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On the Well-Posedness of SPDEs with Singular Drift in Divergence Form Abstract We prove existence and uniqueness of strong solutions for a class of second-order stochastic PDEs with multiplicative Wiener noise and drift of the form ${\mathrm {div}}\gamma (\nabla \cdot )$, where $\gamma $ is a maximal monotone graph in $\mathbb {R}^n \times \mathbb {R}^n$ obtained as the subdifferential of a convex function satisfying very mild assumptions on its behavior at infinity. The well-posedness result complements the corresponding one in our recent work arXiv:1612.08260 where, under the additional assumption that $\gamma $ is single-valued, a solution with better integrability and regularity properties is constructed. The proof given here, however, is self-contained. KeywordsStochastic evolution equations Singular drift Divergence form Multiplicative noise Monotone operators 2010 Mathematics Subject ClassificationPrimary: 60H15 47H06 Secondary: 46N30 Notes Acknowledgements The authors are partially supported by The Royal Society through its International Exchange Scheme. Parts of this chapter were written while the first-named author was visiting the Interdisziplinäres Zentrum für Komplexe Systeme at the University of Bonn, hosted by Prof. S. Albeverio. References 1. 2. 3. 4.Krylov, N.V., Rozovskiĭ, B.L.: Current problems in mathematics. Stochastic Evolution Equations, vol. 14 (Russian), pp. 71–147, 256. Akad. Nauk SSSR, Vsesoyuz. Inst. Nauchn. i Tekhn. Informatsii, Moscow (1979). MR MR570795 (81m:60116)Google Scholar 5. 6. 7.Marinelli, C., Scarpa, L.: Strong solutions to SPDEs with monotone drift in divergence form. arXiv:1612.08260 8.Marinelli, C., Scarpa, L.: A note on doubly nonlinear SPDEs with singular drift in divergence form. arXiv:1712.05595 9.Pardoux, E.: Equations aux derivées partielles stochastiques nonlinéaires monotones, Ph.D. thesis, Université Paris XI (1975)Google Scholar 10.Scarpa, L.: Well-posedness for a class of doubly nonlinear stochastic PDEs of divergence type. J. Differ. Eqn. 263(4), 2113–2156 (2017)Google Scholar 11.
So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions. In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in Introduction to Functions and Graphs, exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions. Derivative of the Exponential Function Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course. First of all, we begin with the assumption that the function $B(x)=b^x,b>0,$ is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined—beginning with the definition of $b^n$, where $n$ is a positive integer—as the product of $b$ multiplied by itself $n$ times. Later, we defined $b^0=1,b^{−n}=\frac{1}{b^n}$, for a positive integer $n$, and $b^{s/t}=(\sqrt[t]{b})^s$ for positive integers $s$ and $t$. These definitions leave open the question of the value of br where r is an arbitrary real number. By assuming the continuity of $B(x)=b^x,b>0$, we may interpret $b^r$ as $\displaystyle \lim_{x→r}b^x$ where the values of $x$ as we take the limit are rational. For example, we may view $4^π$ as the number satisfying \[4^3<4^π<4^4,4^{3.1}<4^π<4^{3.2},4^{3.14}<4^π<4^{3.15},\] \[4^{3.141}<4^{π}<4^{3.142},4^{3.1415}<4^{π}<4^{3.1416},….\] As we see in the following table, $4^π≈77.88.$ $x$ $4^x$ $x$ $4^x$ $4^3$ 64 $4^{3.141593}$ 77.8802710486 $4^{3.1}$ 73.5166947198 $4^{3.1416}$ 77.8810268071 $4^{3.14}$ 77.7084726013 $4^{3.142}$ 77.9242251944 $4^{3.141}$ 77.8162741237 $4^{3.15}$ 78.7932424541 $4^{3.1415}$ 77.8702309526 $4^{3.2}$ 84.4485062895 $4^{3.14159}$ 77.8799471543 $4^{4}$ 256 Approximating a Value of $4^π$ We also assume that for $B(x)=b^x,\, b>0$, the value $B′(0)$ of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function $B(x)$ is differentiable everywhere. We make one final assumption: that there is a unique value of $b>0$ for which $B′(0)=1$. We define e to be this unique value, as we did in Introduction to Functions and Graphs. Figure provides graphs of the functions $y=2^x, \,y=3^x, \,y=2.7^x,$ and $y=2.8^x$. A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of e lies somewhere between 2.7 and 2.8. The function $E(x)=e^x$ is called the natural exponential function. Its inverse, $L(x)=\log_e x=\ln x$ is called the natural logarithmic function. For a better estimate of $e$, we may construct a table of estimates of $B′(0)$ for functions of the form $B(x)=b^x$. Before doing this, recall that \[B′(0)=\lim_{x→0}\frac{b^x−b^0}{x−0}=\lim_{x→0}\frac{b^x−1}{x}≈\frac{b^x−1}{x}\] for values of $x$ very close to zero. For our estimates, we choose $x=0.00001$ and $x=−0.00001$ to obtain the estimate \[\frac{b^{−0.00001}−1}{−0.00001}<B′(0)<\frac{b^{0.00001}−1}{0.00001}.\] See the following table. b $\frac{b^{−0.00001}−1}{−0.00001}<B′(0)<\frac{b^{0.00001}−1}{0.00001}.$ b $\frac{b^{−0.00001}−1}{−0.00001}<B′(0)<\frac{b^{0.00001}−1}{0.00001}.$ 2 $0.693145<B′(0)<0.69315$ 2.718 $1.000002<B′(0)<1.000012$ 2.7 $0.993247<B′(0)<0.993257$ 2.719 $1.000259<B′(0)<1.000269$ 2.71 $0.996944<B′(0)<0.996954$ 2.72 $1.000627<B′(0)<1.000637$ 2.718 $0.999891<B′(0)<0.999901$ 2.8 $1.029614<B′(0)<1.029625$ 2.7182 $0.999965<B′(0)<0.999975$ 3 $1.098606<B′(0)<1.098618$ The evidence from the table suggests that $2.7182<e<2.7183.$ The graph of $E(x)=e^x$ together with the line $y=x+1$ are shown in Figure. This line is tangent to the graph of $E(x)=e^x$ at $x=0$. Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of $B(x)=b^x, \,b>0$. Recall that we have assumed that $B′(0)$ exists. By applying the limit definition to the derivative we conclude that \[B′(0)=\lim_{h→0}\frac{b^{0+h}−b^0}{h}=\lim_{h→0}\frac{b^h−1}{h} \] Turning to $B′(x)$, we obtain the following. $\displaystyle B′(x)=\lim_{h→0}\frac{b^{x+h}−b^x}{h}$ Apply the limit definition of the derivative. $\displaystyle =\lim_{h→0}\frac{b^xb^h−b^x}{h}$ Note that $b^{x+h}=b^xb^h$. $\displaystyle =\lim_{h→0}\frac{b^x(b^h−1)}{h}$ Factor out $b^x$. $\displaystyle =b^x\lim_{h→0}\frac{b^h−1}{h}$ Apply a property of limits. $=b^xB′(0)$ Use $\displaystyle B′(0)=\lim_{h→0}\frac{b^{0+h}−b^0}{h}=\lim_{h→0}\frac{b^h−1}{h}$. We see that on the basis of the assumption that $B(x)=b^x$ is differentiable at $0,B(x)$ is not only differentiable everywhere, but its derivative is \[B′(x)=b^xB′(0).\nonumber\] For $E(x)=e^x, \,E′(0)=1.$ Thus, we have $E′(x)=e^x$. (The value of $B′(0)$ for an arbitrary function of the form $B(x)=b^x, \,b>0,$ will be derived later.) Derivative of the Natural Exponential Function Let $E(x)=e^x$ be the natural exponential function. Then \[E′(x)=e^x.\] In general, \[\frac{d}{dx}(e^{g(x)})=e^{g(x)}g′(x)\] Example $\PageIndex{1}$: Derivative of an Exponential Function Find the derivative of $f(x)=e^{\tan(2x)}$. Solution: Using the derivative formula and the chain rule, \[f′(x)=e^{\tan(2x)}\frac{d}{dx}(\tan(2x))=e^{\tan(2x)}\sec^2(2x)⋅2\] Example $\PageIndex{2}$: Combining Differentiation Rules Find the derivative of $y=\dfrac{e^{x^2}}{x}$. Solution Use the derivative of the natural exponential function, the quotient rule, and the chain rule. $y′=\dfrac{(e^{x^2}⋅2)x⋅x−1⋅e^{x^2}}{x^2}$ Apply the quotient rule. $=\dfrac{e^{x^2}(2x^2−1)}{x^2}$ Simplify. Exercise $\PageIndex{1}$ Find the derivative of $h(x)=xe^{2x}$. Hint Don’t forget to use the product rule. Answer $h′(x)=e^{2x}+2xe^{2x}$ Example $\PageIndex{3}$: Applying the Natural Exponential Function A colony of mosquitoes has an initial population of 1000. After $t$ days, the population is given by $A(t)=1000e^{0.3t}$. Show that the ratio of the rate of change of the population, $A′(t)$, to the population, $A(t)$ is constant. Solution First find $A′(t)$. By using the chain rule, we have $A′(t)=300e^{0.3t}.$ Thus, the ratio of the rate of change of the population to the population is given by \[A′(t)=\frac{300e^{0.3t}}{1000e^{0.3t}}=0.3.\] The ratio of the rate of change of the population to the population is the constant 0.3. Exercise $\PageIndex{2}$ If $A(t)=1000e^{0.3t}$ describes the mosquito population after $t$ days, as in the preceding example, what is the rate of change of $A(t)$ after 4 days? Hint Find $A′(4)$. Answer $996$ Derivative of the Logarithmic Function Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function. Definition: The Derivative of the Natural Logarithmic Function If $x>0$ and $y=\ln x$,then \[\frac{dy}{dx}=\frac{1}{x}.\] More generally, let $g(x)$ be a differentiable function. For all values of $x$ for which $g′(x)>0$, the derivative of $h(x)=\ln(g(x))$ is given by \[h′(x)=\frac{1}{g(x)}g′(x).\] Proof If $x>0$ and $y=\ln x$, then $e^y=x.$ Differentiating both sides of this equation results in the equation \[e^y\frac{dy}{dx}=1.\] Solving for $\dfrac{dy}{dx}$ yields \[\frac{dy}{dx}=\frac{1}{e^y}.\] Finally, we substitute $x=e^y$ to obtain \[\frac{dy}{dx}=\frac{1}{x}.\] We may also derive this result by applying the inverse function theorem, as follows. Since $y=g(x)=\ln x$ is the inverse of $f(x)=e^x$, by applying the inverse function theorem we have \[\frac{dy}{dx}=\frac{1}{f′(g(x))}=\frac{1}{e^{lnx}}=\frac{1}{x}.\] Using this result and applying the chain rule to $h(x)=\ln(g(x))$ yields \[h′(x)=\frac{1}{g(x)}g′(x).\] □ The graph of $y=\ln x$ and its derivative $\dfrac{dy}{dx}=\dfrac{1}{x}$ are shown in Figure. Example $\PageIndex{4}$:Taking a Derivative of a Natural Logarithm Find the derivative of $f(x)=\ln(x^3+3x−4)$. Solution Use Equation directly. $f′(x)=\dfrac{1}{x^3+3x−4}⋅(3x^2+3)$ Use $g(x)=x^3+3x−4$ in $h′(x)=\dfrac{1}{g(x)}g′(x)$. $=\dfrac{3x^2+3}{x^3+3x−4}$ Rewrite. Example $\PageIndex{5}$:Using Properties of Logarithms in a Derivative Find the derivative of $f(x)=\ln(\frac{x^2\sin x}{2x+1})$. Solution At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler. $f(x)=\ln(\frac{x^2\sin x}{2x+1})=2\ln x+\ln(\sin x)−\ln(2x+1)$ Apply properties of logarithms. $f′(x)=\dfrac{2}{x}+\cot x−\dfrac{2}{2x+1}$ Apply sum rule and $h′(x)=\dfrac{1}{g(x)}g′(x)$. Exercise $\PageIndex{3}$ Differentiate: $f(x)=\ln(3x+2)^5$. Hint Use a property of logarithms to simplify before taking the derivative. Answer $f′(x)=\dfrac{15}{3x+2}$ Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of $y=\log_b x$ and $y=b^x$ for $b>0, \,b≠1$. Derivatives of General Exponential and Logarithmic Functions Let $b>0,b≠1,$ and let $g(x)$ be a differentiable function. i. If, $y=\log_b x$, then \[\frac{dy}{dx}=\frac{1}{x\ln b}.\] More generally, if $h(x)=\log_b(g(x))$, then for all values of $x$ for which $g(x)>0$, \[h′(x)=\frac{g′(x)}{g(x)\ln b}.\] ii. If $y=b^x,$ then \[\frac{dy}{dx}=b^x\ln b.\] More generally, if $h(x)=b^{g(x)},$ then \[h′(x)=b^{g(x)}g''(x)\ln b\] Proof If $y=\log_b x,$ then $b^y=x.$ It follows that $\ln(b^y)=\ln x$. Thus $y\ln b=\ln x$. Solving for $y$, we have $y=\dfrac{\ln x}{\ln b}$. Differentiating and keeping in mind that $\ln b$ is a constant, we see that \[\frac{dy}{dx}=\frac{1}{x\ln b}.\] The derivative in Equation now follows from the chain rule. If $y=b^x$. then $\ln y=x\ln b.$ Using implicit differentiation, again keeping in mind that $\ln b$ is constant, it follows that $\dfrac{1}{y}\dfrac{dy}{dx}=\ln b$. Solving for $\dfrac{dy}{dx}$ and substituting $y=b^x$, we see that \[\frac{dy}{dx}=y\ln b=b^x\ln b.\] The more general derivative (Equation) follows from the chain rule. □ Example $\PageIndex{6}$:Applying Derivative Formulas Find the derivative of $h(x)=\dfrac{3^x}{3^x+2}$. Solution Use the quotient rule and Note. $h′(x)=\dfrac{3^x\ln 3(3^x+2)−3^x\ln 3(3^x)}{(3^x+2)^2}$ Apply the quotient rule. $=\dfrac{2⋅3^x\ln 3}{(3x+2)^2}$ Simplify. Example $\PageIndex{7}$: Finding the Slope of a Tangent Line Find the slope of the line tangent to the graph of $y=\log_2 (3x+1)$ at $x=1$. Solution To find the slope, we must evaluate $\dfrac{dy}{dx}$ at $x=1$. Using Equation, we see that \[\frac{dy}{dx}=\frac{3}{(3x+1)\ln 2}.\] By evaluating the derivative at $x=1$, we see that the tangent line has slope \[\frac{dy}{dx}\bigg{\|}_{x=1}=\frac{3}{4\ln 2}=\frac{3}{\ln 16}.\] Exercise $\PageIndex{4}$ Find the slope for the line tangent to $y=3^x$ at $x=2.$ Hint Evaluate the derivative at $x=2.$ Answer $9\ln(3)$ Logarithmic Differentiation At this point, we can take derivatives of functions of the form $y=(g(x))^n$ for certain values of $n$, as well as functions of the form $y=b^{g(x)}$, where $b>0$ and $b≠1$. Unfortunately, we still do not know the derivatives of functions such as $y=x^x$ or $y=x^π$. These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form $h(x)=g(x)^{f(x)}$. It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of $y=\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}$. We outline this technique in the following problem-solving strategy. Problem-Solving Strategy: Using Logarithmic Differentiation To differentiate $y=h(x)$ using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain $\ln y=\ln(h(x)).$ Use properties of logarithms to expand $\ln(h(x))$ as much as possible. Differentiate both sides of the equation. On the left we will have $\dfrac{1}{y}\dfrac{dy}{dx}$. Multiply both sides of the equation by $y$ to solve for $\dfrac{dy}{dx}$. Replace $y$ by $h(x)$. Example $\PageIndex{8}$: Using Logarithmic Differentiation Find the derivative of $y=(2x^4+1)^{\tan x}$. Solution Use logarithmic differentiation to find this derivative. $\ln y=\ln(2x^4+1)^{\tan x}$ Step 1. Take the natural logarithm of both sides. $\ln y=\tan x\ln(2x^4+1)$ Step 2. Expand using properties of logarithms. $\dfrac{1}{y}\dfrac{dy}{dx}=\sec^2 x\ln(2x^4+1)+\dfrac{8x^3}{2x^4+1}⋅\tan x$ Step 3. Differentiate both sides. Use the product rule on the right. $\dfrac{dy}{dx}=y⋅(\sec^2 x\ln(2x4+1)+\dfrac{8x^3}{2x^4+1}⋅\tan x)$ Step 4. Multiply by $y$ on both sides. $\dfrac{dy}{dx}=(2x^4+1)^{\tan x}(\sec^2 x\ln(2x^4+1)+\dfrac{8x^3}{2x^4+1}⋅\tan x)$ Step 5. Substitute $y=(2x^4+1)^{\tan x}$. Example $\PageIndex{9}$: Extending the Power Rule Find the derivative of $y=\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}$. Solution This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter. $\ln y=\ln\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}$ Step 1. Take the natural logarithm of both sides. $\ln y=\ln x+\frac{1}{2}\ln(2x+1)−x\ln e−3\ln \sin x$ Step 2. Expand using properties of logarithms. $\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\dfrac{\cos x}{\sin x}$ Step 3. Differentiate both sides. $\dfrac{dy}{dx}=y(\dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\cot x)$ Step 4. Multiply by $y$ on both sides. $\dfrac{dy}{dx}=\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}(\dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\cot x)$ Step 5. Substitute $y=\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}.$ Exercise $\PageIndex{5}$ Use logarithmic differentiation to find the derivative of $y=x^x$. Hint Follow the problem solving strategy. Answer Solution: $\dfrac{dy}{dx}=x^x(1+\ln x)$ Exercise $\PageIndex{6}$ Find the derivative of $y=(\tan x)^π$. Hint Use the power rule (since the exponent $\pi$ is a constant) and the chain rule. Answer $y′=π(\tan x)^{π−1}\sec^2 x$ Key Concepts On the basis of the assumption that the exponential function $y=b^x, \,b>0$ is continuous everywhere and differentiable at $0$, this function is differentiable everywhere and there is a formula for its derivative. We can use a formula to find the derivative of $y=\ln x$, and the relationship $\log_b x=\dfrac{\ln x}{\ln b}$ allows us to extend our differentiation formulas to include logarithms with arbitrary bases. Logarithmic differentiation allows us to differentiate functions of the form $y=g(x)^{f(x)}$ or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating. Key Equations Derivative of the natural exponential function $\dfrac{d}{dx}(e^{g(x)})=e^{g(x)}g′(x)$ Derivative of the natural logarithmic function $\dfrac{d}{dx}(\ln g(x))=\dfrac{1}{g(x)}g′(x)$ Derivative of the general exponential function $\dfrac{d}{dx}(b^{g(x)})=b^{g(x)}g′(x)\ln b$ Derivative of the general logarithmic function $\dfrac{d}{dx}(\log_b g(x))=\dfrac{g′(x)}{g(x)\ln b}$ Glossary logarithmic differentiation is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.
Closed subschemes and closed immersions of schemes have been causing me a lot of confusion for a while now. I have a few questions that I think might clear things up. Please assume that when I use the term "ring" that I mean "commutative ring with identity" whose morphisms take $0 \mapsto 0$ and $1 \mapsto 1$. Sorry for the long question, but I feel it is necessary to spell out the definitions I am working with rather than expect people to chase them up. My first exposure to this was Hartshorne page 85. There he makes the following definitions: A closed immersion is a morphism of schemes $\iota: Y \longrightarrow X$ such that $\iota$ induces a homeomorphism of sp$(Y)$ onto a closed subset of sp$(X)$, and further that the induced map of sheaves, $\iota^{\#}: \mathcal{O}_{X} \longrightarrow \iota_{}\mathcal{O}_{Y}$ is surjective. A closed subscheme is an equivalence class of closed immersions, where we say that $\iota: Y \longrightarrow X$ and $\iota': Y' \longrightarrow X$ are equivalent if there is an isomorphism $\psi: Y' \longrightarrow Y$ satisfying $\iota' = \iota \circ \psi$. After having a bit of confusion with the closed subscheme part, I consulted Görtz & Wedhorn where, on page 84 (Definition 3.41) they give their own definitions. Their definition for a closed immersion is identical, however their definition for a closed subscheme is as follows: A closed subscheme of a scheme $X$ is given by a closed subset $Y \subseteq X$ (let $\iota: Y \hookrightarrow X$ be the inclusion) and a sheaf $\mathcal{O}_{Y}$ on $Y$ such that $(Y, \mathcal{O}_{Y})$ is a scheme, and such that the sheaf $\iota _{}\mathcal{O}_{Y}$ is isomorphic to $\mathcal{O}_{X}/ \mathcal{I}$ for $\mathcal{I}$ a subsheaf of ideals of $\mathcal{O}_{X}$. With that in place, my question is basically threefold: 1) Aside from the suggestive name, Hartshorne doesn't seem to suggest (at least not to me) that a closed subscheme is actually a scheme. Indeed, how does one even make sense of putting a scheme structure on an equivalence class of morphisms? 2) Görtz & Wedhorn seem to overcome this by simply defining it to be a scheme. How are their definitions equivalent (if they are)? One big problem I am having seeing this equivalence is relating the surjectivity of $\iota^{\#}$ and $\iota^{'\#}$ to the sheaf of ideals in Görtz & Wedhorn. The issue is, since the category of rings is not abelian (indeed, not even additive as far as I know), I can't expect to have kernels and cokernels, and so I can't expect to be able to take $\mathcal{I}$ to be the kernel of the surjective morphisms like I could if they were sheaves of abelian groups. 3) I tried to play around with the affine case in Hartshorne's definition to make sense of things. Let $A$ be a ring with ideals $\mathfrak{a}$ and $\mathfrak{b}$. These give morphisms of schemes Spec$(A /\mathfrak{a}) \longrightarrow$ Spec $A$ and Spec$(A /\mathfrak{b}) \longrightarrow$ Spec $A$ which induce homemorphisms onto the closed sets in the obvious way. My intuition suggests that these should be "equivalent" (for the purposes of defining a closed subscheme) precisely if $\mathfrak{a}$ and $\mathfrak{b}$ have the same radical, since then $V(\mathfrak{a})$ and $V(\mathfrak{b})$ would be the same. Indeed the radical is just the intersection of all primes containing them. However, by Hartshorne's definition, that would require that the isomorphism $i$ be induced by an isomorphism of rings $A / \mathfrak{a} \simeq A/ \mathfrak{b}$ whenever $\mathfrak{a}$ and $\mathfrak{b}$ have the same nilradical, which is obviously nonsense. The long of the short is I am stumped and incredibly confused. Any advice, or references, or answers (partial or full) to all or any of these questions would be greatly appreciated. Bonus question: Is there a way to fix my confusion $\textit{without}$ resorting to quasi-coherent sheaves?
I'm designing a habitable earth-like planet that has oceans, forests and life. Now this planet is a bit larger in diameter than earth but its mass and therefore its gravity is a bit lower. This lower gravity has resulted into large lifeforms. Now, one of this lifeforms resembles in appearance a large jellyfish with wings that glides through the air. However, I wasn't sure if a lower gravity is enough to allow such a huge creature to fly though the air, so I figured out a way to solve this by making the atmosphere thicker. But then it hit, " if a planet has lower gravity, how can it have such a thick atmosphere?" I was looking if there were real life planets that have these traits (lower gravity, thick atmosphere) and realized than Venus is a good match. But then again, Venus isn't really the most hospitable planet there is? So, to sum it up, could a planet have lower gravity and yet a thicker atmosphere? And also could such a planet be Earth-like? Yes. In fact, given your premise of a planet with a larger diameter but lower surface gravity than Earth, I'd expect a thicker atmosphere (both in terms of atmospheric pressure and literal thickness) than Earth. The relevant variables here are gravitational acceleration and escape velocity. You want a planet with less gravity than Earth at its surface, but a higher escape velocity at its upper atmosphere. Is this plausible? Absolutely! The relevant variables that you can directly tune here are the mass and radius of the planet. Note that on the scales of planets (other than gas giants), the thickness of the atmosphere is insignificant in comparison to the radius of the planet itself- so, to simplify things here, I'll just ignore the thickness of the atmosphere entirely. The relevant formulas are gravitational acceleration at the planet's surface: $$ g = {GM\over r^2} $$ and escape velocity: $$ v_e = \sqrt {2GM \over r} $$ where r is the radius of your planet, M is the planet's mass, and G is Newton's gravitational constant. You want to reduce the surface gravity (as per your premise), while increasing the escape velocity (which will make it easier for the planet to hold on to an atmosphere). Surface gravity decreases rapidly with radius, as indicated by that $r^2$ term, so increasing the planet's radius will very quickly decrease its surface gravity. However, in order to keep the escape velocity from decreasing, the mass must be increased by at least as much. As an example, let's take a planet with 9 times Earth's mass and 4 times Earth's radius. $$ g = {G (9M_\oplus) \over (4r_\oplus)^2} = {9GM_\oplus\over 16r_\oplus^2} = {9 \over 16} g_\oplus $$ $$ v_e = \sqrt {2G (9M_\oplus) \over 4r_\oplus} = \sqrt {9 \over 4} \cdot \sqrt {2GM_\oplus \over r_\oplus} = {3 \over 2} {v_e}_\oplus $$ As you can see, this planet has just over half of Earth's gravity, but 1.5 times Earth's escape velocity. You don't need to use these exact numbers, of course- I chose them mainly to avoid getting irrational numbers. But the point is there: Take Earth, increase the radius a bit, increase the mass a bit more (but not too much more), and you're done. Now, there is one more complication for you to consider, handwave, or ignore: The material your planet is made of. Density is proportional to $M/r^3$, so it decreases even faster than surface gravity. This example planet's density is about 14% that of Earth. Which means that it must mostly be made of something less dense than water- more like ethanol or kerosene. Making things worse, larger, more massive planets tend to be denser than smaller planets, since gravity compresses the material in the middle of the planet. Earth is (slightly) denser than Mercury, even though Mercury's iron core makes up over half of its volume. Thus, your planet must be made of materials much less dense than those that make up Earth, yet able to withstand even higher pressures. Perhaps instead of silicon, magnesium, and iron, your planet has a lot of lithium? You have to remember that surface gravity and escape velocity are two different attributes of a planet, and you need one to be as low as possible and the other to be as high as possible. As far as I remember, the escape velocity from the planet, at the uppermost level of the atmosphere where gases escape from the planet, has to be a certain number of times - 5 or 6 I think - greater than the average velocity of gas molecules at that level in order to retain the atmosphere for billions of years. The average velocity of the gas molecules at the top of the atmosphere depends on their average temperature, which may be different from that at ground level. If the ratio between escape velocity and average molecular velocity is not that high, the planet will lose it s atmosphere in less than billions of years and needs to replenish its atmosphere from various sources in order to have one long enough to develop multi celled life, intelligent natives, a breathable for humans atmospheric composition, or other stuff necessary for an interesting planet in most types of science fiction stories. It is believed that the atmosphere on Titan is replenished about as fast as it escapes, for example. Earth seems capable of retaining it's original atmosphere for billions of years. Venus, for example, has a slightly lower escape velocity and a much denser and more massive atmosphere than Earth, so it is hard to imagine that a lot of Venusian atmosphere is constantly being lost and replenished. Another thing that can remove a planetary atmosphere is the solar or stellar wind of particles that knock molecules out of the upper atmosphere and into space. A strong planetary magnetosphere deflects most of those particles and protects the atmosphere. What you desire is for a planet that has as low a surface gravity as possible, in order to make flying easier, and as dense an atmosphere as possible in order to make flying easier, and having a denser atmosphere is made easier if the escape velocity is as high as possible. Mercury has a surface gravity of 3.7 meters per second per second, 0.3772 of Earth's, and an escape velocity of 4.25 kilometers per second, which is 0.3799 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.0071. Venus has a surface gravity of 8.87 meters per second per second, 0.9044 of Earth's and an escape velocity of 10.36 kilometers per second, which is 0.92615 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.0240. Earth has a surface gravity of 9.807 meters per second per second, and an escape velocity of 11.186 kilometers per second, and of course both are exactly 1.000 of Earth's.The ratio of the escape velocity compared to the ratio of the surface gravity is 1.0000. The Moon has a surface gravity of 1.62 meters per second per second, 0.1651 of Earth's, and an escape velocity of 2.38 kilometers per second, which is 0.2127 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.2883. Mars has a surface gravity of 3.711 meters per second per second, 0.3784 of Earth's, and an escape velocity of 5.027 kilometers per second, which is 0.4494 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.1876. Io has a surface gravity of 1.796 meters per second per second, 0.1831 of Earth's, and an escape velocity of 2.558 kilometers per second, which is 0.2286 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.2484. Europa has a surface gravity of 1.314 meters per second per second, 0.1339 of Earth's, and an escape velocity of 2.025 kilometers per second, which is 0.1810 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.3517. Ganymede has a surface gravity of 1.428 meters per second per second, 0.1456 of Earth's, and an escape velocity of 2.741 kilometers per second, which is 0.2450 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.6826. Callisto has a surface gravity of 1.235 meters per second per second, 0.1259 of Earth's, and an escape velocity of 2.440 kilometers per second, which is 0.2181 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.7323. Titan has a surface gravity of 1.352 meters per second per second, 0.1378 of Earth's, and an escape velocity of 2.639 kilometers per second, which is 0.2359 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.7119. Triton has a surface gravity of 0.779 meters per second per second, 0.0794 of Earth's, and an escape velocity of 1.455 kilometers per second, which is 0.1300 of Earth's escape velocity. The ratio of the escape velocity compared to the ratio of the surface gravity is 1.6372. All of these planets and moons have lower diameters, masses, densities, surface gravity, and escape velocities than Earth. And the ratio of their escape velocity to that of Earth is greater than the ratio of their surface gravity to that of Earth, in ratios varying from 1.0071 to 1.7323. So a smaller planet than Earth is likely to have an escape velocity that is not as much lower than Earth's as its surface gravity is lower than Earth's. And that is what you want, a surface gravity as low as possible to help with flying, and an escape velocity as high as possible to help retain as dense an atmosphere as possible. And your planet will have a much denser atmosphere than many other planets with similar escape velocities, because of various factors that created a much denser original atmosphere and helped it retain and/or replenish it much more than other similar planets did. Note that Venus is slightly less able to retain an atmosphere than Earth is, and yet has an atmosphere many times as dense as Earth's. Titan is less able to retain an atmosphere than Ganymede or Callisto but has a much denser atmosphere than they do. Titan is much less able to retain an atmosphere than Earth, but has an atmosphere comparable in density to Earth's. There will be an upper limit to the density of your planet's atmosphere if you want Earth humans to breath it, and a somewhat higher upper limit if you want the lifeforms on the planet to breath it. All of the gases in Earth's atmosphere, nitrogen, carbon dioxide, water vapor, even oxygen, have safe upper pressure limits. Beyond those limits all of them, even oxygen, become toxic. So you will have to add up the safe pressure limits of every gas likely to be found in a planetary atmosphere and the total pressure will be the total limit of atmospheric pressure for humans to breath. Life forms native to the planet may have evolved to tolerate higher pressures.
Exercise $\PageIndex{1}$: Hexagonal numbers (cornered) Consider the hexagonal numbers are the sequence $1,6,15, 28,45,66 \cdots.$ Predict the n th term. Explain your prediction. Answer: $2n^2-n$. Exercise $\PageIndex{2}$: Finite sum For each of the following, find the sum and explain your reasoning. Please do not use any formula. $1+3+5+7+9+\cdots +197+199$ $1+ \displaystyle \frac{1}{2}+ \displaystyle \frac{1}{4}+\cdots + \displaystyle \frac{1}{2^{16}}+\displaystyle \frac{1}{2^{17}}$ Answer: $1+3+5+7+9+···+197+199$ Notice that $1,3,5,7 ,\cdots$ terms of a sequence. This is an Arithmetic Sequence because the difference remains the same between the terms throughout the entire sequence. Hence,$ a = 1 \& \, d = 2$. Consider, $S_ n = 1+3+5+7+9+···+197+199$ $S n = 199+197+195+193+191+···+3+1$ By adding we get, $(2S_n = 200+200+200+200+200+···+200+200$ $2S_n = 100(200)$ $S_n = ((100)/2))(200)$ $S_n = (50)(200)$ $S_n = 10000$ Hence, the sum of the sequence is $10000.$ 2. Exercise $\PageIndex{3}$: Proof by induction Consider the sequence $ 4,10,16, 22, 28,,\dots$, assume that the pattern continues. Show that the $n^{th}$ term of this sequence can be expressed as $6n-2$. Prove by using induction for all integers $ n \geq 1, 4+10+16+\dots+(6n-2)=n(3n+1)$ Answer: 1. Term First difference 4 10 6 16 6 22 6 Notice that the first difference is constant. Hence the $n^{th}$ term is a linear function. Let $t_n = a_n + b.$ Then we need to find $a, b$. First Equation: Let $n = 1$ $t_1 = 4$ $4 = a(1) + b$ $4 = a + b$ Second Equation: Let $n = 2$ $t_2 = 10$ $10 = a(2) + b$ $10 = 2a + b$ To find $a$, we use $10 = 2a + b$ and $- 4 = a + b$. Therfore, $6 = a.$ Now to find $b$, we use $a = 6$ and $4 = a + b$, $4 = (6) + b$ $4 - 6 = b$ $-2 = b$. Hence,$t_n = 6n - 2.$ 2. Step 1: Base Step: Show that this statement is true for the smallest value Check statement is true for n = 1. L.H.S = 4 R.H.S = n(3n + 1) = (1)(3(1) + 1) = (1)(3 + 1) = (1)(4) = 4 Hence, the statement is true for n = 1. Step 2: Induction Assumption: We shall assume that the statement is true for n = k. 4 + 10 + 16 + . . . + (6k − 2) = k(3k + 1) Step 3: Induction: We shall show that the statement is true for n = k + 1. 4 + 10 + 16 + . . . + (6k − 2) + (6 (k + 1) − 2) = (k+1)(3(k + 1) + 1) Consider, L.H.S = 4 + 10 + 16 + . . . + (6k − 2) + (6 (k + 1) − 2) = k(3k + 1) + (6 ( k + 1) - 2) = k (3k + 1) + (6k + 6 - 2) = k (3k + 1) + (6k + 4) = 3k 2 + k + 6k + 4 = 3k 2 + 7k + 4 = (k + 1)(3k + 4) Hence, the statement is true for n = k + 1 Therefore, by induction the statement is true, ∀n ∈ N. Exercise $\PageIndex{4}$: Proof by induction Consider the sequence $ 3,11,19, 27, 35,\dots$, assume that the pattern continues. Show that the $n^{th}$ term of this sequence can be expressed as $8n-5$. Prove by using induction for all integers $ n\geq 1, 3+11+19 \dots + (8n-5)=4n^2-n.$ Exercise $\PageIndex{5}$: Tribonacci Let's start with the numbers $0,0,1,$ and generate future numbers in our sequence by adding up the previous three numbers. Write out the first $15$ terms in this sequence, starting with the first $1$. Exercise $\PageIndex{6}$: Proof by induction The sequence $b_0,b_1,b_2....$ is defined as follows: $b_0=1,b_1=3,b_2=5,$ and for any integer $n \geq 3, \, b_n=3b_{n-2}+2b_{n-3}.$ Find $b_3,b_4,b_5$ and $b_6$. Prove that $b_n < 2^{n+1}$ for all integers $n \geq 1.$ Exercise $\PageIndex{7}$: Quadratic Sequence Find the $n^{th}$ term of the sequence $5,10,17, 26, 37, \cdots$, assume that the pattern continues. Answer: $n^2+2n+2$ Exercise $\PageIndex{8}$: Proof by induction Prove by using induction: for all integers $ n\geq 1, \, 1+4+7 \dots + (3n-2)=\frac{n(3n-1)}{2}.$ Answer: Step 1: Base Step: Show that this statement is true for the smallest value Check statement is true for n = 1. L.H.S = 1 R.H.S = n(3n−1) / (2) = (1)(3(1) − 1) / (2) = (1)(3 − 1) / (2) = (1)(2) / (2) = 1 Hence, the statement is true for n = 1. Step 2: Induction Assumption: We shall assume that the statement is true for n = k. 1+4+7...+(3k−2)= k(3k−1) / (2) Step 3: Induction: We shall show that the statement is true for n = k + 1. 1+4+7...+ (3k - 2) + (3(k + 1) − 2) = (k + 1)(3(k + 1) − 1) / (2) Consider, L.H.S = k(3k − 1) / (2) + (3 (k + 1) − 2) = k (3k − 1) / (2) + (3k + 3) − 2) = k (3k − 1) / (2) + (3k + 1) = (3k 2 + k ) / (2) + (3k + 1) = (3k 2 + k + 3k + 1) / (2) = (3k 2 + 4k + 1) / (2) = ((k + 1)(3k + 1)) / (2) Hence, the statement is true for n = k + 1 Therefore, by induction the statement is true, ∀n ∈ N. Exercise $\PageIndex{9}$: Recognizing sequence Predict $n^{th}$ term of the sequence $\frac{2}{3},\frac{3}{4}, \frac{4}{5}\cdots\,$ assume that the pattern continues. Explain your prediction. Answer: $ \frac{n}{n+1}$. Exercise $\PageIndex{10}$: Recognizing sequence Consider the sequence $ t_1=1, t_2=3+5, t_3=7+9+11, \cdots $. Predict the n th term. Justify your prediction. Exercise $\PageIndex{11}$: Proof by induction Show that the perimeter of the design by joining $n$ hexagons in a row is $8n+4$ cm. Exercise $\PageIndex{13}$: Pentagonal Numbers (cornered) Find the $n^{th}$ term of the sequence $1,5,12, 22, \cdots$,assume that the pattern continues. Exercise $\PageIndex{14}$: Square Pyramidal numbers Find the $n^{th}$ term of the sequence $1,5,14,30 \cdots$., assume that the pattern continues. Exercise $\PageIndex{15}$: Difference Compute the difference of each of the following sequences: $a_n= n^3$ $a_n=n^{\underline{3}}$ $a_n= \displaystyle {n \choose 3 } $ Answer $n^2+2n+1$ $3 n^2 $ $n \choose 2$.
