n^2\to\infty$$. Then, a.a.s. $$\Gamma\in{\mathcal G}(n,p)$$ is in $$\mathcal{AS}$$. □ Proof Let $$\delta>0$$ be a small constant to be specified later (the choice of $$\delta$$ will depend on $$\epsilon$$). By Corollary 4.2 (i) for $$p(n)$$ in the range we are considering, a.a.s. all joint links have size at least $$(1-\delta)p^2(n-2)$$. Denote this event by $$\mathcal{E}_1$$. We henceforth condition on $$\mathcal{E}_1$$ occurring (not this only affects the values of probabilities by an additive factor of $$\mathbb{P}(\mathcal{E}_1^c)=O(n^{-\varepsilon})=o(1)$$). With probability $$1-p^{\binom{n}{2}}=1-o(1)$$, $$\Gamma$$ is not a clique, whence, there there exist non-adjacent vertices in $$\Gamma$$. We henceforth assume $$\Gamma\neq K_n$$, and choose $$v_1, v_2\in \Gamma$$ which are not adjacent. Let $$B$$ be the maximal block associated with the pair $$(v_1,v_2)$$. We separate the range of $$p$$ into three. Case 1: $$\boldsymbol{p}$$ is “far” from both the threshold and $$1$$. Let $$\alpha>0$$ be fixed, and suppose $$\alpha n^{-1/4}\leq p \leq 1- \frac{\log n}{2n}$$. Let $$\mathcal{E}_2$$ be the event that for every vertex $$v\in\Gamma-B$$ the set $$\mathrm{Lk}_{\Gamma}(v)\cap B$$ has size at least $$\frac{1}{2}p^3(n-3)$$. By Corollary 4.2, a.a.s. event $$\mathcal{E}_2$$ occurs, that is, all vertices in $$\Gamma-B$$ have this property. We claim that a.a.s. there is no clique of order at least $$\frac{1}{2}p^3(n-3)$$ in $$\Gamma$$. Indeed, if $$p<1-\eta$$ for some fixed $$\eta>0$$, then by Lemma 4.3 part (ii), a largest clique in $$\Gamma$$ has order $$O(\log n)=o(p^3n)$$. On the other hand, if $$1-\eta <p\leq 1- \frac{\log n}{2n}$$, then by Lemma 4.3 part (i), a largest clique in $$\Gamma$$ has order $$o(n)=o(p^3n)$$. Thus in either case a.a.s. for every$$v\in \Gamma-B$$, $$\mathrm{Lk}_{\Gamma}(v)\cap B$$ is not a clique and hence $$v \in \overline{B}$$, so that a.a.s. $$\overline{B}=\Gamma$$, and $$\Gamma \in \mathcal{AS}$$ as required. Case 2: $$\boldsymbol{p}$$ is “close” to the threshold. Suppose that $$(1+\epsilon)\left(\displaystyle\frac{\log{n}}{n}\right)^{\frac{1}{3}}\le p(n)$$ and $$np^4\to 0$$. Let $$\vert B\vert=m+2$$. By our conditioning, we have $$(1-\delta)(n-2)p^2\leq m \leq (1+\delta)(n-2)p^2$$. The probability that a given vertex $$v\in \Gamma$$ is not in $$\overline{B}$$ is given by: P(v∉B¯|{|B|=m})=(1−p)m+mp(1−p)m+1+∑r=2m(mr)pr(1−p)m−rp(r2). (1) In this equation, the first two terms come from the case where $$v$$ is connected to $$0$$ and $$1$$ vertex in $$B\setminus\{v_1,v_2\}$$ respectively, while the third term comes from the case where the link of $$v$$ in $$B\setminus\{v_1,v_2\}$$ is a clique on $$r\geq 2$$ vertices. As we shall see, in the case $$np^4\to 0$$ which we are considering, the contribution from the first two terms dominates. Let us estimate their order: (1−p)m+mp(1−p)m−1=(1+mp1−p)(1−p)m ≤(1+mp1−p)e−mp ≤(1+(1−δ)(1+ϵ)3logn1−p)n−(1−δ)(1+ϵ)3. Taking $$\delta<1-\frac{1}{(1+\epsilon)^3}$$ this expression is $$o(n^{-1})$$. We now treat the sum making up the remaining terms in Equation 1. To do so, we will analyze the quotient of successive terms in the sum. Fixing $$2\le r\le m-1$$ we see: (mr+1)pr+1(1−p)m−r−1p(r+12)(mr)pr(1−p)m−rp(r2)=m−r−1r+1⋅pr+11−p≤mpr+1≤mp3. Since $$np^4\to 0$$ (by assumption), this also tends to zero as $$n\to\infty$$. The quotients of successive terms in the sum thus tend to zero uniformly as $$n\to \infty$$, and we may bound the sum by a geometric series: ∑r=2m(mr)pr(1−p)m−rp(r2)≤(m2)p3(1−p)m−2∑i=0m−2(mp3)i≤(12+o(1))m2p3(1−p)m−2. Now, $$m^2p^3(1-p)^{m-2}=\frac{mp^2}{1-p}\cdot mp(1-p)^{m-1}$$. The second factor in this expression was already shown to be $$o(n^{-1})$$, while $$mp^2\le(1+\delta)np^4 \to 0$$ by assumption, so the total contribution of the sum is $$o(n^{-1})$$. Thus for any value of $$m$$ between $$(1-\delta)p^2(n-2)$$ and $$(1+\delta)p^2(n-2)$$, the right hand side of Equation (1) is $$o(n^{-1})$$, and we conclude: P(v∉B¯|E1)≤o(n−1). Thus, by Markov’s inequality, P(B¯=Γ)≥P(E1)(1−∑vP(v∉B¯|E1))=1−o(1), establishing that a.a.s. $$\Gamma\in\mathcal{AS}$$, as claimed. Case 3: $$p$$ is “close” to $$1$$. Suppose $$n^{-2}\ll (1-p)\leq \frac{\log n}{2n}$$. Consider the complement of $$\Gamma$$, $$\Gamma^c\in {\mathcal G}(n, 1-p)$$. In the range of the parameter $$\Gamma^c$$ a.a.s. has at least two connected components that contain at least two vertices. In particular, taking complements, we see that $$\Gamma$$ is a.a.s. a join of two subgraphs, neither of which is a clique. It is a simple exercise to see that such as graph is in $$\mathcal{AS}$$, thus a.a.s. $$\Gamma\in\mathcal{AS}$$. ■ As we now show, the bound obtained in the above theorem is actually a sharp threshold. Analogous to the classical proof of the connectivity threshold [21], we consider vertices which are “isolated” from a block to prove that graphs below the threshold strongly fail to be in $$\mathcal{AS}$$. Theorem 4.5. If $$p\le\left(1-\epsilon\right)\left(\displaystyle\frac{\log{n}}{n}\right)^{\frac{1}{3}}$$ for some $$\epsilon>0$$, then $$\Gamma\in{\mathcal G}(n,p)$$ is asymptotically almost surely not in $$\mathcal{AS}$$. □ Proof We will show that, for $$p$$ as hypothesized, every block has a vertex “isolated” from it. Explicitly, let $$\Gamma\in{\mathcal G}(n,p)$$ and consider $$B=B_{v,w}=\mathrm{Lk}(v)\cap \mathrm{Lk}(w)\cup\{v,w\}$$. Let $$X(v,w)$$ be the event that every vertex of $$\Gamma-B$$ is connected by an edge to some vertex of $$B$$. Clearly $$\Gamma\in \mathcal{AS}$$ only if the event $$X(v,w)$$ occurs for some pair of non-adjacent vertices $$\{v,w\}$$. Set $$X=\bigcup_{\{v,w\}} X(v,w)$$. Note that $$X$$ is a monotone event, closed under the addition of edges, so that the probability it occurs in $$\Gamma\in {\mathcal G}(n,p)$$ is a non-decreasing function of $$p$$. We now show that when $$p=\left(1-\epsilon\right)\left(\log{n}/ n\right)^{\frac{1}{3}}$$, a.a.s. $$X$$ does not occur, completing the proof. Consider a pair of vertices $$\{v,w\}$$, and set $$k=\vert B_{v,w}\vert$$. Conditional on $$B_{v,w}$$ having this size and using the standard inequality $$(1-x)\le e^{-x}$$, we have that P(X(v,w))=(1−(1−p)k)n−k≤e−(n−k)(1−p)k. Now, the value of $$k$$ is concentrated around its mean: by Corollary 4.2, for any fixed $$\delta>0$$ and all $$\{v,w\}$$, the order of $$B_{v,w}$$ is a.a.s. at most $$(1+\delta)np^2$$. Conditioning on this event $$\mathcal{E}$$, we have that for any pair of vertices $$v,w$$, P(X(v,w)|E)≤maxk≤(1+δ)np2e−(n−k)(1−p)k=e−(n−(1+δ)np2)(1−p)(1+δ)np2. Now $$(1-p)^{(1+\delta)np^2} = e^{(1+\delta)np^2\log(1-p)}$$ and by Taylor’s theorem $$\log(1-p)=-p+O(p^2)$$, so that: P(X(v,w)|E)≤e−n(1+O(p2))e−(1+δ)np3(1+O(p))=e−n1−(1+δ)(1−ϵ)3+o(1). Choosing $$\delta<\displaystyle\frac{1}{(1-\epsilon)^3}-1$$, the expression above is $$o(n^{-2})$$. Thus P(X) ≤P(Ec)+∑{v,w}P(X(v,w)|E) =o(1)+(n2)o(n−2)=o(1). Thus a.a.s. the monotone event $$X$$ does not occur in $$\Gamma\in{\mathcal G}(n,p)$$ for $$p=\left(1-\epsilon\right)\left(\log{n}/ n\right)^{\frac{1}{3}}$$, and hence a.a.s. the property $$\mathcal{AS}$$ does not hold for $$\Gamma\in{\mathcal G}(n,p)$$ and $$p(n)\leq \left(1-\epsilon\right)\left(\log{n}/ n\right)^{\frac{1}{3}}$$. ■ 5 Genericity of $$\boldsymbol{\mathcal{CFS}}$$ The two main results in this section are upper and lower bounds for inclusion in $$\mathcal{CFS}$$. These results are established in Theorems 5.1 and 5.7. Theorem 5.1. If $$p\colon{\mathbb{N}}\rightarrow(0,1)$$ satisfies $$(1-p)n^2\to\infty$$ and $$p(n)\geq 5\sqrt{\frac{\log n}{n}}$$ for all sufficiently large $$n$$, then a.a.s. $$\Gamma \in {\mathcal G}(n,p)$$ lies in $$\mathcal{CFS}$$. □ The proof of Theorem 5.1 divides naturally into two ranges. First of all for large $$p$$, namely for $$p(n)\geq 2\left(\log{n}/n\right)^{\frac{1}{3}}$$, we appeal to Theorem 4.4 to show that a.a.s. a random graph $$\Gamma\in {\mathcal G}(n,p)$$ is in $$\mathcal{AS}$$ and hence, by Lemma 2.8, in $$\mathcal{CFS}$$. In light of our proof of Theorem 4.4, we may think of this as the case when we can “beam up” every vertex of the graph $$\Gamma$$ to a single block $$B_{x,y}$$ in an appropriate way, and thus obtain a connected component of $$\square(\Gamma)$$ whose support is all of $$V(\Gamma)$$ Secondly we have the case of “small $$p$$” where 5lognn≤p(n)≤2(lognn)13, which is the focus of the remainder of the proof. Here we construct a path of length of order $$n/\log n$$ in $$\square(\Gamma)$$ on to which every vertex $$v\in V(\Gamma)$$ can be “beamed up” by adding a $$4$$–cycle whose support contains $$v$$. This is done in the following manner: we start with an arbitrary pair of non-adjacent vertices contained in a block $$B_{0}$$. We then pick an arbitrary pair of non-adjacent vertices in the block $$B_0$$ and let $$B_1$$ denote the intersection of the block they define with $$V(\Gamma)\setminus B_0$$. We repeat this procedure, to obtain a chain of blocks $$B_0, B_1, B_2, \ldots, B_t$$, with $$t=O(n/\log n)$$, whose union contains a positive proportion of $$V(\Gamma)$$, and which all belong to the same connected component $$C$$ of $$\square(\Gamma)$$. This common component $$C$$ is then large enough that every remaining vertex of $$V(\Gamma)$$ can be attached to it. The main challenge is showing that our process of recording which vertices are included in the support of a component of the square graph does not die out or slow down too much, that is, that the block sizes $$\vert B_i\vert$$ remains relatively large at every stage of the process and that none of the $$B_i$$ form a clique. Having described our strategy, we now fill in the details, beginning with the following upper bound on the probability of $$\Gamma\in {\mathcal G}(n,p)$$ containing a copy of $$K_{10}$$, the complete graph on $$10$$ vertices. The following lemma is a variant of [21, Corollary 4]: Lemma 5.2. Let $$\Gamma\in{\mathcal G}(n,p)$$. If $$p=o(n^{-\frac{1}{4}})$$, then the probability that $$\Gamma\in {\mathcal G}(n,p)$$ contains a clique with at least $$10$$ vertices is at most $$o(n^{-\frac{5}{4}})$$. □ Proof The expected number of copies of $$K_{10}$$ in $$\Gamma$$ is (n10)p(102)≤n10p45=o(n−5/4). The statement of the lemma then follows from Markov’s inequality. ■ Proof of Theorem 5.1. As remarked above, Theorem 4.4 proves Theorem 5.1 for “large” $$p$$, so we only need to deal with the case where 5lognn≤p(n)≤2(lognn)13. We iteratively build a chain of blocks, as follows. Let $$\{x_0,y_0\}$$ be a pair of non-adjacent vertices in $$\Gamma$$, if such a pair exists, and an arbitrary pair of vertices if not. Let $$B_0$$ be the block with ends $$\{x_0, y_0\}$$. Now assume we have already constructed the blocks $$B_0, \ldots, B_i$$, for $$i\geq 0$$. Let $$C_i=\bigcup_i B_i$$ (for convenience we let $$C_{-1}=\emptyset$$). We will terminate the process and set $$t=i$$ if any of the three following conditions occur: $$\vert \mathrm{core}(B_i)\vert\leq 6\log n$$ or $$i\geq n/6\log n$$ or $$\vert V(\Gamma)\setminus C_i\vert \leq n/2$$. Otherwise, we let $$\{x_{i+1}, y_{i+1}\}$$ be a pair of non-adjacent vertices in $$\mathrm{core}(B_i)$$, if such a pair exists, and an arbitrary pair of vertices from $$\mathrm{core}(B_i)$$ otherwise. Let $$B_{i+1}$$ denote the intersection of the block whose ends are $$\{x_{i+1}, y_{i+1}\}$$ and the set $$\left(V(\Gamma)\setminus(C_i)\right)\cup\{x_{i+1}, y_{i+1}\}$$. Repeat. Eventually this process must terminate, resulting in a chain of blocks $$B_0, B_1, \ldots, B_t$$. We claim that a.a.s. both of the following hold for every $$i$$ satisfying $$0 \leq i \leq t$$: (i) $$\vert\mathrm{core}(B_i)\vert > 6\log n$$; and (ii) $$\{x_i,y_i\}$$ is a non-edge in $$\Gamma$$. Part (i) follows from the Chernoff bound given in Lemma 4.1: for each $$i\geq -1$$ the set $$V(\Gamma)\setminus C_i$$ contains at least $$n/2$$ vertices by construction. For each vertex $$v\in V(\Gamma)\setminus C_i$$, let $$X_v$$ be the indicator function of the event that $$v$$ is adjacent to both of $$\{x_{i+1}, y_{i+1}\}$$. The random variables $$(X_v)$$ are independent identically distributed Bernoulli random variables with mean $$p$$. Their sum $$X=\sum_v X_v$$ is exactly the size of the core of $$B_{i+1}$$, and its expectation is at least $$p^2n/2$$. Applying Lemma 4.1, we get that P(X<6logn) ≤P(X≤12EX) ≤2e−(12)225logn6n=2e−2524logn. Thus the probability that $$\vert \mathrm{core}(B_i)\vert <6\log n$$ for some $$i$$ with $$0\leq i \leq t$$ is at most: t2e−2524logn≤4n5logn2e−2524logn=o(1). Part (ii) is a trivial consequence of part (i) and Lemma 5.2: a.a.s. $$\mathrm{core}(B_i)$$ has size at least $$6\log n$$ for every $$i$$ with $$0 \leq i \leq t$$, and a.a.s. $$\Gamma$$ contains no clique on $$10 < \log n$$ vertices, so that a.a.s. at each stage of the process we could choose an non-adjacent pair $$\{x_i,y_i\}$$. From now on we assume that both (i) and (ii) occur, and that $$\Gamma$$ contains no clique of size $$10$$. In addition, we assume that $$\vert \mathrm{core}(B_0)\vert < 8n^{\frac{1}{3}}(\log{n})^{\frac{2}{3}}$$, which occurs a.a.s. by the Chernoff bound. Since $$\mathrm{core}(B_i)\geq 6\log n$$ for every $$i$$, we must have that by time $$0<t\leq n/6\log n$$ the process will have terminated with $$C_t=\bigcup_{i=0}^t B_i$$ supported on at least half of the vertices of $$V(\Gamma)$$. Lemma 5.3. Either one of the assumptions above fails or there exists a connected component $$F$$ of $$\square(\Gamma)$$ such that: (i) for every $$i$$ with $$0 \leq i \leq t$$ and every pair of non-adjacent vertices $$\{v,v'\}\in B_i$$, there is a vertex in $$F$$ whose support in $$\Gamma$$ contains the pair $$\{v,v'\}$$; and (ii) the support in $$\Gamma$$ of the $$4$$–cycles corresponding to vertices of $$F$$ contains all of $$C_t$$ with the exception of at most $$9$$ vertices of $$\mathrm{core}(B_0)$$; moreover, these exceptional vertices are each adjacent to all the vertices of $$\mathrm{core}(B_0)$$. □ Proof By assumption the ends $$\{x_0, y_0\}$$ of $$B_0$$ are non-adjacent. Thus, every pair of non-adjacent vertices $$\{v,v'\}$$ in $$\mathrm{core}(B_0)$$ induces a $$4$$–cycle in $$\Gamma$$ when taken together with $$\{x_0,y_0\}$$, and all of these squares clearly lie in the same component $$F$$ of $$\square(\Gamma)$$. Repeating the argument with the non-adjacent pair $$\{x_1, y_1\}\in \mathrm{core}(B_0)$$ and the block $$B_1$$, and then the non-adjacent pair $$\{x_2, y_2\}\in \mathrm{core}(B_1)$$ and the block $$B_2$$, and so on, we see that there is a connected component $$F$$ in $$\square(\Gamma)$$ such that for every $$0\leq i \leq t$$, every pair of non adjacent vertices $$\{v,v'\} \in B_i$$ lies in a $$4$$–cycle corresponding to a