I was unsure if I should post this because the question has not been active for a year. I decided to go ahead anyway, because people might have the same question in the future. The following paragraph is taken from Structure of Lie Groups and Lie Algebras byV. V. Gorbatsevich, A. L. Onishchik and E. B. Vinberg. Let $A$ be a simply-connected covering group for $SL_2(\mathbb{R})$. As isknown, $Z(A) \simeq \mathbb{Z}$. Let $z_0$ be a generator of the group $Z(A)$, and let $t_0 \in \mathbb{T}=SO_2$ be a rotation through an angle incommensurable with $\pi$. Then $\Gamma = \langle(t_0, z_0)\rangle$ is a discrete normal subgroupof $\mathbb{T} \times A$. Let $G = (\mathbb{T} \times A)/\Gamma$. Then the image $L$ of the subgroup A under the natural homomorphism $\mathbb{T} \times A \rightarrow G$ is a Levi subgroup of G. The factthat the subgroup $\langle t_0 \rangle$ is dense in $\mathbb{T}$ implies that $L$ is dense in $G$, i.e. it is not a Lie subgroup. This shows that question (1), which is true for algebraic groups in characteristic 0 as explained in the other answer, is not generally true for Lie groups, as the Levi factor can fail to be closed.
TIPS FOR SOLVING QUESTIONS RELATED TO AREA: 1. Area of Square = (side) Perimeter of Square = 4 x side 2. Area of Rectangle = Length x Breadth Perimeter of Rectangle = 2(Length + Breadth) 3. Area of 4 walls of a room = 2(Length + Breadth) x Height 4. Triangle: A triangle is a a plane figure with three straight sides and three angles. (i) Right triangle with base and height given. \begin{aligned} \text{Area of a triangle =} \frac{1}{2}BaseHeight \\ \end{aligned} (ii) Triangle with three different sides a, b and c. \begin{aligned} \text{Area of a triangle =}\sqrt{s(s-a)(s-b)(s-c)}, \\ \end{aligned} \begin{aligned} s= \frac{1}{2}(a+b+c) \\ \end{aligned} (iii) Equilateral triangle - A triangle with all three sides of equal length. \begin{aligned} \text{Area of a equilateral triangle =}\frac{\sqrt{3}}{4}(side)^2 \end{aligned} (iv). Radius of incircle of an equilateral triangle of side a = \begin{aligned} 5. 6. Circle: A circle is a round plane figure whose boundary (the circumference) consists of points equidistant from a fixed point (the centre). (i) \begin{aligned} \text{Area of a circle} = \pi R^2 \end{aligned}, where R is radius of the circle (ii) \begin{aligned} \text{Circumference of a circle} = 2 \pi R \end{aligned} (iii) \begin{aligned} \text{Length of a arc} = \frac{2 \pi R \theta }{360} \\ \text{ where } \theta \text{ is the central angle } \end{aligned} (iv) \begin{aligned} \text{Area of sector = } \frac{1}{2}(arc \theta) \\ = \frac{\pi R^2 \theta}{360} \end{aligned} (v) \begin{aligned} \text{Area of a semi circle =} \frac{\pi R^2}{2} \end{aligned} (vi) \begin{aligned}\text{Circumference of a semi circle =} \pi R \end{aligned}
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building) It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore) In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of @TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies \|\| x ( t ) - 0 \|\| < \varepsilon.$$The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $... "If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}\|f(z)\| \leq C_K \\|f\\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed? Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2 Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$ Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight. hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$ for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$ I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything. I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set).theorem ... The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious. (but seriously, the best tactic is over powered...) Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field? Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement? "Infinity exists" comes to mind as a potential candidate statement. Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system @Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity but so far failed Put it in another way, an equivalent formulation of that (possibly open) problem is: > Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object? If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite. My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science... O great, given a transcendental $s$, computing $\min_P(\|P(s)\|)$ is a knapsack problem hmm... By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as: $$P(x) = \prod_{k=0}^n (x - \lambda_k)$$ If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic Thus given $s$ transcendental, to minimise $\|P(s)\|$ will be given as follows: The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases. In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that0<\|x−pq\|<1qn.{\displaystyle 0<\left\|x-{\frac {p}... Do these still exist if the axiom of infinity is blown up? Hmmm... Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum: $$\sum_{k=1}^M \frac{1}{b^{k!}}$$ The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework There's this theorem in Spivak's book of Calculus:Theorem 7Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and$$f'... and neither Rolle nor mean value theorem need the axiom of choice Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set > are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion
The question is that I want to know whether there is difference in the applying of sign bias test in detecting the exist of asymmetric effects and the adequacy of symmetric GARCH model. In the definition of sign bias test, we need to do the t test for the coefficient $\beta$ in the regression equation $\hat{\epsilon}_t^2 = \alpha + \beta S_{t-1}^- + z_t$, where $\hat{\epsilon}_t$ is the estimated residual, $S_{t-1}^-$ is the dummy function that $\hat{\epsilon}_{t-1} < 0 $ and 0 otherwise and $z_t$ is the noise. My quiz is that how do I get the estimated $\hat{\epsilon}$. In the original paper by Engle and Ng http://www.finance.martinsewell.com/stylized-facts/volatility/EngleNg1993.pdf It explained that $\hat{\epsilon}_t$ is from the equation that $\hat{\epsilon}_t = y_t - \mu_t$. My understanding is that, if $\hat{\epsilon}_t$ is from the that equation, then it is only with the purpose to check the exist of asymmetric effects. But some material explain $\hat{\epsilon}_t$ is estimated from a symmetric GARCH model, for example, GARCH(1,1), if we want to show whether a symmetric GARCH model is adequate in describing the asymmetric effects. I am confused about it. Thanks in advance.
Let $P$ be a convex $n$-gon. Suppose that we have an infinite number of $P$s, and that each of them is colored either red or blue. Here, let us consider the following operations : Operation 1 : Place a red $P$ on a plane. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ Operation 2 : Place $n$ blue $P$s around the red $P$ such that each of the blue $P$s and the red $P$ are laid symmetrically with each $E_i\ (i=1,2,\cdots,n)$ where $E_i$ is an edge of the red $P$. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ Operation 3 : Place $n$ red $P$s around every blue $P$ in the same way as operation 2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ Operation 4 : Place $n$ blue $P$s around every red $P$ in the same way as operation 2. Operation 5 : Repeat operation 3 and 4. Here, let us consider the following conditions : Condition 1 : These $P$s are plane tessellation figures. Condition 2 : There exists no place where both red $P$ and blue $P$ are placed. Note that a regular hexagon, for example, does not satisfy the condition 2. See the figure below. The blue $P$ on which a letter $P$ is written, for example, will be colored red by the next operation. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ Then, here is the first question. Question 1: Is the following true? $P$ satisfies these two conditions $\iff$ $P$ is either "a $45–45–90$ triangle", "a $30–60–90$ triangle", "an equilateral triangle" or "a rectangle". I reached this conjecture by considering the inner angles of $P$. The followings are what I've thought : Every inner angle, say $\alpha$, of $P$ has to satisfy $2m\alpha=360^{\circ}$ where $m\ge 2\in\mathbb N$. Hence, $\alpha$ has to be any of the positive divisors of $180$ except $180$. This leads $n\le 4$ and so on. Then, here is the second question. Question 2: Letting $P$ be a convexpolyhedron, how about the three dimensional version of this question? Suppose that we consider the plane symmetry instead of the line symmetry. In the two dimensional version, I think we can consider the inner angles of $P$. However, I don't have any good idea for the higher dimensional version. Can anyone help?
Systematics on lifetime measurments We propose the following text (latex format). latex abbreviations are defined using the standard lhcb latex defintion file. The particle decay times are measured from the distance between the primary vertex and secondary decay vertex in the VELO. The accuracy with which this distance is known is dependent on the precision with which the relative position along the beam line ($z$ axis) of the LHCb modules is determined.\\ There are two contributions to this systematic uncertainty. First there is the precision with which the VELO modules were assembled. This has been determined during a survey at the time of assembly to be 100~\mum~ of the measurement of the baseplate over the whole length of the VELO \cite{bib:VELOPerformance}. \begin{equation} \sigma_{\text{survey}} = \frac{100\times10^{-3}~\text{mm}}{1000~\text{mm}} = 0.01 \%. \end{equation} The second contribution originates from track-based alignment \cite{bib:alignKalman, bib:alignVELO, bib:alignVELOResult}. This is mostly determined by the first two modules on the track since the following modules are weighted down due to multiple scattering effects. So in principle the $z$-scale uncertainty is obtained comparing the $z$ module position from the track-based alignment with the metrology (20~\mum) divided by the spacing between two modules (30~mm). However since the signal tracks have some spread in $z$ within the VELO, they do not all hit the same module first (see Fig.\ref{fig_zpos}). The RMS of this distributions (100~mm) is a measure for the effective spread of the tracks. Therefore the resulting uncertainty from tack-based alignment is given by \begin{equation} \sigma_{\text{track}} = \frac{20\times10^{-3}~\text{mm}}{100~\text{mm}} = 0.02 \%. \end{equation} \begin{figure} \centering \includegraphics[width=0.6\linewidth]{./plots/ct/zpos_first_hit.eps} \caption{$z$-position of the first hit on each track used in this analysis.\label{fig_zpos}} \end{figure} For the overall $z$-scale systematic we add the two contributions in quadrature and end up with \begin{equation} \sigma_{z\text{-scale}} = 0.022 \%. \end{equation} This is directly translated into a relative uncertainty on \Dms. Therefore the systematic uncertainty on \Dms that we assign to the $z$-scale is $\pm 0.004$ ps$^{-1}$. \bibitem{bib:alignKalman}{W.~Hulsbergen, ``The global covariance matrix of tracks fitted with a Kalman filter and an application in detector alignment'', Nucl.\ Instrum.\ Meth.\ A {\bf 600} (2009) 471 [arXiv:0810.2241 [physics.ins-det]]. %%CITATION = NUIMA,A600,471;%% } \bibitem{bib:alignVELO}{ S.~Viret, C.~Parkes and M.~Gersabeck, ``Alignment procedure of the LHCb Vertex Detector'', Nucl.\ Instrum.\ Meth.\ A {\bf 596} (2008) 157 [arXiv:0807.5067 [physics.ins-det]]. %%CITATION = NUIMA,A596,157;%% } \bibitem{bib:alignVELOResult}{ S.~Borghi {\it et al.}, ``First spatial alignment of the LHCb VELO and analysis of beam absorber collision data,'', Nucl.\ Instrum.\ Meth.\ A {\bf 618} (2010) 108. %%CITATION = NUIMA,A618,108;%% } \bibitem{bib:VELOPerformance}{LHCb VELO grou},``Performance of the LHCb VELO, to be submitted to JINST''}
A tetrahedral snake, sometimes called a Steinhaus snake, is a collection of tetrahedra, linked face to face. Steinhaus showed in 1956 that the last tetrahedron in the snake can never be a translation of the first one. This is a consequence of the fact that the group generated by the four reflexions in the faces of a tetrahedron form the free product $C_2 \ast C_2 \ast C_2 \ast C_2$. For a proof of this, see Stan Wagon’s book The Banach-Tarski paradox, starting at page 68. The thread $(3\|3)$ is the spine of the $(9\|1)$-snake which involves the following lattices \[ \xymatrix{& & 1 \frac{1}{3} \ar@[red]@{-}[dd] & & \\ & & & & \\ 1 \ar@[red]@{-}[rr] & & 3 \ar@[red]@{-}[rr] \ar@[red]@{-}[dd] & & 1 \frac{2}{3} \\ & & & & \\ & & 9 & &} \] It is best to look at the four extremal lattices as the vertices of a tetrahedron with the lattice $3$ corresponding to its point of gravity. The congruence subgroup $\Gamma_0(9)$ fixes each of these lattices, and the arithmetic group $\Gamma_0(3\|3)$ is the conjugate of $\Gamma_0(1)$ \[ \Gamma_0(3\|3) = \{ \begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}.\begin{bmatrix} a & b \\ c & d \end{bmatrix}.\begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & \frac{b}{3} \\ 3c & 1 \end{bmatrix}~\|~ad-bc=1 \} \] We know that $\Gamma_0(3\|3)$ normalizes the subgroup $\Gamma_0(9)$ and we need to find the moonshine group $(3\|3)$ which should have index $3$ in $\Gamma_0(3\|3)$ and contain $\Gamma_0(9)$. So, it is natural to consider the finite group $A=\Gamma_0(3\|3)/\Gamma_9(0)$ which is generated by the co-sets of \[ x = \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix} \qquad \text{and} \qquad y = \begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix} \] To determine this group we look at the action of it on the lattices in the $(9\|1)$-snake. It will fix the central lattice $3$ but will move the other lattices. Recall that it is best to associate to the lattice $M.\frac{g}{h}$ the matrix \[ \alpha_{M,\frac{g}{h}} = \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} \] and then the action is given by right-multiplication. \[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.x = \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix}.x = \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 1 \end{bmatrix}.x=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] That is, $x$ corresponds to a $3$-cycle $1 \rightarrow 1 \frac{1}{3} \rightarrow 1 \frac{2}{3} \rightarrow 1$ and fixes the lattice $9$ (so is rotation around the axis through the vertex $9$). To compute the action of $y$ it is best to use an alternative description of the lattice, replacing the roles of the base-vectors $\vec{e}_1$ and $\vec{e}_2$. These latices are projectively equivalent \[ \mathbb{Z} (M \vec{e}_1 + \frac{g}{h} \vec{e}_2) \oplus \mathbb{Z} \vec{e}_2 \quad \text{and} \quad \mathbb{Z} \vec{e}_1 \oplus \mathbb{Z} (\frac{g’}{h} \vec{e}_1 + \frac{1}{h^2M} \vec{e}_2) \] where $g.g’ \equiv~1~(mod~h)$. So, we have equivalent descriptions of the lattices \[ M,\frac{g}{h} = (\frac{g’}{h},\frac{1}{h^2M}) \quad \text{and} \quad M,0 = (0,\frac{1}{M}) \] and we associate to the lattice in the second normal form the matrix \[ \beta_{M,\frac{g}{h}} = \begin{bmatrix} 1 & 0 \\ \frac{g’}{h} & \frac{1}{h^2M} \end{bmatrix} \] and then the action is again given by right-multiplication. In the tetrahedral example we have \[ 1 = (0,\frac{1}{3}), \quad 1\frac{1}{3}=(\frac{1}{3},\frac{1}{9}), \quad 1\frac {2}{3}=(\frac{2}{3},\frac{1}{9}), \quad 9 = (0,\frac{1}{9}) \] and \[ \begin{bmatrix} 1 & 0 \\ \frac{1}{3} & \frac{1}{9} \end{bmatrix}.y = \begin{bmatrix} 1 & 0 \\ \frac{2}{3} & \frac{1}{9} \end{bmatrix},\quad \begin{bmatrix} 1 & 0 \\ \frac{2}{3} & \frac{1}{9} \end{bmatrix}. y = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{9} \end{bmatrix}, \quad \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{9} \end{bmatrix}. y = \begin{bmatrix} 1 & 0 \\ \frac{1}{3} & \frac{1}{9} \end{bmatrix} \] That is, $y$ corresponds to the $3$-cycle $9 \rightarrow 1 \frac{1}{3} \rightarrow 1 \frac{2}{3} \rightarrow 9$ and fixes the lattice $1$ so is a rotation around the axis through $1$. Clearly, these two rotations generate the full rotation-symmetry group of the tetrahedron \[ \Gamma_0(3\|3)/\Gamma_0(9) \simeq A_4 \] which has a unique subgroup of index $3$ generated by the reflexions (rotations with angle $180^o$ around axis through midpoints of edges), generated by $x.y$ and $y.x$. The moonshine group $(3\|3)$ is therefore the subgroup generated by \[ (3\|3) = \langle \Gamma_0(9),\begin{bmatrix} 2 & \frac{1}{3} \\ 3 & 1 \end{bmatrix},\begin{bmatrix} 1 & \frac{1}{3} \\ 3 & 2 \end{bmatrix} \rangle \] Similar Posts: Snakes, spines, threads and all that Roots of unity and the Big Picture The defining property of 24 Monstrous dessins 2 looking for the moonshine picture A forgotten type and roots of unity (again) The Big Picture is non-commutative nc-geometry and moonshine? the Bost-Connes Hecke algebra Conway’s big picture
Given a simple complex Lie algebra $\mathfrak{g}$, recall the Springer resolution of its nilpotent cone $\widetilde{\mathcal{N}}\to \mathcal{N}$. Several times I have seen someone explaining Springer theory in terms of perverse sheaves, and each time the existence of the Grothendieck-Springer alteration $\pi: \widetilde{\mathfrak{g}}\to\mathfrak{g}$ along with the diagram $$\require{AMScd}\begin{CD} \widetilde{\mathcal{N}} @>>> \widetilde{\mathfrak{g}} @>>>\mathfrak{t}\\ @VVV @VVV @VVV\\ \mathcal{N} @>>> \mathfrak{g}@>>>\mathfrak{t}/W \end{CD}$$ is magically pulled out of a hat (eg: There also exist this other thing that...), where $\mathfrak{t}$ is the universal Cartan and $W$ is the Weyl group. Then one uses the fact that $\pi$ is a small map, giving an IC sheaf $\pi_\ast\underline{\mathbb{Q}}_\widetilde{\mathfrak{g}}$ with a natural $W$-action, which in turn induces a $W$-action on the Springer sheaf by some functoriality. I find this unsatisfying because it seems like $\widetilde{\mathfrak{g}}$ is kept mysterious. My vague question is how should I think about the Grothendieck-Springer resolution and what is its role in modern representation theory? I know this is not a good question, so let me try to refine it by asking the two following questions. 1) Is there a broader theoretical context to fit the above diagram into where I am given a resolution of singularities $X_0\to Y_0$ (maybe with $X$ symplectic?), and can find a smooth family $X\to T$ and a proper map of smooth varieties $X\to Y$ fitting into the diagram $$\require{AMScd}\begin{CD} X_0 @>>> X \\ @VVV @VVV\\ Y_0 @>>> Y, \end{CD}$$ or is the Springer map special in a sense I don't understand? Aside from applications to proving a generalized Springer correspondence, are there other examples where the existence and properties of this remarkable space are used in representation theory? 2) What are other applications of the Grothendieck-Springer resolution? For example, the Springer resolution can be interpreted as a moment map, and David Ben-Zvi's answer to this question shows how this may be interpreted as the semiclassical shadow to Beilinson-Bernstein localization. Is there an analogous quantization of $\widetilde{\mathfrak{g}}\to \mathfrak{g}$? EDIT: I would be particularly interested in applications which are not so closely connected with the Springer resolution. I'll stop here, since I have probably already asked too many questions. I would greatly appreciate any references to a modern understanding of $\widetilde{\mathfrak{g}}$.
Given two categories $I$ and $J$ we say that colimits of shape $I$ commute with limits of shape $J$ in the category of sets, if for any functor $F : I \times J \to \text{Set}$ the canonical map $$\textrm{colim}_{i\in I} \text{lim}_{j\in J} F(i,j) \to \textrm{lim}_{j\in J} \text{colim}_{i\in I} F(i,j)$$ is an isomorphism. The standard examples are a) filtered colimits commute with finite limits and b) sifted colimits commute with finite products. (Those statements can be regarded as definitions of which categories $I$ are filtered or sifted respectively, but both terms have independent definitions for which these commutation results are propositions.) A third, less known example is to take $I$ a finite group and $J$ a cofiltered category, in other words, if $G$ is a finite group and $X_j$ is an inverse system of $G$-sets, then the canonical map $$(\varprojlim_{j\in J} X_j)/G \to \varprojlim_{j \in J}(X_j/G)$$ is an isomorphism. Now, all of these examples are easy to prove separately (here's a proof of the $G$-set result, for example) but I see no unifying pattern. Is there a simple criterion for when $I$-colimits and $J$-colimits commute in the category of sets? [Note: It's true that $I$ is filtered (resp. sifted) if and only if for all finite (resp. finite discrete) $J$ the diagonal functor $I \to I^J$ is final; but I don't think that for arbitrary $I$ and $J$, if the diagonal $I \to I^J$ is final then $I$-colimits commute with $J$-limits. If I'm wrong and that condition on the diagonal actually is sufficient for commutation: why? and is it also necessary?]
I need the angular parts ($d\vartheta^2 + \sin^2 \vartheta \; d\varphi^2$ ) to be perfectly aligned, for aesthetical reasons only. Of course, the alignment problem is caused by the hyperbolic functions which don't have exactly the same number of letters (and same letters shape) as the trigonometric functions. Here's a preview of what the code above is doing, and the problem indicated in red : So how should I achieve the alignment I'm looking for ? Why on Earth can't I use simple LaTeX commands on a LaTeX board here !? The LaTeX symbols should be displayed correctly, instead of showing the commands ! EDIT : Inserting the command \phantom{h} inside the second equation doesn't do the trick very well, since it gives an exagerated visible space after the sin^2 {\chi} function. I tried adding \, and \; here and there, but the result isn't satisfying (too approximate). I guess the solution is to center the \sin^2 {\chi} by adding more space at its left and at its right, to balance the \sinh^2 {\chi} above it.
Absolutely!In fact, in my opinion, the most important "math skill" that should be taught in conjunction with, and using, word problems is checking whether the answers make sense. This is an absolutely invaluable part of making any practical use of mathematics, as opposed to just blindly applying formulas for the sake of passing an exam.There are several ... Consider these 5 problems:What is $ 35 \div 10 $ ?We are in the store and want to buy packages of party plates for a birthday party. The plates come in packages of 10. There will be 35 children at the party. How many packages should we buy so that everyone will have a plate?10 friends go out to get ice cream. The bill comes to $35. If the friends ... The answers to a word problem should in my opinion make sense (within reasonable limits).The goal you mentioned that students should check their answers is one shared by many, as also witnessed by our recent question How to award points for sense-making at the end of a problem?Now, if it is not a given any more that the solution does indeed make sense, ... I think there's a countervailing issue that the book you're describing is trying to deal with. I teach college students, so I don't know what the particular approach it's taking is age appropriate for a 6th grader, but it is trying to solve a real problem.What I would do about this is introduce them as separate topics, probably spaced far in time. Let ... Here are some suggestions for problem sources in English. Some of them are appropriate for very bright students studying geometry or Algebra II, but might nonetheless prove too difficult for students accelerated to this extent.-Mathematical Circles, Fomin et al.-Mathematical Problems: An Anthology, Dynkin et al.-Problems in Elementary Mathematics, ... Does your colleague have any information about these students' reading comprehension? The reason I ask is that, in my opinion, difficulty with word problems is not always a matter of taking the content of the word problem and reformulating it as equations. Rather, it is often a matter of getting the content of the problem in the first place. In other words,... DASL (pronounced "dazzle" and short for Data And Story Library) is an online collection of stories with matching data sets to be used for educational purposes. They are real data from real research. Searchable by statistics concept and by theme of the story.OzDASL is similar, but most of the data has an Australian or New Zealandish source. Personally I ... Professional work in general, and scientific and mathematical work in particular, is done principally in writing. Requirements, specifications, orders, field reports, case studies, journals, grants, etc., are all disseminated and documented in writing. Symbolic mathematical notation is inherently a specialized system of concise writing (arguably, it is nigh-... (edited)a. For all of these, I would think to expand the category into subcategories. So not just "bank account" but time value of money and NPV, bond details, etc. You learn something when doing this finer description.b. Maybe it fits under your (2) but radioactive decay is a huge one you did not mention. Applications include carbon (and other) ... Yes, there is evidence for the claim; for example, consult the following:Abedi, J., & Lord, C. (2001). The language factor in mathematics tests. Applied Measurement in Education, 14(3), 219-234. Link (no paywall).You will find in this piece that the authors, as you considered in your "soft evidence" proposal, modified mathematical questions (from ... I suggest you break it down into smaller parts. In fact, don't start on equations. Start on translation of terms:Double of a number, translation: $2x$Sum of two numbers, translation: $x+y$The triple of a number after adding five units, translation: $3(x+5)$Next combine translations into equations:The double of a number is equal to the triple of ... I feel like I always post the same thing in these threads, but this again sounds like an issue of blocking vs interleaving. In this case, the textbook may have started interleaving different problems a little bit too early, but generally it should be introduced earlier than feels intuitively right.That's because the algorithms for $\frac{1}{6}\times 90$ ... Physically (or even economically) unreasonable answers are confusing. The student needs to be shown that math is a tool that can solve practical problems. Answers that are unreasonable send the wrong message. (That math is a logic game versus a commercially relevant skill.) Why do that, when there are so many easy problems to construct that don't make ... Have your students covered division and multiplication yet? To me, these problems can all be solved using the "meaning of division" and the "meaning of multiplication"."Meaning of multiplication": If I have n groups of equal size m, then I have n*m total objects."Meaning of division": There are actually two!(a) If I have m objects which are ... I think you have a very different understanding of what "mathematics" is than I do.Consider the following questions:What kinds of geometric properties are held by the figure you get when you join the midpoints of adjacent sides of a polygon?Under what conditions on the parameters $a,b$ does an equation of the form $a^x = x^b$ have a rational solution?... I do avoid using this phrase in all of my math classes. Not that I've ever thought of it as a particular goal, but I would want to reserve the word "term" to specify an addend.In your first example, I would specify, "Write a formula using the variable $x$". In your second example, I specify, "Solve the following equation for $y$" (end of direction), ... Recently, I got to teach a new high school class called "Essentials of College Math." It sounds like hard content, but it was basically just a transition course from Algebra II to Precal. Anyways, this is the teacher's manual for the course (which also includes all the student handouts). The first unit was called "algebraic expressions" and dealt with this ... I only have anecdotal evidence but since you said it was welcome, I am adding my 2 cents from both the perspective of a teacher and an item writer, hired to write questions for standardized tests.I taught math to gifted elementary students. A small fraction of them were gifted in math and had reading levels below grade level. Students with poor reading ... I work with gifted elementary school students, but one of my favorite sites, nrich has challenging problems that you could use for older gifted students.Try looking at secondary problems for stages 4 and 5.Here are some suggested problems to see if you'll like the site:You can also look at the Post-16 Curriculum on nrich here. See description below.... One of my colleagues introduced me to a strategy called "3 Reads" which he details in his blog Misteristhisright. This strategy has two major features:You remove the question from the problem. The problem only gives the context. This can be done with Dan Meyer's 3 act problems as well.You read the problem 3 different times in order to become familiar ... I would say that this book is trying to get students to think about what the calculations mean, rather than simply execute an algorithm, and I strongly believe that is something that should be done right from the beginning, not saved for later or for more advanced students.I cannot tell you how many students I have worked with as a private tutor who, if ... (I came to mention DASL, but since it's already been mentioned, I'll give some other resources.)opendata.stackexchange.com often has mentions of useful sources of data, some of which are small (and others of which might be sampled to generate smaller data sets).It's also a good place to ask about data setsThe datasets subreddit http://www.reddit.com/r/... This is not a good answer, but I'll mention anyway that Mathematica now connects to vast "curated" data sources. It would takesome effort to master these sources, but they could provide endless streams of real data. Here is one example I just ran (NB: Up-to-date in that no Pluto!). I think there's a very basic answer to this. It's the very origin of math and numbers. Try to teach a child that 1+1=2 without the visual (or tactile) example of using fingers to show the concept of 1 and 2, and how the oneness of a single finger has something in common with the oneness of a single block. It's only when this sinks in, do we get comfortable ... There are three direct benefits, as far as I can tell.Word problems answer questions like "why do I need to know this". If you have a student who thinks that learning math is pointless then you can use word problems to help them understand why math is importantFor some students math is very difficult when's it's just abstract numbers. For more ... One basic model of atmospheric pressure has it decaying exponentially as a function of height above sea level. Two places to look for this arehttps://people.clas.ufl.edu/kees/files/AtmosphericPressure.pdf andhttp://nova.stanford.edu/projects/mod-x/ad-expatm.html. The website mathschallenge.net has pdfs of tiered problems with solutions. They are nicely posed problems, often with some humour, and are based on different stages of mathematical maturity. To begin with they have no real prerequisites, but eventually they require some of the 'canonical' mathematical knowledge. I just found them to be well organised, ... Perhaps an example would clarify what you mean.Question 1:What number is obtained after reducing 40 by 5%?Question 2:Farmer John has lately been having problems selling his corn. He needs to start selling more of them soon, or they will start to go bad. It seems that the retail price of 40 cents is too high, so he decides to start offering a 5% ... Catenary curves are the sum of two exponentials -- One concave upward but decreasing, and the other concave upward and increasing. They are a good model for the curve of suspension bridge cables. (The key assumptions are that the weight of the cables is small compared to the weight of the bridge deck, and the weight of the bridge deck is constant along the ... I think trying to make "real-world" word problems is often posing more problem than it solves. One expects using maths in real-world problem would make them more appealing and show their usefulness, but honestly in most cases the problems are either ridiculous, or artificial, or boring to students, or all of these. Then the whole idea backfires, and what you ...
[Improved version!!] Suppose $\mathsf{Prov}_R$ is a Rosserized proof predicate for PA (or some other suitable theory). To fix ideas, suppose $$\mathsf{Prov}_R(\overline{\ulcorner\varphi\urcorner}) = \mathsf{\exists x(Prf(x, \overline{\ulcorner\varphi\urcorner}) \land (\forall y \leq x)\neg Prf(y, \overline{\ulcorner\bot\urcorner}))}$$ where $\mathsf{Prf}$ represents the relation that $m$ has to $n$ when $m$ numbers a proof of the sentence numbered $n$. We can construct a $\Delta_0$ wff $\mathsf{Prf}$, making $\mathsf{Prov}_R$ $\Sigma_1$. Then, as is familiar, PA $\vdash \neg\mathsf{Prov}_R(\overline{\ulcorner\bot\urcorner})$ So the HBL derivability conditions can't all hold for $\mathsf{Prov}_R$. The first condition still holds. And it is reasonably easy to see why the second condition might fail (suppose that ordered by Gödel numbers the proof of $A$ and $A \to C$ precede the first proof of $\bot$ which precedes the first proof of $C$). But we also know that using Jereslow's trick we can get a version of the unprovability of consistency without appeal to the second condition, so the third condition should be the crucial failure. Well the usual proof of the third condition runs by showing PA $\vdash \psi \to \mathsf{Prov}(\overline{\ulcorner\psi\urcorner})$ for any $\Sigma_1$ $\psi$, and then remarking that $\mathsf{Prov}(\overline{\ulcorner\varphi\urcorner})$ is itself $\Sigma_1$. But this line of argument is presumably going to be blocked at the first stage when we turn to the Rosserized predicate $\mathsf{Prov}_R$: i.e. we won't have PA $\vdash \psi \to \mathsf{Prov}_R(\overline{\ulcorner\psi\urcorner})$ for every $\Sigma_1$ $\psi$. [And it is plausible this should fail, I guess.] So we know the third condition fails. But -- and here at last is the questions: Is there a simple counterexample to PA $\vdash \psi \to \mathsf{Prov}_R(\overline{\ulcorner\psi\urcorner})$ for $\Sigma_1$ $\psi$, and if that counterexample doesn't already involve $\mathsf{Prov}_R$, Are there known constructions of suitable wffs $\theta$ such that we have a nice illustration of a case where PA $\nvdash \mathsf{Prov}_R(\overline{\ulcorner\theta\urcorner}) \to \mathsf{Prov}_R(\overline{\ulcorner\mathsf{Prov}_R(\overline{\ulcorner\theta\urcorner})\urcorner})$ either for the given Rosserized predicate, or some cousin?
We report a new measurement of the pseudorapidity (eta) and transverse-energy (Et) dependence of the inclusive jet production cross section in pbar b collisions at sqrt(s) = 1.8 TeV using 95 pb*-1 of data collected with the DZero detector at the Fermilab Tevatron. The differential cross section d^2sigma/dEt deta is presented up to \|eta\| = 3, significantly extending previous measurements. The results are in good overall agreement with next-to-leading order predictions from QCD and indicate a preference for certain parton distribution functions. We have searched for the pair production of first generation scalar leptoquarks in the eejj channel using the full data set (123 pb-1) collected with the D0 detector at the Fermilab Tevatron during 1992--1996. We observe no candidates with an expected background of approximately 0.4 events. Comparing the experimental 95% confidence level upper limit to theoretical calculations of the cross section with the assumption of a 100% branching fraction to eq, we set a lower limit on the mass of a first generation scalar leptoquark of 225 GeV/c^2. The results of this analysis rule out the interpretation of the excess of high Q^2 events at HERA as leptoquarks which decay exclusively to eq. This paper presents the first measurement of the inclusive J/Psi production cross section in the forward pseudorapidity region 2.5<\|eta\|<3.7 in ppbar collisions at sqrt(s)=1.8TeV. The results are based on 9.8 pb-1 of data collected using the D0 detector at the Fermilab Tevatron Collider. The inclusive J/Psi cross section for transverse momenta between 1 and 16 GeV/c is compared with theoretical models of charmonium production. We have searched for second generation leptoquark (LQ) pairs in the \mu\mu+jets channel using 94+-5 pb^{-1} of pbar-p collider data collected by the D0 experiment at the Fermilab Tevatron during 1993-1996. No evidence for a signal is observed. These results are combined with those from the \mu\nu+jets and \nu\nu+jets channels to obtain 95% confidence level (C.L.) upper limits on the LQ pair production cross section as a function of mass and $beta, the branching fraction of a LQ decay into a charged lepton and a quark. Lower limits of 200(180) GeV/c^2 for \beta=1(1/2) are set at the 95% C.L. on the mass of scalar LQ. Mass limits are also set on vector leptoquarks as a function of \beta. Azimuthal decorrelations between the two central jets with the largest transverse momenta are sensitive to the dynamics of events with multiple jets. We present a measurement of the normalized differential cross section based on the full dataset (L=36/pb) acquired by the ATLAS detector during the 2010 sqrt(s)=7 TeV proton-proton run of the LHC. The measured distributions include jets with transverse momenta up to 1.3 TeV, probing perturbative QCD in a high energy regime. This letter describes the observation of the light-by-light scattering process, $\gamma\gamma\rightarrow\gamma\gamma$, in Pb+Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV. The analysis is conducted using a data sample corresponding to an integrated luminosity of 1.73 nb$^{-1}$, collected in November 2018 by the ATLAS experiment at the LHC. Light-by-light scattering candidates are selected in events with two photons produced exclusively, each with transverse energy $E_{\textrm{T}}^{\gamma} > 3$ GeV and pseudorapidity $\|\eta_{\gamma}\| < 2.37$, diphoton invariant mass above 6 GeV, and small diphoton transverse momentum and acoplanarity. After applying all selection criteria, 59 candidate events are observed for a background expectation of 12 $\pm$ 3 events. The observed excess of events over the expected background has a significance of 8.2 standard deviations. The measured fiducial cross section is 78 $\pm$ 13 (stat.) $\pm$ 7 (syst.) $\pm$ 3 (lumi.) nb. We report on a search for second generation leptoquarks with the D\O\ detector at the Fermilab Tevatron $p\overline{p}$ collider at $\sqrt{s}$ = 1.8 TeV. This search is based on 12.7 pb$~{-1}$ of data. Second generation leptoquarks are assumed to be produced in pairs and to decay into a muon and quark with branching ratio $\beta$ or to neutrino and quark with branching ratio $(1-\beta)$. We obtain cross section times branching ratio limits as a function of leptoquark mass and set a lower limit on the leptoquark mass of 111 GeV/c$~{2}$ for $\beta = 1 $ and 89 GeV/c$~{2}$ for $\beta = 0.5 $ at the 95\%\ confidence level. A study of the particle multiplicity between jets with large rapidity separation has been performed using the D\O\ detector at the Fermilab Tevatron $p\bar{p}$ Collider operating at $\sqrt{s}=1.8$\,TeV. A significant excess of low-multiplicity events is observed above the expectation for color-exchange processes. The measured fractional excess is $1.07 \pm 0.10({\rm stat})~{ + 0.25}_{- 0.13}({\rm syst})\%$, which is consistent with a strongly-interacting color-singlet (colorless) exchange process and cannot be explained by electroweak exchange alone. A lower limit of $0.80\%$ (95\% C.L.) is obtained on the fraction of dijet events with color-singlet exchange, independent of the rapidity gap survival probability. The inclusive cross sections times leptonic branching ratios for W and Z boson production in PbarP collisions at Sqrt(s)=1.8 TeV were measured using the D0 detector at the Fermilab Tevatron collider: Sigma_WB(W->e, nu) = 2.36 +/- 0.07 +/- 0.13 nb, Sigma_WB(W->mu,nu) = 2.09 +/- 0.23 +/- 0.11 nb, Sigma_ZB(Z-> e, e) = 0.218 +/- 0.011 +/- 0.012 nb, Sigma_ZB(Z->mu,mu) = 0.178 +/- 0.030 +/- 0.009 nb. The first error is the combined statistical and systematic uncertainty, and the second reflects the uncertainty in the luminosity. For the combined electron and muon analyses we find: [Sigma_WB(W->l,nu)]/[Sigma_Z*B(Z->l,l)] = 10.90 +/- 0.49. Assuming Standard Model couplings, this result is used to determine the width of the W boson: Gamma(W) = 2.044 +/- 0.093 GeV. Bottom quark production in pbar-p collisions at sqrt(s)=1.8 TeV is studied with 5 inverse picobarns of data collected in 1995 by the DO detector at the Fermilab Tevatron Collider. The differential production cross section for b jets in the central rapidity region (\|y(b)\| < 1) as a function of jet transverse energy is extracted from a muon-tagged jet sample. Within experimental and theoretical uncertainties, DO results are found to be higher than, but compatible with, next-to-leading-order QCD predictions. We report a new measurement of the cross section for the production of isolated photons, with transverse energies (ET) above 10 GeV and pseudorapidities \|eta\| < 2.5, in p pbar collisions at sqrt{s} = 1.8 TeV. The results are based on a data sample of 107.6 pb-1 recorded during 1992--1995 with the D0 detector at the Fermilab Tevatron collider. The background, predominantly from jets which fragment to neutral mesons, was estimated using the longitudinal shower shape of photon candidates in the calorimeter. The measured cross section is in good agreement with the next-to-leading order (NLO) QCD calculation for ET > 36 GeV. We have made a precise measurement of the central inclusive jet cross section at sqrt(s) = 1.8 TeV. The measurement is based on an integrated luminosity of 92 pb-1 collected at the Fermilab Tevatron pbar-p Collider with the D-Zero detector. The cross section, reported as a function of jet transverse energy (ET >= 60 GeV) in the pseudorapidity interval \|eta\| <= 0.5, is in good agreement with predictions from next-to-leading order quantum chromodynamics. Using 1.8 fb-1 of pp collisions at a center-of-mass energy of 7 TeV recorded by the ATLAS detector at the Large Hadron Collider, we present measurements of the production cross sections of Upsilon(1S,2S,3S) mesons. Upsilon mesons are reconstructed using the di-muon decay mode. Total production cross sections for p_T<70 GeV and in the rapidity interval \|Upsilon\|<2.25 are measured to be 8.01+-0.02+-0.36+-0.31 nb, 2.05+-0.01+-0.12+-0.08 nb, 0.92+-0.01+-0.07+-0.04 nb respectively, with uncertainties separated into statistical, systematic, and luminosity measurement effects. In addition, differential cross section times di-muon branching fractions for Upsilon(1S), Upsilon(2S), and Upsilon(3S) as a function of Upsilon transverse momentum p_T and rapidity are presented. These cross sections are obtained assuming unpolarized production. If the production polarization is fully transverse or longitudinal with no azimuthal dependence in the helicity frame the cross section may vary by approximately +-20%. If a non-trivial azimuthal dependence is considered, integrated cross sections may be significantly enhanced by a factor of two or more. We compare our results to several theoretical models of Upsilon meson production, finding that none provide an accurate description of our data over the full range of Upsilon transverse momenta accessible with this dataset. We have searched for first generation scalar leptoquark (LQ) pairs in the enu+jets channel using ppbar collider data (integrated luminosity= 115 pb^-1) collected by the DZero experiment at the Fermilab Tevatron during 1992-96. The analysis yields no candidate events. We combine the results with those from the ee+jets and nunu+jets channels to obtain 95% confidence level (CL) upper limits on the LQ pair production cross section as a function of mass and of beta, the branching fraction to a charged lepton. Comparing with the next-to-leading order theory, we set 95% CL lower limits on the LQ mass of 225, 204, and 79 GeV/c^2 for beta=1, 1/2, and 0, respectively. Using the DZero detector at the 1.8 TeV pbarp Fermilab Tevatron collider, we have measured the inclusive dijet mass spectrum in the central pseudorapidity region \|eta_jet\| < 1.0 for dijet masses greater than 200 Gev/c^2. We have also measured the ratio of spectra sigma(\|eta_jet\| < 0.5)/sigma(0.5 < \|eta_jet\| < 1.0). The order alpha_s^3 QCD predictions are in good agreement with the data and we rule out models of quark compositeness with a contact interaction scale < 2.4 TeV at the 95% confidence level. This Letter describes a measurement of the muon cross section originating from b quark decay in the forward rapidity range 2.4 < y(mu) < 3.2 in pbarp collisions at sqrt(s) = 1.8 TeV. The data used in this analysis were collected by the D0 experiment at the Fermilab Tevatron. We find that NLO QCD calculations underestimate b quark production by a factor of four in the forward rapidity region. A cross section measurement using muon+jet data has been included in this version of the paper. We measure the ttbar production cross section in ppbar collisions at sqrt{s}=1.96 TeV in the lepton+jets channel. Two complementary methods discriminate between signal and background, b-tagging and a kinematic likelihood discriminant. Based on 0.9 fb-1 of data collected by the D0 detector at the Fermilab Tevatron Collider, we measure sigma_ttbar=7.62+/-0.85 pb, assuming the current world average m_t=172.6 GeV. We compare our cross section measurement with theory predictions to determine a value for the top quark mass of 170+/-7 GeV.
Defining parameters Level: $ N $ = $ 4003 $ Weight: $ k $ = $ 2 $ Nonzero newspaces: $ 8 $ Sturm bound: $2670668$ Dimensions The following table gives the dimensions of various subspaces of $M_{2}(\Gamma_1(4003))$. Total New Old Modular forms 669668 669668 0 Cusp forms 665667 665667 0 Eisenstein series 4001 4001 0 Decomposition of $S_{2}^{\mathrm{new}}(\Gamma_1(4003))$ We only show spaces with even parity, since no modular forms exist when this condition is not satisfied. Within each space $ S_k^{\mathrm{new}}(N, \chi) $ we list the newforms together with their dimension. Label $\chi$ Newforms Dimension $\chi$ degree 4003.2.a $\chi_{4003}(1, \cdot)$ 4003.2.a.a 2 1 4003.2.a.b 152 4003.2.a.c 179 4003.2.c $\chi_{4003}(822, \cdot)$ n/a 666 2 4003.2.e $\chi_{4003}(551, \cdot)$ n/a 7304 22 4003.2.f $\chi_{4003}(102, \cdot)$ n/a 9296 28 4003.2.i $\chi_{4003}(129, \cdot)$ n/a 14652 44 4003.2.j $\chi_{4003}(10, \cdot)$ n/a 18648 56 4003.2.m $\chi_{4003}(6, \cdot)$ n/a 204512 616 4003.2.o $\chi_{4003}(4, \cdot)$ n/a 410256 1232 "n/a" means that newforms for that character have not been added to the database yet
For the better part of the 30ties, Ernst Witt (1) did hang out with the rest of the ‘Noetherknaben’, the group of young mathematicians around Emmy Noether (3) in Gottingen. In 1934 Witt became Helmut Hasse‘s assistent in Gottingen, where he qualified as a university lecturer in 1936. By 1938 he has made enough of a name for himself to be offered a lecturer position in Hamburg and soon became an associate professor, the down-graded position held by Emil Artin (2) until he was forced to emigrate in 1937. A former fellow student of him in Gottingen, Erna Bannow (4), had gone earlier to Hamburg to work with Artin. She continued her studies with Witt and finished her Ph.D. in 1939. In 1940 Erna Bannow and Witt married. So, life was smiling on Ernst Witt that sunday january 28th 1940, both professionally and personally. There was just one cloud on the horizon, and a rather menacing one. He was called up by the Wehrmacht and knew he had to enter service in february. For all he knew, he was spending the last week-end with his future wife… (later in february 1940, Blaschke helped him to defer his military service by one year). Still, he desperately wanted to finish his paper before entering the army, so he spend most of that week-end going through the final version and submitted it on monday, as the published paper shows. In the 70ties, Witt suddenly claimed he did discover the Leech lattice $ {\Lambda} $ that sunday. Last time we have seen that the only written evidence for Witt’s claim is one sentence in his 1941-paper Eine Identität zwischen Modulformen zweiten Grades. “Bei dem Versuch, eine Form aus einer solchen Klassen wirklich anzugeben, fand ich mehr als 10 verschiedene Klassen in $ {\Gamma_{24}} $.” But then, why didn’t Witt include more details of this sensational lattice in his paper? Ina Kersten recalls on page 328 of Witt’s collected papers : “In his colloquium talk “Gitter und Mathieu-Gruppen” in Hamburg on January 27, 1970, Witt said that in 1938, he had found nine lattices in $ {\Gamma_{24}} $ and that later on January 28, 1940, while studying the Steiner system $ {S(5,8,24)} $, he had found two additional lattices $ {M} $ and $ {\Lambda} $ in $ {\Gamma_{24}} $. He continued saying that he had then given up the tedious investigation of $ {\Gamma_{24}} $ because of the surprisingly low contribution $ \displaystyle \| Aut(\Lambda) \|^{-1} < 10^{-18} $ to the Minkowski density and that he had consented himself with a short note on page 324 in his 1941 paper.” In the last sentence he refers to the fact that the sum of the inverse orders of the automorphism groups of all even unimodular lattices of a given dimension is a fixed rational number, the Minkowski-Siegel mass constant. In dimension 24 this constant is $ \displaystyle \sum_{L} \frac{1}{\| Aut(L) \|} = \frac {1027637932586061520960267}{129477933340026851560636148613120000000} \approx 7.937 \times 10^{-15} $ That is, Witt was disappointed by the low contribution of the Leech lattice to the total constant and concluded that there might be thousands of new even 24-dimensional unimodular lattices out there, and dropped the problem. If true, the story gets even better : not only claims Witt to have found the lattices $ {A_1^{24}=M} $ and $ {\Lambda} $, but also enough information on the Leech lattice in order to compute the order of its automorphism group $ {Aut(\Lambda)} $, aka the Conway group $ {Co_0 = .0} $ the dotto-group! Is this possible? Well fortunately, the difficulties one encounters when trying to compute the order of the automorphism group of the Leech lattice from scratch, is one of the better documented mathematical stories around. The books From Error-Correcting Codes through Sphere Packings to Simple Groups by Thomas Thompson, Symmetry and the monster by Mark Ronan, and Finding moonshine by Marcus du Sautoy tell the story in minute detail. It took John Conway 12 hours on a 1968 saturday in Cambridge to compute the order of the dotto group, using the knowledge of Leech and McKay on the properties of the Leech lattice and with considerable help offered by John Thompson via telephone. But then, John Conway is one of the fastest mathematicians the world has known. The prologue of his book On numbers and games begins with : “Just over a quarter of a century ago, for seven consecutive days I sat down and typed from 8:30 am until midnight, with just an hour for lunch, and ever since have described this book as “having been written in a week”.” Conway may have written a book in one week, Ernst Witt did complete his entire Ph.D. in just one week! In a letter of August 1933, his sister told her parents : “He did not have a thesis topic until July 1, and the thesis was to be submitted by July 7. He did not want to have a topic assigned to him, and when he finally had the idea, he started working day and night, and eventually managed to finish in time.” So, if someone might have beaten John Conway in fast-computing the dottos order, it may very well have been Witt. Sadly enough, there is a lot of circumstantial evidence to make Witt’s claim highly unlikely. For starters, psychology. Would you spend your last week-end together with your wife to be before going to war performing an horrendous calculation? Secondly, mathematical breakthroughs often arise from newly found insight. At that time, Witt was also working on his paper on root lattices “Spiegelungsgrupen and Aufzähling halbeinfacher Liescher Ringe” which he eventually submitted in january 1941. Contained in that paper is what we know as Witt’s lemma which tells us that for any integral lattice the sublattice generated by vectors of norms 1 and 2 is a direct sum of root lattices. This leads to the trick of trying to construct unimodular lattices by starting with a direct sum of root lattices and ‘adding glue’. Although this gluing-method was introduced by Kneser as late as 1967, Witt must have been aware of it as his 16-dimensional lattice $ {D_{16}^+} $ is constructed this way. If Witt wanted to construct new 24-dimensional even unimodular lattices in 1940, it would be natural for him to start off with direct sums of root lattices and trying to add vectors to them until he got what he was after. Now, all of the Niemeier-lattices are constructed this way, except for the Leech lattice! I’m far from an expert on the Niemeier lattices but I would say that Witt definitely knew of the existence of $ {D_{24}^+} $, $ {E_8^3} $ and $ {A_{24}^+} $ and that it is quite likely he also constructed $ {(D_{16}E_8)^+, (D_{12}^2)^+, (A_{12}^2)^+, (D_8^3)^+} $ and possibly $ {(A_{17}E_7)^+} $ and $ {(A_{15}D_9)^+} $. I’d rate it far more likely Witt constructed another two such lattices on sunday january 28th 1940, rather than discovering the Leech lattice. Finally, wouldn’t it be natural for him to include a remark, in his 1941 paper on root lattices, that not every even unimodular lattices can be obtained from sums of root lattices by adding glue, the Leech lattice being the minimal counter-example? If it is true he was playing around with the Steiner systems that sunday, it would still be a pretty good story he discovered the lattices $ {(A_2^{12})^+} $ and $ {(A_1^{24})^+} $, for this would mean he discovered the Golay codes in the process! Which brings us to our next question : who discovered the Golay code?
Data on the mean multiplicity of strange hadrons produced in minimum bias proton--proton and central nucleus--nucleus collisions at momenta between 2.8 and 400 GeV/c per nucleon have been compiled. The multiplicities for nucleon--nucleon interactions were constructed. The ratios of strange particle multiplicity to participant nucleon as well as to pion multiplicity are larger for central nucleus--nucleus collisions than for nucleon--nucleon interactions at all studied energies. The data at AGS energies suggest that the latter ratio saturates with increasing masses of the colliding nuclei. The strangeness to pion multiplicity ratio observed in nucleon--nucleon interactions increases with collision energy in the whole energy range studied. A qualitatively different behaviour is observed for central nucleus--nucleus collisions: the ratio rapidly increases when going from Dubna to AGS energies and changes little between AGS and SPS energies. This change in the behaviour can be related to the increase in the entropy production observed in central nucleus-nucleus collisions at the same energy range. The results are interpreted within a statistical approach. They are consistent with the hypothesis that the Quark Gluon Plasma is created at SPS energies, the critical collision energy being between AGS and SPS energies. This publication describes the methods used to measure the centrality of inelastic Pb-Pb collisions at a center-of-mass energy of 2.76 TeV per colliding nucleon pair with ALICE. The centrality is a key parameter in the study of the properties of QCD matter at extreme temperature and energy density, because it is directly related to the initial overlap region of the colliding nuclei. Geometrical properties of the collision, such as the number of participating nucleons and the number of binary nucleon-nucleon collisions, are deduced from a Glauber model with a sharp impact parameter selection and shown to be consistent with those extracted from the data. The centrality determination provides a tool to compare ALICE measurements with those of other experiments and with theoretical calculations. The ALICE Collaboration has studied J/psi production in pp collisions at $\sqrt{s}$=7 TeV at the LHC through its muon pair decay. The polar and azimuthal angle distributions of the decay muons were measured, and results on the J/$\psi$ polarization parameters lambda_theta and lambda_phi were obtained. The study was performed in the kinematic region 2.5 < y < 4, 2 < $p_t$ < 8 GeV/c, in the helicity and Collins-Soper reference frames. The results for lambda_theta in the helicity frame show a longitudinal polarization at low $p_t$, vanishing with increasing transverse momentum. All the other polarization parameters are consistent with zero, within uncertainties. The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (\|y\|<0.8) in the transverse momentum range 1<p_T<8 GeV/c with the ALICE experiment at the CERN LHC in pp collisions at a center of mass energy s=7 TeV using an integrated luminosity of 2.2 nb^-^1. Electrons from beauty hadron decays were selected based on the displacement of the decay vertex from the collision vertex. A perturbative QCD calculation agrees with the measurement within uncertainties. The data were extrapolated to the full phase space to determine the total cross section for the production of beauty quark-antiquark pairs. The ALICE experiment has measured the inclusive J/psi production in Pb-Pb collisions at sqrt(sNN) = 2.76 TeV down to zero transverse momentum in the rapidity range 2.5 < y < 4. A suppression of the inclusive J/psi yield in Pb-Pb is observed with respect to the one measured in pp collisions scaled by the number of binary nucleon-nucleon collisions. The nuclear modification factor, integrated over the 0%-80% most central collisions, is 0.545+/-0.032(stat)+/-0.083(syst) and does not exhibit a significant dependence on the collision centrality. These features appear significantly different from measurements at lower collision energies. Models including J/psi production from charm quarks in a deconfined partonic phase can describe our data. A measurement of the multi-strange Xi- and Omega- baryons and their antiparticles by the ALICE experiment at the CERN Large Hadron Collider (LHC) is presented for inelastic proton-proton collisions at centre of mass energy of 7 TeV. The transverse momentum (pt) distributions were studied at mid-rapidity (\|y\| < 0.5) in the range of 0.6 < pt < 8.5 GeV/c for Xi- and Xi+ baryons, and in the range of 0.8 < pt < 5 GeV/c for Omega- and Omega+. Baryons and antibaryons were measured as separate particles and we find that the baryon to antibaryon ratio of both particle species is consistent with unity over the entire range of the measurement. The statistical precision of the current LHC data has allowed us to measure a difference between the mean pt of Xi- (Xi+) and Omega- (Omega+). Particle yields, mean pt, and the spectra in the intermediate pt range are not well described by the PYTHIA Perugia 2011 tune Monte Carlo event generator, which has been tuned to reproduce the early LHC data. The discrepancy is largest for Omega- (Omega+). This PYTHIA tune approaches the pt spectra of Xi- and Xi+ baryons below pt < 0.85 GeV/c and describes the Xi- and Xi+ spectra above pt > 6.0 GeV/c. We also illustrate the difference between the experimental data and model by comparing the corresponding ratios of (Omega-+Omega+)/(Xi-+Xi+) as a function of transverse mass. The transverse momentum (pT) distribution of primary charged particles is measured in minimum bias (non-single-diffractive) p+Pb collisions at sNN=5.02 TeV with the ALICE detector at the LHC. The pT spectra measured near central rapidity in the range 0.5<pT<20 GeV/c exhibit a weak pseudorapidity dependence. The nuclear modification factor RpPb is consistent with unity for pT above 2 GeV/c. This measurement indicates that the strong suppression of hadron production at high pT observed in Pb+Pb collisions at the LHC is not due to an initial-state effect. The measurement is compared to theoretical calculations. We present measurements of Underlying Event observables in pp collisions at sqrt(s) = 0.9 and 7 TeV. The analysis is performed as a function of the highest charged-particle transverse momentum pT,LT in the event. Different regions are defined with respect to the azimuthal direction of the leading (highest transverse momentum) track: Toward, Transverse and Away. The Toward and Away regions collect the fragmentation products of the hardest partonic interaction. The Transverse region is expected to be most sensitive to the Underlying Event activity. The study is performed with charged particles above three different pT thresholds: 0.15, 0.5 and 1.0 GeV/c. In the Transverse region we observe an increase in the multiplicity of a factor 2-3 between the lower and higher collision energies, depending on the track pT threshold considered. Data are compared to Pythia 6.4, Pythia 8.1 and Phojet. On average, all models considered underestimate the multiplicity and summed pT in the Transverse region by about 10-30%. Measurements of the sphericity of primary charged particles in minimum bias proton--proton collisions at sqrt(s)=0.9, 2.76 and 7 TeV with the ALICE detector at the LHC are presented. The observable is linearized to be collinear safe and is measured in the plane perpendicular to the beam direction using primary charged tracks with $p_{\rm T}\geq0.5$ GeV/c in $\|\eta\|\leq0.8$. The mean sphericity as a function of the charged particle multiplicity at mid-rapidity ($N_{\rm ch}$) is reported for events with different $p_{\rm T}$ scales ("soft" and "hard") defined by the transverse momentum of the leading particle. In addition, the mean charged particle transverse momentum versus multiplicity is presented for the different event classes, and the sphericity distributions in bins of multiplicity are presented. The data are compared with calculations of standard Monte Carlo event generators. The transverse sphericity is found to grow with multiplicity at all collision energies, with a steeper rise at low $N_{\rm ch}$, whereas the event generators show the opposite tendency. The combined study of the sphericity and the mean $p_{\rm T}$ with multiplicity indicates that most of the tested event generators produce events with higher multiplicity by generating more back-to-back jets resulting in decreased sphericity (and isotropy). The PYTHIA6 generator with tune PERUGIA-2011 exhibits a noticeable improvement in describing the data, compared to the other tested generators. Measurements of cross sections of inelastic and diffractive processes in proton--proton collisions at LHC energies were carried out with the ALICE detector. The fractions of diffractive processes in inelastic collisions were determined from a study of gaps in charged particle pseudorapidity distributions: for single diffraction (diffractive mass $M_X < 200$ GeV/$c^2$) $\sigma_{\rm SD}/\sigma_{\rm INEL} = 0.21 \pm 0.03, 0.20^{+0.07}_{-0.08}$, and $0.20^{+0.04}_{-0.07}$, respectively at centre-of-mass energies $\sqrt{s} = 0.9, 2.76$, and 7 TeV; for double diffraction (for a pseudorapidity gap $\Delta\eta > 3$) $\sigma_{\rm DD}/\sigma_{\rm INEL} = 0.11 \pm 0.03, 0.12 \pm 0.05$, and $0.12^{+0.05}_{-0.04}$, respectively at $\sqrt{s} = 0.9, 2.76$, and 7 TeV. To measure the inelastic cross section, beam properties were determined with van der Meer scans, and, using a simulation of diffraction adjusted to data, the following values were obtained: $\sigma_{\rm INEL} = 62.8^{+2.4}_{-4.0} (model) \pm 1.2 (lumi)$ mb at $\sqrt{s} =$ 2.76 TeV and $73.2^{+2.0}_{-4.6} (model) \pm 2.6 (lumi)$ mb at $\sqrt{s}$ = 7 TeV. The single- and double-diffractive cross sections were calculated combining relative rates of diffraction with inelastic cross sections. The results are compared to previous measurements at proton--antiproton and proton--proton colliders at lower energies, to measurements by other experiments at the LHC, and to theoretical models. The production of muons from heavy flavour decays is measured at forward rapidity in proton--proton collisions at \sqrt(s) = 7 TeV collected with the ALICE experiment at the LHC. The analysis is carried out on a data sample corresponding to an integrated luminosity L_{int} = 16.5 nb^{-1}. The transverse momentum and rapidity differential production cross sections of muons from heavy flavour decays are measured in the rapidity range 2.5 < y < 4, over the transverse momentum range 2 < p_{t} < 12 GeV/c. The results are compared to predictions based on perturbative QCD calculations.
These mock exams are provided to help you prepare for Term/Final tests. The best way to use these practice tests is to try the problems as if you were taking the test. Please don't look at the solution until you have attempted the question(s). Only reading through the answers or studying them, will typically not be helpful in preparing since it is too easy to convince yourself that you understand it. Mock Exam (Test 2) You can try timing yourself for 50 minutes. Exercise $\PageIndex{1}$: definition of the derivative a) Use the definition of the derivative to find $f'(x)$ for $$\displaystyle{f(x)=\frac{1}{3x+1}}$$ Hint: The definition of the derivative $f'(x)= \lim_{h \rightarrow 0} \displaystyle\frac{f(x+h)-f(x)}{h}$. Answer: $\displaystyle{f(x)=\frac{-3}{(3x+1)^2}}$. Solution: $f'(x)= \lim_{h \rightarrow 0} \displaystyle\frac{f(x+h)-f(x)}{h}$. b) Find the slope of the tangent line to the graph at the point $x=1.$ Answer $\displaystyle{f(x)=\frac{-3}{16}}$. Exercise $\PageIndex{2}$: L'Hopital's Find the following limits: $\displaystyle \lim_{x→0^+}x^2lnx$ $ \displaystyle \lim_{x→π}\frac{1+cosx}{sinx}$ $\displaystyle \lim_{x\rightarrow 0 }\left(e^x + x\right)^{1/x}.$ Answer: $0. 0, e^2$ Exercise $\PageIndex{3}$: derivatives Find the derivative $\displaystyle{\frac{dy}{dx}}$ of the following functions/relations: $y(x)=\displaystyle \frac{ln(x^2+1)}{x^2}$. Note that $y(x)$ is $y$ as a function of $x$. $\displaystyle{y=\sin^{-1}(x^3)}$. $x^2y+xy+y^2=1$ $y(x)= x^{\sin x}$ Answer: $ \displaystyle\frac{2}{x^2+x} -\displaystyle \frac{2 \ln(x^2+1)}{x^3}, \displaystyle\frac{3x^2}{\sqrt{1-x^6}}, \displaystyle\frac{-(2xy+y)}{x^2+x+2y}, x^{\sin x-1}(\sin(x)+x \ln(x) \cos(x))$. Solution: 1. $\displaystyle \frac{dy}{dx}=\displaystyle \frac{\displaystyle \frac{2x}{x^2+1}-(2x)ln(x^2+1)x^2 }{x^4},$ by quotient rule, $= \displaystyle \frac{2}{x(x^2+1)}-\displaystyle \frac{2ln(x^2+1)}{x^3},$ $= \displaystyle\frac{2}{x^2+x}- \displaystyle \frac{2 \ln(x^2+1)}{x^3}.$ 4. Let $y=x^{\sin x}$. Then $ ln(y)= ln \left(x^{\sin x} \right)$. Which implies, $ ln(y)= \sin(x) ln \left(x\right)$. Differentiate with respect to $x$ both sides, $ \displaystyle \frac{1}{y} (\displaystyle \frac{dy}{dx}= \cos(x) \ln(x)+ \sin(x) \displaystyle \frac{1}{x}$, $ (\displaystyle \frac{dy}{dx}= y\left(\cos(x) \ln(x)+ \sin(x) \displaystyle \frac{1}{x} \right)$, $ (\displaystyle \frac{dy}{dx}= x^{\sin x}\left(\cos(x) \ln(x)+ \sin(x) \displaystyle \frac{1}{x} \right)$, Hence the result. Exercise $\PageIndex{4}$: Related rates A spherical balloon is being inflated at a rate of $2 m^3/min.$ Find how fast the surface area of the balloon is increasing when $r=5m.$ (The surface area of the sphere is $4\pi r^2$ and the volume $v=\displaystyle \frac{4}{3} \pi r^3,$ where $r$ is the radius of the sphere.) Answer $\displaystyle\frac{4}{5} m^2/min$. Exercise $\PageIndex{5}$: Consider the function $f(x)=x^5 - 5x^4$. Find the intervals on which $f$ is increasing. Find the intervals on which $f$ is decreasing. Find the value of the relative minima(s)( if any) of the function. Find the value of the relative maxima(s)( if any) of the function. Find the open intervals on which $f$ is concave up. Find the open intervals on which $f$ is concave down. Find the $x-$ coordinates of all inflection points. Answer $(-\infty, 0)\cup (4, \infty), (0,4), x=4, x=0,(3,\infty), (-\infty,3). x=3$ Contributors Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)
Home Integration by PartsIntegration by Parts Examples Integration by Parts with a definite integral Going in Circles Tricks of the Trade Integrals of Trig FunctionsAntiderivatives of Basic Trigonometric Functions Product of Sines and Cosines (mixed even and odd powers or only odd powers) Product of Sines and Cosines (only even powers) Product of Secants and Tangents Other Cases Trig SubstitutionsHow Trig Substitution Works Summary of trig substitution options Examples Completing the Square Partial FractionsIntroduction to Partial Fractions Linear Factors Irreducible Quadratic Factors Improper Rational Functions and Long Division Summary Strategies of IntegrationSubstitution Integration by Parts Trig Integrals Trig Substitutions Partial Fractions Improper IntegralsType 1 - Improper Integrals with Infinite Intervals of Integration Type 2 - Improper Integrals with Discontinuous Integrands Comparison Tests for Convergence Modeling with Differential EquationsIntroduction Separable Equations A Second Order Problem Euler's Method and Direction FieldsEuler's Method (follow your nose) Direction Fields Euler's method revisited Separable EquationsThe Simplest Differential Equations Separable differential equations Mixing and Dilution Models of GrowthExponential Growth and Decay The Zombie Apocalypse (Logistic Growth) Linear EquationsLinear ODEs: Working an Example The Solution in General Saving for Retirement Parametrized CurvesThree kinds of functions, three kinds of curves The Cycloid Visualizing Parametrized Curves Tracing Circles and Ellipses Lissajous Figures Calculus with Parametrized CurvesVideo: Slope and Area Video: Arclength and Surface Area Summary and Simplifications Higher Derivatives Polar CoordinatesDefinitions of Polar Coordinates Graphing polar functions Video: Computing Slopes of Tangent Lines Areas and Lengths of Polar CurvesArea Inside a Polar Curve Area Between Polar Curves Arc Length of Polar Curves Conic sectionsSlicing a Cone Ellipses Hyperbolas Parabolas and Directrices Shifting the Center by Completing the Square Conic Sections in Polar CoordinatesFoci and Directrices Visualizing Eccentricity Astronomy and Equations in Polar Coordinates Infinite SequencesApproximate Versus Exact Answers Examples of Infinite Sequences Limit Laws for Sequences Theorems for and Examples of Computing Limits of Sequences Monotonic Covergence Infinite SeriesIntroduction Geometric Series Limit Laws for Series Test for Divergence and Other Theorems Telescoping Sums Integral TestPreview of Coming Attractions The Integral Test Estimates for the Value of the Series Comparison TestsThe Basic Comparison Test The Limit Comparison Test Convergence of Series with Negative TermsIntroduction, Alternating Series,and the AS Test Absolute Convergence Rearrangements The Ratio and Root TestsThe Ratio Test The Root Test Examples Strategies for testing SeriesStrategy to Test Series and a Review of Tests Examples, Part 1 Examples, Part 2 Power SeriesRadius and Interval of Convergence Finding the Interval of Convergence Power Series Centered at $x=a$ Representing Functions as Power SeriesFunctions as Power Series Derivatives and Integrals of Power Series Applications and Examples Taylor and Maclaurin SeriesThe Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor PolynomialsTaylor Polynomials When Functions Are Equal to Their Taylor Series When a Function Does Not Equal Its Taylor Series Other Uses of Taylor Polynomials Functions of 2 and 3 variablesFunctions of several variables Limits and continuity Partial DerivativesOne variable at a time (yet again) Definitions and Examples An Example from DNA Geometry of partial derivatives Higher Derivatives Differentials and Taylor Expansions Differentiability and the Chain RuleDifferentiability The First Case of the Chain Rule Chain Rule, General Case Video: Worked problems Multiple IntegralsGeneral Setup and Review of 1D Integrals What is a Double Integral? Volumes as Double Integrals Iterated Integrals over RectanglesHow To Compute Iterated Integrals Examples of Iterated Integrals Fubini's Theorem Summary and an Important Example Double Integrals over General RegionsType I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinatesdA = r dr (d theta) Examples Multiple integrals in physicsDouble integrals in physics Triple integrals in physics Integrals in Probability and StatisticsSingle integrals in probability Double integrals in probability Change of VariablesReview: Change of variables in 1 dimension Mappings in 2 dimensions Jacobians Examples Bonus: Cylindrical and spherical coordinates Substitution The most basic trick for doing integrals is $u$-substitution. The simplest uses involve taking $u$ to be the inside part of a composition of functions, and have $du$ or a multiple of $du$ appearing in the integral as well. More sophisticated uses involve clever choices of $u$. $\displaystyle\int x \sin\left(x^2+3\right) \,dx \overset{\fbox{$\,\,\,\, u\,=\,x^2+3\\\,\,\, du\,=\,2x\,dx\\ \frac{1}{2}du\,=\,x\,dx$}\\}{=} \frac{1}{2} \int\sin u\,du = -\frac{1}{2}\cos u + C = -\frac{1}{2}\cos\left(x^2+3\right) + C$
Since my question relates directly to a part of the text from a 2004 book, Logic in Computer Science: Modelling and Reasoning about Systems (2nd Edition) by Michael Huth and Mark Ryan, in order to provide context for the following discussion, I'm partially quoting the book verbatim: The decision problem of validity in predicate logic is undecidable: no program exists which, given any $\varphi$, decides whether $\varphi$. PROOF: As said before, we pretend that validity is decidable for predicate logic and thereby solve the (insoluble) Post correspondence problem. Given a correspondence problem instance $C$: $$s_1 s_2 ... s_k$$ $$t_1 t_2 ... t_k$$ we need to be able to construct, within finite space and time and uniformly so for all instances, some formula $\varphi$ of predicate logic such that $\varphi$ holds iff the correspondence problem instance $C$ above has a solution. As function symbols, we choose a constant $e$ and two function symbols $f_0$ and $f_1$ each of which requires one argument. We think of $e$ as the empty string, or word, and $f_0$ and $f_1$ symbolically stand for concatenation with 0, respectively 1. So if $b_1 b_2 ... b_l$ is a binary string of bits, we can code that up as the term $f_{b_l}(f_{b_{l−1}}...(f_{b_2}(f_{b_1}(e)))...)$. Note that this coding spells that word backwards. To facilitate reading those formulas, we abbreviate terms like $f_{b_l}(f_{b_{l−1}}...(f_{b_2}(f_{b_1}(t)))...)$ by $f_{{b_1}{b_2}...{b_l}}(t)$. We also require a predicate symbol $P$ which expects two arguments. The intended meaning of $P(s,t)$ is that there is some sequence of indices $(i_1,i_2,...,i_m)$ such that $s$ is the term representing $s_{i_1} s_{i_2}...s_{i_m}$ and $t$ represents $t_{i_1} t_{i_2}...t_{i_m}$. Thus, $s$ constructs a string using the same sequence of indices as does $t$; only $s$ uses the $s_i$ whereas $t$ uses the $t_i$. Our sentence $\varphi$ has the coarse structure $\varphi_1 \wedge \varphi_2 \implies \varphi_3$ where we set $$\varphi_1 \stackrel{def}{=} \bigwedge\limits_{i=1}^k P\left(f_{s_i}(e),f_{t_i}(e)\right)$$ $$\varphi_2 \stackrel{def}{=} \forall v,w \hspace{1mm} P(v,w)\rightarrow\bigwedge\limits_{i=1}^kP(f_{s_i}(v),f_{t_i}(w))$$ $$\varphi_3 \stackrel{def}{=} \exists z\hspace{1mm} P(z,z)$$. Our claim is $\varphi$ holds iff the Post correspondence problem $C$ has a solution. In proving PCP ⟹ Validity: Conversely, let us assume that the Post correspondence problem C has some solution, [...] The way we proceed here is by interpretingfinite, binary strings in the domain of values $A′$ of the model $M′$. This is not unlike the coding of an interpreter for one programming language in another. The interpretation is done by a function interpretwhich is defined inductively on the data structure of finite, binary strings: $$\text{interpret}(\epsilon) \stackrel{def}{=} e^{M′}$$ $$\text{interpret}(s0) \stackrel{def}{=} {f_0}^{M′}(\text{interpret}(s))$$ $$\text{interpret}(s1) \stackrel{def}{=} {f_1}^{M′}(\text{interpret}(s))$$. [...] Using [$\text{interpret}(b_1 b_2...b_l) = f_{b_l}^{M′}(f_{b_{l-1}}^{M′}(...(f_{b_1}^{M′}(e{M′})...)))$] and the fact that $M\models\varphi_1$, we conclude that $(\text{interpret}(s_i), \text{interpret}(t_i)) \in P^{M′}$ for $i = 1,2,...,k$. [...] since $M′ \models \varphi_2$, we know that for all $(s,t) \in P^{M′}$ we have that $(\text{interpret}(ss_i),\text{interpret}(tt_i)) \in P^{M′}$ for $i=1,2,...,k$. Using these two facts, starting with $(s, t) = (s_{i_1}, t_{i_1})$, we repeatedly use the latter observation to obtain (2.9) $(\text{interpret}(s_{i_1}s_{i_2}...s_{i_n}),\text{interpret}(t_{i_1}t_{i_2}...t_{i_n})) \in P^{M′}$. [...] Hence (2.9) verifies $\exists{z} P(z,z)$ in $M′$ and thus $M′ \models \varphi_3$. In proving that the validity of predicate logic is undecidable, according to the approach I learned from school, which is based on that of the Huth & Ryan book (2nd edition, page 135), when constructing the reduction of PCP to Validity problem, the "finite binary strings" of the universe are interpreted with a " interpret function", which encodes binary strings into composites of functions of the model. Then it goes on to show that, using the fact that the antecedent of $\varphi$ must hold for it to be non-trivial, both sub-formulae of the antecedent can be expressed with the said " interpret function". From there, it follows that the consequence holds, too, since it can also be expressed in a way with the interpret function that follows from the previous expressions with interpret. My question is: what is the purpose of this " interpret function"? Why can't we just use the previously devised φ and get the same result? What do we get out of using interpret to express our elements? And also, what if our universe contains some arbitrary elements; that is, what if they are not binary strings? Do we just construct some mapping of the two?