vertex of $$F$$. This establishes (i). We now show that the support of $$F$$ contains all of $$C_t$$ except possibly some vertices in $$B_0$$. We already established that every pair $$\{x_i,y_i\}$$ is in the support of some vertex of $$F$$. Suppose $$v\in \mathrm{core}(B_i)$$ for some $$i>0$$. By construction, $$v$$ is not adjacent to at least one of $$\{x_{i-1}, y_{i-1}\}$$, say $$x_{i-1}$$. Thus, $$\{x_{i-1},x_i, y_i, v\}$$ induces a $$4$$–cycle which contains $$v$$ and is associated to a vertex of $$F$$. Finally, suppose $$v\in \mathrm{core}(B_0)$$. By (i), $$v$$ fails to be in the support of $$F$$ only if $$v$$ is adjacent to all other vertices of $$\mathrm{core}(B_0)$$. Since, by assumption, $$\Gamma$$ does not contain any clique of size $$10$$, there are at most $$9$$ vertices not in the support of $$F$$, proving (ii). ■ Lemma 5.3, shows that a.a.s. we have a “large” component $$F$$ in $$\square(\Gamma)$$ whose support contains “many” pairs of non-adjacent vertices. In the last part of the proof, we use these pairs to prove that the remaining vertices of $$V(\Gamma)$$ are also supported on our connected component. For each $$i$$ satisfying $$0 \leq i \leq t$$, consider a a maximal collection, $$M_{i}$$, of pairwise-disjoint pairs of vertices in $$\mathrm{core}(B_i)\setminus\{x_{i+1},y_{i+1}\}$$. Set $$M=\bigcup_i M_i$$, and let $$M'$$ be the subset of $$M$$ consisting of pairs, $$\{v,v'\}$$, for which $$v$$ and $$v'$$ are not adjacent in $$\Gamma$$. We have |M|=∑i=1t(⌊12|core(Bi)|⌋−1)≥|Ct|2−2t≥n4(1−o(1)). The expected size of $$M'$$ is thus $$(1-p)n(\frac{1}{4}-o(1))=\frac{n}{4}(1-o(1))$$, and by the Chernoff bound from Lemma 4.1 we have P(|M′|≤n5) ≤2e−(15+o(1))2(1−p)n12≤e−(1300+o(1))n, which is $$o(1)$$. Thus a.a.s. $$M'$$ contains at least $$n/5$$ pairs, and by Lemma 5.3 each of these lies in some $$4$$–cycle of $$F$$. We now show that we can “beam up” every vertex not yet supported on $$F$$ by a $$4$$–cycle using a pair from $$M'$$. By construction we have at most $$n/2$$ unsupported vertices from $$V(\Gamma)\setminus C_t$$ and at most $$9$$ unsupported vertices from $$\mathrm{core}(B_0)$$. Assume that $$\vert M' \vert \geq n/5$$. Fix a vertex $$w\in V(\Gamma)\setminus C_t$$. For each pair $$\{v,v'\}\in M'$$, let $$X_{v,v'}$$ be the event that $$w$$ is adjacent to both $$v$$ and $$v'$$. We now observe that if $$X_{v,v'}$$ occurs for some pair $$\{v,v'\}\in M'\cap \mathrm{core}(B_i)$$, then $$w$$ is supported on $$F$$. By construction, $$w$$ is not adjacent to at least one of $$\{x_i, y_i\}$$, let us say without loss of generality $$x_i$$. Hence, $$\{x_i,v,v',w\}$$ is an induced $$4$$–cycle in $$\Gamma$$ which contains $$w$$ and which corresponds to a vertex of $$F$$. The probability that $$X_{v,v'}$$ fails to happen for every pair $$\{v,v'\}\in M'$$ is exactly (1−p2)|M′|≤(1−p2)n/5≤e−p2n5=e−5logn. Thus the expected number of vertices $$w \in V(\Gamma)\setminus C_t$$ which fail to be in the support of $$F$$ is at most $$\frac{n}{2}e^{-5\log n}=o(1)$$, whence by Markov’s inequality a.a.s. no such bad vertex $$w$$ exists. Finally, we deal with the possible $$9$$ left-over vertices $$b_1, b_2, \ldots b_9$$ from $$\mathrm{core}(B_0)$$ we have not yet supported. We observe that since $$\mathrm{core}(B_0)$$ contains at most $$8n^{\frac{1}{3}}(\log{n})^{\frac{2}{3}}$$ vertices (as we are assuming and as occurs a.a.s. , see the discussion before Lemma 5.3 ), we do not stop the process with $$B_0$$, $$\mathrm{core}(B_1)$$ is non-empty and contains at least $$6\log n$$ vertices. As stated in Lemma 5.3, each unsupported vertex $$b_i$$ is adjacent to all other vertices in $$\mathrm{core}(B_0)$$, and in particular to both of $$\{x_1, y_1\}$$. If $$b_i$$ fails to be adjacent to some vertex $$v\in \mathrm{core}(B_1)$$, then the set $$\{b_i, x_1, y_1, v\}$$ induces a $$4$$–cycle corresponding to a vertex of $$F$$ and whose support contains $$b_i$$. The probability that there is some $$b_i$$ not supported in this way is at most 9P(bi adjacent to all of core(B1)) =9p6logn=o(1). Thus a.a.s. we can “beam up” each of the vertices $$b_1, \ldots b_9$$ to $$F$$ using a vertex $$v\in \mathrm{core}(B_1)$$, and the support of the component $$F$$ in the square graph $$\square(\Gamma)$$ contains all vertices of $$\Gamma$$. This shows that a.a.s. $$\Gamma \in \mathcal{CFS}$$, and concludes the proof of the theorem. ■ Remark 5.4. The constant $$5$$ in Theorem 5.1 is not optimal, and indeed it is not hard to improve on it slightly, albeit at the expense of some tedious calculations. We do not try to obtain a better constant, as we believe that the order of the upper bound we have obtained is not sharp. We conjecture that the actual threshold for $$\mathcal{CFS}$$ occurs when $$p(n)$$ is of order $$n^{-1/2}$$ (see the discussion below Theorem 5.7), but a proof of this is likely to require significantly more involved and sophisticated arguments than the present article. □ A simple lower bound for the emergence of the $$\mathcal{CFS}$$ property can be obtained from the fact that if $$\Gamma\in\mathcal{CFS}$$, then $$\Gamma$$ must contain at least $$n-3$$ squares; if $$p(n)\ll n^{-\frac{3}{4}}$$, then by Markov’s inequality a.a.s. a graph in $${\mathcal G}(n,p)$$ contains fewer than $$o(n)$$ squares and thus cannot be in $$\mathcal{CFS}$$. Below, in Theorem 5.7, we prove a better lower bound, showing that the order of the upper bound we proved in Theorem 5.1 is not off by a factor of more than $$(\log n)^{3/2}$$. Lemma 5.5. Let $$\Gamma$$ be a graph and let $$C$$ be the subgraph of $$\Gamma$$ supported on a given connected component of $$\square(\Gamma)$$. Then there exists an ordering $$v_1<v_2< \cdots <v_{\vert C\vert}$$ of the vertices of $$C$$ such that for all $$i\ge3$$, $$v_i$$ is adjacent in $$\Gamma$$ to at least two vertices preceding it in the order. □ Proof As $$C$$ is a component of $$\square(\Gamma)$$, it contains at least one induced $$4$$–cycle. Let $$v_1, v_2$$ be a pair of non-adjacent vertices from such an induced $$4$$–cycle. Then the two other vertices $$\{v_3,v_4\}$$ of the $$4$$–cycle are both adjacent in $$\Gamma$$ to both of $$v_{1}$$ and $$v_{2}$$. If this is all of $$C$$, then we are done. Otherwise, we know that each $$4$$–cycle in $$C$$ is “connected” to the cycle $$F=\{v_{1},v_{2},v_{3},v_{4}\}$$ via a sequence of induced $$4$$–cycles pairwise intersecting in pairwise non-adjacent vertices. In particular, there is some such $$4$$–cycle whose intersection with $$F$$ is either a pair of non-adjacent vertices in $$F$$ or three vertices of $$F$$; either way, we may add the new vertex next in the order. Continuing in this way and using the fact that the number of vertices not yet reached is a monotonically decreasing set of positive integers, the lemma follows. ■ Proposition 5.6. Let $$\delta>0$$. Suppose $$p\leq\frac{1}{\sqrt{n}\log n}$$. Then a.a.s. for $$\Gamma\in{\mathcal G}(n,p)$$, no component of $$\square(\Gamma)$$ has support containing more than $$4\log n$$ vertices of $$\Gamma$$. □ Proof Let $$\delta>0$$. Let $$m=\left\lceil \min\left(4\log n, 4\log \left(\frac{1}{p}\right) \right)\right\rceil$$, with $$p\leq 1/\left(\sqrt{n}\log n\right)$$. We shall show that a.a.s. there is no ordered $$m$$–tuple of vertices $$v_1<v_2<\cdots <v_m$$ from $$\Gamma$$ such that for every $$i\ge 2$$ each vertex $$v_i$$ is adjacent to at least two vertices from $$\{v_j: 1\leq j <i\}$$. By Lemma 5.5, this is enough to establish our claim. Let $$v_1<v_2<\cdots < v_m$$ be an arbitrary ordered $$m$$–tuple of vertices from $$V(\Gamma)$$. For $$i\geq 2$$, let $$A_i$$ be the event that $$v_i$$ is adjacent to at least two vertices in the set $$\{v_j: 1\leq j <i\}$$. We have: Pr(Ai)=∑j=2i−1(i−1j)pj(1−p)i−j−1. (2) As in the proof of Theorem 4.5 we consider the quotients of successive terms in the sum to show that its order is given by the term $$j=2$$. To see this, observe: (i−1j+1)pj+1(1−p)i−j−2(i−1j)pj(1−p)i−j−1≤i−j−1j+1⋅p1−p<mp where the final inequality holds for $$n$$ sufficiently large and $$p=p(n)$$ satisfying our assumption. Since $$m=O\left(\log n\right)$$ and $$p=o(n^{-1/2})$$ we have, again for $$n$$ large enough, that $$mp=o(1)$$, and we may bound the sum in equation (2) by a geometric series to obtain the bound: Pr(Ai) =(i−12)p2(1−p)i−3(1+O(mp)) ≤(i−1)22p2(1+O(mp)). Now let $$A=\bigcap_{i=1}^m A_i$$. Note that the events $$A_i$$ are mutually independent, since they are determined by disjoint edge-sets. Thus we have: Pr(A)=∏i=3mPr(Ai) ≤∏i=3m((i−1)22p2(1+O(mp))) =((m−1)!)2p2m−42m−2(1+O(m2p)), where in the last line we used the equality $$(1+O(mp))^{m-2} =1+O(m^2p)$$ to bound the error term. Thus we have that the expected number $$X$$ of ordered $$m$$–tuples of vertices from $$\Gamma$$ for which $$A$$ holds is at most: E(X)=n!(n−m)!P(A) ≤nm4e2(m2p2−4me22)m(1+O(m2p)) =4e2(nm2p2−4/m2e2)m(1+O(m2p)), where in the first line we used the inequality $$(m-1)!\leq e (m/e)^{m}$$. We now consider the quantity f(n,m,p)=nm2p2−4/m2e2 which is raised to the $$m^{\textrm{th}}$$ power in the inequality above. We claim that $$f(n,m,p)\leq e^{-1+\log 2+o(1)}$$. We have two cases to consider: Case 1: $$m=\lceil 4\log n\rceil$$. Since $$4\log n \leq 4\log(1/p)$$, we deduce that $$p\leq n^{-1}$$. Then f(n,m,p) =n(4logn)2p2−o(1)2e2≤n−1+o(1)≤e−1+log2+o(1). Case 2: $$m=\lceil 4\log(1/p)\rceil$$. First, note that $$p^{-4/m}=\exp\left(\frac{4\log(1/p)}{\lceil 4 \log 1/p\rceil}\right)\leq e$$. Also, for $$p$$ in the range $$[0, n^{-1/2}(\log n)^{-1}]$$ and $$n$$ large enough, $$p^2(\log(1/p))^2$$ is an increasing function of $$p$$ and is thus at most: 1n(logn)2(12logn)2(1+O(loglognlogn))=14n−1(1+o(1)). Plugging this into the expression for $$f(n,m,p)$$, we obtain: f(n,m,p) =(1+o(1))16n(log(1/p))2p2−4/m2e2 ≤(1+o(1))2e=e−1+log2+o(1). Thus, in both cases (1) and (2) we have $$f(n,m,p)\leq e^{-1 +\log 2+o(1)}$$, as claimed, whence E(X) ≤4e2(f(n,m,p))m(1+O(m2p))≤4e2e−(1−log2)m+o(m)(1+O(m2p))=o(1). It follows from Markov’s inequality that the non-negative, integer-valued random variable $$X$$ is a.a.s. equal to $$0$$. In other words, a.a.s. there is no ordered $$m$$–tuple of vertices in $$\Gamma$$ for which the event $$A$$ holds and, hence by Lemma 5.5, no component in $$\square(\Gamma)$$ covering more than $$m\leq 4\log n$$ vertices of $$\Gamma$$. ■ Theorem 5.7. Suppose $$p\leq\frac{1}{\sqrt{n} \log n}$$. Then a.a.s. $$\Gamma\in{\mathcal G}(n,p)$$ is not in $$\mathcal{CFS}$$. □ Proof To show that $$\Gamma\not\in\mathcal{CFS}$$, we first show that, for $$p\leq \frac{1}{\sqrt{n} \log n}$$, a.a.s. there is no non-empty clique $$K$$ such that $$\Gamma=\Gamma' \star K$$. Indeed the standard Chernoff bound guarantees that we have a.a.s. no vertex in $$\Gamma$$ with degree greater than $$\sqrt{n}$$. Thus to prove the theorem, it is enough to show that a.a.s. there is no connected component $$C$$ in $$\square(\Gamma)$$ containing all the vertices in $$\Gamma$$. Theorem 5.6 does this by establishing the stronger bound that a.a.s. there is no connected component $$C$$ covering more than $$4\log n$$ vertices. ■ While Theorem 5.7 improves on the trivial lower bound of $$n^{-3/4}$$, it is still off from the upper bound for the emergence of the $$\mathcal{CFS}$$ property established in Theorem 5.1. It is a natural question to ask where the correct threshold is located. Remark 5.8. We strongly believe that there is a sharp threshold for the $$\mathcal{CFS}$$ property analogous to the one we established for the $$\mathcal{AS}$$ property. What is more, we believe this threshold should essentially coincide with the threshold for the emergence of a giant component in the auxiliary square graph $$\square(\Gamma)$$. Indeed, our arguments in Proposition 5.6 and Theorems 5.1 both focus on bounding the growth of a component in $$\square(\Gamma)$$. Heuristically, we would expect a giant component to emerge in $$\square(\Gamma)$$ to emerge for $$p(n)=cn^{-1/2}$$, for some constant $$c>0$$, when the expected number of common neighbors of a pair of non-adjacent vertices in $$\Gamma$$ is $$c^2$$, and thus the expected number of distinct vertices in $$4$$–cycles which meet a fixed $$4$$–cycle in a non-edge is $$2c^2$$. What the precise value of $$c$$ should be is not entirely clear (a branching process heuristics suggests $$\sqrt{\displaystyle\frac{\sqrt{17}-3}{2}}$$ as a possible value, see Remark 6.2), however, and the dependencies in the square graph make its determination a delicate matter. □ 6 Experiments Theorems 4.4 and 4.5 show that $$\left(\log n/n\right)^{\frac{1}{3}}$$ is a sharp threshold for the family $$\mathcal{AS}$$ and Theorem 5.1 shows that $$n^{-\frac{1}{2}}$$ is the right order of magnitude of the threshold for $$\mathcal{CFS}$$. Below we provide some empirical results on the behavior of random graphs near the threshold for $$\mathcal{AS}$$ and the conjectured threshold for $$\mathcal{CFS}$$. We also compare our experimental data with analogous data at the connectivity threshold. Our experiments are based on various algorithms that we implemented in $$\texttt{C++}$$; the source code is available from the authors. At www.wescac.net/research.html or math.columbia.edu/~jason, all source code and data can be downloaded. We begin with the observation that computer simulations of $$\mathcal{AS}$$ and $$\mathcal{CFS}$$ are tractable. Indeed, it is easily seen that there are polynomial-time algorithms for deciding whether a given graph is in $$\mathcal{AS}$$ and/or $$\mathcal{CFS}$$. Testing for $$\mathcal{AS}$$ by examining each block and determining whether it witnesses $$\mathcal{AS}$$ takes $$O(n^5)$$ steps, where $$n$$ is the number of vertices. The $$\mathcal{CFS}$$ property is harder to detect, but essentially reduces to determining the component structure of the square graph. The square graph can be produced in polynomial time and, in polynomial time, one can find its connected components and check the support of these components in the original graph. Using our software, we tested random graphs in $${\mathcal G}(n,p)$$ for membership in $$\mathcal{AS}$$ for $$n\in\{300+100k\mid0\leq k\leq12\}$$ and $$\{p(n)=\alpha\left(\log n/n\right)^{\frac{1}{3}}\mid \alpha=0.80+0.1k,\,0\leq k\leq 9\}$$. For each pair $$(n,p)$$ of this type, we generated $$400$$ random graphs and tested each for membership in $$\mathcal{AS}$$. (This number of tests ensures that, with probability approximately $$95\%$$, the measured proportion of $$\mathcal{AS}$$ graphs is within at most $$0.05$$ of the actual proportion.) The results are summarized in Figure 2. The data suggest that, fixing $$n$$, the probability that a random graph is in $$\mathcal{AS}$$ increases monotonically