Home Integration by PartsIntegration by Parts Examples Integration by Parts with a definite integral Going in Circles Tricks of the Trade Integrals of Trig FunctionsAntiderivatives of Basic Trigonometric Functions Product of Sines and Cosines (mixed even and odd powers or only odd powers) Product of Sines and Cosines (only even powers) Product of Secants and Tangents Other Cases Trig SubstitutionsHow Trig Substitution Works Summary of trig substitution options Examples Completing the Square Partial FractionsIntroduction to Partial Fractions Linear Factors Irreducible Quadratic Factors Improper Rational Functions and Long Division Summary Strategies of IntegrationSubstitution Integration by Parts Trig Integrals Trig Substitutions Partial Fractions Improper IntegralsType 1 - Improper Integrals with Infinite Intervals of Integration Type 2 - Improper Integrals with Discontinuous Integrands Comparison Tests for Convergence Modeling with Differential EquationsIntroduction Separable Equations A Second Order Problem Euler's Method and Direction FieldsEuler's Method (follow your nose) Direction Fields Euler's method revisited Separable EquationsThe Simplest Differential Equations Separable differential equations Mixing and Dilution Models of GrowthExponential Growth and Decay The Zombie Apocalypse (Logistic Growth) Linear EquationsLinear ODEs: Working an Example The Solution in General Saving for Retirement Parametrized CurvesThree kinds of functions, three kinds of curves The Cycloid Visualizing Parametrized Curves Tracing Circles and Ellipses Lissajous Figures Calculus with Parametrized CurvesVideo: Slope and Area Video: Arclength and Surface Area Summary and Simplifications Higher Derivatives Polar CoordinatesDefinitions of Polar Coordinates Graphing polar functions Video: Computing Slopes of Tangent Lines Areas and Lengths of Polar CurvesArea Inside a Polar Curve Area Between Polar Curves Arc Length of Polar Curves Conic sectionsSlicing a Cone Ellipses Hyperbolas Parabolas and Directrices Shifting the Center by Completing the Square Conic Sections in Polar CoordinatesFoci and Directrices Visualizing Eccentricity Astronomy and Equations in Polar Coordinates Infinite SequencesApproximate Versus Exact Answers Examples of Infinite Sequences Limit Laws for Sequences Theorems for and Examples of Computing Limits of Sequences Monotonic Covergence Infinite SeriesIntroduction Geometric Series Limit Laws for Series Test for Divergence and Other Theorems Telescoping Sums Integral TestPreview of Coming Attractions The Integral Test Estimates for the Value of the Series Comparison TestsThe Basic Comparison Test The Limit Comparison Test Convergence of Series with Negative TermsIntroduction, Alternating Series,and the AS Test Absolute Convergence Rearrangements The Ratio and Root TestsThe Ratio Test The Root Test Examples Strategies for testing SeriesStrategy to Test Series and a Review of Tests Examples, Part 1 Examples, Part 2 Power SeriesRadius and Interval of Convergence Finding the Interval of Convergence Power Series Centered at $x=a$ Representing Functions as Power SeriesFunctions as Power Series Derivatives and Integrals of Power Series Applications and Examples Taylor and Maclaurin SeriesThe Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor PolynomialsTaylor Polynomials When Functions Are Equal to Their Taylor Series When a Function Does Not Equal Its Taylor Series Other Uses of Taylor Polynomials Functions of 2 and 3 variablesFunctions of several variables Limits and continuity Partial DerivativesOne variable at a time (yet again) Definitions and Examples An Example from DNA Geometry of partial derivatives Higher Derivatives Differentials and Taylor Expansions Differentiability and the Chain RuleDifferentiability The First Case of the Chain Rule Chain Rule, General Case Video: Worked problems Multiple IntegralsGeneral Setup and Review of 1D Integrals What is a Double Integral? Volumes as Double Integrals Iterated Integrals over RectanglesHow To Compute Iterated Integrals Examples of Iterated Integrals Fubini's Theorem Summary and an Important Example Double Integrals over General RegionsType I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinatesdA = r dr (d theta) Examples Multiple integrals in physicsDouble integrals in physics Triple integrals in physics Integrals in Probability and StatisticsSingle integrals in probability Double integrals in probability Change of VariablesReview: Change of variables in 1 dimension Mappings in 2 dimensions Jacobians Examples Bonus: Cylindrical and spherical coordinates Finding the slope of the tangent line to a graph $y=f(x)$ is easy -- just compute $f'(x)$. Likewise, the area under the curve between $x=a$ and $x=b$ is just $\int_a^b f(x) dx$. But how do we compute slopes and areas with parametrized curves? The answer is to express everything in terms of the parameter $t$. By computing derivatives with respect to $t$ and integrals with respect to $t$, we can compute everything we want. For slopes, we are looking for $dy/dx$. This is a limit \begin{eqnarray} \frac{dy}{dx} &=& \lim \frac{\Delta y}{\Delta x} \cr &=& \lim \frac{\Delta y/\Delta t}{\Delta x/\Delta t} \cr &=& \frac{\lim (\Delta y/\Delta t)}{\lim (\Delta x/\Delta t)} \cr &=& \frac {dy/dt}{dx/dt}, \end{eqnarray}where the limits are as $\Delta t$ and $\Delta x$ and $\Delta y$ all go to zero. As long as we can take the derivatives of $x$ and $y$, we can compute $dy/dx$. To find the area under a curve between $x=a$ and $x=b$, we need to compute $$\int_a^b y\; dx = \int_{t_1}^{t_2} y(t) \frac{dx(t)}{dt} dt,$$ where $x(t_1)=a$ and $x(t_2)=b$. To find the area, we need to both compute a derivative as well as an integral. In the following video, we derive these formulas, work out the tangent line to a cycloid, and compute the area under one span of the cycloid.
tl;dr;You can teach your machine to break arbitrary Caesar cipher by observing enough training examples using Trusted Region Policy Optimization for Policy Gradients: Full textImagine the world where a hammer was introduced to the public just couple a years ago. Everyone is running around trying to apply the hammer to anything that even resembles a nail. This is the world we are living in and the hammer is deep learning. Today I will be applying it to a task that can be much easier solved by other means but hey, it's Deep Learning Age! Specifically, I will teach my machine to break a simple cipher like Caesar cipher just by looking at several (actually, a lot) examples of English text and corresponding encoded strings. You may have heard that machines are getting pretty good at playing games so I decided to formulate this code breaking challenge as a game. Fortunately there is this OpenAI Gym toolkit that can be used "for developing and comparing reinforcement learning algorithms". It provides some great abstractions that help us define games in terms that computer can understand. For instance, they have a game (or environment) called "Copy-v0" with the following setup and rules: There is an input tape with some characters. You can move cursor one step left or right along this tape. You can read symbols under the cursor and output characters one at a time to the output tape. You need to copy input tape characters to output tape to win. Now let's talk a bit about the hammer itself. The hottest thing on the Reinforcement Learning market right now is Policy Gradients and specifically this flavorTrust Region Policy Optimization. There is an amazing article from Andrej Karpathy on Policy Gradients so I will not give here an introduction. If you are new to Reinforcement Learning you just stop reading this post and go read that one. Seriously, it's so much better! Still here? Ok, I will tell you about TRPO then. TRPO is a technique for Policy Gradients optimization that produces much better results than vanilla gradient descent and even guarantees (theoretically, of course) that you can get an improved policy network on every iteration. With vanilla PG you start by defining a policy network that produces scores for the actions given the current state. You then simulate hundreds and thousands of games taking actions suggested by the network and note which actions produced better results. Having this data available you can then use backpropagation to update your policy network and start all over again. The only thing that TRPO adds to this is that you solve a constrained optimization problem instead of an unconstrained one: $$ \textrm{maximize } L(\theta) \textrm{ subject to } \bar{D}_{KL}(\theta_{\textrm{old}},\theta)<\delta$$ Here $L(\theta)$ is a loss that we are trying to optimize. It is defined as $$E_{a \sim q}[\frac{\pi_\theta(a\|s_n)}{q(a\|s_n)} A_{\theta_{\textrm{old}}}(s_n,a)],$$ where $\theta$ is our weights vector, $\pi_\theta(a\|s_n)$ is a probability (score) of the selected action $a$ in state $s_n$ according to the policy network, $q(a\|s_n)$ is a corresponding score using the policy network from the iteration before and $A_{\theta_{\textrm{old}}}(s_n,a)$ is an advantage (more on it later). Running simple gradient descent on this is the vanilla Policy Gradients approach. TRPO approach doesn't blindly descend along the gradient but takes into account the $\bar{D}_{KL}(\theta_{\textrm{old}},\theta)<\delta$ constraint. To make sure the constraint is satisfied we do the following. First, we approximately solve the following equation to find a search direction: $$Ax = g,$$ where A is the Fisher information matrix, $A_{\textrm{ij}} = \frac{\partial}{\partial \theta_i}\frac{\partial}{\partial \theta_j}\bar{D}_{KL}(\theta_{\textrm{old}},\theta)$ and $g$ is the gradient that you can get from the loss using backpropagation. This is done using conjugate gradients algorithm. Once we have a search direction we can easily find a maximum step along this direction that still satisfies the constraint. One thing that I promised to get back to is the advantage. It is defined as $$A_\pi(s,a)= Q_\pi(s,a)−V_\pi(s),$$ where $Q_\pi(s,a)$ is a state-action value function (actual reward of taking an action in this state, it usually includes discounted rewards for all upcoming states) and $V_\pi(s)$ is a value function (in our case it's just a separate network that we train to predict the value of the state). Bored enough already? I promise, it's not that scary in code. You can find the full implementation here: tilarids/reinforcement_learning_playground. Specifically, look at trpo_agent.py. You can reproduce the Caesar cipher breaking by running trpo_caesar.py. wojzaremba's implementation a lot - you are right. I was copying some TRPO code from there and then rewriting it to make it more readable and also to make sure it follows the paper closely.
Harald Grosse, Alexander Hock, Raimar Wulkenhaar March 29, 2019 Let $F_g(t)$ be the generating function of intersection numbers on the moduli spaces $\overline{\mathcal{M}}_{g,n}$ of complex curves of genus $g$. As by-product of a complete solution of all non-planar correlation functions of the renormalised $\Phi^3$-matrical QFT model, we explicitly construct a Laplacian $\Delta_t$ on the space of formal parameters $t_i$ satisfying $\exp(\sum_{g\geq 2} N^{2-2g}F_g(t))=\exp((-\Delta_t+F_2(t))/N^2)1$ for any $N>0$. The result is achieved via Dyson-Schwinger equations from noncommutative quantum field theory combined with residue techniques from topological recursion. The genus-$g$ correlation functions of the $\Phi^3$-matricial QFT model are obtained by repeated application of another differential operator to $F_g(t)$ and taking for $t_i$ the renormalised moments of a measure constructed from the covariance of the model. open access link @article{Grosse:2019nes, author = "Grosse, Harald and Hock, Alexander and Wulkenhaar, Raimar", title = "{A Laplacian to compute intersection numbers on $\overline{\mathcal{M}}_{g,n}$ and correlation functions in NCQFT}", year = "2019", eprint = "1903.12526", archivePrefix = "arXiv", primaryClass = "math-ph", SLACcitation = "%%CITATION = ARXIV:1903.12526;%%" } Keywords: none
Neural Networks From Ufldl Line 11: Line 11: This "neuron" is a computational unit that takes as input <math>x_1, x_2, x_3</math> (and a +1 intercept term), and This "neuron" is a computational unit that takes as input <math>x_1, x_2, x_3</math> (and a +1 intercept term), and - outputs <math>h_{W,b}(x) = f(W^Tx) = f(\sum_{i=1}^3 W_{i}x_i +b)</math>, where <math>f : \Re \mapsto \Re</math> is + outputs <math>h_{W,b}(x) = f(W^Tx) = f(\sum_{i=1}^3 W_{i}x_i +b)</math>, where <math>f : \Re \mapsto \Re</math> is called the '''activation function'''. In these notes, we will choose called the '''activation function'''. In these notes, we will choose <math>f(\cdot)</math> to be the sigmoid function: <math>f(\cdot)</math> to be the sigmoid function: Line 31: Line 31: - + - [[Image:Sigmoid_Function.png\|400px\| + [[Image:Sigmoid_Function.png\|400px\|\|Sigmoid activation function.]] - [[Image:Tanh_Function.png\|400px\| + [[Image:Tanh_Function.png\|400px\|\|Tanh activation function.]] + The <math>\tanh(z)</math> function is a rescaled version of the sigmoid, and its output range is The <math>\tanh(z)</math> function is a rescaled version of the sigmoid, and its output range is <math>[-1,1]</math> instead of <math>[0,1]</math>. <math>[-1,1]</math> instead of <math>[0,1]</math>. - Note that unlike + Note that unlike (parts of CS229, we are not using the convention here of <math>x_0=1</math>. Instead, the intercept term is handled separately by the parameter <math>b</math>. here of <math>x_0=1</math>. Instead, the intercept term is handled separately by the parameter <math>b</math>. Line 52: Line 53: A neural network is put together by hooking together many of our simple A neural network is put together by hooking together many of our simple - + neurons,so that the output of a neuron can be the input of another. For example, here is a small neural network: example, here is a small neural network: Line 58: Line 59: In this figure, we have used circles to also denote the inputs to the network. The circles In this figure, we have used circles to also denote the inputs to the network. The circles - labeled + labeled +1are called '''bias units''', and correspond to the intercept term. The leftmost layer of the network is called the '''input layer''', and the The leftmost layer of the network is called the '''input layer''', and the rightmost layer the '''output layer''' (which, in this example, has only one rightmost layer the '''output layer''' (which, in this example, has only one Line 94: Line 95: In the sequel, we also let <math>z^{(l)}_i</math> denote the total weighted sum of inputs to unit <math>i</math> in layer <math>l</math>, In the sequel, we also let <math>z^{(l)}_i</math> denote the total weighted sum of inputs to unit <math>i</math> in layer <math>l</math>, - including the bias term (e.g., <math>z_i^{(2)} = \sum_{j=1}^n W^{(1)}_{ij} x_j + b^{(1)}_i</math>), so that + including the bias term (e.g., <math>z_i^{(2)} = \sum_{j=1}^n W^{(1)}_{ij} x_j + b^{(1)}_i</math>), so that <math>a^{(l)}_i = f(z^{(l)}_i)</math>. <math>a^{(l)}_i = f(z^{(l)}_i)</math>. Line 101: Line 102: to apply to vectors in an element-wise fashion (i.e., to apply to vectors in an element-wise fashion (i.e., <math>f([z_1, z_2, z_3]) = [f(z_1), f(z_2), f(z_3)]</math>), then we can write <math>f([z_1, z_2, z_3]) = [f(z_1), f(z_2), f(z_3)]</math>), then we can write - + more compactly as: compactly as: :<math>\begin{align} :<math>\begin{align} Line 109: Line 110: h_{W,b}(x) &= a^{(3)} = f(z^{(3)}) h_{W,b}(x) &= a^{(3)} = f(z^{(3)}) \end{align}</math> \end{align}</math> - More generally, recalling that we also use <math>a^{(1)} = x</math> to also denote the values from the input layer, + More generally, recalling that we also use <math>a^{(1)} = x</math> to also denote the values from the input layer, then given layer <math>l</math>'s activations <math>a^{(l)}</math>, we can compute layer <math>l+1</math>'s activations <math>a^{(l+1)}</math> as: then given layer <math>l</math>'s activations <math>a^{(l)}</math>, we can compute layer <math>l+1</math>'s activations <math>a^{(l+1)}</math> as: :<math>\begin{align} :<math>\begin{align} Line 120: Line 121: We have so far focused on one example neural network, but one can also build neural We have so far focused on one example neural network, but one can also build neural - networks with other + networks with other architectures(meaning patterns of connectivity between neurons), including ones with multiple hidden layers. - architectures + The most common choice is a <math>\textstyle n_l</math>-layered network The most common choice is a <math>\textstyle n_l</math>-layered network where layer <math>\textstyle 1</math> is the input layer, layer <math>\textstyle n_l</math> is the output layer, and each where layer <math>\textstyle 1</math> is the input layer, layer <math>\textstyle n_l</math> is the output layer, and each layer <math>\textstyle l</math> is densely connected to layer <math>\textstyle l+1</math>. In this setting, to compute the layer <math>\textstyle l</math> is densely connected to layer <math>\textstyle l+1</math>. In this setting, to compute the output of the network, we can successively compute all the activations in layer output of the network, we can successively compute all the activations in layer - <math>\textstyle L_2</math>, then layer <math>\textstyle L_3</math>, and so on, up to layer <math>\textstyle L_{n_l}</math>, using + <math>\textstyle L_2</math>, then layer <math>\textstyle L_3</math>, and so on, up to layer <math>\textstyle L_{n_l}</math>, using . This is one example of a '''feedforward''' neural network, since the connectivity graph example of a '''feedforward''' neural network, since the connectivity graph does not have any directed loops or cycles. does not have any directed loops or cycles. - - Line 144: Line 142: patient, and the different outputs <math>y_i</math>'s might indicate presence or absence patient, and the different outputs <math>y_i</math>'s might indicate presence or absence of different diseases.) of different diseases.) + + + + + +
From Wikipedia on 24: “$24$ is the only number whose divisors, namely $1, 2, 3, 4, 6, 8, 12, 24$, are exactly those numbers $n$ for which every invertible element of the commutative ring $\mathbb{Z}/n\mathbb{Z}$ is a square root of $1$. It follows that the multiplicative group $(\mathbb{Z}/24\mathbb{Z})^* = \{ \pm 1, \pm 5, \pm 7, \pm 11 \}$ is isomorphic to the additive group $(\mathbb{Z}/2\mathbb{Z})^3$. This fact plays a role in monstrous moonshine.” Where did that come from? In the original “Monstrous Moonshine” paper by John Conway and Simon Norton, section 3 starts with: “It is a curious fact that the divisors $h$ of $24$ are precisely those numbers $h$ for which $x.y \equiv 1~(mod~h)$ implies $x \equiv y~(mod~h)$.” and a bit further they even call this fact: “our ‘defining property of $24$'”. The proof is pretty straightforward. We want all $h$ such that every unit in $\mathbb{Z}/h \mathbb{Z}$ has order two. By the Chinese remainder theorem we only have to check this for prime powers dividing $h$. $5$ is a unit of order $4$ in $\mathbb{Z}/16 \mathbb{Z}$. $2$ is a unit of order $6$ in $\mathbb{Z}/ 9 \mathbb{Z}$. A generator of the cyclic group $(\mathbb{Z}/p\mathbb{Z})^*$ is a unit of order $p-1 > 2$ in $\mathbb{Z}/p \mathbb{Z}$, for any prime number $p \geq 5$. This only leaves those $h$ dividing $2^3.3=24$. But, what does it have to do with monstrous moonshine? Moonshine assigns to elements of the Monster group $\mathbb{M}$ a specific subgroup of $SL_2(\mathbb{Q})$ containing a cofinite congruence subgroup \[ \Gamma_0(N) = \{ \begin{bmatrix} a & b \\ cN & d \end{bmatrix}~\|~a,b,c,d \in \mathbb{Z}, ad-Nbc = 1 \} \] for some natural number $N = h.n$ where $n$ is the order of the monster-element, $h^2$ divides $N$ and … $h$ is a divisor of $24$. To begin to understand how the defining property of $24$ is relevant in this, take any strictly positive rational number $M$ and any pair of coprime natural numbers $g < h$ and associate to $M \frac{g}{h}$ the matrix\[\alpha_{M\frac{g}{h}} = \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} \]We say that $\Gamma_0(N)$ fixes $M \frac{g}{h}$ if we have that \[ \alpha_{M\frac{g}{h}} \Gamma_0(N) \alpha_{M\frac{g}{h}}^{-1} \subset SL_2(\mathbb{Z}) \] For those in the know, $M \frac{g}{h}$ stands for the $2$-dimensional integral lattice \[ \mathbb{Z} (M \vec{e}_1 + \frac{g}{h} \vec{e}_2) \oplus \mathbb{Z} \vec{e}_2 \] and the condition tells that $\Gamma_0(N)$ preserves this lattice under base-change (right-multiplication). In “Understanding groups like $\Gamma_0(N)$” Conway describes the groups appearing in monstrous moonshine as preserving specific finite sets of these lattices. For this, it is crucial to determine all $M\frac{g}{h}$ fixed by $\Gamma_0(N)$. \[ \alpha_{M\frac{g}{h}}.\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.\alpha_{M\frac{g}{h}}^{-1} = \begin{bmatrix} 1 & M \\ 0 & 1 \end{bmatrix} \] so we must have that $M$ is a natural number, or that $M\frac{g}{h}$ is a number-like lattice, in Conway-speak. \[ \alpha_{M\frac{g}{h}}.\begin{bmatrix} 1 & 0 \\ N & 1 \end{bmatrix}.\alpha_{M\frac{g}{h}}^{-1} = \begin{bmatrix} 1 + \frac{Ng}{Mh} & – \frac{Ng^2}{Mh^2} \\ \frac{N}{M} & 1 – \frac{Ng}{Mh} \end{bmatrix} \] so $M$ divides $N$, $Mh$ divides $Ng$ and $Mh^2$ divides $Ng^2$. As $g$ and $h$ are coprime it follows that $Mh^2$ must divide $N$. Now, for an arbitrary element of $\Gamma_0(N)$ we have \[ \alpha_{M\frac{g}{h}}.\begin{bmatrix} a & b \\ cN & d \end{bmatrix}.\alpha_{M\frac{g}{h}}^{-1} = \begin{bmatrix} a + c \frac{Ng}{Mh} & Mb – c \frac{Ng^2}{Mh^2} – (a-d) \frac{g}{h} \\ c \frac{N}{M} & d – c \frac{Ng}{Mh} \end{bmatrix} \] and using our divisibility requirements it follows that this matrix belongs to $SL_2(\mathbb{Z})$ if $a-d$ is divisible by $h$, that is if $a \equiv d~(mod~h)$. We know that $ad-Nbc=1$ and that $h$ divides $N$, so $a.d \equiv 1~(mod~h)$, which implies $a \equiv d~(mod~h)$ if $h$ satisfies the defining property of $24$, that is, if $h$ divides $24$. Concluding, $\Gamma_0(N)$ preserves exactly those lattices $M\frac{g}{h}$ for which \[ 1~\|~M~\|~\frac{N}{h^2}~\quad~\text{and}~\quad~h~\|~24 \] A first step towards figuring out the Moonshine Picture.
After completing this reading you should be able to: Construct, apply, and interpret hypothesis tests and confidence intervals for a single coefficient in a multiple regression. Construct, apply, and interpret joint hypothesis tests and confidence intervals for multiple coefficients in a multiple regression. Interpret the $F$-statistic. Interpret tests of a single restriction involving multiple coefficients. Interpret confidence sets for multiple coefficients. Identify examples of omitted variable bias in multiple regressions. Interpret the ${ R }^{ 2 }$ and adjusted ${ R }^{ 2 }$ in a multiple regression. Hypothesis Tests and Confidence Intervals for a Single Coefficient This section is about the calculation of the standard error, hypotheses testing, and confidence interval construction for a single regression in a multiple regression equation. Introduction In a previous chapter, we looked at simple linear regression where we deal with just one regressor (independent variable). The response (dependent variable) is assumed to be affected by just one independent variable. M ultiple regression, on the other hand , simultaneously considers the influence of multiple explanatory variables on a response variable Y. We may want to establish the confidence interval of one of the independent variables. We may want to evaluate whether any particular independent variable has a significant effect on the dependent variable. Finally, We may also want to establish whether the independent variables as a group have a significant effect on the dependent variable. In this chapter, we delve into ways all this can be achieved. Hypothesis Tests for a single coefficient Suppose that we are testing the hypothesis that the true coefficient ${ \beta }_{ j }$ on the $j$th regressor takes on some specific value ${ \beta }_{ j,0 }$. Let the alternative hypothesis be two-sided. Therefore, the following is the mathematical expression of the two hypotheses: $$ { H }_{ 0 }:{ \beta }_{ j }={ \beta }_{ j,0 }\quad vs.\quad { H }_{ 1 }:{ \beta }_{ j }\neq { \beta }_{ j,0 } $$ This expression represents the two-sided alternative. The following are the steps to follow while testing the null hypothesis: Computing the coefficient’s standard error. Computing the $t$-statistic, as previously described: Computing the test’s $p-value$ as previously described: $$ p-value=2\Phi \left( -\|{ t }^{ act }\| \right) $$ Also, the $t$-statistic can be compared to the critical value corresponding to the significance level that is desired for the test. Confidence Intervals for a Single Coefficient The confidence interval for a regression coefficient in multiple regression is calculated and interpreted the same way as it is in simple linear regression. The t-statistic has degrees of freedom where k = number of independents n – k – 1 Supposing that an interval contains the true value of ${ \beta }_{ j }$ with a probability of 95%. This is simply the 95% two-sided confidence interval for ${ \beta }_{ j }$. The implication here is that the true value of ${ \beta }_{ j }$ is contained in 95% of all possible randomly drawn variables. Alternatively, the 95% two-sided confidence interval for ${ \beta }_{ j }$ is the set of values that are impossible to reject when a two-sided hypothesis test of 5% is applied. Therefore, with a large sample size: $$ 95\%\quad confidence\quad interval\quad for\quad { \beta }_{ j }=\left[ { \hat { \beta } }_{ j }-1.96SE\left( { \hat { \beta } }_{ j } \right) ,{ \hat { \beta } }_{ j }+1.96SE\left( { \hat { \beta } }_{ j } \right) \right] $$ Tests of Joint Hypotheses In this section, we consider the formulation of the joint hypotheses on multiple regression coefficients. We will further study the application of an $F$-statistic in their testing. Hypotheses Testing on Two or More Coefficients Joint Null Hypothesis In multiple regression, we canno test the null hypothesis that t allslope coefficients are equal 0 based on t-tests that each individualslope coefficient equals 0. Why? individual t-tests do not account for the effects of among the independent variables. interactions For this reason, we conduct the The F-test tests the null hypothesis that all of the slope coefficients in the multiple regression model are jointly equal to 0, .i.e., F-test which uses the F-statistic. $F$-Statistic The F-statistic, which is always a one-tailed test , is calculated as: To determine whether at least one of the coefficients is statistically significant, the calculated F-statistic is compared with the one-tailed critical F-value, at the appropriate level of significance. Decision rule: Rejection of the null hypothesis at a stated level of significance indicates that at least one of the coefficients is significantly different than zero, i.e, at least one of the independent variables in the regression model makes a significant contribution to the dependent variable. Example An analyst runs a regression of monthly value-stock returns on four independent variables over 48 months. The total sum of squares for the regression is 360, and the sum of squared errors is 120. Test the null hypothesis at the 5% significance level (95% confidence) that all the four independent variables are equal to zero. Solution Solution ${ H }_{ 0 }:{ \beta }_{ 1 }=0,{ \beta }_{ 2 }=0,\dots ,{ \beta }_{ 4 }=0 $ Versus ${ H }_{ 1 }:{ \beta }_{ j }\neq 0$ (at least one j is not equal to zero, j=1,2… k ) ESS = TSS – SSR = 360 – 120 = 240 The calculated test statistic = (ESS/k)/(SSR/(n-k-1)) =(240/4)/(120/43) = 21.5 ${ F }_{ 43 }^{ 4 }$ is approximately 2.44 at 5% significance level. Decision: Reject H 0. Conclusion: at least one of the 4 independents is significantly different than zero. Omitted Variable Bias in Multiple Regression This is the bias in the OLS estimator arising when at least one included regressor gets collaborated with an omitted variable. The following conditions must be satisfied for an omitted variable bias to occur: There must be a correlation between at least one of the included regressors and the omitted variable. The dependent variable $Y$ must be determined by the omitted variable. Practical Interpretation of the ${ R }^{ 2 }$ and the adjusted ${ R }^{ 2 }$, ${ \bar { R } }^{ 2 }$ To determine the accuracy within which the OLS regression line fits the data, we apply the coefficient of determination and the regression’s standard error. The coefficient of determination, represented by ${ R }^{ 2 }$, is a measure of the “goodness of fit” of the regression. It is interpreted as the percentage of variation in the dependent variable explained by the independent variables ${ R }^{ 2 }$ is not a reliable indicator of the explanatory power of a multiple regression model.Why? ${ R }^{ 2 }$ increases as new independent variables are added to the model, even if the marginal contribution of the new variable is not statistically significant. Thus, a high ${ R }^{ 2 }$ may reflect the impact of a large set of independents rather than how well the set explains the dependent.This problem is solved by the use of the adjusted ${ R }^{ 2 }$ (extensively covered in chapter 8) almost always The following are the factors to watch out when guarding against applying the ${ R }^{ 2 }$ or the ${ \bar { R } }^{ 2 }$: An added variable doesn’t have to be statistically significant just because the ${ R }^{ 2 }$ or the ${ \bar { R } }^{ 2 }$ has increased. It is not always true that the regressors are a true cause of the dependent variable, just because there is a high ${ R }^{ 2 }$ or ${ \bar { R } }^{ 2 }$. It is not necessary that there is no omitted variable bias just because we have a high ${ R }^{ 2 }$ or ${ \bar { R } }^{ 2 }$. It is not necessarily true that we have the most appropriate set of regressors just because we have a high ${ R }^{ 2 }$ or ${ \bar { R } }^{ 2 }$. It is not necessarily true that we have an inappropriate set of regressors just because we have a low ${ R }^{ 2 }$ or ${ \bar { R } }^{ 2 }$. Question 1 An economist tests the hypothesis that GDP growth in a certain country can be explained by interest rates and inflation. Using some 30 observations, the analyst formulates the following regression equation: $$ GDP growth = { \hat { \beta } }_{ 0 } + { \hat { \beta } }_{ 1 } Interest+ { \hat { \beta } }_{ 2 } Inflation $$ Regression estimates are as follows: Intercept 0.10 0.5% Interest rates 0.20 0.05 Inflation 0.15 0.03 Is the coefficient for interest rates significant at 5%? Since the test statistic < t-critical, we accept H 0; the interest rate coefficient is significant at the 5% level. not Since the test statistic > t-critical, we reject H 0; the interest rate coefficient is significant at the 5% level. not Since the test statistic > t-critical, we reject H 0; the interest rate coefficient is significant at the 5% level. Since the test statistic < t-critical, we accept H 1; the interest rate coefficient is significant at the 5% level. The correct answer is C. We have GDP growth = 0.10 + 0.20(Int) + 0.15(Inf) Hypothesis: $$ { H }_{ 0 }:{ \hat { \beta } }_{ 1 } = 0 \quad vs \quad { H }_{ 1 }:{ \hat { \beta } }_{ 1 }≠0 $$ The test statistic is: $$ t = \left( \frac { 0.20 – 0 }{ 0.05 } \right) = 4 $$ The critical value is t (α/2, n-k-1) = t 0.025,27 = 2.052 (which can be found on the t-table). Decision: Since test statistic > t-critical, we reject H 0. Conclusion: The interest rate coefficient is significant at the 5% level.
The magnitude of the force of gravity between two bodies is proportional to the product of their masses:$$F=G\frac{m_1m_2}{r^2}$$This doesn't change depending on which body you're applying the force to, i.e. if you interchange the masses. The magnitude is the same. What does change is the direction of the force. Force is a vector quantity, denoted as $\vec{F}$ or $\mathbf{F}$. If we write the equation for gravity using proper vector notation, we have$$\mathbf{F}=G\frac{m_1m_2}{\|\mathbf{r}_1-\mathbf{r_2}\|^2}\frac{\mathbf{r_1}-\mathbf{r_2}}{\|\mathbf{r_1}-\mathbf{r_2}\|}$$Here, the positions of the objects are represented by vectors, $\mathbf{r}_1$ and $\mathbf{r_2}$. Additionally, $\|\mathbf{x}\|$ denotes the norm of a vector $\mathbf{x}$ - its magnitude. Now, if you interchange the masses, the direction of the force changes, although $\|\mathbf{r_1}-\mathbf{r_2}\|=\|\mathbf{r_2}-\mathbf{r_1}\|$, because this refers to the magnitude of the vectors. So the force applied on one object is the opposite of the force applied on the other object. This is Newton's third law. The acceleration is more interesting. The force on object $1$ due to gravity is$$F_1=m_1g_1$$Here,$$g_1=\frac{Gm_2}{r^2}$$where $m_2$ is the other mass. This should tell you that $g_1\neq g_2$, except when $m_1=m_2$. I'm not an expert in general relativity, but I do know that it describes how spacetime curves due to the presence of one body. The solution to the Einstein Field Equations, the metric, is different for different bodies, because one piece of it, the stress-energy tensor, is different for objects of different mass/energy/etc.
Hey guys! I built the voltage multiplier with alternating square wave from a 555 timer as a source (which is measured 4.5V by my multimeter) but the voltage multiplier doesn't seem to work. I tried first making a voltage doubler and it showed 9V (which is correct I suppose) but when I try a quadrupler for example and the voltage starts from like 6V and starts to go down around 0.1V per second. Oh! I found a mistake in my wiring and fixed it. Now it seems to show 12V and instantly starts to go down by 0.1V per sec. But you really should ask the people in Electrical Engineering. I just had a quick peek, and there was a recent conversation about voltage multipliers. I assume there are people there who've made high voltage stuff, like rail guns, which need a lot of current, so a low current circuit like yours should be simple for them. So what did the guys in the EE chat say... The voltage multiplier should be ok on a capacitive load. It will drop the voltage on a resistive load, as mentioned in various Electrical Engineering links on the topic. I assume you have thoroughly explored the links I have been posting for you... A multimeter is basically an ammeter. To measure voltage, it puts a stable resistor into the circuit and measures the current running through it. Hi all! There is theorem that links the imaginary and the real part in a time dependent analytic function. I forgot its name. Its named after some dutch(?) scientist and is used in solid state physics, who can help? The Kramers–Kronig relations are bidirectional mathematical relations, connecting the real and imaginary parts of any complex function that is analytic in the upper half-plane. These relations are often used to calculate the real part from the imaginary part (or vice versa) of response functions in physical systems, because for stable systems, causality implies the analyticity condition, and conversely, analyticity implies causality of the corresponding stable physical system. The relation is named in honor of Ralph Kronig and Hans Kramers. In mathematics these relations are known under the names... I have a weird question: The output on an astable multivibrator will be shown on a multimeter as half the input voltage (for example we have 9V-0V-9V-0V...and the multimeter averages it out and displays 4.5V). But then if I put that output to a voltage doubler, the voltage should be 18V, not 9V right? Since the voltage doubler will output in DC. I've tried hooking up a transformer (9V to 230V, 0.5A) to an astable multivibrator (which operates at 671Hz) but something starts to smell burnt and the components of the astable multivibrator get hot. How do I fix this? I check it after that and the astable multivibrator works. I searched the whole god damn internet, asked every god damn forum and I can't find a single schematic that converts 9V DC to 1500V DC without using giant transformers and power stage devices that weight 1 billion tons.... something so "simple" turns out to be hard as duck In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it? If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum. @AaronStevens Yeah, I had a good laugh to myself when he responded back with "Yeah, maybe they considered it and it was just too complicated". I can't even be mad at people like that. They are clearly fairly new to physics and don't quite grasp yet that most "novel" ideas have been thought of to death by someone; likely 100+ years ago if it's classical physics I have recently come up with a design of a conceptual electromagntic field propulsion system which should not violate any conservation laws, particularly the Law of Conservation of Momentum and the Law of Conservation of Energy. In fact, this system should work in conjunction with these two laws ... I rememeber that Gordon Freeman's thesis was "Observation of Einstein-Podolsky-Rosen Entanglement on Supraquantum Structures by Induction Through Nonlinear Transuranic Crystal of Extremely Long Wavelength (ELW) Pulse from Mode-Locked Source Array " In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it? If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum. @ACuriousMind What confuses me is the interpretation of Peskin to this infinite c-number and the experimental fact He said, the second term is the sum over zero point energy modes which is infnite as you mentioned. He added," fortunately, this energy cannot be detected experm., since the experiments measure only the difference between from the ground state of H". @ACuriousMind Thank you, I understood your explanations clearly. However, regarding what Peskin mentioned in his book, there is a contradiction between what he said about the infinity of the zero point energy/ground state energy, and the fact that this energy is not detectable experimentally because the measurable quantity is the difference in energy between the ground state (which is infinite and this is the confusion) and a higher level. It's just the first encounter with something that needs to be renormalized. Renormalizable theories are not "incomplete", even though you can take the Wilsonian standpoint that renormalized QFTs are effective theories cut off at a scale. according to the author, the energy differenc is always infinite according to two fact. the first is, the ground state energy is infnite, secondly, the energy differenc is defined by substituting a higher level energy from the ground state one. @enumaris That is an unfairly pithy way of putting it. There are finite, rigorous frameworks for renormalized perturbation theories following the work of Epstein and Glaser (buzzword: Causal perturbation theory). Just like in many other areas, the physicist's math sweeps a lot of subtlety under the rug, but that is far from unique to QFT or renormalization The classical electrostatics formula $H = \int \frac{\mathbf{E}^2}{8 \pi} dV = \frac{1}{2} \sum_a e_a \phi(\mathbf{r}_a)$ with $\phi_a = \sum_b \frac{e_b}{R_{ab}}$ allows for $R_{aa} = 0$ terms i.e. dividing by zero to get infinities also, the problem stems from the fact that $R_{aa}$ can be zero due to using point particles, overall it's an infinite constant added to the particle that we throw away just as in QFT @bolbteppa I understand the idea that we need to drop such terms to be in consistency with experiments. But i cannot understand why the experiment didn't predict such infinities that arose in the theory? These $e_a/R_{aa}$ terms in the big sum are called self-energy terms, and are infinite, which means a relativistic electron would also have to have infinite mass if taken seriously, and relativity forbids the notion of a rigid body so we have to model them as point particles and can't avoid these $R_{aa} = 0$ values.
Geometry Category : 6th Class Geometry Learning Objectives Geometry The branch of mathematics which deals with mathematical objects like points, lines, panes and space is called geometry. Point: A point is to be thought of as a location in space. In other words, a point determines location in a space. Line segment: Let A and B be two point on a plane. Then the straight path between points a and B is known as Sine segment AB. The above line segment is denoted as\[\overline{AB}\]. Ray: A line segment extended endlessly in one direction Is. called a ray. The above ray Is denoted as \[\overrightarrow{AB}\]. Line: A line is a straight path that extends on and on in both directions endlessly. The above ray Is denoted as \[\overrightarrow{AB}\]. Line: A line is a straight path that extends on and on in both directions endlessly. Parts of a circle Centre: The fixed point in the plane which is equidistant from every point lying on the boundary of the circle is called centre of the circle. Radius : The line segment joining the centre of the circle to any point on the circle is the radius of that circle. Chord: A line segment joining any two points on a circle is called chord of the circle. Diameter: A chord which passes through the centre of a circle is called a diameter of that circle. Secant: A line which intersects the circles at two distinct points is called a secant. Arc: An arc is a part of a circle. Segment: A region in the interior of the circle enclosed by a chord and an arc is called a segment of the circle. Sector: A region in the interior of the circle enclosed by two radii and an arc is called sector of the circle. Semicircle: A diameter divides a circle into two equal parts. Each part is called a semicircle. Circumference: The length of the boundary of the interior of a circle is called circumference of the circle. Concentric circles: Two or more circles with the same centre are called concentric circles. Three Dimensional Shapes Cuboid: Cuboid is a three dimensional solid shape. A rectangular wooden box is an example of a cuboid. A cuboid has 6 rectangular faces, 12 edges and 8 vertices. Cube: A cuboid whose length, breadth and height are equal is called a cube, A cube has 5 square faces, 12 edges and 8 vertices. Cylinder: A cylinder is three dimensional geometric figure that has two congruent and parallel bases. It has no vertex, two circular faces, one curved face and two curved edges. Sphere: An object which is in the shape of a ball is said to have the shape of sphere. It has a curved surface but no vertex and no edge. Cone: A cone is a three dimensional geometric shape that tapers smoothly from a fiat to a point called the apex or vertex. It has one vertex, one curved edge, one curved face and one flat face. Pyramid: A pyramid is a solid whose base is a plane rectilinear figure and whose side faces are triangles having a common vertex called the vertex of the pyramid. Symmetry A figure drawn on the paper is symmetric, if it can be folded in such a way that the two halves of the figure exactly cover each other. The line of fold is called axis of the symmetry, Symmetry of some geometrical shapes Line segment: A line segment is symmetrical about its perpendicular bisector. Angle: An angle with equal arms is symmetrical about the bisector of the angle, Scalene triangle: A scalene triangle has no line of symmetry. Parallelogram: A parallelogram has no line of symmetry. Table of Symmetry of Geometrical Shapes Shape Figure Number of lines of symmetry Equilateral triangle 3 Isosceles triangle 1 Square 4 Rectangle 2 Rhombus 2 Semicircle 1 Circle Infinite Commonly Asked Questions Which one of the following is the complementary angle of \[\mathbf{30}{}^\circ \] ? (a) \[30{}^\circ \] (b) \[60{}^\circ \] (c) \[90{}^\circ \] (d) All of these (e) None of these Answer (b) Explanation: Complementary angle of \[30{}^\circ =90{}^\circ -30{}^\circ =60{}^\circ .\] In the options given below the pair of angles are given. Find the complementary pair. (a)\[51{}^\circ ,25{}^\circ \] (b) \[61{}^\circ ,29{}^\circ \] (c) \[45{}^\circ ,55{}^\circ \] (d) All of these (d) None of these Answer: (b) Explanation: If the sum of pair of angles is \[90{}^\circ \] then the pair is called cornplementary pair of angles. The sum of angles \[61{}^\circ +29{}^\circ =90{}^\circ .\] Hence the pair of angles \[61{}^\circ ,29{}^\circ \] is a complementary pair of angles, Choose the pair of supplementary angle from the options given below? (a) \[100{}^\circ ,200{}^\circ \] (b) \[105{}^\circ ,75{}^\circ \] (c) \[30{}^\circ ,180{}^\circ \] (d) All of these (e) None of these Answer: (b) Explanation: The sum of angles of the pair \[105{}^\circ ,75{}^\circ =105{}^\circ +75{}^\circ =180{}^\circ \] Hence, the pair of angles \[105{}^\circ ,75{}^\circ \] is a supplementary pair of angles. Find the measurement of A if B and C are given when A, B and C are the angle of the triangle. (a)\[A=180{}^\circ -\left( B+C \right)\] (b) \[A=180{}^\circ +\left( B-C \right)\] (c) \[A=180{}^\circ +C-B\] (d) All of these (e) None of these Answer (a) Explanation: The sum of angles of the triangle \[=180{}^\circ \]. Hence, the angle \[A=180{}^\circ -\left( B+C \right).\] If the base of a triangle is 18 cm and distance between base and opposite vertex is 10 cm, what will be the area of triangle? (a) \[206\text{ }c{{m}^{2}}\] (b) \[10\text{ }c{{m}^{2}}\] (c) \[90\text{ }c{{m}^{2}}\] (d) All of these None of these Answer (c) Explanation: The distance between base and opposite vertex is height, \[The\,area={{\frac{1}{2}}^{~}}\times base\times height\] \[=\frac{1}{2}18\times 10=90c{{m}^{2}}.\] You need to login to perform this action. You will be redirected in 3 sec
Home Integration by PartsIntegration by Parts Examples Integration by Parts with a definite integral Going in Circles Tricks of the Trade Integrals of Trig FunctionsAntiderivatives of Basic Trigonometric Functions Product of Sines and Cosines (mixed even and odd powers or only odd powers) Product of Sines and Cosines (only even powers) Product of Secants and Tangents Other Cases Trig SubstitutionsHow Trig Substitution Works Summary of trig substitution options Examples Completing the Square Partial FractionsIntroduction to Partial Fractions Linear Factors Irreducible Quadratic Factors Improper Rational Functions and Long Division Summary Strategies of IntegrationSubstitution Integration by Parts Trig Integrals Trig Substitutions Partial Fractions Improper IntegralsType 1 - Improper Integrals with Infinite Intervals of Integration Type 2 - Improper Integrals with Discontinuous Integrands Comparison Tests for Convergence Modeling with Differential EquationsIntroduction Separable Equations A Second Order Problem Euler's Method and Direction FieldsEuler's Method (follow your nose) Direction Fields Euler's method revisited Separable EquationsThe Simplest Differential Equations Separable differential equations Mixing and Dilution Models of GrowthExponential Growth and Decay The Zombie Apocalypse (Logistic Growth) Linear EquationsLinear ODEs: Working an Example The Solution in General Saving for Retirement Parametrized CurvesThree kinds of functions, three kinds of curves The Cycloid Visualizing Parametrized Curves Tracing Circles and Ellipses Lissajous Figures Calculus with Parametrized CurvesVideo: Slope and Area Video: Arclength and Surface Area Summary and Simplifications Higher Derivatives Polar CoordinatesDefinitions of Polar Coordinates Graphing polar functions Video: Computing Slopes of Tangent Lines Areas and Lengths of Polar CurvesArea Inside a Polar Curve Area Between Polar Curves Arc Length of Polar Curves Conic sectionsSlicing a Cone Ellipses Hyperbolas Parabolas and Directrices Shifting the Center by Completing the Square Conic Sections in Polar CoordinatesFoci and Directrices Visualizing Eccentricity Astronomy and Equations in Polar Coordinates Infinite SequencesApproximate Versus Exact Answers Examples of Infinite Sequences Limit Laws for Sequences Theorems for and Examples of Computing Limits of Sequences Monotonic Covergence Infinite SeriesIntroduction Geometric Series Limit Laws for Series Test for Divergence and Other Theorems Telescoping Sums Integral TestPreview of Coming Attractions The Integral Test Estimates for the Value of the Series Comparison TestsThe Basic Comparison Test The Limit Comparison Test Convergence of Series with Negative TermsIntroduction, Alternating Series,and the AS Test Absolute Convergence Rearrangements The Ratio and Root TestsThe Ratio Test The Root Test Examples Strategies for testing SeriesStrategy to Test Series and a Review of Tests Examples, Part 1 Examples, Part 2 Power SeriesRadius and Interval of Convergence Finding the Interval of Convergence Power Series Centered at $x=a$ Representing Functions as Power SeriesFunctions as Power Series Derivatives and Integrals of Power Series Applications and Examples Taylor and Maclaurin SeriesThe Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor PolynomialsTaylor Polynomials When Functions Are Equal to Their Taylor Series When a Function Does Not Equal Its Taylor Series Other Uses of Taylor Polynomials Functions of 2 and 3 variablesFunctions of several variables Limits and continuity Partial DerivativesOne variable at a time (yet again) Definitions and Examples An Example from DNA Geometry of partial derivatives Higher Derivatives Differentials and Taylor Expansions Differentiability and the Chain RuleDifferentiability The First Case of the Chain Rule Chain Rule, General Case Video: Worked problems Multiple IntegralsGeneral Setup and Review of 1D Integrals What is a Double Integral? Volumes as Double Integrals Iterated Integrals over RectanglesHow To Compute Iterated Integrals Examples of Iterated Integrals Fubini's Theorem Summary and an Important Example Double Integrals over General RegionsType I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinatesdA = r dr (d theta) Examples Multiple integrals in physicsDouble integrals in physics Triple integrals in physics Integrals in Probability and StatisticsSingle integrals in probability Double integrals in probability Change of VariablesReview: Change of variables in 1 dimension Mappings in 2 dimensions Jacobians Examples Bonus: Cylindrical and spherical coordinates On the last slide, we learned how to work with 2-dimensionallaminas. Of course, the world isn't 2 dimensional! In the3-dimensional world, density is mass per unit This is usually computed in rectangular coordinates, with $dV = dx\, dy\, dz$, but sometimes it's easier to use cylindrical coordinates, with $dV = r \,dr \,d\theta\, dz$, or spherical coordinates. Center of mass works almost the same as in 2 dimensions: You have to be careful with moment of inertia, since that depends on which axis you are rotating around. For rotations around the $z$ axis, the moment of inertia is $$I_3 = \iiint_S (x^2+y^2) \rho(x,y,z) dV,$$ since the distance from $(x,y,z)$ to the $z$ axis is $r=\sqrt{x^2+y^2}$. However, if you are rotating around the $x$ axis, then the moment of inertia is $I_1=\iiint_S (y^2+z^2) \rho(x,y,z) dV$, and if you are rotating around the $y$ axis, then the moment of inertia is $I_2=\iiint_S (x^2+z^2) \rho(x,y,z) dV.$ Rotating around diagonal axes is even more complicated. To dealwith such cases, physicists define a moment ofinertia
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building) It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore) In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of @TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies \|\| x ( t ) - 0 \|\| < \varepsilon.$$The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $... "If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}\|f(z)\| \leq C_K \\|f\\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed? Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2 Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$ Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight. hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$ for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$ I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything. I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set).theorem ... The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious. (but seriously, the best tactic is over powered...) Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field? Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement? "Infinity exists" comes to mind as a potential candidate statement. Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system @Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity but so far failed Put it in another way, an equivalent formulation of that (possibly open) problem is: > Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object? If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite. My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science... O great, given a transcendental $s$, computing $\min_P(\|P(s)\|)$ is a knapsack problem hmm... By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as: $$P(x) = \prod_{k=0}^n (x - \lambda_k)$$ If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic Thus given $s$ transcendental, to minimise $\|P(s)\|$ will be given as follows: The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases. In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that0<\|x−pq\|<1qn.{\displaystyle 0<\left\|x-{\frac {p}... Do these still exist if the axiom of infinity is blown up? Hmmm... Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum: $$\sum_{k=1}^M \frac{1}{b^{k!}}$$ The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework There's this theorem in Spivak's book of Calculus:Theorem 7Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and$$f'... and neither Rolle nor mean value theorem need the axiom of choice Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set > are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion
section 6.5 exercises In each of the following triangles, solve for the unknown side and angles. 1. 2. 3. 4. Find a possible formula for the trigonometric function whose values are in the following tables. 5. $x$ 0 1 2 3 4 5 6 $y$ -2 4 10 4 -2 4 10 6. $x$ 0 1 2 3 4 5 6 $y$ 1 -3 -7 -3 1 -3 -7 7. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the high temperature for the day is 63 degrees and the low temperature of 37 degrees occurs at 5 AM. Assuming $t$ is the number of hours since midnight, find an equation for the temperature, $D$, in terms of $t$. 8. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the high temperature for the day is 92 degrees and the low temperature of 78 degrees occurs at 4 AM. Assuming $t$ is the number of hours since midnight, find an equation for the temperature, $D$, in terms of $t$. 9. A population of rabbits oscillates 25 above and below an average of 129 during the year, hitting the lowest value in January ($t = 0$). a. Find an equation for the population, $P$, in terms of the months since January, $t$. b. What if the lowest value of the rabbit population occurred in April instead? 10. A population of elk oscillates 150 above and below an average of 720 during the year, hitting the lowest value in January ($t = 0$). a. Find an equation for the population, $P$, in terms of the months since January, $t$. b. What if the lowest value of the rabbit population occurred in March instead? 11. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 105 degrees occurs at 5 PM and the average temperature for the day is 85 degrees. Find the temperature, to the nearest degree, at 9 AM. 12. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 84 degrees occurs at 6 PM and the average temperature for the day is 70 degrees. Find the temperature, to the nearest degree, at 7 AM. 13. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 47 and 63 degrees during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight does the temperature first reach 51 degrees? 14. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 64 and 86 degrees during the day and the average daily temperature first occurs at 12 AM. How many hours after midnight does the temperature first reach 70 degrees? 15. A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. How many minutes of the ride are spent higher than 13 meters above the ground? 16. A Ferris wheel is 45 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 27 meters above the ground? 17. The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square kilometers in March. Assuming sinusoidal fluctuation, during how many months are there less than 9 million square kilometers of sea ice? 18. The sea ice area around the South Pole fluctuates between about 18 million square kilometers in September to 3 million square kilometers in March. Assuming sinusoidal fluctuation, during how many months are there more than 15 million square kilometers of sea ice? 19. A respiratory ailment called “Cheyne-Stokes Respiration” causes the volume per breath to increase and decrease in a sinusoidal manner, as a function of time. For one particular patient with this condition, a machine begins recording a plot of volume per breath versus time (in seconds). Let $b(t)$ be a function of time t that tells us the volume (in liters) of a breath that starts at time $t$. During the test, the smallest volume per breath is 0.6 liters and this first occurs for a breath that starts 5 seconds into the test. The largest volume per breath is 1.8 liters and this first occurs for a breath beginning 55 seconds into the test. [UW] a. Find a formula for the function $b(t)$ whose graph will model the test data for this patient. b. If the patient begins a breath every 5 seconds, what are the breath volumes during the first minute of the test? 20. Suppose the high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m, at which time the water is 10 feet above the height of low tide. Low tides occur 6 hours after high tides. Suppose there are two high tides and two low tides every day and the height of the tide varies sinusoidally. [UW] a. Find a formula for the function $y = h(t)$ that computes the height of the tide above low tide at time $t$. (In other words, $y = 0$ corresponds to low tide.) b. What is the tide height at 11:00 a.m.? 21. A communications satellite orbits the earth $t$ miles above the surface. Assume the radius of the earth is 3,960 miles. The satellite can only “see” a portion of the earth’s surface, bounded by what is called a horizon circle. This leads to a two-dimensional cross-sectional picture we can use to study the size of the horizon slice: [UW] a. Find a formula for $\alpha$ in terms of $t$. b. If $t = 30,000$ miles, what is $\alpha$? What percentage of the circumference of the earth is covered by the satellite? What would be the minimum number of such satellites required to cover the circumference? c. If $t = 1,000$ miles, what is $\alpha$? What percentage of the circumference of the earth is covered by the satellite? What would be the minimum number of such satellites required to cover the circumference? d. Suppose you wish to place a satellite into orbit so that 20% of the circumference is covered by the satellite. What is the required distance $t$? 22. Tiffany is a model rocket enthusiast. She has been working on a pressurized rocket filled with nitrous oxide. According to her design, if the atmospheric pressure exerted on the rocket is less than 10 pounds/sq.in., the nitrous oxide chamber inside the rocket will explode. Tiff worked from a formula $p=14.7e^{-h/10}$ pounds/sq.in. for the atmospheric pressure $h$ miles above sea level. Assume that the rocket is launched at an angle of $\alpha$ above level ground at sea level with an initial speed of 1400 feet/sec. Also, assume the height (in feet) of the rocket at time $t$ seconds is given by the equation $y\left(t\right)=-16t^{2} +1400\sin \left(\alpha \right)t$. [UW] a. At what altitude will the rocket explode? b. If the angle of launch is $\alpha$ = 12$\mathrm{{}^\circ}$, determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? c. If the angle of launch is $\alpha$ = 82$\mathrm{{}^\circ}$, determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? d. Find the largest launch angle $\alpha$ so that the rocket will not explode. Answer 1. $c = \sqrt{89}$, $A = 57.9946^{\circ}$, $B = 32.0054^{\circ}$ 3. $b = \sqrt{176}$, $A = 27.8181^{\circ}$, $B = 62.1819^{\circ}$ 5. $y(x) = 6 \sin (\dfrac{\pi}{2}(x - 1)) + 4$ 7. $D(t) = 50 - 13 \cos(\dfrac{\pi}{12}(t - 5))$ 9. a. $P(t) = 129 - 25 \cos(\dfrac{\pi}{6} t)$ b. $P(t) = 129 - 25 \cos(\dfrac{\pi}{6} (t - 3))$ 11. 75 degrees 13. 8 15. 2.80869431742 17. 5.035 months
Home Integration by PartsIntegration by Parts Examples Integration by Parts with a definite integral Going in Circles Tricks of the Trade Integrals of Trig FunctionsAntiderivatives of Basic Trigonometric Functions Product of Sines and Cosines (mixed even and odd powers or only odd powers) Product of Sines and Cosines (only even powers) Product of Secants and Tangents Other Cases Trig SubstitutionsHow Trig Substitution Works Summary of trig substitution options Examples Completing the Square Partial FractionsIntroduction to Partial Fractions Linear Factors Irreducible Quadratic Factors Improper Rational Functions and Long Division Summary Strategies of IntegrationSubstitution Integration by Parts Trig Integrals Trig Substitutions Partial Fractions Improper IntegralsType 1 - Improper Integrals with Infinite Intervals of Integration Type 2 - Improper Integrals with Discontinuous Integrands Comparison Tests for Convergence Modeling with Differential EquationsIntroduction Separable Equations A Second Order Problem Euler's Method and Direction FieldsEuler's Method (follow your nose) Direction Fields Euler's method revisited Separable EquationsThe Simplest Differential Equations Separable differential equations Mixing and Dilution Models of GrowthExponential Growth and Decay The Zombie Apocalypse (Logistic Growth) Linear EquationsLinear ODEs: Working an Example The Solution in General Saving for Retirement Parametrized CurvesThree kinds of functions, three kinds of curves The Cycloid Visualizing Parametrized Curves Tracing Circles and Ellipses Lissajous Figures Calculus with Parametrized CurvesVideo: Slope and Area Video: Arclength and Surface Area Summary and Simplifications Higher Derivatives Polar CoordinatesDefinitions of Polar Coordinates Graphing polar functions Video: Computing Slopes of Tangent Lines Areas and Lengths of Polar CurvesArea Inside a Polar Curve Area Between Polar Curves Arc Length of Polar Curves Conic sectionsSlicing a Cone Ellipses Hyperbolas Parabolas and Directrices Shifting the Center by Completing the Square Conic Sections in Polar CoordinatesFoci and Directrices Visualizing Eccentricity Astronomy and Equations in Polar Coordinates Infinite SequencesApproximate Versus Exact Answers Examples of Infinite Sequences Limit Laws for Sequences Theorems for and Examples of Computing Limits of Sequences Monotonic Covergence Infinite SeriesIntroduction Geometric Series Limit Laws for Series Test for Divergence and Other Theorems Telescoping Sums Integral TestPreview of Coming Attractions The Integral Test Estimates for the Value of the Series Comparison TestsThe Basic Comparison Test The Limit Comparison Test Convergence of Series with Negative TermsIntroduction, Alternating Series,and the AS Test Absolute Convergence Rearrangements The Ratio and Root TestsThe Ratio Test The Root Test Examples Strategies for testing SeriesStrategy to Test Series and a Review of Tests Examples, Part 1 Examples, Part 2 Power SeriesRadius and Interval of Convergence Finding the Interval of Convergence Power Series Centered at $x=a$ Representing Functions as Power SeriesFunctions as Power Series Derivatives and Integrals of Power Series Applications and Examples Taylor and Maclaurin SeriesThe Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor PolynomialsTaylor Polynomials When Functions Are Equal to Their Taylor Series When a Function Does Not Equal Its Taylor Series Other Uses of Taylor Polynomials Functions of 2 and 3 variablesFunctions of several variables Limits and continuity Partial DerivativesOne variable at a time (yet again) Definitions and Examples An Example from DNA Geometry of partial derivatives Higher Derivatives Differentials and Taylor Expansions Differentiability and the Chain RuleDifferentiability The First Case of the Chain Rule Chain Rule, General Case Video: Worked problems Multiple IntegralsGeneral Setup and Review of 1D Integrals What is a Double Integral? Volumes as Double Integrals Iterated Integrals over RectanglesHow To Compute Iterated Integrals Examples of Iterated Integrals Fubini's Theorem Summary and an Important Example Double Integrals over General RegionsType I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinatesdA = r dr (d theta) Examples Multiple integrals in physicsDouble integrals in physics Triple integrals in physics Integrals in Probability and StatisticsSingle integrals in probability Double integrals in probability Change of VariablesReview: Change of variables in 1 dimension Mappings in 2 dimensions Jacobians Examples Bonus: Cylindrical and spherical coordinates Of all the techniques of integration that we have learned, the most powerful, and the simplest, is $u$-substitution. If $u = g(x)$, then $$\int_a^b f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(u) du.$$ You probably learned this method in terms of anti-derivatives, as a way to turn the chain rule inside out. That sort of reasoning doesn't generalize very well to multiple dimensions. Instead, let's analyze $u$-substitution from the definition of an integral. By definition, $\int_a^b f(g(x)) g'(x) dx$ is the limit of a sum$$\sum_{i=1}^N f(g(x_i^)) g'(x_i^) \Delta_i x,$$ where we have broken the interval from $a$ to $b$ into $N$ pieces, $x_i^$ is a point in the $i$-th piece, and $\Delta_i x$is the length of the $i$-th piece. If welet $u_i^ = g(x_i^)$, then $\Delta_i u \approx g'(x_i^) \Delta_i x$, andour sum is approximately $$\sum_{i=1}^N f(u_i^*) \Delta_i u.$$The limit of this sum is, by definition, the definite integral of $f(u) du$from the starting value of $u$ (namely $g(a)$) to the ending value of $u$($g(b)$). We have converted an integral in $x$-space into an integral in $u$-space,but the functions being integrated are not the same. In one case we integrate$f(u)$. In the other case we integrate $f(g(x))$ We also learned about
I just posted a short essay that defines a function with some curious properties. The function is defined by \[f(x) = \begin{cases}\frac{1}{b} & x = \frac{a}{b}\in\mathbf{Q}\text{ in lowest terms}\ \0 & \text{if } x \in \mathbf{R}\setminus\mathbf{Q}.\end{cases}\] In the essay I prove that $f(x)$ is continuous on $\mathbf{R}\setminus\mathbf{Q}$ $f(x)$ is discontinuous on $\mathbf{Q}$ $f(x)$ is integrable, and in particular $\int_0^1 f(x), dx = 0$ At first blush, these properties may seem counterintuitive. The theory of Lebesgue measure completely characterizes Riemann integrable functions: they are functions which are almost everywhere continuous. The above function is almost everywhere continuous because its set of discontinuities ($\mathbf{Q}$) is countably infinite, and therefore has Lebesgue measure $0$.
XVIIIth Century Math, Chuck Norris and Updating Risks of Default We have seen previously how machine learning and statistics could help predicting how much money a loan would pay back. Here we’ll use another branch of statistics, probabilities, to determine how likely a loan is to default knowing it has already made some payments. Very Basic Notions In probabilities, the chance or risk of an event A to occur is written $P(A) $. $P(A) $ is always between 0 an 1, where 0 means it will never occur and 1 that it is absolutely certain, and can be calculated as: \[ P(A) = \frac{\text{Number of ways ‘A’ can happen}}{\text{Total number of possible outcomes}} \]. For instance, the probability to get a ‘Head’ when flipping a coin (provided the coin is fair) is: \[ P(H) = \frac{\text{Head}}{\text{Head or Tail}} = \frac{1}{2} = 0.5 \]. Things get more interesting when we combine the probabilities of multiple events. For instance, the probability of drawing two heads in a row is: \[ P(H) \times P(H) = 0.25 \]. While the probability of tossing Head at least once when doing 2 throws is equal to the probability of tossing Head at the first throw, plus the probability of tossing head at the second throw when the first one was Tail, or: \[ P(H) + P(H) \times P(T) = 0.5 + 0.25 = 0.75 \]. But this is only valid if the two events are independent. The probability of a fair coin to fall on ‘head’ is not influenced by the previous toss (which is somehow counter-intuitive: if you toss a fair coin and get ‘Head’ 99 times in a row, most would believe that is time for ‘Tail’. It’s not, at least not more that during the very first throw). In real life, many events are dependent upon each others. Chuck Who? Let’s consider $P(A) $ your probability to be in awe of someone your encounter in general to be 0.05, and$ P(C) $ your probability of bumping into Chuck Norris to be .000000003 (1 out of 320 million people in the U.S). Your probability to bump into Chuck Norris and to be in awe is NOT the probability of bumping into Chuck Norris times the probability of being in awe in general because… well, he’s Chuck Norris, so anybody would be utterly impressed. In probabilities, $P(A\|C) $ means the probability of event A knowing that C happens. Here, $P(A\|C) $ means the probability to be in awe knowing you bumped into Chuck Norris, which we’ll put very conservatively at only 0.99 [1]. Therefore your probability of bumping into Chuck Norris and being in awe is: \[ P(A\|C) \times P(C) = 0.99 \times 0.000000003 \] England a long time ago The probability for being in front of Chuck Norris and in awe being the same that the probability of being in awe and in front of Chuck Norris, we can write: \[P(A\|C) \times P(C) = P(C\|A) \times P(A) \] And therefore, the probability of being in front of Chuck Norris knowing that you’re in awe can be written: \[ P(C\|A) = \frac{P(A\|C) \times P(C)}{P(A)} \] Here, we have $P(C\|A) = 0.99 \times 0.000000003 \div 0.05 = 0.000000059 $. Which makes even sense intuitively: as you’re difficult to impress (only 5% chance), the fact that you’re in awe right now significantly increases the probability that you’re in front of Chuck Norris himself. Baye’s Theorem has countless uses in medicine, finance or physics. Its main interest is that it allows to update the probability of an event as new evidence is acquired, a process named Bayesian inference. Back to the XXIst Century A central issue when lending money is to estimate the probability that a borrower will fail to make all the due payments, an event known as a loan defaulting. As more and more payments are made, that probability shall decrease. After all, when the very last payment is made, the default probability decreases to 0. If D is the event of a loan defaulting, and N the event of a loan to have made at least N payments already, then Bayesian inference allows use to calculate $ P(D\|N) $ the updated probability that it will default knowing it has already made N payments, as: \[ P(D\|N) = \frac{P(N\|D) \times P(D)}{P(N)} \] Luckily, all the terms on the right side of the equation are relatively easy to measure. For instance, we analyze the historical data of 218,480 Prosper loans. Amongst those, 31,924 are 36-months loans that were issued more than 3 years ago, and are therefore old enough to have reached maturity. Let’s consider the loans with the grade ‘D’. 1,551 of them have defaulted, out of a total of 6,754, which gives an overall probability of default of: \[ P(D) = 1551 \div 6754 = 0.2296 \] Nota Bene: This average default rate may seem high, but one must consider that ‘D’ is a risky grade, and that defaulting loans usually make many payments before defaulting, therefore reducing the losses. Thanks to interest rates around 22% on average, even with such a high probability of default, net annual returns for investors exceeded 7.17%. Prosper’s data do not tell how many payments each of the loans has made. But the data show for each loan how much was paid back. Knowing the loan amount $ A $, yearly interest rate $ r $ and term in months $ T $, we can calculate the monthly installment $ p $ for each loan, using the formula: \[ p = A \times \frac{\frac{r}{12}}{1 – (1 + \frac{r}{12)})^-T} \] Once we calculated the monthly installment, we can estimate the number of monthly payments made by a loan by dividing the total amount paid by the monthly installment. By counting the proportion of loans that made at least N payments, both for all the 6,754 grade D loans, and for the 1,551 loans of them that we know will default, we obtain respectively $ P(N) $ and $ P(N\|D) $: Month P(N) P(N\|D) 1 0.990820254664 0.960025789813 2 0.982973053006 0.925854287556 3 0.975273911756 0.892327530625 4 0.966834468463 0.855577047066 5 0.957506662718 0.814958091554 … … … 15 0.873852531833 0.450676982592 16 0.867337873853 0.422308188266 17 0.860823215872 0.393939393939 18 0.85667752443 0.375886524823 … … … 34 0.780574474386 0.0522243713733 35 0.773319514362 0.0348162475822 This kind of information is enough to update the probability of default of a loan knowing it has made N payments already. For instance, the probability of default for a 36-months, grade D loan that has made already 17 payments is: \[ P(D\|17) = \frac{P(17\|D) \times P(D)}{P(17)} = \frac{0.3939\times 0.2296}{0.8608} = 10.50\% \] This simple formula allows to make two interesting calculations. First, to estimate the probability of default of ongoing loans in an investor’s portfolio. Second, by comparing historical default rates and updated rates, it shows how healthy a marketplace is. For instance, by applying this Bayesian inference on Prosper’s 12,270 non-mature loans with the same grade ‘D’, we can calculate than on average, the default rate is now down to 15.9%, indicating significant improvements from the platform; all the more impressive since, as mentioned previously, it was already return over 7% per year for that grade. Thomas Bayes and Chuck Norris should invest in Marketplace Lending. Although it’s probably too late for one of them. Emmanuel Marot May 26, 2015 2 Comment
Energy conservation does work perfectly in general relativity. The overall Lagrangian is invariant under time translations and Noether's Theorem can be used to derive a non-trivial and exact conserved current for energy. The only thing that makes general relativity a little different from electromagnetism is that the time translation symmetry is part of a larger gauge symmetry so time is not absolute and can be chosen in many ways. However there is no problem with the derivation of conserved energy with respect to any given choice of time translation. There is a long and interesting history to this problem. Einstein gave a valid formula for the energy in the gravitational field shortly after publishing general relativity. The mathematicians Hilbert and Klein did not like the coordinate dependence in Einstein's formulation and claimed it reduced to a trivial identity. They enlisted Noether to work out a general formalism for conservation laws and claimed that her work supported their view. The debate continued for many years especially in the context of gravitational waves which some people claimed did not exist. They thought that the linearised solutions for gravitational waves were equivalent to flat space via co-ordinate transformations and that they carried no energy. At one point even Einstein doubted his own formalism, but later he returned to his original view that energy conservation holds up. The issue was finally resolved when exact non-linear gravitational wave solutions were found and it was shown that they do carry energy. Since then this has even been verified empirically to very high precision with the observation of the slowing down of binary pulsars in exact agreement with the predicted radiation of gravitational energy from the system. The formula for energy in general relativity is usually given in terms of pseudo tensors such as those proposed by Laundau & Lifshitz, Dirac, Weinberg or Einstein himself. Wikipedia has a good article on these and how they confirm energy conservation. Although pseudotensors are mathematically rigorous objects which can be understood as sections of jet bundles, some people don't like their apparent co-ordinate dependence. There are other covariant approaches such as the Komar Superpotential or a more general formula of mine which gives the energy current in terms of the time translation vector $k^{\mu}$ as $ J^{\mu}_G = \frac{1}{16\pi G} (k^{\mu}R - 2k^{\mu}\Lambda - 2{{k^{\alpha}}_{;\alpha}}^{\mu} + {{k^{\alpha}}_{;}}^{\mu}_{\alpha}+ {{k^{\mu}}_{;}}^{\alpha}_{\alpha})$ Despite these general formulations of energy conservation in general relativity there are some cosmologists who still take the view that energy conservation is only approximate or that it only works in special cases or that it reduces to a trivial identity. In each case these claims can be refuted either by studying the formulations I have referenced or by comparing the arguments given by these cosmologists with analogous situations in other gauge theories where conservation laws are accepted and follow analogous rules. One area of particular contention is energy conservation in a homogeneous cosmology with cosmic radiation and a cosmological constant. Despite all the contrary claims, a valid formula for energy conservation in this case can be derived from the general methods and is given by this equation. $ E = Mc^2 + \frac{\Gamma}{a} + \frac{\Lambda c^2}{\kappa}a^3 - \frac{3}{\kappa}\dot{a}^2a - Ka = 0$ $a(t)$ is the universal expansion factor as a funcrtion of time normalised to 1 at the current epoch. $E$ is the total energy in an expanding region of volume $a(t)^3$. This always comes to zero in a perfectly homogeneous cosmology. $M$ is the total mass of matter in the region $c$ is the speed of light $\Gamma$ is the density of cosmic radiation normalised to the current epoch $\Lambda$ is the cosmological constant, thought to be positive. $\kappa$ is the gravitational coupling constant $K$ is a constant that is positive for spherical closed space, negative for hyperbolic space and zero for flat space. The first two terms describe the energy in matter and radiation with the matter energy not changing and the radiation decreasing as the universe expands. Both are positive.The third term is "dark energy" which is currently though to be positive and contributing about 75% of the non-gravitational energy, but this increases with time.The final two terms represent the gravitational energy which is negative to balance the other terms. This equation holds as a consequence of the well-known Friedmann cosmological equations, that come from the Einstein field equations, so it is in no sense trivial as some people have claimed it must be.