in the range of $$p$$ we are considering, rising sharply from almost zero to almost one. Fig. 2. View largeDownload slide Experimental prevalence of $$\mathcal{AS}$$ at density $$\alpha\left(\frac{\log n}{n}\right)^{\frac{1}{3}}$$. (A colour version of this figure is available from the journal’s website.) Fig. 2. View largeDownload slide Experimental prevalence of $$\mathcal{AS}$$ at density $$\alpha\left(\frac{\log n}{n}\right)^{\frac{1}{3}}$$. (A colour version of this figure is available from the journal’s website.) In Figure 3, we display the results of testing random graphs for membership in $$\mathcal{CFS}$$ for $$n\in\{100+100k\mid 0\leq k\leq 15\}$$ and $$\{p(n)=\alpha n^{-\frac{1}{2}} \mid \alpha=0.700+0.025k,\,0\leq k\leq 8\}$$. For each pair $$(n,p)$$ of this type, we generated $$400$$ random graphs and tested each for membership in $$\mathcal{CFS}$$. (This number of tests ensures that, with probability approximately $$95\%$$, the measured proportion of $$\mathcal{AS}$$ graphs is within at most $$0.05$$ of the actual proportion.) The data suggest that, fixing $$n$$, the probability that a random graph is in $$\mathcal{CFS}$$ increases monotonically in considered range of $$p$$: rising sharply from almost zero to almost one inside a narrow window. Fig. 3. View largeDownload slide Experimental prevalence of $$\mathcal{CFS}$$ membership at density $$\alpha n^{-\frac{1}{2}}$$. (A colour version of this figure is available from the journal’s website.) Fig. 3. View largeDownload slide Experimental prevalence of $$\mathcal{CFS}$$ membership at density $$\alpha n^{-\frac{1}{2}}$$. (A colour version of this figure is available from the journal’s website.) Remark 6.1 (Block and core sizes). For each graph $$\Gamma$$ tested, the $$\mathcal{AS}$$ software also keeps track of how many nonadjacent pairs $$\{x,y\}$$ — that is, how many blocks — were tested before finding one sufficient to verify membership in $$\mathcal{AS}$$; if no such block is found, then all non-adjacent pairs have been tested and the graph is not in $$\mathcal{AS}$$. ({A set of such data comes with the source code, and more is available upon request.}) At densities near the threshold, this number of blocks is generally very large compared to the number of blocks tested at densities above the threshold. For example, in one instance with $$(n,\alpha)=(1000,0.89)$$, verifying that the graph was in $$\mathcal{AS}$$ was accomplished after testing just $$86$$ blocks, while at $$(1000,0.80)$$, a typical test checked all $$422961$$ blocks (expected number: $$423397$$) before concluding that the graph is not $$\mathcal{AS}$$. At the same $$(n,\alpha)$$, another test found that the graph was in $$\mathcal{AS}$$, but only after $$281332$$ tests. These data are consonant with the spirit of our proofs of Theorems 4.5 and 4.4: in the former case, we showed that no “good” block exists, while in the latter we show that every block is good. We believe that right at the threshold we should have some intermediate behavior, with the expected number of “good” blocks increasing continuously from $$0$$ to $$(1-p)\binom{n}{2}(1-o(1))$$. What is more, we expect that the more precise threshold for the $$\mathcal{AS}$$ property, coinciding with the appearance of a single “good” block, should be located “closer” to our lower bound than to our upper one, that is, at $$p(n)= \left((1-\epsilon)\log n/n\right)^{1/3}$$, where $$\epsilon(n)$$ is a sequence of strictly positive real numbers tending to $$0$$ as $$n\rightarrow \infty$$ (most likely decaying at a rate just faster than $$(\log n)^{-2}$$, see below). Our experimental data, which exhibit a steep rise in $$\Pr(\mathcal{AS})$$ strictly before the value $$\alpha$$ hits one, gives some support to this guess. Finally, our observations on the number of blocks suggests a natural way to understand the influence of higher-order terms on the emergence of the $$\mathcal{AS}$$ property: at exactly the threshold for $$\mathcal{AS}$$, the event $$E(v,w)$$ that a pair of non-adjacent vertices $$\{v,w\}$$ gives rise to a “good” block is rare and, despite some mild dependencies, the number $$N$$ of pairs $$\{v,w\}$$ for which $$E(v,w)$$ occurs is very likely to be distributed approximatively like a Poisson random variable. The probability $$\Pr(N\geq 1)$$ would then be a very good approximation for $$\Pr(\mathcal{AS})$$. “Good” blocks would thus play a role for the emergence of the $$\mathcal{AS}$$ property in random graphs analogous to that of isolated vertices for connectivity in random graphs. When $$p=\left((1-\epsilon) \log n/n\right)^{\frac{1}{3}}$$, the expectation of $$N$$ is roughly $$ne^{-n^{\epsilon}(1-\epsilon)\log n}$$. This expectation is $$o(1)$$ when $$0<\epsilon(n)=\Omega(1/n)$$ and is $$1/2$$ when $$\epsilon(n)=(1+o(1)) \log 2/(\log n)^2$$. This suggest that the emergence of $$\mathcal{AS}$$ should occur when $$\epsilon(n)$$ decays just a little faster than $$(\log n)^2$$. □ Remark 6.2. Our data suggest that the prevalence of $$\mathcal{CFS}$$ is closely related to the emergence of a giant component in the square graph. Indeed, below the experimentally observed threshold for $$\mathcal{CFS}$$, not only is the support of the largest component in the square graph not all of $$\Gamma \in {\mathcal G}(n,p)$$, but in fact the size of the support of the largest component is an extremely small proportion of the vertices (see Figure 4). In the Erdős–Rényi random graph, a giant component emerges when $$p$$ is around $$1/n$$, that is, when vertices begin to expect at least one neighbor; this corresponds to a paradigmatic condition of expecting at least one child for survival of a Galton–Watson process (see [9] for a modern treatment of the topic). The heuristic observation that when $$p=cn^{-1/2}$$ the vertices of a diagonal $$e$$ in a fixed $$4$$-cycle $$F$$ expect to be adjacent to $$cn^2$$ vertices outside the $$4$$-cycle, giving rise to an expected $$\binom{cn^2+2}{2}-1$$ new $$4$$-cycles connected to $$F$$ through $$e$$ in $$\square(\Gamma)$$ suggests that $$c=\sqrt{\displaystyle\frac{\sqrt{17}-3}{2}}\approx 0.7494$$ could be a reasonable guess for the location of the threshold for the $$\mathcal{CFS}$$ property. 2 Our data, although not definitive, appears somewhat supportive of this guess: see Figure 4 which is based on the same underlying data set as Figure 3. We note that unlike an Erdős–Rényi random graph, the square graph $$\square(\Gamma)$$ exhibits some strong local dependencies, which may make the determination of the exact location of its phase transition a much more delicate affair. □ Fig. 4. View largeDownload slide Fraction of vertices supporting the largest $$\mathcal{CFS}$$–subgraph at density $$\alpha n^{-\frac{1}{2}}$$. (A colour version of this figure is available from the journal’s website.) Fig. 4. View largeDownload slide Fraction of vertices supporting the largest $$\mathcal{CFS}$$–subgraph at density $$\alpha n^{-\frac{1}{2}}$$. (A colour version of this figure is available from the journal’s website.) Remark 6.3. For comparison with the threshold data for $$\mathcal{AS}$$ and $$\mathcal{CFS}$$, we include below a similar figure of experimental data for connectivity for $$\alpha$$ from $$0.8$$ to $$1.4$$, where $$p=\frac{\alpha\log n}{n}$$. Given, what we know about the thresholds for connectivity and the $$\mathcal{AS}$$ property, this last set of data together with Figure 2 should serve as a warning not to draw overly strong conclusions: the graphs we tested are sufficiently large for the broader picture to emerge, but probably not large enough to allow us to pinpoint the exact location of the threshold for $$\mathcal{CFS}$$. □ Fig. 5. View largeDownload slide Experimental prevalence of connectedness at density $$\alpha\frac{\log n}{n}$$. (A colour version of this figure is available from the journal’s website.) Fig. 5. View largeDownload slide Experimental prevalence of connectedness at density $$\alpha\frac{\log n}{n}$$. (A colour version of this figure is available from the journal’s website.) Funding This work was supported by National Science Foundation [DMS-0739392 to J.B, 1045119 to M.F.H.] and also supported by a Simons Fellowship to J.B. Acknowledgments J.B. thanks the Barnard/Columbia Mathematics Department for their hospitality during the writing of this article. J.B. thanks Noah Zaitlen for introducing him to the joy of cluster computing and for his generous time spent answering remedial questions about programming. Also, thanks to Elchanan Mossel for several interesting conversations about random graphs. M.F.H. and T.S. thank the organizers of the 2015 Geometric Groups on the Gulf Coast conference, at which some of this work was completed. This work benefited from several pieces of software, the results of one of which is discussed in Section 6. Some of the software written by the authors incorporates components previously written by J.B. and M.F.H. jointly with Alessandro Sisto. Another related useful program was written by Robbie Lyman under the supervision of J.B. and T.S. during an REU program supported by NSF Grant DMS-0739392, see [30]. 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This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/about_us/legal/notices) For permissions, please e-mail: journals. [email protected]
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https://mc-stan.org/docs/2_28/functions-reference/math-functions.html | # 28 Mathematical Functions
This appendix provides the definition of several mathematical functions used throughout the manual. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9931914210319519, "perplexity": 1837.907759803501}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00385.warc.gz"} |
https://mathoverflow.net/questions/221128/probability-of-paths-to-the-boundary-of-a-tree | Probability of paths to the boundary of a tree
Let $G_n$ be the $4$-regular tree of depth $n$, that is to say the finite graph given by the ball of radius $n$ in the Cayley graph of the free group on two generators. By the root I mean the vertex at the center. If I choose vertices uniformly at random from $G_n$, what is the probability that when I choose the root it will be connected to the boundary of the graph by a path through vertices I have chosen already? Equivalently, if I take a random ordering of the vertices of $G_n$, what is the probability that the segment preceding the root will contain a path to the boundary? More precisely, I would like to know whether this probability approaches $0$ as $n$ goes to infinity.
Perhaps easiest to think about it in this way; assign every vertex an independent time which is uniform on $[0,1]$. If the vertices of $G_n$ are added in increasing order of their times, then this is equivalent to adding them one by one uniformly as you describe. But this way we can easily think about all the vertices in the infinite graph simultaneously.
Now condition on the time of the root. Given that this time is $p$, the set of vertices preceding the root contains each other vertex independently with probability $p$. Call this the set of open vertices. Effectively this is percolation with a random probability $p$ (itself chosen uniformly on $[0,1]$).
If we do percolation with probability $p$, then there exists an infinite open path starting from the root with positive probability iff $p>1/3$. This is almost the same as the probability of survival of a Galton-Watson branching process whose offspring distribution is Binomial($3,p$); the difference is that here the root has $4$ possible offspring, while each other vertex has only $3$. The probability (as a fuction of $p$) can be quite easily obtained as the solution of a recursive equation. To get the probability that the set of open vertices contains an infinite path from the root, integrate over $p$ from $1/3$ to $1$.
The limit as $n\to\infty$ of the probability that the set of open vertices contains a path from the root to distance $n$ is then just this probability that the set of open vertices contains an infinite path from the root. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9746314883232117, "perplexity": 58.80431288242358}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141723602.80/warc/CC-MAIN-20201203062440-20201203092440-00143.warc.gz"} |
https://mathoverflow.net/questions/378016/integration-by-parts-on-a-k%C3%A4hler-manifold/378022 | # Integration by parts on a Kähler manifold
I am trying to make sense of integration by parts on a Kähler manifold $$X$$ equipped with a Kähler metric $$\omega$$. Given two smooth real functions $$f$$ and $$h$$ on $$X$$, I want to write down the integration by parts formula for the following: $$\int_{X} h \Delta_{\omega} f \omega^n.$$ In local coordinates $$\Delta_{\omega} f = \sum_{i, j} g^{i \bar\jmath} \partial_{\bar\jmath} \partial_{i}f$$. My guess is that $$\int_{X} h \Delta_{\omega} f \omega^n = -\int_{X} g^{i \bar\jmath} \partial_{\bar\jmath} h \partial_{i} f \omega^n.$$ However, this is not quite right since the LHS is real while the right hand side is not necessarily so. What is the correct formula? Is there a general strategy for thinking of such things in the complex case?
• Try to write in terms of the exterior derivative and the Hodge star, since integration by parts is Stokes' theorem applied to an exact form. – Ben McKay Dec 3 '20 at 9:16
• The formula that you wrote is correct, and the RHS is real! The integrand is not always real-valued, but the integral is. – YangMills Dec 3 '20 at 13:55
• Is it obvious that the integral has to be real?@YangMills – penny Dec 3 '20 at 14:05
• It follows from the equation that you wrote, since the LHS is real. Proving your formula is a simple exercise using the divergence theorem (and the definition of covariant derivatives for a Kahler metric). – YangMills Dec 3 '20 at 14:36
Assume $$(X, d = \partial + \bar{\partial})$$ to be a compact Kähler manifold. The Kähler metric $$g$$ induces a metric on all differential forms, which we will also call $$g$$. It follows that $$\omega^n$$ defines a Hilbert space of $$i$$-forms on $$X$$ by $$\langle u, v\rangle = \int_X g(u,v) \omega^n.$$ For functions $$u, v$$, this is $$\langle u, v\rangle = \int_X u \bar{v} \omega^n.$$
The adjoints $$d^*$$ and $$\bar{\partial}^*$$ are the operators such that for all $$(i-1)$$-form $$u$$ and all $$i$$-form $$v$$, $$\langle du,v\rangle = \langle u,d^*v\rangle,\quad\langle\bar{\partial}u,v\rangle = \langle u, \bar{\partial}^* v\rangle.$$ The expression $$\bar{\partial}^*$$ equals $$-* \bar{\partial} *$$, where $$*$$ is the Hodge-$$*$$ and it involves the Kähler metric $$g$$.