Dijet angular distributions from the first LHC pp collisions at center-of-mass energy sqrt(s) = 7 TeV have been measured with the ATLAS detector. The dataset used for this analysis represents an integrated luminosity of 3.1 pb-1. Dijet $\chi$ distributions and centrality ratios have been measured up to dijet masses of 2.8 TeV, and found to be in good agreement with Standard Model predictions. Analysis of the $\chi$ distributions excludes quark contact interactions with a compositeness scale $\Lambda$ below 3.4 TeV, at 95% confidence level, significantly exceeding previous limits. The results of a search for pair production of light top squarks are presented, using 4.7 fb^-1 of sqrt(s) = 7 TeV proton-proton collisions collected with the ATLAS detector at the Large Hadron Collider. This search targets top squarks with masses similar to, or lighter than, the top quark mass. Final states containing exclusively one or two leptons (e, mu), large missing transverse momentum, light-jets and b-jets are used to reconstruct the top squark pair system. Global mass scale variables are used to separate the signal from a large ttbar background. No excess over the Standard Model expectations is found. The results are interpreted in the framework of the Minimal Supersymmetric Standard Model, assuming the top squark decays exclusively to a chargino and a b-quark. Light top squarks with masses between 123-167 GeV are excluded for neutralino masses around 55 GeV. A search is presented for direct top squark pair production in final states with one isolated electron or muon, jets, and missing transverse momentum in proton-proton collisions at sqrt(s) = 7 TeV. The measurement is based on 4.7 fb-1 of data collected with the ATLAS detector at the LHC. Each top squark is assumed to decay to a top quark and the lightest supersymmetric particle (LSP). The data are found to be consistent with Standard Model expectations. Top squark masses between 230 GeV and 440 GeV are excluded with 95% confidence for massless LSPs, and top squark masses around 400 GeV are excluded for LSP masses up to 125 GeV. Measurements are presented of differential cross-sections for top quark pair production in pp collisions at relative to the total inclusive top quark pair production cross-section. A data sample of 2.05 fb(−1) recorded by the ATLAS detector at the Large Hadron Collider is used. Relative differential cross-sections are derived as a function of the invariant mass, the transverse momentum and the rapidity of the top quark pair system. Events are selected in the lepton (electron or muon) + jets channel. The background-subtracted differential distributions are corrected for detector effects, normalized to the total inclusive top quark pair production cross-section and compared to theoretical predictions. The measurement uncertainties range typically between 10 % and 20 % and are generally dominated by systematic effects. No significant deviations from the Standard Model expectations are observed. Jet cross sections have been measured for the first time in proton-proton collisions at a centre-of-mass energy of 7 TeV using the ATLAS detector. The measurement uses an integrated luminosity of 17 nb-1 recorded at the Large Hadron Collider. The anti-kt algorithm is used to identify jets, with two jet resolution parameters, R = 0.4 and 0.6. The dominant uncertainty comes from the jet energy scale, which is determined to within 7% for central jets above 60 GeV transverse momentum. Inclusive single-jet differential cross sections are presented as functions of jet transverse momentum and rapidity. Dijet cross sections are presented as functions of dijet mass and the angular variable $\chi$. The results are compared to expectations based on next-to-leading-order QCD, which agree with the data, providing a validation of the theory in a new kinematic regime. This paper describes measurements of the sum of the transverse energy of particles as a function of particle pseudorapidity, eta, in proton-proton collisions at a centre-of-mass energy, sqrt(s) = 7 TeV using the ATLAS detector at the Large Hadron Collider. The measurements are performed in the region \|eta\| < 4.8 for two event classes: those requiring the presence of particles with a low transverse momentum and those requiring particles with a significant transverse momentum. In the second dataset measurements are made in the region transverse to the hard scatter. The distributions are compared to the predictions of various Monte Carlo event generators, which generally tend to underestimate the amount of transverse energy at high \|eta\|. Results are presented of a search for new particles decaying to large numbers of jets in association with missing transverse momentum, using 4.7 fb^-1 of pp collision data at sqrt(s) = 7 TeV collected by the ATLAS experiment at the Large Hadron Collider in 2011. The event selection requires missing transverse momentum, no isolated electrons or muons, and from >=6 to >=9 jets. No evidence is found for physics beyond the Standard Model. The results are interpreted in the context of a MSUGRA/CMSSM supersymmetric model, where, for large universal scalar mass m_0, gluino masses smaller than 840 GeV are excluded at the 95% confidence level, extending previously published limits. Within a simplified model containing only a gluino octet and a neutralino, gluino masses smaller than 870 GeV are similarly excluded for neutralino masses below 100 GeV. Combined ATLAS and CMS measurements of the Higgs boson production and decay rates, as well as constraints on its couplings to vector bosons and fermions, are presented. The combination is based on the analysis of five production processes, namely gluon fusion, vector boson fusion, and associated production with a W or a Z boson or a pair of top quarks, and of the six decay modes H → ZZ, W W , γγ, ττ, bb, and μμ. All results are reported assuming a value of 125.09 GeV for the Higgs boson mass, the result of the combined measurement by the ATLAS and CMS experiments. The analysis uses the CERN LHC proton-proton collision data recorded by the ATLAS and CMS experiments in 2011 and 2012, corresponding to integrated luminosities per experiment of approximately 5 fb$^{−1}$ at $ \sqrt{s}=7 $ TeV and 20 fb$^{−1}$ at $ \sqrt{s}=8 $ TeV. The Higgs boson production and decay rates measured by the two experiments are combined within the context of three generic parameterisations: two based on cross sections and branching fractions, and one on ratios of coupling modifiers. Several interpretations of the measurements with more model-dependent parameterisations are also given. The combined signal yield relative to the Standard Model prediction is measured to be 1.09 ± 0.11. The combined measurements lead to observed significances for the vector boson fusion production process and for the H → ττ decay of 5.4 and 5.5 standard deviations, respectively. The data are consistent with the Standard Model predictions for all parameterisations considered. A study of WZ production in proton-proton collisions at sqrt(s) = 7 TeV is presented using data corresponding to an integrated luminosity of 4.6 fb^-1 collected with the ATLAS detector at the Large Hadron Collider in 2011. In total, 317 candidates, with a background expectation of 68+/-10 events, are observed in double-leptonic decay final states with electrons, muons and missing transverse momentum. The total cross-section is determined to be sigma_WZ(tot) = 19.0+1.4/-1.3(stat.)+/-0.9(syst.)+/-0.4(lumi.) pb, consistent with the Standard Model expectation of 17.6+1.1/-1.0 pb. Limits on anomalous triple gauge boson couplings are derived using the transverse momentum spectrum of Z bosons in the selected events. The cross section is also presented as a function of Z boson transverse momentum and diboson invariant mass. This letter reports the results of a search for top and bottom squarks from gluino pair production in 4.7 fb^-1 of pp collisions at sqrt(s) = 7 TeV using the ATLAS detector at the LHC. The search is performed in events with large missing transverse momentum and at least three jets identified as originating from a b-quark. Exclusion limits are presented for a variety of gluino-mediated models with gluino masses up to 1 TeV excluded. First measurements of the W -> lnu and Z/gamma* -> ll (l = e, mu) production cross sections in proton-proton collisions at sqrt(s) = 7 TeV are presented using data recorded by the ATLAS experiment at the LHC. The results are based on 2250 W -> lnu and 179 Z/gamma* -> ll candidate events selected from a data set corresponding to an integrated luminosity of approximately 320 nb-1. The measured total W and Z/gamma-boson production cross sections times the respective leptonic branching ratios for the combined electron and muon channels are $\stotW$ BR(W -> lnu) = 9.96 +- 0.23(stat) +- 0.50(syst) +- 1.10(lumi) nb and $\stotZg$ * BR(Z/gamma* -> ll) = 0.82 +- 0.06(stat) +- 0.05(syst) +- 0.09(lumi) nb (within the invariant mass window 66 < m_ll < 116 GeV). The W/Z cross-section ratio is measured to be 11.7 +- 0.9(stat) +- 0.4(syst). In addition, measurements of the W+ and W- production cross sections and of the lepton charge asymmetry are reported. Theoretical predictions based on NNLO QCD calculations are found to agree with the measurements. This paper describes a measurement of the flavour composition of dijet events produced in pp collisions at sqrt{s}=7 TeV using the ATLAS detector. The measurement uses the full 2010 data sample, corresponding to an integrated luminosity of 39 pb^-1. Six possible combinations of light, charm and bottom jets are identified in the dijet events, where the jet flavour is defined by the presence of bottom, charm or solely light flavour hadrons in the jet. Kinematic variables, based on the properties of displaced decay vertices and optimised for jet flavour identification, are used in a multidimensional template fit to measure the fractions of these dijet flavour states as functions of the leading jet transverse momentum in the range 40 GeV to 500 GeV and jet rapidity \|y\| < 2.1. The fit results agree with the predictions of leading- and next-to-leading-order calculations, with the exception of the dijet fraction composed of bottom and light flavour jets, which is underestimated by all models at large transverse jet momenta. The ability to identify jets containing two b-hadrons, originating from e.g. gluon splitting, is demonstrated. The difference between bottom jet production rates in leading and subleading jets is consistent with the next-to-leading-order predictions.
I started by giving a brief history of the problem including the Nash-Tognoli Theorem. I also review some of the major publications concerning the algebraic realization problem written by Prof. Heiner Dovermann and others. I really enjoy the Notices of the AMS “What is…?” articles. They are generally brief perfunctory definitions and material aimed at mathematicians outside of the fields described. I enjoy them for gaining some breadth before diving into articles outside of my repertoire. Here are a list of highlights relating to AT, because they’re a fun place to start reading about these topics. Equivariant Cohomology– Loring W. Tu an Infinity Category– Jacob Lurie a Linear Algebraic Group– Skip Garibaldi Persistent Homology– Shmuel Weinberger an Operad– Jim Stasheff a Projective Structure– William M. Goldman a Legendrian Knot– Joshua M. Sabloff a Perverse Sheaf– Mark Andrea de Cataldo and Luca Migliorini a Stack– Dan Edidin a Derived Stack– Gabriele Vezzosi So far we’ve swapped stories over wine and cheese, Indian food, Thai food, and Crème brûlée. You don’t need a community of women to get through graduate school, but it certainly helps! Aside from a snorkelling break (see picture above), I spent the weekend reading a 2001 paper by the Kyoto group. Classification of equivariant complex vector bundles over a circle Jin-Hwan Cho, Sung Sook Kim, Mikiya Masuda, and Dong Youp Suh J. Math. Kyoto Univ. Volume 41, Number 3 (2001), 517-534. They wrote a series of papers about classification of vector bundles and the algebraic realization problem during the late 1990’s and early 2000’s. Jin-Hwan Cho was a student of Dong Youp Suh’s and he is now at Suwon University. A key tool that is used in this and their subsequent complex classification paper is in section 2 when they decompose the G-vector bundles using a result based on a functorial property of a previously defined induced bundle. In August of 2013, I finally presented on Milnor’s Characteristic Classes for my final comprehensive exam. I am officially a PhD candidate now! I focused on his proof of the polynomial ring structure of the cohomology ring of the complex Grassmann manifold. I assumed everyone already had working definitions of the Euler class and Stiefel-Whitney classes. I presented Milnor’s definition of Chern classes based on the Gysin sequence, and then powered through the notationally heavy proof. I gave a talk during the June 2013 algebraic topology summer course at MSRI. Can you see how nervous I am? notes I spoke about the Becker Gottlieb transfer map as described in J. Becker, D. Gottlieb, “The transfer map and fiber bundles” Topology , 14 (1975) pp. 1–12 link In Gottlieb’s 1975 paper, he proved the existence of a transfer homomorphism involving the total space and the base space of a fiber bundle. The transfer homomorphism is between the nth cohomology of the total space, and the nth cohomology of the base space. The transfer map is defined s.t. the composition of the induced pullback of the projection and the transfer map is just multiplication by the Euler characteristic of the fiber. $\hat{\tau} p: H(B) \rightarrow H*(B)$ A natural question to pose is whether or not this transfer homomorphism was induced by some map from $B$ to $E$. This would make the transfer homomorphism geometric. The paper ends with a proof of the Adams conjecture. It is an important result dealing with real vector bundle isomorphism classes and cohomology operations in K-Theory. The original proof used algebraic geometry. Quillen then proved the Adams Conjecture using the Brauer Induction Theorem. Becker and Gottlieb’s proof is considered exceptional, because it only uses algebraic topology. Their proof reduces the problem to elements of $KO(X)$ which are $2n$-dimensional vector bundles. It uses the transfer map by applying a splitting principal derived from the transfer map. During my first two years of grad school, I was an NSF graduate fellow through the GK-12 grant held here at U. Hawaii. I worked with both a Hawaiian immersion school Halau Lokahi and a more traditional public high school on island, Kailua High School. My favorite projects included teaching middle schoolers how to turn squares into spheres, mobius bands, and tori. I had high schoolers explore surfaces with connected and disconnected boundaries by cutting up large quantities of paper. Super M Website Mahalo nui loa NSF!
I'm trying to work with Dermott's Law to develop a generalized "formula" for assigning major moons to gas- and ice-giant planets, but it doesn't seem to work. If I use the values specified for Jupiter: $T_0 = 0.444$ and $C = 2.03$, and I assume that Ganymede would be considered the third major moon of Jupiter, the orbital period I get out of the equation is: $$ T(3) = 0.444 \times 2.03^3 \\ \space \\ T(3) = 0.444 \times 8.365 \\ \space \\ T(3) = 3.714 $$ ... which is just over half the correct value of 7.155 days. If I use the known orbital period for Ganymede (7.155) days, and determine the value for $n$, I get 3.930: $$ T(n) = T_0C^n \\ \space \\ 7.155 = 0.444 \times 2.03^n \\ \space \\ \frac{7.155}{0.444} = 2.03^n \\ \space \\ 16.115 = 2.03^n \\ \space \\ \log_{10}{16.115} = n \times \log_{10}{2.03} \\ \space \\ n = \frac{\log_{10}{16.115}}{\log_{10}{2.03}} \\ \space \\ n = \frac{1.028}{0.307} \\ \space \\ n = 3.930 $$ ... which is not even an integer, let alone the 3.0 I was expecting. I find a similar problem with Io, Europa, and Callisto, which come out as: $n_\text{Io} = 1.956$ $n_\text{Europa} = 2.941$ $n_\text{Callisto} = 5.128$ ... where I would expect the values to be 1, 2, and 4 (or maybe 0, 1, and 3). Has anybody else worked with this? Can you tell me what I'm not understanding about it?
Let $A$ be an integrally closed domain, with quotient field $K_A$. My main question is the following: Question. Does any non constant irreducible polynomial of $A[X]$ stays irreducible in $K_A[X]$ ? Of course, this is true for monic non constant polynomials, so my first idea what to reduce to this case. This leads to the following subquestion: Subquestion 1. Let $A$ be a domain, and let $P\in A[X]$ be an irreducible polynomial of degree $n\geq 1$, with leading coefficient $a_n$. Is $a_n^{n-1}P(X/a_n)\in A[X]$ irreducible ? If not, is this the case if $A$ is an integral domain ? For the moment, I've tried to prove it without any success, but I came across another subquestion, which would imply a positive answer to the previous one, and thus a positive one to the main question (if I am not mistaken...). Subquestion 2 Let $A$ be an integral domain, and let be a non constant irreducible polynomial $P\in A[X]$. Finally , let $a\in A\setminus\{0\}$. Is is true that the divisors of $aP$ have the form , $b\in A$, or $cP$, $b,c\in A\setminus\{0\}$. If not, is is true if $A$ is an integrally closed domain ? Greg
I'm studying the Simple Harmonic Motion, and I am hesitant about, how to get mass values for infinite period? When mass is 0. When mass is infinite. With $\tau=2\pi/\sqrt{k/m}$. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.Sign up to join this community Because if you have zero mass you can't use the Newtonian $F=m a$ to find force, you'll have a hard time making a physical argument for having zero mass. Likewise, you break all physical barriers when you go to an infinite mass. The best one can do is say that the mass approaches zero or approaches infinity. As $m\to \infty$, $\frac{m}{k}\to \infty$, and so the period goes to infinity, which can be physically interpreted as the spring's $k$ constant is just not strong enough to accelerate the mass at all. As $m\to 0$, $\frac{m}{k}\to 0$, and so the period goes to zero, which can be physically interpreted as very very fast vibrations. The trick used here is the same one used in $\varepsilon$ - $\delta$ limit/derivative proofs in pure math. To avoid the problem of calculating $\infty-\infty$ or $\infty/\infty$, we just calculate behavior "arbitrarily large" or "arbitrarily small", but not actually infinite nor zero.
Science Advisor Homework Helper 1,200 400 Part 1 of (9) ##\phi: R \to S## is a ring epimorphism. Define ##I:= \phi^{-1}(J)##. It is well known that the inverse image of an ideal is an ideal, thus ##I## is an ideal. Define ##\psi: R/I \to S/J: [r] \mapsto [\phi(r)]## This is well defined: If ##r \in I##, then ##\phi(r) \in J##. Clearly, this is also a ring morphism. For injectivity, assume ##[\phi(r)] = 0##, then ##\phi(r) \in J##, and ##r \in \phi^{-1}(J) = I##, thus ##[r] = 0##. The kernel is trivial and the map is injective. Surjectivity follows immediately by surjectivity of ##\phi##. It follows that ##\psi## is an isomorphism, and thus ##R/I \cong S/J##. Define ##\psi: R/I \to S/J: [r] \mapsto [\phi(r)]## This is well defined: If ##r \in I##, then ##\phi(r) \in J##. Clearly, this is also a ring morphism. For injectivity, assume ##[\phi(r)] = 0##, then ##\phi(r) \in J##, and ##r \in \phi^{-1}(J) = I##, thus ##[r] = 0##. The kernel is trivial and the map is injective. Surjectivity follows immediately by surjectivity of ##\phi##. It follows that ##\psi## is an isomorphism, and thus ##R/I \cong S/J##. Last edited:
I am reading a magazine for acoustics, one article is about the choice of speaker wire. The article said the ideal speaker wire should has no resistance, capacitance and inductance. I understand that for the low resistance in ideal case but doubts on the capacitance and inductance. We are driving in the speaker with AC signal, says ideal sinusoidal wave, in the text, it is given that the voltage drop on the capacitor which is connected to the AC power should be $$v_c = I_{max}X_C\sin\omega t $$ where $X_C=1/(\omega C)$. My understanding is for ideal wire, the voltage potential on a short section on the wire should be zero, so we should have $v_c=0$ so does it mean $C\to \infty$ when $\omega\neq 0$? Similar reasoning but on inductance, we have $$v_L = -I_{max}X_L\sin\omega t $$ so for ideal wire, $X_L$ should be zero. Are those reasoning corret?
Before the question, I need to mention some necessary definitions. The rapidity is defined as: $$y=\frac{1}{2}\ln\frac{E+p_z}{E-p_z}=\frac{1}{2}\ln\frac{1+v_z}{1-v_z}=\tanh^{-1}(v_z)$$ where $v_z=p_z/E$ is the velocity along $z$ direction. $v_z=\tanh y$ We have defined the transverse mass $m_T$ and the longitudinal boost factor $\gamma_z$: $$m^2_T=m^2+p^2_T$$ $$\gamma_z=\frac{1}{\sqrt{1-v^2_z}}=\frac{E}{\sqrt{E^2-p^2_z}}=\frac{E}{\sqrt{m^2+p^2_T}}=\frac{E}{m_T}=\cosh y$$ It is easy to show that: $$E=m_T\gamma_z=m_T\cosh y$$ $$p_z=m_T\gamma_zv_z=m_T\sinh y$$ We note that under longitudinal boost, both $p_T$ and $m_T$ remain constant. In high energy physics, one usually uses the Lorentz-invariant particle spectrum $EdN/d^3p$.$$p_z = m_T\sinh y \Rightarrow dp_z=m_T\cosh ydy=Edy\Rightarrow\frac{dp_z}{E}=dy$$Therefore,$$\frac{d^3p}{E}=\frac{dp_zd^2p_T}{E}=dyd^2p_T=dyp_Tdp_Td\phi_p$$The above Lorentz-invariant spectrum is often written as$$E\frac{dN}{d^3p}=\frac{dN}{dyd^2p_T}=\frac{dN}{dyp_Tdp_Td\phi_p}=\frac{dN}{dym_Tdm_Td\phi_p}$$One can see that under Lorentz boost, $d^2p_T$ and $dy$ remain invariant, therefore, $d^3p/E = d^2p_Tdy$ is Lorentz invariant quantity. My question is that is there another different method to show that $E\frac{dN}{d^3p}$ is Lorentz-invariant?
Numerical solution of a non-local elliptic problem modeling a thermistor with a finite element and a finite volume method 1. Department of Mathematics, University of the Aegean, GR-832 00 Karlovassi, Samos, Greece, Greece − $w''(x) = \lambda (f(w(x)))/((\eq^1_(-1) f(w(z)) dz)^2) \all x \in$ (−1, 1), $w'(1) + \alpha w(1) = 0$, $w'$(−1) − $\alpha w$(−1)$ = $0, where $\alpha$ and $\lambda$ are positive constants and $f$ is a function satisfying $f(s)$ > 0, $f'(s) < 0, f''(s) > 0$ for $s > 0, \eq^\infty_0 f(s)ds < \infty.$ The solution of the equation represents the steady state of a thermistor device. The problem has a unique solution for a critical value $\lambda$* of the parameter $\lambda$, at least two solutions for $\lambda < \lambda$* and has no solution for $\lambda > \lambda$. We apply a finite element and a finite volume method in order to find a numerical approximation of the solution of the problem from the space of continuous piecewise quadratic functions, for the case that $\lambda < \lambda$ and for the stable branch of the bifurcation diagram. A comparison of these two methods is made regarding their order of convergence for $f(s) = e^( - s)$ and $f(s) = (1 + s)^( - 2)$. Also, for the same equation but with Dirichlet boundary conditions, a situation where the solution is unique for $\lambda < \lambda$*, a similar comparison of the finite element and the finite volume method is presented. Keywords:finite element method, quadratic finite elements., finite volume method, Non-local elliptic equation. Mathematics Subject Classification:Primary: 65N30, 65N99; Secondary: 80M10, 80M9. Citation:Christos V. Nikolopoulos, Georgios E. Zouraris. Numerical solution of a non-local elliptic problem modeling a thermistor with a finite element and a finite volume method. Conference Publications, 2007, 2007 (Special) : 768-778. doi: 10.3934/proc.2007.2007.768 [1] Caterina Calgaro, Meriem Ezzoug, Ezzeddine Zahrouni. Stability and convergence of an hybrid finite volume-finite element method for a multiphasic incompressible fluid model. [2] Paola Goatin, Sheila Scialanga. Well-posedness and finite volume approximations of the LWR traffic flow model with non-local velocity. [3] [4] [5] Assyr Abdulle, Yun Bai, Gilles Vilmart. Reduced basis finite element heterogeneous multiscale method for quasilinear elliptic homogenization problems. [6] Ming Yan, Lili Chang, Ningning Yan. Finite element method for constrained optimal control problems governed by nonlinear elliptic PDEs. [7] Yinnian He, Yanping Lin, Weiwei Sun. Stabilized finite element method for the non-stationary Navier-Stokes problem. [8] Nora Aïssiouene, Marie-Odile Bristeau, Edwige Godlewski, Jacques Sainte-Marie. A combined finite volume - finite element scheme for a dispersive shallow water system. [9] Alexander Zlotnik, Ilya Zlotnik. Finite element method with discrete transparent boundary conditions for the time-dependent 1D Schrödinger equation. [10] Xufeng Xiao, Xinlong Feng, Jinyun Yuan. The stabilized semi-implicit finite element method for the surface Allen-Cahn equation. [11] Runchang Lin, Huiqing Zhu. A discontinuous Galerkin least-squares finite element method for solving Fisher's equation. [12] Binjie Li, Xiaoping Xie, Shiquan Zhang. New convergence analysis for assumed stress hybrid quadrilateral finite element method. [13] Kun Wang, Yinnian He, Yueqiang Shang. Fully discrete finite element method for the viscoelastic fluid motion equations. [14] Junjiang Lai, Jianguo Huang. A finite element method for vibration analysis of elastic plate-plate structures. [15] [16] Donald L. Brown, Vasilena Taralova. A multiscale finite element method for Neumann problems in porous microstructures. [17] [18] [19] [20] Impact Factor: Tools Metrics Other articles by authors [Back to Top]
Forgot password? New user? Sign up Existing user? Log in Consider the integral, I=∫1xdx\displaystyle I =\int \dfrac{1}{x}dx I=∫x1dx We know that I=logx I= \log x I=logx But, I=∫1x(1)dx=(x)1x−∫(−1x2)xdx\displaystyle I =\int \dfrac{1}{x}(1)dx =(x)\dfrac{1}{x}-\int \bigg(-\dfrac{1}{x^2} \bigg) xdxI=∫x1(1)dx=(x)x1−∫(−x21)xdx 1+I=I⇒1=01+I=I \Rightarrow 1=01+I=I⇒1=0 What is the mistake? I do not know the mistake Note by Sparsh Sarode 2 years, 9 months ago Easy Math Editor This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: italics _italics_ bold __bold__ - bulleted- list 1. numbered2. list paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org) > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines# 4 spaces, and now they show# up as a code block.print "hello world" \[ \] 2 \times 3 2^{34} a_{i-1} \frac{2}{3} \sqrt{2} \sum_{i=1}^3 \sin \theta \boxed{123} Sort by: you actually got ∫1x−∫1x=1\int \frac{1}{x} -\int \frac{1}{x}=1 ∫x1−∫x1=1. There is absolutely nothing wrong with that. Those are indefinite integrals. Log in to reply What if I add limits from 1 to 2..you still get 1=0 I should have added limits. It is xx∣ab\frac{x}{x} \|_{a}^{b}xx∣ab. That yeilds zero. 1 there has limits. @Kushal Patankar – I didn't quite get u, u meant xx=b−ab−a≠1 \dfrac{x}{x} =\dfrac{b-a}{b-a} \neq 1xx=b−ab−a=1 @Sparsh Sarode – I can't believe you just stated that... @Kushal Patankar – Correct me if I am wrong, isn't 2−12−1=11=1 \dfrac{2-1}{2-1}=\dfrac{1}{1}=12−12−1=11=1 ? @Sparsh Sarode – its bb−aa\frac{b}{b}-\frac{a}{a}bb−aa and still 1∣ab=1a−1b=1−1=01\|_a^b = 1_a-1_b=1-1=01∣ab=1a−1b=1−1=0. @Kushal Patankar – Oops! I can't believe I made such a mistake... Anyways thx.. @Sparsh Sarode – Sure... Glad to help. @Sparsh Sarode – Can you post this as a problem so others can learn from this misconception? @Calvin Lin – Ok, surely @Kushal Patankar – Exactly... Then he should get correctly. How is LHS not equal to 0 Integration constants need not be equal. @Kushal Patankar – K.. I agree on that And what happens without limits?∫dxx=xx+∫dxx\displaystyle \int \frac{dx}{x} =\frac{x}{x}+\displaystyle \int \frac{dx}{x}∫xdx=xx+∫xdx It's still correct.. Integration constants need not be same Oh, right, thank you @Hjalmar Orellana Soto – Welcome.. :) You cannot cancel the integrals on both sides because each function has infinitely many anti-derivatives off by a constant. That's why, for good reason, you should always include "+C" in your answers. Problem Loading... Note Loading... Set Loading...
A Linear Energy Stable Scheme for a Thin Film Model Without Slope Selection 497 Downloads Citations Abstract We present a linear numerical scheme for a model of epitaxial thin film growth without slope selection. The PDE, which is a nonlinear, fourth-order parabolic equation, is the L 2 gradient flow of the energy $\int_{\Omega}( - \frac{1}{2} \ln(1 + \|\nabla\phi\|^{2} ) + \frac{\epsilon^{2}}{2}\|\Delta\phi(\mathbf{x})\|^{2})\,\mathrm{d}\mathbf{x}$. The idea of convex-concave decomposition of the energy functional is applied, which results in a numerical scheme that is unconditionally energy stable, i.e., energy dissipative. The particular decomposition used here places the nonlinear term in the concave part of the energy, in contrast to a previous convexity splitting scheme. As a result, the numerical scheme is fully linear at each time step and unconditionally solvable. Collocation Fourier spectral differentiation is used in the spatial discretization, and the unconditional energy stability is established in the fully discrete setting using a detailed energy estimate. We present numerical simulation results for a sequence of ϵ values ranging from 0.02 to 0.1. In particular, the long time simulations show the −log( t) decay law for the energy and the t 1/2 growth law for the surface roughness, in agreement with theoretical analysis and experimental/numerical observations in earlier works. KeywordsEpitaxial thin film growth Slope selection Energy stability Convexity splitting Fourier collocation spectral Preview Unable to display preview. Download preview PDF. References 1.Ehrlich, G., Hudda, F.G.: Atomic view of surface diffusion: tungsten on tungsten. J. Chem. Phys. 44, 1036–1099 (1966) Google Scholar 2. 3.Eyre, D.J.: Unconditionally gradient stable time marching the Cahn-Hilliard equation. In: Bullard, J.W., Kalia, R., Stoneham, M., Chen, L.Q. (eds.) Computational and Mathematical Models of Microstructural Evolution, vol. 53, pp. 1686–1712. Materials Research Society, Warrendale (1998) Google Scholar 4.Guan, Z., Wang, C., Wise, S.M.: Convergence of a convex splitting scheme for the nonlocal Cahn-Hilliard Equation (in preparation) Google Scholar 5. 6. 7.Kohn, R.V.: Energy-driven pattern formation. In: Sanz-Solé, M., Soria, J., Varona, J.L., Verdera, J. (eds.) Proceedings of the International Congress of Mathematicians, vol. 1, pp. 359–383. European Mathematical Society Publishing House, Madrid (2006) Google Scholar 8. 9. 10. 11. 12.Moldovan, D., Golubovic, L.: Interfacial coarsening dynamics in epitaxial growth with slope selection. Phys. Rev. E 61, 6190–6214 (2000) Google Scholar 13. 14. 15. 16.
I've been stuck on what I'm pretty sure is a simple part of a larger question. It's a cylinder (radius a) spinning in a viscous fluid. It's rotating at rate $\Omega$ .During this question we get that the motion of a uniform axisymetric flow is: $$\displaystyle\frac{\partial u_\theta}{\partial t} = \nu \left ( \frac{\partial ^2u_\theta}{\partial r^2} + \frac{1}{r} \frac{\partial u_\theta}{ \partial r} - \frac{u_\theta}{r^2} \right )$$ We are told that the specific axial angular momentum is given by: $$m = ru_\theta$$ Solving for steady state I then find that: $$\displaystyle u_\theta = \frac{\Omega a^2}{r}$$ I'm then asked to "hence" obtain an expression for the viscous torque (per unit length in the axial direction) on this cylinder neglecting end effects and assuming the flow to be axismmetric (in terms of $\nu$, $\Omega$, $\rho$ and $a$. I can see physically that there must be a toque driving this. I just can't for the life of me figure out what I'm meant to do in the next step (as by being in "steady state" there's no acceleration as such). If I've done anything horribly wrong or you can see the next step help would be appreciated.