The Laplacian is then $$\Delta_d = d^* d + d d^* = 2 (\bar{\partial} \bar{\partial}^* +\bar{\partial}^* \bar{\partial}) = 2 \Delta_{\bar{\partial}}.$$ Since $$h,f$$ are functions, we have $$\langle h, \Delta_{\bar{\partial}} f\rangle = \langle h, (\bar{\partial} \bar{\partial}^* +\bar{\partial}^* \bar{\partial}) f\rangle = \langle\bar{\partial}^* h, \bar{\partial}^* f\rangle + \langle\bar{\partial} h, \bar{\partial} f\rangle = \langle\bar{\partial} h, \bar{\partial} f\rangle.$$ In terms of your notation, this means $$\int_X h \Delta_\omega f \omega^n = \int_X g(\bar{\partial} h, \bar{\partial} f) \omega^n.$$
• Generally people use $\langle u, v\rangle$ rather than $<u, v>$. The former is given by $\langle u, v\rangle$. – Michael Albanese Dec 3 '20 at 12:10
• I made the edit @MichaelAlbanese mentions. Note also that \quad or \qquad or one of the other spacing commands is probably preferable to repeated \ \ \ . – LSpice Dec 3 '20 at 13:58
• Sorry I am a bit confused. Usually $\bar \partial^{\ast}$ would take a $(1, 1)$ form then gives back a smooth function. Here you seem to be acting it on a function directly. – penny Dec 3 '20 at 14:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.980088472366333, "perplexity": 162.8145589174161}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178369721.76/warc/CC-MAIN-20210305030131-20210305060131-00023.warc.gz"} |
http://adas-fusion.eu/element/detail/adf38/nrb05%5D%5Bn/nrb05%5D%5Bn_v16ic1-2.dat | nrb05#n_v16ic1-2.dat
Photoexcitation-autoionisation Rate Coefficients
Ion
V16+
Filename
nrb05#n_v16ic1-2.dat
Full Path
Parent states
1s2 2s2 2p3 4S1.5
1s2 2s2 2p3 2D1.5
1s2 2s2 2p3 2D2.5
1s2 2s2 2p3 2P0.5
1s2 2s2 2p3 2P1.5
1s2 2s1 2p4 4P2.5
1s2 2s1 2p4 4P1.5
1s2 2s1 2p4 4P0.5
1s2 2s1 2p4 2D1.5
1s2 2s1 2p4 2D2.5
1s2 2s1 2p4 2S0.5
1s2 2s1 2p4 2P1.5
1s2 2s1 2p4 2P0.5
1s2 2p5 2P1.5
1s2 2p5 2P0.5
Recombined states
1s2 2s2 2p4 3P2.0
1s2 2s2 2p4 3P0.0
1s2 2s2 2p4 3P1.0
1s2 2s2 2p4 1D2.0
1s2 2s2 2p4 1S0.0
1s2 2s1 2p5 3P2.0
1s2 2s1 2p5 3P1.0
1s2 2s1 2p5 3P0.0
1s2 2s1 2p5 1P1.0
1s2 2p6 1S0.0
1s1 2s2 2p5 3P2.0
1s1 2s2 2p5 3P1.0
1s1 2s2 2p5 3P0.0
1s1 2s2 2p5 1P1.0
1s1 2s1 2p6 3S1.0
1s1 2s1 2p6 1S0.0
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
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https://www.physicsforums.com/threads/categorizing-a-physics-demonstration.799338/ | # Categorizing a Physics Demonstration
1. Feb 22, 2015
### Richardj1701
ABSTRACT:
I have a collection of springs and disks (masses) to choose from. I have a solid rod fixed to the ground. I slide a piece of sheet metal into the rod to act as a base. I now slide a disk into the rod. Then a spring. Then another disk. Then another spring. And one final disk. I now lift the system (from the sheet metal base) to a desired height. After releasing the system, it drops and we can observe conservation of momentum since the top-most disk will shoot upwards.
GOAL:
Under what category of physics would you place this? I want to know what I have to research in order to analyze the physics behind this reaction. (My end goal is to maximize the speed at which that top most disk flies off.)
ATTEMPTS:
I've looked into coupled, spring-mass systems, but I don't think that this will help me since my system is not so much coupled but sitting on each other.
2. Feb 22, 2015
### Bystander
First quick dirty impression? Newton's cradle.
3. Feb 23, 2015
### CWatters
Perhaps I misunderstand the description but how does dropping the system cause the top disc "shoot upwards"?
Draft saved Draft deleted
Similar Discussions: Categorizing a Physics Demonstration | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8764632344245911, "perplexity": 1550.5251440572736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948567785.59/warc/CC-MAIN-20171215075536-20171215095536-00598.warc.gz"} |
https://plainmath.net/12172/x-2-plus-y-2-equal-25-what-is-dydx-dx-dy | # x^{2} +y^{2} =25 What is dydx dx/dy?
Question
Differential equations
$$x^{2} +y^{2} =25$$
What is dydx dx/dy?
2021-02-10
$$\frac{dx}{dy} =−\frac{x}{y}$$
Since y is not on it's own side, we will have to differentiate implicitly. We will have to take the derivative with respect to x, since the question is asking for dydx dx/dy . After taking the derivative we get: $$2x+2y(dydx)=02x+2y\frac{dx}{dy}=0$$
Since we want to solve for dy/dx we must isolate it. First, we will subtract 2x from both sides to get: $$2y(dydx)=−2x2y\frac{dx}{dy}=−2x$$
Then we will divide both sides by 2y to get: $$dy/dx=−\frac{x}{y}$$
$$\frac{dx}{dy} =−\frac{x}{y}$$
### Relevant Questions
Solve the equation:
$$\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{2}}+{1}\right)}$$
Solve the equation:
$$\displaystyle{\left({a}-{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}$$
The graph of y = f(x) contains the point (0,2), $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{-{x}}}{{{y}{e}^{{{x}^{{2}}}}}}}$$, and f(x) is greater than 0 for all x, then f(x)=
A) $$\displaystyle{3}+{e}^{{-{x}^{{2}}}}$$
B) $$\displaystyle\sqrt{{{3}}}+{e}^{{-{x}}}$$
C) $$\displaystyle{1}+{e}^{{-{x}}}$$
D) $$\displaystyle\sqrt{{{3}+{e}^{{-{x}^{{2}}}}}}$$
E) $$\displaystyle\sqrt{{{3}+{e}^{{{x}^{{2}}}}}}$$
Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
Q. 2# $$(x+1)\frac{dy}{dx}=x(y^{2}+1)$$
Q. 2# $$\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{{2}}}+{1}\right)}$$
$$\displaystyle{y}={\frac{{{1}}}{{{2}+{\sin{{x}}}}}}$$
find general solution in semi homogenous method of $$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={x}-{y}+\frac{{1}}{{x}}+{y}-{1}$$
Deterrmine the first derivative $$\displaystyle{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}$$ :
$$\displaystyle{y}={2}{e}^{{2}}{x}+{I}{n}{x}^{{3}}-{2}{e}^{{x}}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9570276737213135, "perplexity": 893.3449493214381}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00485.warc.gz"} |
https://kg15.herokuapp.com/abstracts/315 | # Zigzags on Interval Greedoids
### Yulia Kempner Holon Institute of Technology
#### Vadim E. Levit Ariel University
PDF
Minisymposium: GENERAL SESSION TALKS
Content: The pivot operation, i.e., exchanging exactly one element in a basis, is one of the fundamental algorithmic tools in linear algebra. Korte and Lov\'{a}sz introduced a combinatorial analog of this operation for bases of greedoids. Let $X,Y$ be two bases of a greedoid $(U,\mathcal{F})$ such that $\left\vert X-Y\right\vert =1$ and $X\cap Y\in\mathcal{F}$. Then $X$ can be obtained from $Y$ by \textit{a pivot operation} where the element $y\in Y-X$ is pivoted out and the element $x\in X-Y$ is pivoted in. We extend this definition to all feasible sets of the same cardinality and introduce \textit{lower and upper zigzags} comprising these sets. A \textit{zigzag} is a sequence of feasible sets $P_{0},P_{1},...,P_{2m}$ such that: \emph{(i)} these sets have only two different cardinalities; \emph{(ii)} any two consecutive sets in this sequence differ by a single element. Zigzag structures allow us to give new metric characterizations of some subclasses of interval greedoids including antimatroids and matroids.
Back to all abstracts | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9503217339515686, "perplexity": 1144.0523171703194}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057973.90/warc/CC-MAIN-20210926205414-20210926235414-00108.warc.gz"} |
http://pixel-druid.com/frobenius-kernel.html | § Frobenius Kernel
§ Some facts about conjugates of a subgroup
Let $H$ be a subgroup of $G$. Define $H_g \equiv \{ g h g^{-1} : h \in H \}$.
• We will always have $e \in H_g$ since $geg^{-1} = e \in H_g$.
• Pick $k_1 k_2 \in H_g$. This gives us $k_i = gh_ig^{-1}$. So, $k_1 k_2 = g h_1 g^{-1} g h_2 g^{-1} = g (h_1 h_2) g^{-1} \in H_g$.
• Thus, the conjugates of a subgroup is going to be another subgroup that has nontrivial intersection with the original subgroup.
• For inverse, send $k = ghg^{-1}$ to $k^{-1} = g h^{-1} g^{-1}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9975124001502991, "perplexity": 516.5145068829011}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662584398.89/warc/CC-MAIN-20220525085552-20220525115552-00572.warc.gz"} |
https://brilliant.org/problems/brilliant-elasticity/ | # Brilliant Elasticity
Brilliant Industries creates a consumer good called problems. Suppose that the demand for problems is relatively elastic, and the supply of problems is relatively inelastic. Which one of these scenarios will see the greatest relative increase in the quantity of problems at the equilibrium point? (Assume that each scenario will affect either the supply of or the demand for the product.)
× | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8883296847343445, "perplexity": 1153.8485420714194}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912205163.72/warc/CC-MAIN-20190326115319-20190326141319-00183.warc.gz"} |
http://wikimechanics.org/momentum | Momentum
Sir Isaac Newton. Painted by G Kneller 1689.
Momentum is the modern English word used for translating the phrase "quantity of motion" that uses on the very first page of his great book, the Principia.1 So to understand motion WikiMechanics starts by using sensation to define the momentum as follows. Consider some particle P characterized by its wavevector $\overline{ \kappa }$ and the total number of quarks it contains $N$. Report on any changes relative to a frame of reference F which is characterized using $\tilde{ \kappa }$ the average wavevector of the quarks in F. Definition: the momentum of particle P in reference frame F is the ordered set of three numbers
\begin{align} \overline{p} \equiv \frac{ h }{ 2 \pi } \left( \overline{ \kappa }^{ \sf{P}} \! - N^{ \sf{P}} \, \tilde{ \kappa }^{ \sf{F}} \right) \end{align}
where $h$ and $\pi$ are constants. The norm of the momentum is marked without an overline
$p \equiv \left\| \, \overline{p} \, \right\|$
If $p=0$ we say that P is stationary or at rest in the F-frame. Alternatively, if $p \ne 0$ then we say that P is in motion.
Sensory interpretation: The momentum is defined by a difference between the wavevector of P and a scaled-down version of the frame's wavevector. Recall that the wavevector has previously been interpreted as a mathematical representation of visual sensation. So momentum can be understood as the visual contrast between a particle and its reference frame.
Next step: conservation of momentum.
page revision: 677, last edited: 26 Sep 2015 19:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9868066310882568, "perplexity": 877.3917049920931}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948520042.35/warc/CC-MAIN-20171212231544-20171213011544-00323.warc.gz"} |
http://ito.wias-berlin.de/publications/wias-publ/run.jsp?template=abstract&type=Preprint&year=2013&number=1802 | WIAS Preprint No. 1802, (2013)
# Closed-loop optimal experiment design: Solution via moment extension
Authors
• Hildebrand, Roland
• Gevers, Michel
• Solari, Gabriel
2010 Mathematics Subject Classification
• 93E12
Keywords
• Optimal experiment design, Closed-loop identification, Convex programming, Power spectral density, Moment method
Abstract
We consider optimal experiment design for parametric prediction error system identification of linear time-invariant multiple-input multiple-output (MIMO) systems in closed-loop when the true system is in the model set. The optimization is performed jointly over the controller and the spectrum of the external excitation, which can be reparametrized as a joint spectral density matrix. We have shown in [18] that the optimal solution consists of first computing a finite set of generalized moments of this spectrum as the solution of a semi-definite program. A second step then consists of constructing a spectrum that matches this finite set of optimal moments and satisfies some constraints due to the particular closed-loop nature of the optimization problem. This problem can be seen as a moment extension problem under constraints. Here we first show that the so-called central extension always satisfies these constraints, leading to a constructive procedure for the optimal controller and excitation spectrum.We then show that, using this central extension, one can construct a broader set of parametrized optimal solutions that also satisfy the constraints; the additional degrees of freedom can then be used to achieve additional objectives. Finally, our new solution method for the MIMO case allows us to considerably simplify the proofs given in [18] for the single-input single-output case.
Appeared in
• IEEE Trans. Autom. Control, 60 (2015) pp. 1731--1744. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9496574401855469, "perplexity": 643.4079098474473}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986705411.60/warc/CC-MAIN-20191020081806-20191020105306-00131.warc.gz"} |
https://www.physicsforums.com/threads/probability-of-finding-a-particle-in-a-box.671124/ | # Probability of finding a particle in a box
1. Feb 11, 2013
### ahhppull
1. The problem statement, all variables and given/known data
Consider ψ (x) for a particle in a box:
ψn(x) = (2/L)1/2sin(n∏x/L)
Calculate the probability of finding the particle in the middle half of the box (i.e., L/4 ≤ x ≤ 3L/4). Also, using this solution show that as ''n'' goes to infinity you get the classical solution of 0.5.
2. Relevant equations
3. The attempt at a solution
I integrated and figured out the probability for n=1,2,3. For n=1 I got 1/2 + 1/∏ which is about 0.818. For n = 1 I got 1/2 and for n=3, I got 0.430.
I don't understand where the problem asks "Also, using this solution show that as ''n'' goes to infinity you get the classical solution of 0.5." From my calculations, as n goes to infinity, it does not approach a value of 0.5.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Feb 11, 2013
### TSny
It will help if you can evaluate the integral for a general (unspecified) values of n and then look at the result as n goes to infinity.
From the 3 values you have obtained, you can't tell whether or not the probability is approaching any specific value as n gets large. (By the way, I agree with your answers for n = 1 and 2, but not for n = 3.)
3. Feb 11, 2013
### clamtrox
Drawing a picture of some of the solutions might give you some insight into what kind of answer you are looking for.
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http://www.investopedia.com/exam-guide/cfa-level-1/quantitative-methods/probability-distribution-properties.asp | # CFA Level 1
AAA
## Quantitative Methods - Common Probability Distribution Properties
Normal Distribution
The normal distribution is a continuous probability distribution that, when graphed as a probability density, takes the form of the so-called bell-shaped curve. The bell shape results from the fact that, while the range of possible outcomes is infinite (negative infinity to positive infinity), most of the potential outcomes tend to be clustered relatively close to the distribution's mean value. Just how close they are clustered is given by the standard deviation. In other words, a normal distribution is described completely by two parameters: its mean (μ) and its standard deviation (σ).
Here are other defining characteristics of the normal distribution: it is symmetric, meaning the mean value divides the distribution in half and one side is the exact mirror image of the other -that is, skewness = 0. Symmetry also requires that mean = median = mode. Its kurtosis (measure of peakedness) is 3 and its excess kurtosis (kurtosis - 3) equals 0. Also, if given 2 or more normally distributed random variables, the linear combination must also be normally distributed.
While any normal distribution will share these defining characteristics, the mean and standard deviation will be unique to the random variable, and these differences will affect the shape of the distribution. On the following page are two normal distributions, each with the same mean, but the distribution with the dotted line has a higher standard deviation.
Univariate vs. Multivariate Distributions
A univariate distribution specifies probabilities for a single random variable, while a multivariate distribution combines the outcomes of a group of random variables and summarizes probabilities for the group. For example, a stock will have a distribution of possible return outcomes; those outcomes when summarized would be in a univariable distribution. A portfolio of 20 stocks could have return outcomes described in terms of 20 separate univariate distributions, or as one multivariate distribution.
Earlier we indicated that a normal distribution is completely described by two parameters: its mean and standard deviation. This statement is true of a univariate distribution. For models of multivariate returns, the mean and standard deviation of each variable do not completely describe the multivariate set. A third parameter is required, the correlation, or co-movement, between each pair of variables in the set. For example, if a multivariate return distribution was being assembled for a portfolio of stocks, and a number of pairs were found to be inversely related (i.e. one increases at the same time the other decreases), then we must consider the overall effect on portfolio variance. For a group of assets that are not completely positively related, there is the opportunity to reduce overall risk (variance) as a result of the interrelationships.
For a portfolio distribution with n stocks, the multivariate distribution is completely described by the n mean returns, the n standard deviations and the n*(n - 1)/2 correlations. For a 20-stock portfolio, that's 20 lists of returns, 20 lists of variances of return and 20*19/2, or 190 correlations.
Confidence Intervals
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