I have a problem deriving the conservation of energy from time translation invariance. The invariance of the Lagrangian under infinitesimal time displacements $t \rightarrow t' = t + \epsilon$ can be written as \begin{equation} \delta L = L\left( q(t),\frac{dq(t)}{dt},t\right) - L\left( q(t+ \epsilon),\frac{dq(t+ \epsilon)}{dt},t+\epsilon \right) = 0. \end{equation} Using Taylor series, keeping only first order terms this gives \begin{equation}\rightarrow \delta L =- \frac{\partial L }{\partial q} \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon - \frac{\partial L }{\partial t} \epsilon = 0. \end{equation} Using the Euler-Lagrange equation and assuming that the Lagrangian does not depend explicitly on time we get \begin{equation}\rightarrow \delta L =- \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \right) \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon =0. \end{equation} Which we can write as \begin{equation}\rightarrow \delta L = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \frac{\partial q}{\partial t} \right) \epsilon = - \frac{d}{d t} \left(p \frac{\partial q}{\partial t} \right) \epsilon = 0. \end{equation} But unfortunatly this is not the Hamiltonian. This computation should yield \begin{equation} \rightarrow \frac{d}{dt} \left( p \dot{q} - L \right) = 0. \end{equation} But I can't find no reason why and how the the extra $-L$ should emerge. I can see that this term can be written at the place where it is written because we have $\delta L = - \frac{d L}{dt } \epsilon$ and therefore \begin{equation} \rightarrow \delta L = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \frac{\partial q}{\partial t} \right) \epsilon = - \frac{d L}{dt } \epsilon. \end{equation} And then the desired equation would only say $0-0=0$. Any idea where i did a mistake would be much appreciated. Reiterating pppqqq's answer, your error is right at the beginning where you set $\delta L = 0$. The Lagrangian is not a constant of motion, so this equation is fallacious. Instead, you want $\frac{dL}{dt} = \frac{\partial L}{\partial q} \dot{q} + \frac{\partial L}{\partial \dot{q}}\ddot{q}$ which assumes $\frac{\partial L}{\partial t} = 0$. When you apply the Euler-Lagrange equation, you get $\frac{dL}{dt} = \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}} \dot{q})$ which is just a short algebra step from showing that the Hamiltonian is conserved. Your original derivation simply shows that if the Lagrangian is time-independent and if also it is a constant of motion, then $p \dot{q}$ is also a constant of motion. I) Firstly, we mention that Noether's Theorem (in its original form) concerns a symmetry of the action $S$, not necessarily the Lagrangian $L$. The relevant notion for the Lagrangian is quasi-symmetry, cf. this Phys.SE answer. II) Secondly, we make the assumption that $$\tag{1} \text{The Lagrangian } L=L(q,\dot{q}) \text{ has no }{\it explicit} \text{ time dependence.} $$ We would like to use Noether's theorem to prove that the energy function$^1$ $$\tag{2} h~:=~p_i\dot{q}^i-L,\qquad p_i ~:=~\frac{\partial L}{\partial \dot{q}^i }, $$ is then conserved on-shell $$ \tag{3} \frac{dh}{dt}~\approx~0. $$ Hence we should identify the relevant symmetry. (Here the $\approx$ symbol means equality modulo eom. Observe btw that we will not use eom for the remainder of this answer. This is because the assumptions of Noether's theorem demand that the symmetry holds also for virtual off-shell configurations which violate eom.) III) It is apparent from OP's first equation that he is considering an infinitesimal pure time translation $$ t^{\prime} - t ~=:~\delta t ~=~-\varepsilon, \qquad \text{(horizontal variation)}\tag{A}$$ $$ q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~0, \qquad \text{(no vertical variation)}\tag{B}$$ $$ q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~-\varepsilon\dot{q}^i. \qquad \text{(full variation)}\tag{C}$$ (The words horizontal and vertical refer to translation in the $t$ direction and the $q^i$ directions, respectively). Also note that we have changed the sign in front of $\varepsilon$ for later convenience. A pure time translation (A) is in general not a symmetry of the Lagrangian $$ \delta L ~=~ \frac{dL}{dt}\delta t ~=~ -\varepsilon \frac{dL}{dt}~\neq~0.\tag{D} $$ The full explanation why the pure horizontal transformation (A)-(C) cannot be used to prove energy conservation is given in Section VI below. But first we show two other transformations that do work in the next Sections IV and V. IV) If we change time (A), the values of $q^{i}$ and $\dot{q}^{i}$ will in general also change. In other words, we must introduce a compensating vertical variation (B'), so that the full variation (C') of the generalized positions are zero: $$ t^{\prime} - t ~=:~\delta t ~=~-\varepsilon, \qquad \text{(horizontal variation)}\tag{A'}$$ $$ q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~\varepsilon\dot{q}^i, \qquad \text{(vertical variation)}\tag{B'}$$ $$ q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~0. \qquad \text{(full variation)}\tag{C'}$$ The transformation (A') - (C') is a symmetry of the Lagrangian: $$ \delta L ~=~\frac{\partial L}{\partial q^i }\delta_0 q^i + \frac{\partial L}{\partial \dot{q}^i }\delta_0 \dot{q}^i + \frac{dL}{dt}\delta t ~=~-\varepsilon\frac{\partial L}{\partial t }~=~0,\tag{D'} $$ where we in the last equality used that the Lagrangian $L$ has no explicit time dependence. Using the standard formula mentioned on Wikipedia, the (bare) Noether current (multiplied with $\varepsilon$) becomes the energy (multiplied with $\varepsilon$) $$ \varepsilon j ~:=~ p_i \delta_0 q^i + L \delta t~=~ p_i \delta q^i - h \delta t~=~ \varepsilon h ,\tag{E'}$$ as we wanted to show. V) Alternatively, as is done in Example 1 on Wikipedia, we can consider a purely vertical infinitesimal transformation $$ t^{\prime} - t ~=:~\delta t ~=~0, \qquad \text{(no horizontal variation)}\tag{A''}$$ $$ q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~\varepsilon\dot{q}^i, \qquad \text{(vertical variation)}\tag{B''}$$ $$ q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~\varepsilon\dot{q}^i. \qquad \text{(full variation)}\tag{C''}$$ The transformation (A'') - (C'') is a quasi-symmetry of the Lagrangian: $$ \delta L ~=~\frac{\partial L}{\partial q^i }\delta_0 q^i + \frac{\partial L}{\partial \dot{q}^i }\delta_0 \dot{q}^i ~=~\varepsilon\frac{\partial L}{\partial q^i }\dot{q}^i + \varepsilon\frac{\partial L}{\partial \dot{q}^i } \ddot{q}^i ~=~ \varepsilon\frac{dL}{dt}-\varepsilon\frac{\partial L}{\partial t}~=~ \varepsilon\frac{dL}{dt}, \tag{D''}$$ where we in the last equality used that the Lagrangian $L$ has no explicit time dependence. The (bare) Noether current (multiplied with $\varepsilon$) becomes $$ \varepsilon j ~:=~ p_i \delta_0 q^i + L \delta t~=~ \varepsilon p_i\dot{q}^i.\tag{E''}$$ The Noether current must be corrected because of the appearance of the total time derivative in eq. (D''). The full Noether current becomes the energy function $$ J~=~j-L~=~p_i\dot{q}^i-L~=~h,\tag{F''}$$ as we wanted to show. VI) Finally, let us return to OP's pure horizontal transformation (A)-(C). While not a symmetry, it is still a quasi-symmetry of the Lagrangian $L$, cf. eq. (D). The (bare) Noether current (multiplied with $\varepsilon$) becomes $$ \varepsilon j ~:=~ p_i \delta_0 q^i + L \delta t~=~ -\varepsilon L .\tag{E}$$ The Noether current must be corrected because of the appearance of the total time derivative in eq. (D). The full Noether current becomes zero: $$ J~=~j-(-L)~=~-L+L~=~0.\tag{F}$$ In other words, the corresponding conservation law is a triviality! This is because we never used in eq. (D) the non-trivial fact (1) that the Lagrangian $L$ has no explicit time dependence. -- Here's the right way to understand this (not that I'm biased or anything). Let me begin that I agree with others who point out that $\delta L \neq 0$ in this case, but I'd like to demonstrate why in a convincing manner. Hopefully the way I present the resolution will be clear. I'll be mathematically precise, but I won't worry about certain technical assumptions such as degrees of differentiability of the functions involved. Generalities. So that we can be absolutely sure that there is no confusion, let me review some notation and definitions. Let a path $q:[t_a, t_b]\to \mathbb R$ in configuration space be given. Let $\hat q:[t_a, t_b]\times (\epsilon_a, \epsilon_b)\to \mathbb R$ be a one-parameter deformation of $q$ with $\epsilon_a<0<\epsilon_b$. We define the variation of $q$ and its derivative $\dot q$ with respect to this deformation as follows:\begin{align} \delta q(t) = \frac{\partial \hat q}{\partial\epsilon}(t,0) , \qquad \delta\dot q(t) = \frac{\partial^2\hat q}{\partial \epsilon\partial t}(t,0)\end{align}By the way, to get some intuition for this (and especially my notation), you might find the following post useful: Now, suppose that a lagrangian $L$ that is local in $q$ and $\dot q$ is given, then for a given path $q$ we define its variation with respect to the deformation $\hat q$ as follows:\begin{align} \delta L(q(t), \dot q(t), t) = \frac{\partial}{\partial\epsilon}L\left(\hat q(t,\epsilon), \frac{\partial\hat q}{\partial t}(t,\epsilon), t\right)\Big\|_{\epsilon=0}\end{align}From these two definitions, we find the following expression for the variation of the Lagrangian (where we suppress the arguments of functions for notational compactness)\begin{align} \delta L = \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot q}\delta\dot q\end{align}We call a given deformation a symmetry of $L$ provided there exists a function $F$ that is local in paths $q$ such that\begin{align} \delta L(q(t), \dot q(t), t) = \frac{dF_q}{dt}(t) \end{align}for any $q$. In other words, a symmetry is a deformation that, to first order in the deformation parameter $\epsilon$, only changes the Lagrangian by at most a total time derivative. These definitions allows us to compactly write the following Lagrangian version of Noether's theorem For every symmetry of the Lagrangian, the quantity \begin{align} Q_q(t) = \frac{\partial L}{\partial \dot q}(q(t), \dot q(t), t) \delta q(t) - F_q(t) \end{align} is conserved for all $q$ satisfying the Euler-Lagrange equations. Time translation symmetry. We consider the deformation\begin{align} \hat q(t,\epsilon) = q(t+\epsilon).\end{align}which, of course, we call time translation. Now, a short computation shows that under this deformation, one has the following variations:\begin{align} \delta q(t) = \dot q(t), \qquad \delta \dot q(t) = \ddot q(t)\end{align}It follows that for any Lagrangian (not just one that has time-translation symmetry) a short computation gives\begin{align} \delta L(q(t), \dot q(t) t) = \frac{d}{dt}L(q(t), \dot q(t), t) - \frac{\partial L}{\partial t}(q(t), \dot q(t), t),\end{align}and we immediately get the following result: If $\partial L/\partial t = 0$, then time-translation is a symmetry of $L$ where the function $F$ is simply given by the Lagrangian itself. Noether's theorem then tells us that there is a conserved charge; \begin{align} Q_q(t) = \frac{\partial L}{\partial \dot q}(q(t), \dot q(t), t)\dot q(t) - L(q(t), \dot q(t), t) \end{align} which is precisely the Hamiltonian. I think the problem is in the first line: invariance for finite time displacement is $$L(q,\dot q ,t+h)-L(q,\dot q ,t)=0.$$In the infinitesimal case this should become: $$L (q,\dot q, t+h)-L (q,\dot q,t)=O(h^2) \iff \partial _t L(q,\dot q ,t)=0$$(note that $q$ and $\dot q $ here are not functions of time).With this and Lagrange's equation of motion, you should be able to prove that $H=p\dot q-L$ is conserved along solutions. I'm not sure about what does the term “infinitesimal time displacement” mean. If $g^{\varepsilon}\colon M \to M$ is a one parameter transformation of the configuration space, then the condition $$\dfrac {\partial}{\partial \varepsilon} \|_{\varepsilon =0}L(g^\varepsilon _(\dot q),t)=0,$$ that I believe expresses symmetry under infinitesimal displacement, is different from $$L(g^\varepsilon _(\dot q),t)=L(\dot q, t)$$ which is (according to Arnold) the usual definition of (finite displacement) symmetry. If we look at the special case where $g^\varepsilon$ (which involves a slight generalization of the precedent discourse) is the time translation, then it's obvious that the finite and infinitesimal displacement symmetry conditions are the same. I'll try to answer the question “how can we see energy naturally emerge from time translation symmetry” in the only sense that I can understand it, that is, “can energy be seen as a Noether's charge?”. Alert: the proof is messy. Recall the definition of the Noether's charge associated to a 1 parameter group of symmetries $g^{\varepsilon}$: $$I=\dfrac{\partial L}{\partial \dot q}\dfrac{\partial }{\partial \varepsilon}\|_{\varepsilon = 0}(g^\varepsilon q).$$ Noether's theorem states that $I$ is conserved along solutions if $\partial _\varepsilon \|_{\varepsilon =0}L(g_* ^\varepsilon \dot q)=0$. As it is, the theorem is stated for an autonomous lagrangian, that is, not time dependent Lagrangian. In order to see the energy naturally emerge as a Noether's charge, one approach is indicated in Arnold's book and is as follows. If $M$ is the configuration space and $L$ is the spurious (i.e. non autonomous) Lagrangian, define the generalized configuration space as $M'=M\times \mathbb R $. Define the Lagrangian on $TM'$: $$\tilde L(q,\dot q,\tau ,\dot \tau)=L(q,\frac{\dot q}{\dot \tau},\tau)\dot \tau.$$ If $q\colon \mathbb [\tau _1 ,\tau _2] \to M$ and $\tau \colon [t_1,t_2] \to [\tau _1,\tau _2] $, note that the action: $$\tilde S[q,\tau]=\int _{t_1} ^{t_2}\tilde L(q(\tau(t)),\dot q (\tau (t)),\tau (t),\dot \tau (t))\text d t=\int _{\tau _1}^{\tau _2}L(q(\tau),\dot q (\tau),\tau)\text d \tau=S[q]$$ doesn't depend on $\tau$. So if $q$ is an extremal of $S$, then $(q\circ \tau,\tau)$ is an extremal of $\tilde S$ and satisfies Euler-Lagrange equations. So we can apply Noether's theorem to $\tilde L$. Note that $\partial _\tau \tilde L(q,\dot q ,\tau , \dot \tau)=\partial _\tau L (q,\dot q/\dot \tau,\tau) \dot \tau$, so $\tilde L$ admits time translations if $L$ does. Finally, Noethers charge related to the time translation is: $$\dfrac{\partial \tilde L}{\partial \dot \tau}=L-\dfrac{\partial L}{\partial \dot q}\frac{\dot q}{\dot \tau},$$ that ìis minus the energy. Ok, so from your comments I understand that you already know how to derive Noether's theorem(?), which means that the Noether's current:$$ j = \left( L- \frac{\partial L}{\partial \dot{q}}\dot{q}(t) \right) \epsilon(t) + \frac{\partial L}{\partial \dot{q}} \delta q (t) \tag{1} $$is conserved:$$ \frac{d j }{dt} = 0 $$if the action of a given system is invariant under the following infinitesimal transformations:\begin{equation}t \rightarrow t' = t + \delta t = t + \epsilon (t)\end{equation}\begin{equation}q(t) \rightarrow q'(t')=q(t) + \delta q (t)\end{equation} Now, note that the Hamiltonian is defined as: $$H = \frac{\partial L}{\partial \dot{q}}\dot{q} - L$$ which means that equation $(1)$ can be written as: $$j = - H \epsilon(t) + \frac{\partial L}{\partial \dot{q}} \delta q (t) $$ Now, let us consider a Lagrangian that does not explicitly depend on time, i.e. $L=L(q,\dot{q})$. Subsequently, we consider a time translation:$$t \rightarrow t' = t + \delta t = t + \epsilon$$where $\epsilon$ is a constant (i.e. $\epsilon\neq \epsilon (t)$). If $S$ is invariant ($\delta S = 0$) under time translations, then the Noether current is given by:\begin{equation}j = -H\epsilon\end{equation}(because the path is not affected by a time tranlation, that is $\delta q (t)=0$) and so the Hamiltonian is a constant of motion.
A blog by Sebastian Liem and Erwin Poeze – ViriCiti Labs ViriCiti provides insights in electric, CNG, diesel, hybrid, and hydrogen vehicles, as well as their charging infrastructure. We can, therefore, provide full insight in energy management, maintenance, route operations, and flexible charging for mixed fleets. Currently, we are working on the service ‘Smart Driving’, a tool assisting drivers to drive in an energy efficient, comfortable, and safe way. One of the core inputs of Smart Driving is the measurements of the vehicle acceleration which is done via the DataHub — ViriCiti’s onboard hardware unit — that has a built-in 3D accelerometer. Before the acceleration signals can be useful to us, however, the coordinate system of the DataHub must be aligned with the vehicle. Only then can we interpret the acceleration in the x-direction (driving direction) as braking or accelerating, and only then can we interpret rotation around the z-direction (downward direction) as the vehicle turning. The DataHub is installed by our customers and we at ViriCiti do not necessarily know the orientation of the DataHub in relation to the vehicle. The coordinate systems of the DataHub and the vehicle are therefore not necessarily aligned, as depicted in Figure 1. Figure 1: Vehicle’s coordinate system ($x_v$, $y_v$, $z_v$) is not aligned with that of the DataHub ($x_d$, $y_d$, $z_d$) To overcome this misalignment, we have developed a method to automatically determine the DataHub’s orientation enabling us to align its coordinate system with the vehicle’s. This method has two steps; in the first step, we align the z-axes of the DataHub and vehicle. Secondly, we align the x– and y-axes. The principle is the same for both steps: We measure the acceleration in a known vehicle state where we know forces acting on the vehicle, meaning that we know the acceleration vector in the vehicle’s coordinate system. In the first step, the vehicle must be stationary and positioned on a level surface. In this situation, only gravity affects the vehicle and acceleration is $a_v = (0, 0, g)$ in the vehicle’s coordinate system. As the DataHub is now misaligned, it measures some other acceleration $a_d$. Knowing how the $a_v$ is expressed in the coordinate system of the DataHub, we can find a coordinate transformation that aligns the z-axes of the vehicle’s and DataHub’s coordinate systems. In the z-aligned coordinate system $x’_d, y’_d, z’_d$ (where $z’_d = z_v$), the x- and y-axes can still be misaligned but this is solved in the second step. In this step, the vehicle should be braking while driving in a straight line. We then derive that $a_v = (-x, 0, g)$, which we use to find the coordinate transformation from the z-aligned DataHub’s coordinate system to the vehicle’s coordinate system. Combining the coordinate transformations from step 1 and 2, we have a coordinate transformation that aligns the measurements from the DataHub with the vehicle. This allows us to use the acceleration to describe the movements of the vehicle. In the two following sections, we provide details on how each step works. Figure 2: Step 1, rotation to align the z-axes of the vehicle and DataHub ($z_v$ and $z_d$). Step 2, rotation around aligned z-axis to align the x- and y-axes. Step 1: Using Gravity In this first step, we need to determine when the vehicle is stationary and standing on a level surface. Determining if the vehicle is stationary is quite straightforward as we have direct access to the vehicle speed. Determining if the vehicle is on a level surface, however, is more difficult, as there is no sensor reading readily available. Instead, we rely on statistics taken from many samples — no city is all uphill after all and with a sufficient number of samples the average models a level surface. The measured acceleration $a_d$ in the DataHub’s coordinate system expressed in the vehicle’s coordinate system should be $a_v = (0, 0, g)$. We can now find a transformation matrix $R$ so that $$a_v = R \cdot a_d.$$ We use $R$ to denote the matrix because we know it should be a rotation matrix, it should preserve the origin, the norm, as well as the orientation of the acceleration signal. Note that $R$ doesn’t fully transform the vector from the DataHub’s coordinate system to that of the vehicle. It does, however, align their z-axes or, equivalently, the xy-planes. There is more than one way to construct a rotation matrix – Euler angles and quaternions are two popular approaches. We found these two methods to be unnecessarily complex and numerically error-prone for our purposes. Instead we took a step back and used the theory of rotations: $SO(3)$ the group of 3D Euclidean rotations. We will sketch, with no claims of rigor, the method we settled on. $SO(3)$ is a Lie group which has corresponding Lie algebra $\mathcal{so}(3)$. If we can express our rotation in this algebra we can generate the rotation matrix. To do this we chose the basis $L = [L_x, L_y, L_z]$ for $\mathcal{so}(3)$ $$L_x =\begin{pmatrix}0 & 0 & 0 \\0 & 0 & -1 \\0 & 1 & 0 \\\end{pmatrix},\quad L_y =\begin{pmatrix}0 & 0 & 1 \\0 & 0 & 0 \\-1 & 0 & 0 \\\end{pmatrix},\quad L_z =\begin{pmatrix}0 & -1 & 0 \\1 & 0 & 0 \\0 & 0 & 0 \\\end{pmatrix}$$ With this basis we can identify angle rotations $\theta$ around some unit vector $u$ as element $\theta u \cdot L \in \mathcal{so}(3)$. With this description of the rotation in the Lie algebra we use the exponential map to generate the actual rotation matrix. The map is defined using the matrix exponential series $$\mathcal{so}(3) \to SO(3); \quad \theta u \cdot L \to R = e^{\theta u \cdot L } = I + \theta u \cdot L + \frac{1}{2!}(\theta u \cdot L )^2 + \ldots.$$ The infinite series has an analytical solution because $u \cdot L $ is skew-symmetric, meaning $(u \cdot L )^3 = -u \cdot L $. The higher order terms simplifies to one $u \cdot L $ term and one $(u \cdot L )^2$ term. We get $$R = I + \left[\theta – \frac{\theta^3}{3!} + \frac{\theta^5}{5!} – \ldots\right] u \cdot L+ \left[ {\theta^2}{2!} – {\theta^4}{4!} + {\theta^2}{6!} – \ldots \right](u \cdot L)^2,$$ which, after remembering our trigonometric expansions, let us write $$R = I + \sin \theta u \cdot L + (1 – \cos \theta) (u \cdot L)^2.$$ With this formula we can return to our problem of rotating $a_d$ onto $a_v$. We rotate in the plane spanned by the two vectors, i.e. around the cross-product $v = a_d x a_v$. The $\cos \theta$ and $\sin \theta$ can be found using the geometric meaning of the dot and cross-products respectively. We identify $$u = \frac{v}{\|\|v\|\|}, \quad \cos \theta = \frac{a_v \cdot a_d}{\|\|a_v\|\| \|\|a_d\|\|}, \quad \sin \theta = \frac{\|\|v\|\|}{ \|\|a_v\|\| \|\|a_d\|\|}.$$ And with $a_d$ measured and $a_v$ assumed we have our rotation matrix $R$ which aligns the z-axes of the DataHub and that of the vehicle. Step 2: Braking With the z-axes aligned the next step is to align the x- and y-axes. We do this by identifying a scenario where the acceleration in the xy-plane is known to be only in the x-direction. When the vehicle is braking in a straight line, the acceleration is $a_v = (-a_{x;v}, 0, g)$ in the vehicle coordinate system. Having measured the acceleration in the DataHub’s coordinate system and rotating this acceleration vector to the aligned xy-planes of the vehicle and DataHub during the braking event, we can then find the rotation matrix to completely align the DataHub with its vehicle. To detect a braking event we simply observe if the speed is rapidly diminishing. As it’s braking, the vehicle must maintain the same driving direction (within some tolerance) and to enforce this we use the circular dispersion of the acceleration sampled during the braking event. Only samples that meet a maximum dispersion threshold are accepted so we can be sure that the vehicle is indeed braking in a sufficiently straight line. We combine samples for a number of braking events for our final measurement $a_d$ with which we find the rotation matrix using the method outlined in step 1. Results By composing the rotation matrices from step 1 and step 2 we achieve the alignment of the DataHub and the vehicle. In Figure 3 we see the difference between the acceleration signals of the unaligned DataHub and one that is aligned properly. As one would expect, the x– and y components of the acceleration (blue and orange) are nearly zero when the vehicle is stationary, between timesteps 750 and 1100. Figure 3: x, y, z acceleration in the coordinate system of the DataHub (left) and the same signals transformed to the vehicle coordinate system (right). The process of DataHub alignment is fully automated and currently runs on a selection of vehicles. In the near future we will roll out this new feature to all vehicles equipped with a DataHub and the users will have access to the acceleration signals, useful for, e.g., brake tests of new buses. In the meantime, we continue to develop Smart Driving using the acceleration signals.
In Brian Day's thesis, he gives a definition of a pro-monoidal structure based on a particular motivating example: Suppose $M:\cal{A}\hookrightarrow \cal{B}$ is the inclusion of a small full subcategory, and suppose $\cal{B}$ is equipped with a monoidal structure $1\in \cal{B}$ and $\otimes:\cal{B} \times \cal{B} \to \cal{B}$. Summarizing the discussion, he essentially gives a condition for the monoidal structure on $\cal{B}$ to restrict to a pro-monoidal structure on $\cal{A}$, which amounts to showing that the four induced maps on coends: $$\int^X B(MX,1) \times B(M(-), MX\otimes MA) \\ \downarrow \\ \int^Y B(Y,1) \times B(M(-), Y\otimes MA) $$ $$\int^X B(MX,1) \times B(M(-), MA\otimes M(X)) \\ \downarrow \\ \int^Y B(Y,1) \times B(M(-), MA\otimes Y) $$ $$\int^X B(MX,MA\otimes MA') \times B(M(-), MX\otimes MA'') \\ \downarrow \\ \int^Y B(Y,MA\otimes MA') \times B(M(-), Y\otimes MA'') $$ $$\int^X B(MX,MA'\otimes MA'') \times B(M(-), MA\otimes MX) \\ \downarrow \\ \int^Y B(Y,MA'\otimes MA'') \times B(M(-), MA\otimes Y) $$ are isomorphisms for all $A,A',A''\in \cal{A}$. (I've used the opposite variance for the Yoneda reduction to the one in Day's paper). Morally this seems to say that when we restrict to tensor products of objects of $\cal{A}$, it is enough to compute the coend over $\cal{A}$ rather than all of $\cal{B}$. In the situation that I care about, $\cal{A}$ is actually also dense in $\cal{B}$, the monoidal structure on $\cal{B}$ is biclosed, and the unit of the monoidal structure lies in $\cal{A}$ and is the terminal object of both $\cal{A}$ and $\cal{B}$, so the first two maps seem to end up working out to be isomorphisms pretty easily (I haven't proven it, but I've convinced myself that it works). I have two questions then: 1.) If $\cal{A}$ is also dense in $\cal{B}$, can these conditions be simplified in any way? What about if any of the other conditions relevant to my case of interest hold ($\cal{B}$ biclosed or $1\in \cal{A}$ and terminal in both categories)? 2.) Are there any examples in the literature where the existence of the induced pro-monoidal structure on $\cal{A}$ has been proven by using these coend maps or a related technique (specifically in the nontrivial case where $\cal{B}$ is not just the category of presheaves on $\cal{A}$ and $\cal{A}$'s induced pro-monoidal structure is not representable).
(Disclaimer : I know very well that $SO(N)$ has a Lie algebra of dimension $N(N-1)/2$ etc. This absolutely not the point of my question.) To make my problem more understandable, I start with the example of $SO(2)$. All $SO(2)$ matrices $M$ can be written as ($\theta\in [0,2\pi[$) $$ M=\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}. $$ Using the basis of $2\times2$ real matrices $\sigma_0=\begin{pmatrix}1 & 0\\0&1\end{pmatrix}$, $\sigma_1=\begin{pmatrix}0 & 1\\1&0\end{pmatrix}$, $\sigma_2=\begin{pmatrix}0 & 1\\-1&0\end{pmatrix}$, $\sigma_3=\begin{pmatrix}1 & 0\\0&-1\end{pmatrix}$, one find that $$M=\cos\theta\;\sigma_0+\sin\theta\;\sigma_2.$$ Clearly, $M$ does not have components along $\sigma_1$ and $\sigma_3$, so the dimension of the smallest linear subspace of $\mathrm{M}_2(\mathbb{R})$ that contains $SO(2)$ is $2$. How to articulate the reasoning (for the cases $N>2$ in particular) is not completely clear. I guess that we can say that the component along $\sigma_0$ and $\sigma_2$ are independent because $\cos\theta$ and $\sin\theta$ are independent functions (in a functional analysis sense). Assuming that made sense, we can try to increase $N$. For example, an $SO(3)$ matrix can be written as $$ M=\left(\begin{matrix} \cos\varphi\cos\psi - \cos\theta\sin\varphi\sin\psi & -\cos\varphi\sin\psi - \cos\theta\sin\varphi\cos\psi & \sin\varphi\sin\theta\\ \sin\varphi\cos\psi + \cos\theta\cos\varphi\sin\psi & -\sin\varphi\sin\psi + \cos\theta\cos\varphi\cos\psi & -\cos\varphi\sin\theta\\ \sin\psi\sin\theta & \cos\psi\sin\theta & \cos\theta \end{matrix}\right)\, $$ with $(\phi,\psi)\in [0,2\pi[^2$ and $\theta\in [0,\pi[$. Now, if I look at each matrix element one by one, they are all independent in a functional sense.$^$ Does that mean that the ''dimension of the matrix space'' that $SO(3)$ matrices live in is $9$ ? Is there a way to generalize that to arbitrary $N$ ? In the end, does any of what I wrote above make any sense ? $^$ It is slightly more subtle than that for the $\theta$ dependence, because in the end I am interested in doing integral over the Haar measure, which means that one should look at $x=\cos(\theta)\in[-1,1]$. But $x$ and $\sqrt{1-x^2}$ are orthogonal, so all should be fine.
I am trying to understand the structure of the fermions in non-abelian gauge theories. Disclaimer: my question might be very trivial (I suspect the answer could simply be "a change of basis"), but I would be grateful is someone could shed some light if there's something deeper lurking around the corner. Let's consider the Yang-Mills Lagrangian $$ \mathcal{L} = -\frac{1}{4} (F_{\mu \nu}^a)^2 + \bar{\psi}(i \gamma^\mu D_\mu -m)\psi $$ where $ D_\mu = \mathbf{1} \partial_\mu - i g A_\mu^a t^a$, then $\psi$ needs to have both Dirac ($\mu$) and colour (a) degrees of freedom. I get confused when I change the regular approach to the problem and I am not sure if it is a real issue or if I'm just overcomplicating/overthinking this. For instance, let's consider $SU(N)$ YM where the generators $t^a$ are $N \times N$ matrices, and there are $N^2-1$ of them. Therefore, the covariant derivative $D_\mu$ is an $N \times N$ matrix, and the index $a$ above runs over $N^2-1$ value. When we start by contracting $D_\mu$ with the $\gamma$-matrices, $$ (\gamma^\mu D_\mu)_{ij} = \delta_{ij} \gamma^\mu \partial_\mu - i g (\gamma^\mu A_\mu^a) (t^a)_{ij} $$ one gets an $N \times N$ matrix of $4 \times 4$ matrices. The corresponding $4N$-component object this matrix acts upon is the $N$ Dirac spinors arranged in a column. However, notice that if we start in the following way instead: $$ (D_\mu \psi)_i = \partial_\mu \psi_i - i g A_\mu^a (t^a)_{ij} \psi_j $$ we get that the covariant derivative acts on $\Psi \equiv (\psi_1, \cdots, \psi_N)$. For all we know, $\Psi$ has no spinor structure since we haven't contracted with the gamma matrices yet. Contracting with the $\gamma^\mu$, one gets a $4 \times 4$ matrix of $N \times N$ matrices that encodes the same information as before. This time, it would seem that we only have one Dirac spinor, where each component is a $N$-valued singlet. However, it seems that the matrix $\gamma^\mu D_\mu$ and the $4N$-component spinor look different, even though all we did was to change the order in which we constructed things. In the first case, we get $N$ spinors corresponding to the $N$ colors of the adjoint representation. In $SU(3)$, this would be like saying we effectively have 3 spinors which correspond to the red/blue/green colors as $(\psi_R, \psi_B, \psi_G)$. In the second case, this identification fails since we only have one big complicated object. What went wrong? Is this difference simply a change of basis for $\psi$? Is there something relevant we can learn from looking at the YM Lagrangian in these two different ways? Also, in the first case, when we get $N$ spinors, I am confused about their signification. I always assumed that in QCD, $\psi$ would correspond to a quark, which is a fermion by itself. Does it mean that quarks are fermions that can be described by fermionic fields/degrees of freedom that we call color?
Let us consider to the example of a mass on a spring. We now examine the case of forced oscillations, which we did not yet handle. That is, we consider the equation \[ mx'' + cx' + kx = F(t)\] for some nonzero $F(t) $. The setup is again: $m$ is mass, $c$ is friction, $k$ is the spring constant, and $F(t)$ is an external force acting on the mass. What we are interested in is periodic forcing, such as noncentered rotating parts, or perhaps loud sounds, or other sources of periodic force. Once we learn about Fourier series in Chapter 4, we will see that we cover all periodic functions by simply considering $F(t) = F_0 \cos (\omega t)$ (or sine instead of cosine, the calculations are essentially the same). 2.6.1 Undamped forced motion and resonance First let us consider undamped $c = 0$ motion for simplicity. We have the equation \[ mx'' + kx = F_0 \cos (\omega t)\] This equation has the complementary solution (solution to the associated homogeneous equation) \[x_c = C_1 \cos ( \omega_0t) + C_2 \sin (\omega_0t)\] where $\omega_0 = \sqrt { \frac {k}{m}}$ is the natural frequency (angular), which is the frequency at which the system “wants to oscillate” without external interference. Let us suppose that $\omega_0 \neq \omega $. We try the solution $x_p = A \cos (\omega t) $ and solve for $A$. Note that we need not have sine in our trial solution as on the left hand side we will only get cosines anyway. If you include a sine it is fine; you will find that its coefficient will be zero. We solve using the method of undetermined coefficients. We find that \[x_p = \dfrac {F_0}{m(\omega^2_0 - \omega^2)} \cos (\omega t) \] We leave it as an exercise to do the algebra required. The general solution is \[ x = C_1 \cos (\omega_0t) + C_2 \sin (\omega_0t) + \frac {F_0}{m(\omega^2_0 - \omega^2)} \cos (\omega t) \] or written another way \[ x = C \cos (\omega_0t - y ) + \frac {F_0}{m(\omega^2_0 - \omega^2)} \cos (\omega t) \] Hence it is a superposition of two cosine waves at different frequencies. Example $\PageIndex{1}$: Take \[ 0.5 x'' + 8 x = 10 \cos (\pi t), ~~~ x(0) = 0, ~~~ x' (0) = 0 \] Let us compute. First we read off the parameters: $ \omega = \pi, \omega_0 = \sqrt { \frac {8}{0.5}} = 4, F_0 = 10, m = 0.5 $. The general solution is \[ x = C_1 \cos (4t) + C_2 \sin (4t) + \frac {20}{16 - {\pi }^2} \cos ( \pi t) \] Solve for $C_1$ and $C_2$ using the initial conditions. It is easy to see that $ C_1 = \frac {-20}{16 - {\pi}^2}$ and $ C_2 = 0 $. Hence \[ x = \frac {20}{16 - {\pi}^2} ( \cos (\pi t) - \cos ( 4t ) ) \] Notice the “beating” behavior in Figure 2.5. First use the trigonometric identity \[ 2 \sin ( \frac {A - B}{2}) \sin ( \frac {A + B}{2} ) = \cos B - \cos A \] to get that \[ x = \frac {20}{16 - {\pi}^2} ( 2 \sin ( \frac {4 - \pi}{2}t) \sin ( \frac {4 + \pi}{2} t)) \] Notice that $x$ is a high frequency wave modulated by a low frequency wave. Now suppose that $ \omega_0 = \omega $. Obviously, we cannot try the solution $ A \cos (\omega t) $ and then use the method of undetermined coefficients. We notice that $ \cos (\omega t) $ solves the associated homogeneous equation. Therefore, we need to try $ x_p = At \cos (\omega t) + Bt \sin (\omega t) $. This time we do need the sine term since the second derivative of $ t \cos (\omega t) $ does contain sines. We write the equation \[ x'' + \omega^2 x = \frac {F_0}{m} \cos (\omega t) \] Plugging $ x_p$ into the left hand side we get \[ 2B \omega \cos (\omega t) - 2A \omega \sin (\omega t) = \frac {F_0}{m} \cos (\omega t) \] Hence $ A = 0 $ and $ B = \frac {F_0}{2m \omega } $. Our particular solution is $ \frac {F_0}{2m \omega } t \sin (\omega t) $ and our general solution is \[ x = C_1 \cos (\omega t) + C_2 \sin (\omega t) + \frac {F_0}{2m \omega } t \sin (\omega t) \] The important term is the last one (the particular solution we found). We can see that this term grows without bound as $ t \rightarrow \infty $. In fact it oscillates between $ \frac {F_0t}{2m \omega } $ and $ \frac {-F_0t}{2m \omega } $. The first two terms only oscillate between $ \pm \sqrt { C^2_1 + C^2_2} $, which becomes smaller and smaller in proportion to the oscillations of the last term as $t$ gets larger. In Figure 2.6 we see the graph with $C_1 = C_2 = 0, F_0 = 2, m = 1, \omega = \pi $. By forcing the system in just the right frequency we produce very wild oscillations. This kind of behavior is called resonance or perhaps pure resonance. Sometimes resonance is desired. For example, remember when as a kid you could start swinging by just moving back and forth on the swing seat in the “correct frequency”? You were trying to achieve resonance. The force of each one of your moves was small, but after a while it produced large swings. On the other hand resonance can be destructive. In an earthquake some buildings collapse while others may be relatively undamaged. This is due to different buildings having different resonance frequencies. So figuring out the resonance frequency can be very important. A common (but wrong) example of destructive force of resonance is the Tacoma Narrows bridge failure. It turns out there was a different phenomenon at play 1. 2.6.2 Damped forced motion and practical resonance In real life things are not as simple as they were above. There is, of course, some damping. Our equation becomes \[ mx'' + cx' + kx = F_0 \cos (\omega t), \] for some $ c > 0 $. We have solved the homogeneous problem before. We let \[ p = \frac {c}{2m} ~~~ \omega_0 = \sqrt { \frac {k}{m} }\] We replace equation (2.6.15) with \[ x'' + 2px' + \omega^2_0x = \frac {F_0}{m} \cos (\omega t) \] The roots of the characteristic equation of the associated homogeneous problem are $r_1, r_2 = -p \pm \sqrt {p^2 - \omega_0^2} $. The form of the general solution of the associated homogeneous equation depends on the sign of $ p^2 - \omega^2_0 $, or equivalently on the sign of $ c^2 - 4km $, as we have seen before. That is, \[ x_c = \begin {cases} C_1e^{r_1t} + C_2e^{r_2t} & if c^2 > 4km, \\ C_1e^{pt} + C_2te^{-pt} & if c^2 = 4km, \\ e^{-pt} ( C_1 \cos (\omega_1t) + C_2 \sin (\omega_1t)) & if c^2 < 4km, \end {cases} \] where $ \omega_1 = \sqrt {\omega^2_0 - p^2 } $. In any case, we can see that $ x_c(t) \rightarrow 0 $ as $ t \rightarrow \infty $. Furthermore, there can be no conflicts when trying to solve for the undetermined coefficients by trying $ x_p = A \cos (\omega t) + B \sin (\omega t) $. Let us plug in and solve for $ A$ and $B$. We get (the tedious details are left to reader) \[ ((\omega^2_0 - \omega^2) B - 2 \omega pA ) \sin (\omega t) + ((\omega^2_0 - \omega^2) A + 2 \omega pB ) \cos (\omega t) = \frac {F_0}{m} \cos (\omega t) \] We get that \[ A = \frac { (\omega^2_0 - \omega^2) F_0}{m{(2 \omega p)}^2 + m{(\omega^2_0 - \omega^2)}^2} \] \[ B = \frac { 2 \omega pF_0}{m{(2 \omega p)}^2 + m{(\omega^2_0 - \omega^2)}^2} \] \[ C = \frac {F_0}{m \sqrt { {(2 \omega p)}^2 + {(\omega^2_0 - \omega^2)}^2}} \] Thus our particular solution is \[ C = \frac {F_0}{m \sqrt { {(2 \omega p)}^2 + {(\omega^2_0 - \omega^2)}^2}} \] \[ x_P = \frac {(\omega^2_0 - \omega^2)F_0}{m {(2 \omega p)}^2 + m {(\omega^2_0 - \omega^2)}^2} \cos (\omega t) + \frac {2 \omega pF_0}{m {(2 \omega p)}^2 + m{(\omega^2_0 - \omega^2)}^2} \sin (\omega t) \] Or in the alternative notation we have amplitude $ C$ and phase shift $ \gamma $ where (if $ \omega \ne \omega_0 $) \[ \tan \gamma = \frac {B}{A} = \frac {2 \omega p}{\omega^2_0 - \omega^2} \] Hence we have \[ x_p = \frac {F_0}{m \sqrt { {(2 \omega p)}^2 + {(\omega^2_0 - \omega ^2)}^2}} \cos (\omega t - \gamma) \] If $ \omega = \omega_0$ we see that $ A = 0, B = C = \frac {F_0}{2m \omega p}, ~\rm{and} ~ \gamma = \frac {\pi}{2} $. The exact formula is not as important as the idea. Do not memorize the above formula, you should instead remember the ideas involved. For different forcing function $ F$, you will get a different formula for $ x_p$. So there is no point in memorizing this specific formula. You can always recompute it later or look it up if you really need it. For reasons we will explain in a moment, we call $x_c$the transient solution and denote it by $ x_{tr} $. We call the $ x_p$ we found above the steady periodic solution and denote it by $ x_{sp}$. The general solution to our problem is \[ x = x_c + x_p = x_{tr} + x_ {sp} \] We note that $ x_c = x_{tr} $ goes to zero as $ t \rightarrow \infty $, as all the terms involve an exponential with a negative exponent. Hence for large $t$, the effect of $ x_{tr} $ is negligible and we will essentially only see $x_{sp}$. Hence the name transient. Notice that $ x_{sp}$ involves no arbitrary constants, and the initial conditions will only affect $x_{tr} $. This means that the effect of the initial conditions will be negligible after some period of time. Because of this behavior, we might as well focus on the steady periodic solution and ignore the transient solution. See Figure 2.7 for a graph of different initial conditions. Notice that the speed at which $x_{tr}$ goes to zero depends on $P$ (and hence $c$). The bigger $P$ is (the bigger $c$ is), the “faster” $x_{tr}$ becomes negligible. So the smaller the damping, the longer the “transient region.” This agrees with the observation that when $ c = 0 $, the initial conditions affect the behavior for all time (i.e. an infinite “transient region”). Let us describe what we mean by resonance when damping is present. Since there were no conflicts when solving with undetermined coefficient, there is no term that goes to infinity. What we will look at however is the maximum value of the amplitude of the steady periodic solution. Let $C$ be the amplitude of $x_{sp}$. If we plot $C$ as a function of $\omega $ (with all other parameters fixed) we can find its maximum. We call the $\omega $ that achieves this maximum the practical resonance frequency. We call the maximal amplitude $C(\omega )$ the practical resonance amplitude. Thus when damping is present we talk of practical resonance rather than pure resonance. A sample plot for three different values of $c$ is given in Figure 2.8. As you can see the practical resonance amplitude grows as damping gets smaller, and any practical resonance can disappear when damping is large. To find the maximum we need to find the derivative $ C' (\omega ) $. Computation shows \[ C' (\omega ) = \frac {-4 \omega (2p^2 + \omega^2 - \omega^2_0)F_0}{m {( {(2 \omega p)}^2 + {(\omega^2_0 - \omega^2)})}^{3/2}} \] This is zero either when $ \omega = 0 $ or when $ 2p^2 + \omega^2 - \omega^2_0 = 0 $. In other words, $ C' (\omega ) = 0 $ when \[ \omega = \sqrt { \omega^2_0 - 2p^2} \rm{~or~} \omega = 0 \] It can be shown that if $ \omega^2_0 - 2p^2 $ is positive, then $ \sqrt {\omega^2_0 - 2p^2} $ is the practical resonance frequency (that is the point where $ C(\omega ) $ is maximal, note that in this case $ C' (\omega ) > 0 $ for small $\omega $). If $\omega = 0 $ is the maximum, then essentially there is no practical resonance since we assume that $ \omega > 0 $ in our system. In this case the amplitude gets larger as the forcing frequency gets smaller. If practical resonance occurs, the frequency is smaller than $ \omega_0$. As the damping $c$ (and hence $P$) becomes smaller, the practical resonance frequency goes to $ \omega_0$. So when damping is very small, $ \omega_0$ is a good estimate of the resonance frequency. This behavior agrees with the observation that when $ c = 0 $, then $ \omega_0$ is the resonance frequency. The behavior is more complicated if the forcing function is not an exact cosine wave, but for example a square wave. It will be good to come back to this section once we have learned about the Fourier series.
Volumes of Revolution Suppose you wanted to make a clay vase. It is made by shaping the clay into a curve and spinning it along an axis. If we want to determine how much water it will hold, we can consider the cross sections that are perpendicular to the axis of rotation, and add up all the volumes of the small cross sections. We have the following definition: \[ \text{Volume} = \int_a^b A(x) dx \] where $A(x)$ is the area the cross section at a point $x$. Example 1 Find the volume of the solid that is produced when the region bounded by the curve \[ y = x^2, \; y = 0\; ,\text{ and } x = 2\] is revolved around the x-axis. Solution Since we are revolving around the x-axis, we have that the cross section is in the shape of a disk with radius equal to the y-coordinate of the point. Hence \[ A(x) = \pi r^2 = \pi (x^2)^2. \] We have \[ \begin{align} \text{ Volume} &= \int_{0}^{2} \pi (x^2)^2 dx \\ &= \left( \dfrac{\pi}{5}x^5 \right]_{0}^{2} \\ &= \dfrac{32 \pi}{5}. \end{align}\] Example 2 Find the volume of the solid formed be revolving the region between the curves \[ y = x^2 \text{ and } y = \sqrt{x}\] about the x-axis. Solution We draw the picture and revolve a cross section about the x-axis and come up with a washer. The area of the washer is equal to the area of the outer disk minus the area of the inner disk. \[ A = \pi (R^2 - r^2) \] We have that $R$ is the y-coordinate of the top curve and $r$ is the y-coordinate of the bottom curve $ x^2$. We have \[ A = \pi([\sqrt{x}]^2 - [x^2]^2) = \pi[x - x^4] \] Hence \[ \begin{align} Volume &= \int _0^1 \pi \left( x-x^2 \right) \,dx \\ &= \pi \left[\dfrac{x^2}{2} - \dfrac{x^3}{3} \right]_0^1 \\ &= \dfrac{\pi}{6}. \end{align} \] Example 3: Revolving about the y-axis Find the volume of the solid that is formed by revolving the curve bounded by \[ y = x \text{ and } y = \sqrt{x}. \] about the y-axis. Solution This time our cross section is perpendicular to the y-axis. When we revolve, we get a washer with $R$ equal to the x-coordinate of the$ y = \sqrt{x} $ curve and $r$ equal to the x-coordinate of the $y = x$ curve. Hence \[ A = \pi\left(\left(y\right)^2 - \left(y^2\right)^2\right) = \pi \left(y^2 - y^4\right). \] We get \[ \begin{align} \text{Volume} &= \int_0^1 \pi(y^2-y^4) dy \\ &= \pi\left(\dfrac{y^3}{3}-\dfrac{y^5}{5}\right]_0^1\\ &= \dfrac{2\pi}{15}. \end{align} \] Revolving About a Non-axis Line Example 4 Find the volume of the region formed by revolving the curve \[y=x^3 \;\;\; 0<x<2 \] about the line $y=-2 $ Solution This time we revolve the cross-section and obtain a disk. The radius of the disk is 2 plus the y-coordinate of the curve. Hence \[ A = \pi (2 + x^3)^2 \] so that \[ \text{Volume} = \int _0^2 \pi (2 + x^3)^2 \, dx. \] This integral can be evaluated by FOIL-ing out the binomial and then integrating each monomial. We get a value of approximately 133. Example 5 Try revolving the curve \[ y =x^2 \] from 0 to 2 about the line $ x = 5 $. We have \[ A = \pi \left[\left(5 - \sqrt{y} \right)^2 - 9\right] \] so that \[ \text{Volume} = \int_0^4 \pi \left[ \left( 5 - \sqrt{y} \right)^2 -9 \right] dy. \] This integral works out to be approximately 59. Applications of Volume Example 6: The Volume of the Khufu Pyramid The base of the Khufu pyramid is a square with wide length 736 feet and the angle that the base makes with the ground is 50.8597 degrees. Find the volume of the Khufu pyramid. Solution First note that we need to change to radian measure. \[ 50.8597^{\circ} = 0.88767 \text{ radians}\] The height of the pyramid is \[ 736 \tan(0.88767) = 904.348.\] We have that the area of a cross section is $s^2$where $s$ is the side length of the square. Placing the y-axis through the top of the pyramid and the origin at the middle of the base, we have that \[ s = \sqrt{2} x. \] Hence \[ A(x) = 2x^2. \] Since the axis perpendicular to the cross sections is the y-axis, we need $A$ in terms of $y$. We set up similar triangles: \[ \dfrac{x}{736} = \dfrac{y}{904.348}\] \[ x = 0.8138\, y. \] Hence \[ A(y) = 1.3247\, x^2 \] We calculate \[ \int _0^{904.35} 1.3247\, x^2\, dx \] \[ = 361,131\; \text{ cubic feet}. \] Example 7 A sphere is formed by rotation the curve \[ y= \sqrt{r^2 - x^2}. \] We have \[ \begin{align} \text{Volume} &= \int_{-r}^{r} \pi (\sqrt{r^2-x^2})^2 dx \\ &= \int_{-r}^{r} \pi (r^2-x^2) dx \\ &= \pi \left( r^2x-\dfrac{x^2}{3} \right]_{-r}^r \\ &= \pi \Big[ \big( r^3-\dfrac{r^3}{3}\big) - \big( -r^3+\dfrac{r^3}{3} \big) \Big] \\ &= \dfrac{4}{3} \pi r^3\end{align}.\] Exercise Two cylinders of radius $r$ intersect each other at right angles. Find the volume of their intersection. Hint: Consider cross sections parallel to both axes of rotation. These cross sections are squares. Then show that the side length is $ 2\sqrt{r^2 - x^2} $ Larry Green (Lake Tahoe Community College) Integrated by Justin Marshall.
I have this confusion about Armijo rule used in line search. I was reading back tracking line search but didn't get what this Armijo rule is all about. Can anyone elaborate what Armijo rule is? The wikipedia doesn't seem to explain well. Thanks Once you obtain a descent direction $p$ for your objective function $f(x)$, you need to pick a "good" step length. You don't want to take a step that is too large such that the function at your new point is larger than your current point. At the same time, you don't want to make your step too small such that it takes forever to get to converge. Armijo's condition basically suggests that a "good" step length is such that you have "sufficient decrease" in $f$ at your new point. The condition is mathematically stated as $$f(x_k+\alpha p_k)\leq f(x_k)+\beta\alpha\nabla f(x_k)^Tp_k$$ where $p_k$ is a descent direction at $x_k$ and $\beta\in(0,1)$. The intuition behind this is that the function value at the new point $f(x_k+\alpha p_k)$ should be under the reduced "tangent line" at $x_k$ in the direction of $p_k$. See Nocedal & Wright's book "Numerical Optimization". In chapter 3, there's an excellent graphical description of armijo's sufficient decrease condition. Five years later, this question is still valid. Here (pages 16 and 17) you can find a great explanation, including an Algorithm.
Datum in ura / Date and time: 21.3.19 (10:00-11:00) Predavalnica / Location: FAMNIT-MP4 Predavatelj / Lecturer: Slobodan Filipovski (University of Primorska) Naslov / Title: A connection between a question of Bermond and Bollobas and Ramanujan graphs Vsebina / Abstract: If we let n(k, d) denote the order of the largest undirected graphs of maximum degree k and diameter d, and let M(k,d) denote the corresponding Moore bound, then n(k,d)\leq M(k,d), for all k\geq 3, d\geq 2. Whilethe inequality has been proved strict for all but very few pairs k and d, the exact relation between the values n(k,d) and M(k,d) is unknown, and the uncertainty of the situation is captured by an open question of Bermond and Bollobas who asked whether it is true that for any positive integer c>0 there exist a pair k and d, such that n(k,d)\leq M(k,d)-c. In this talk we present a connection of this question to the value 2\sqrt{k-1}, which is also essential in the definition of the Ramanujan graphs defined as k-regular graphs whose second largest eigenvalue (in modulus) does not exceed 2\sqrt{k-1}. We further reinforce this surprising connection by showing that if the answer to the question of Bermond and Bollobas were negative and there existed a c > 0 such that n(k,d) \geq M(k,d) - c, for all k\geq 3, d\geq 2, then, for any fixed k and all sufficiently large even d's, the largest undirected graphs of degree k and diameter d would have to be Ramanujan graphs. This would imply a positive answer to the open question whether infinitely many non-bipartite k-regular Ramanujan graphs exist for any degree k.
Acta Mathematica Acta Math. Volume 204, Number 2 (2010), 273-292. Astala’s conjecture on distortion of Hausdorff measures under quasiconformal maps in the plane Abstract Let $ E \subset \mathbb{C} $ be a compact set, $ g:\mathbb{C} \to \mathbb{C} $ be a K-quasiconformal map, and let 0 < t < 2. Let $ {\mathcal{H}^t} $ denote t-dimensional Hausdorff measure. Then$ {\mathcal{H}^t}(E) = 0\quad \Rightarrow \quad {\mathcal{H}^{t'}}\left( {gE} \right) = 0,\quad t' = \frac{{2Kt}}{{2 + \left( {K - 1} \right)t}}. $ This is a refinement of a set of inequalities on the distortion of Hausdorff dimensions by quasiconformal maps proved by K. Astala in [2] and answers in the positive a conjecture of K. Astala in op. cit. Note M.T. Lacey was supported in part by a grant from the NSF. E. T. Sawyer was supported in part by a grant from the NSERC. Article information Source Acta Math., Volume 204, Number 2 (2010), 273-292. Dates Received: 11 June 2008 First available in Project Euclid: 31 January 2017 Permanent link to this document https://projecteuclid.org/euclid.acta/1485892470 Digital Object Identifier doi:10.1007/s11511-010-0048-5 Mathematical Reviews number (MathSciNet) MR2653055 Zentralblatt MATH identifier 1211.30036 Rights 2010 © Institut Mittag-Leffler Citation Lacey, Michael T.; Sawyer, Eric T.; Uriarte-Tuero, Ignacio. Astala’s conjecture on distortion of Hausdorff measures under quasiconformal maps in the plane. Acta Math. 204 (2010), no. 2, 273--292. doi:10.1007/s11511-010-0048-5. https://projecteuclid.org/euclid.acta/1485892470
Cistern is also known as tank / vessel / sometimes reservoir. Outlet: The pipe connected to the cistern, which helps to empty the tank is called an outlet. Inlet: The pipe which is connected with the tank, that fills it is called an inlet. NOTE: Very important point is that the inlet is always positive since it fills the tank and outlet is negative work since it empties the tank. Pipes and Cisterns are somewhat similar to the concepts of Work and Wages: The problems of pipes and cisterns usually have two kinds of pipes, Inlet pipe and Outlet pipe / Leak. Inlet pipe is the pipe that fills the tank / reservoir / cistern and Outlet pipe / Leak is the one that empties it. If a pipe can fill a tank in ‘n’ hours, then in 1 hour, it will fill ‘$\frac{1}{n}$‘ parts. For example, if a pipe takes 6 hours to fill a tank completely, say of 12 liters, then in 1 hour, it will fill $(\frac{1}{6})^{th}$ of the tank, i.e., 2 liters. If a pipe can empty a tank in ‘n’ hours, then in 1 hour, it will empty ‘$\frac{1}{n}$‘ parts. For example, if a pipe takes 6 hours to empty a tank completely, say of 18 liters, then in 1 hour, it will empty $(\frac{1}{6})^{th}$ of the tank, i.e., 3 liters. If we have a number of pipes such that some fill the tank and some empty it, and we open all of them together, then in one hour, part of the tank filled/emptied = $\sum(\frac{1}{m_{i}})$ – $\sum(\frac{1}{n_{j}})$ where ‘$m_{i}$‘ is the time taken by inlet pipe ‘i’ to fill the tank completely if only it were open and ‘$n_{j}$‘ is the time taken by outlet pipe ‘j’ to empty the tank completely if only it were open. If the sign of this equation is positive, the tank would be filled and if the sign is negative, the tank would be emptied. Example 1: Two pipes A and B can fill a tank separately in 12 and 16 hours respectively. If both of them are opened together when the tank is initially empty, how much time will it take to completely fill the tank? Solution: Method 1: Part of tank filled by pipe A in one hour working alone = $\frac{1}{12}$ Part of tank filled by pipe B in one hour working alone = $\frac{1}{16}$ ⇒ Part of tank filled by pipe A and pipe B in one hour working together = ($\frac{1}{12}$) + ($\frac{1}{16}$) = $\frac{7}{48}$ Therefore, time taken to completely fill the tank if both A and B work together = $\frac{48}{7}$ hours Method 2: Let the capacity of tank be LCM (12, 16) = 48 units ⇒ Efficiency of pipe A = $\frac{48}{12}$ = 4 units / hour ⇒ Efficiency of pipe B = $\frac{48}{16}$ = 3 units / hour ⇒ Combined efficiency of pipes A and B = 7 units / hour Therefore, time taken to completely fill the tank = $\frac{48}{7}$ hours Example 2: Three pipes A, B and C are connected to a tank. Out of the three, A is the inlet pipe and B and C are the outlet pipes. If opened separately, A fills the tank in 10 hours, B empties the tank in 12 hours and C empties the tank in 30 hours. If all three are opened simultaneously, how much time does it take to fill / empty the tank? Solution: Method 1: Part of tank filled by pipe A in one hour working alone = $\frac{1}{10}$ Part of tank emptied by pipe B in one hour working alone = $\frac{1}{12}$ Part of tank emptied by pipe B in one hour working alone = $\frac{1}{30}$ ⇒ Part of tank filled by pipes A, B and C in one hour working together = ($\frac{1}{10}$) – ($\frac{1}{12}$) – ($\frac{1}{30}$) = –$\frac{1}{60}$ Therefore, time taken to completely empty the tank if all pipes are opened simultaneously = $\frac{1}{60}$ hours = 60 hours Method 2: Let the capacity of tank be LCM (10, 12, 30) = 60 units ⇒ Efficiency of pipe A = $\frac{60}{10}$ = 6 units / hour ⇒ Efficiency of pipe B = – $\frac{60}{10}$ = – 5 units / hour (Here, ‘-‘ represents outlet pipe) ⇒ Efficiency of pipe C = – $\frac{60}{30}$ = – 2 units / hour (Here, ‘-‘ represents outlet pipe) ⇒ Combined efficiency of pipes A, B and C = 6 – 5 – 2 = – 1 units / hour (Here, ‘-‘represents outlet pipe) Therefore, time taken to completely empty the tank = $\frac{60}{(1)}$ = 60 hours Example 3: A cistern has two pipes. Both working together can fill the cistern in 12 minutes. First pipe is 10 minutes faster than the second pipe. How much time would it take to fill the cistern if only second pipe is used? Solution: Method 1: Let the time taken by first pipe working alone be ‘t’ minutes. ⇒ Time taken by second pipe working alone = t + 10 minutes. Part of tank filled by pipe A in one hour working alone = $\frac{1}{t}$ Part of tank filled by pipe B in one hour working alone = $\frac{1}{t + 10}$ => Part of tank filled by pipe A and B in one hour working together = $\frac{1}{t}$ + $\frac{1}{t + 10}$ = $\frac{2t + 10}{t \times (t + 10)}$ But we are given that it takes 12 minutes to completely fill the cistern if both pipes are working together. ⇒ $\frac{2t + 10}{t \times (t + 10)}$ = $\frac{1}{12}$ ⇒ $\frac{t \times (t + 10)}{2t + 10}$ = 12 ⇒ $t^{2}$ + 10t = 24t +120 ⇒ $t^{2}$ – 14t – 120 = 0 ⇒ (t -20)(t + 6) = 0 ⇒ t = 20 minutes (Time cannot be negative) Therefore, time taken by second pipe working alone = 20 + 10 = 30 minutes Method 2: Let the time taken by first pipe working alone be ‘t’ minutes. => Time taken by second pipe working alone = t + 10 minutes. Let the capacity of cistern be t x (t + 10) units. => Efficiency of first pipe = $\frac{t \times (t + 10)}{t}$ = (t + 10) units / minute => Efficiency of second pipe = $\frac{t \times (t + 10)}{(t + 10)}$ = t units / minute => Combined efficiency of pipes = (2t + 10) units / minute => Time taken to fill the cistern completely = $\frac{t \times (t + 10)}{(2t + 10)}$ But we are given that it takes 12 minutes to completely fill the cistern if both pipes are working together. $\frac{t \times (t + 10)}{(2t + 10)}$ = 12 $t^{2}$ + 10t = 24t +120 $t^{2}$ – 14t – 120 = 0 (t – 20)(t + 6) = 0 t = 20 minutes (Time cannot be negative) Therefore, time taken by second pipe working alone = 20 + 10 = 30 minutes
A long while ago I promised to take you from the action by the modular group $\Gamma=PSL_2(\mathbb{Z})$ on the lattices at hyperdistance $n$ from the standard orthogonal laatice $L_1$ to the corresponding ‘monstrous’ Grothendieck dessin d’enfant. Speaking of dessins d’enfant, let me point you to the latest intriguing paper by Yuri I. Manin and Matilde Marcolli, ArXived a few days ago Quantum Statistical Mechanics of the Absolute Galois Group, on how to build a quantum system for the absolute Galois group from dessins d’enfant (more on this, I promise, later). Where were we? We’ve seen natural one-to-one correspondences between (a) points on the projective line over $\mathbb{Z}/n\mathbb{Z}$, (b) lattices at hyperdistance $n$ from $L_1$, and (c) coset classes of the congruence subgroup $\Gamma_0(n)$ in $\Gamma$. How to get from there to a dessin d’enfant? The short answer is: it’s all in Ravi S. Kulkarni’s paper, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math 113 (1991) 1053-1135. It is a complete mystery to me why Tatitscheff, He and McKay don’t mention Kulkarni’s paper in “Cusps, congruence groups and monstrous dessins”. Because all they do (and much more) is in Kulkarni. I’ve blogged about Kulkarni’s paper years ago: – In the Dedekind tessalation it was all about assigning special polygons to subgroups of finite index of $\Gamma$. – In Modular quilts and cuboid tree diagram it did go on assigning (multiple) cuboid trees to a (conjugacy class) of such finite index subgroup. – In Hyperbolic Mathieu polygons the story continued on a finite-to-one connection between special hyperbolic polygons and cuboid trees. – In Farey codes it was shown how to encode such polygons by a Farey-sequence. – In Generators of modular subgroups it was shown how to get generators of the finite index subgroups from this Farey sequence. The modular group is a free product \[ \Gamma = C_2 \ast C_2 = \langle s,u~\|~s^2=1=u^3 \rangle \] with lifts of $s$ and $u$ to $SL_2(\mathbb{Z})$ given by the matrices \[ S=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},~\qquad U= \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \] As a result, any permutation representation of $\Gamma$ on a set $E$ can be represented by a $2$-coloured graph (with black and white vertices) and edges corresponding to the elements of the set $E$. Each white vertex has two (or one) edges connected to it and every black vertex has three (or one). These edges are the elements of $E$ permuted by $s$ (for white vertices) and $u$ (for black ones), the order of the 3-cycle determined by going counterclockwise round the vertex. Clearly, if there’s just one edge connected to a vertex, it gives a fixed point (or 1-cycle) in the corresponding permutation. The ‘monstrous dessin’ for the congruence subgroup $\Gamma_0(n)$ is the picture one gets from the permutation $\Gamma$-action on the points of $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$, or equivalently, on the coset classes or on the lattices at hyperdistance $n$. Kulkarni’s paper (or the blogposts above) tell you how to get at this picture starting from a fundamental domain of $\Gamma_0(n)$ acting on teh upper half-plane by Moebius transformations. Sage gives a nice image of this fundamental domain via the command FareySymbol(Gamma0(n)).fundamental_domain() Here’s the image for $n=6$: The boundary points (on the halflines through $0$ and $1$ and the $4$ half-circles need to be identified which is indicaed by matching colours. So the 2 halflines are identified as are the two blue (and green) half-circles (in opposite direction). To get the dessin from this, let’s first look at the interior points. A white vertex is a point in the interior where two black and two white tiles meet, a black vertex corresponds to an interior points where three black and three white tiles meet. Points on the boundary where tiles meet are coloured red, and after identification two of these reds give one white or black vertex. Here’s the intermediate picture The two top red points are identified giving a white vertex as do the two reds on the blue half-circles and the two reds on the green half-circles, because after identification two black and two white tiles meet there. This then gives us the ‘monstrous’ modular dessin for $n=6$ of the Tatitscheff, He and McKay paper: Let’s try a more difficult example: $n=12$. Sage gives us as fundamental domain giving us the intermediate picture and spotting the correct identifications, this gives us the ‘monstrous’ dessin for $\Gamma_0(12)$ from the THM-paper: In general there are several of these 2-coloured graphs giving the same permutation representation, so the obtained ‘monstrous dessin’ depends on the choice of fundamental domain. You’ll have noticed that the domain for $\Gamma_0(6)$ was symmetric, whereas the one Sage provides for $\Gamma_0(12)$ is not. This is caused by Sage using the Farey-code \[ \xymatrix{ 0 \ar@{-}[r]_1 & \frac{1}{6} \ar@{-}[r]_1 & \frac{1}{5} \ar@{-}[r]_2 & \frac{1}{4} \ar@{-}[r]_3 & \frac{1}{3} \ar@{-}[r]_4 & \frac{1}{2} \ar@{-}[r]_4 & \frac{2}{3} \ar@{-}[r]_3 & \frac{3}{4} \ar@{-}[r]_2 & 1} \] One of the nice results from Kulkarni’s paper is that for any $n$ there is a symmetric Farey-code, giving a perfectly symmetric fundamental domain for $\Gamma_0(n)$. For $n=12$ this symmetric code is \[ \xymatrix{ 0 \ar@{-}[r]_1 & \frac{1}{6} \ar@{-}[r]_2 & \frac{1}{4} \ar@{-}[r]_3 & \frac{1}{3} \ar@{-}[r]_4 & \frac{1}{2} \ar@{-}[r]_4 & \frac{2}{3} \ar@{-}[r]_3 & \frac{3}{4} \ar@{-}[r]_2 & \frac{5}{6} \ar@{-}[r]_1 & 1} \] It would be nice to see whether using these symmetric Farey-codes gives other ‘monstrous dessins’ than in the THM-paper. Remains to identify the edges in the dessin with the lattices at hyperdistance $n$ from $L_1$. Using the tricks from the previous post it is quite easy to check that for any $n$ the monstrous dessin for $\Gamma_0(n)$ starts off with the lattices $L_{M,\frac{g}{h}} = M,\frac{g}{h}$ as below Let’s do a sample computation showing that the action of $s$ on $L_n$ gives $L_{\frac{1}{n}}$: \[ L_n.s = \begin{bmatrix} n & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -n \\ 1 & 0 \end{bmatrix} \] and then, as last time, to determine the class of the lattice spanned by the rows of this matrix we have to compute \[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -n \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -n \end{bmatrix} \] which is class $L_{\frac{1}{n}}$. And similarly for the other edges.2 Comments
Closure under intersection with regular sets is natural in the sense that most machine models for language classes are finite-state based with some additional control mechanism, and by some standard constructions a finite automaton describing some other regular language could be coded into this finite-state mechanism to get closure under intersection. This is how most of the proofs that show this closure property work. I will give a more detailed and formal answer under some very specific framework below. But before some remarks on the boolean operations. 1) Note the curious fact that for a trio, closure under intersection is the same as closure under shuffle. The shuffle of two words is the language of all possible interleavings of two words, i.e., if $u,v \in X^$ for some alphabet $X$, then $$ u \diamond v := \{ u_1 v_1 u_2 v_2 \cdots u_k v_k \mid k \ge 1, u_i, v_i \in X \cup \{\varepsilon\}\}.$$ For languages $U, V \subseteq X^$ we set $U \diamond V = \bigcup_{u \in U, v \in V} u\diamond v$. We can write the shuffle with intersection and the trio operations. Let $\overline X = \{ \overline x : x \in X \}$ be some disjoint copy of $X$ and $h : (X \cup \overline X)^* \to X^$ the homomorphism given by $h(x) = h(\overline x) = x$ and $h_1, h_2 : (X \cup \overline X)^ \to X$ two homomorphisms given by $h_1(x) = h_2(\overline x) = \varepsilon$, $h_1(\overline x) = x$ and $h_2(x) = x$. Then$$ U \diamond V = h(h_1^{-1}(U) \cap h_2^{-1}(V)).$$This formula also gives that we always have closure under shuffle with regular languages in any trio. Conversely if a trio is closed under shuffle then$$ U \cap V = h((U \diamond \overline V) \cap (X\overline X)^*)$$where $\overline U = \{ \overline u : u \in U \}$ is the image of U under the homomorphism given by $x \mapsto \overline x$. 2) If we allow complementation as already mentioned many interesting and natural language classes are excluded. But this could be stated more formal as was done in Georg Zetzsche's dissertation Monoids as Storage Mechanisms. If a trio containing only recursively enumerable languages is closed under complementation, then it is precisely the class of all regular languages. So, if we also want some meaningful way to talk about computability, there is not trio closed under complementation except the regular languages. 3) Let me make my intuitive remarks why closure under intersection with regular languages is a natural operation more precise, or represents a structural property as noted in @babou answer. As said it has to do with the fact that most machine models are finite state based with some additional control (a stack, an infinite tape, a linear bounded tape etc). One way to make this more precise and unify these models is the notion of valence automata, which is also a central topic in the above mentioned dissertation. A valence automaton is essentially a finite automaton enriched with some ''counting mechanism/or storage mechanism'' in form of a monoid. Imagine you have some additional register that can save an element from some monoid, and you can alter it by multiplication with other monoid elements. The following monoids correspond to the following language classes. Monoid \| Language class ------------------------------- B \| Context-Free F_2 \| Context-Free B x B \| Recursively Enumerable 1 \| Regular Z^n \| Blind counter languages N^n \| Partially blind counter languages where B denotes the bicyclic monoid, F_2 the free group of rank $2$, 1 the trivial monoid, Z the integers, and N the natural numbers, see the mentioned dissertation for more details. By a slight modification of the product automaton construction we can show that if $U$ and $V$ are accepted by some valence automata over monoids $M$ and $N$, then $U \cap V$ is accepted by some valence automaton over $M \times N$. But if $V$ is regular, we can choose $N = 1$, the trivial monoid. Hence in this case $M \times N \cong M$, so that we have not left our language class, or said differently every language described by valence automata is closed under intersection with regular languages (and in fact they are semi AFL's).
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It turns out that if this second answer is yes, the series behaves much like a finite sum, i.e. it behaves well. Consider the alternating harmonic series $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac12 + \frac13 - \frac14+\cdots.$$It Example: converges (we saw this previously by using the AST). The series with the absolute values of its terms, which is the harmonic series $\sum \frac{1}{n}$, diverges ($p$-series with $p\le 1$). Since th e series converges, but not in absolute value, we say it is conditionally convergent. Example: Consider the alternating $p$-series, with $p=2$, $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} = 1 - \frac14 + \frac19 - \frac1{16}+\cdots.$$Since the series with the absolute values of the terms of our series, $\sum\frac{1}{n^2}$, is a convergent $p$-series, our series is absolutely convergent. By the fact above, this means it is also convergent. It is not conditionally convergent. Be careful with these terms Conditional convergence of a series means it is convergent but not absolutely convergent. If we are told that a series is convergent, we do not know a priori whether it is conditionally convergent or absolutely convergent. It is one or the other, but not both. Every series is either divergent, conditionally convergent, or absolutely convergent, but it is only one of these things. Justification of the fact above, and some examples, are discussed in the video.
Help me please, how can I show that Poincaré inequality doesn't hold in an unbounded domain? Thanks a lot! If $\Omega$ is a bounded domain and $u \in H_{0}^{1}(\Omega)$ the following inequality holds : $\left \\| u \right \\|_{L^{2}(\Omega) }\leq c \left \\| \nabla u \right \\|_{L^{2}(\Omega) } $ where $c$ depends only on $\Omega$ and not on $u$.
Here's how I would have approached the problem. (1) Since this matrix is diagonal, we can calculate $$\begin{bmatrix}1&0\\0&-1\end{bmatrix}^4=\begin{bmatrix}1^4&0\\0&(-1)^4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$so this matrix satisfies $A^4=I$. (2) For this matrix, I notice that $\sqrt{2}/2$ appears a lot, so my brain goes to the fact that $\sin(\pi/4)=\cos(\pi/4)=\sqrt{2}/2$. Thinking of this matrix as corresponding to a linear transformation in the plane, I took a look at the form for a rotation matrix:$$\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}.$$This tells me that our matrix corresponds to a rotation by $-\pi/4$, so $A^4$ corresponds to composing this rotation by itself four times, resulting in rotating by $-\pi$. Since, in general, rotating a vector by $-\pi$ doesn't result in the same vector, it follows that $A^4\neq I$. (3) Here I noticed that the bottom row is a scalar multiple of the top row, so the matrix cannot be invertible. You can justify this by realizing this means the rows are not linearly independent, or simply by computing the determinant and seeing that it's zero. Since the matrix is not invertible, it is impossible that $A^4=I$. We can see this in a few ways: first, the properties of determinants tell me that $A^4=I$ implies that $\det(A)^4=\det(I)=1$ and so $\det(A)\neq{0}$, which we know isn't true for our $A$ since it's not invertible; second, $A^4=I$ can be written as $A(A^3)=I$, and this would imply that $A$ is invertible with $A^{-1}=A^3$, but $A$ isn't invertible. (4) Like with (2), I see $1/2$ and $\sqrt{3}/2$, so I think of rotation matrices and see that this matrix corresponds to a rotation by $\pi/3$. Again, since for a vector $v\in\mathbb{R}^2$, $A^4v$ is an application of $A$ to the vector $v$ four times, it follows that $A^4$ corresponds to a rotation by $4\pi/3$. Thus, $A^4v\neq v$ for a general $v$. (5) Correcting my answer thanks to the comments below, if one did not wish to do the matrix multiplication, one could again realize that the matrix is a rotation matrix corresponding to $\theta=\pi/2.$ As applying this rotation four times to any vector results in a rotation by $2\pi$, or, equivalently, no rotation at all, we have that $A^4=I$.
I just have a question for the beginning of a proof: Suppose $\frac{dS_{t}}{S_{t}}=(r_{t}-q_{t})dt+\sigma(t,S_{t})dW_{t}$ with $r,q,S$ stochastic. In the book I read, it is written: We define the Arrow-Debreu price $\psi(x',y',z',t)$ as the present value of a derivative that pays off $\delta([S_{t},r_{t},q_{t}]-[x',y',z'])$ at time $t$. This is related to the $t$-forward measure probability density of $(x,y,z)$, $\phi(x,y,z,t)$ by: $$\psi(x,y,z,t)=B(0,t)\ \phi(x,y,z,t)$$ as can be seen from the defining equation for $\psi$ and $\phi$: $V(S_{0},r_{0},q_{0},t=0)=\iiint V(x,y,z,t)\ \psi(x,y,z,t)\ dx\ dy\ dz$ and $V(S_{0},r_{0},q_{0},t=0)=B(0,t)\ \iiint V(x,y,z,t)\ \phi(x,y,z,t)\ dx\ dy\ dz$ With these 2 last equations I understand why: $\psi(x,y,z,t)=B(0,t)\ \phi(x,y,z,t)$. But I don't understand why $V(S_{0},r_{0},q_{0},t=0)=\iiint V(x,y,z,t)\ \psi(x,y,z,t)\ dx\ dy\ dz$. Because for me we have $V(S_{0},r_{0},q_{0},t=0)=\mathbb{E}^{Q}[e^{-\int_{0}^{t}r_{s}ds}\ V(S_{t},r_{t},q_{t},t)]$ so where is the discount term $e^{-\int_{0}^{t}r_{s}ds}$ gone? Thanks
Solid-state T. Sefzik, J. B. Houseknecht, T. M. Clark, S. Prasad, T. L. Lowary, Z. Gan, and P. J. Grandinetti 17O NMR in Carbohydrates 055 - J. Magn. Reson., 185(2), 326-330 (2007) Abstract Solid-state $^{17}$O magic-angle spinning nuclear magnetic resonance measurements at 19.5 Tesla were performed on $^{17}$O-enriched methyl $\alpha$-D-galactopyranoside (4-$^{17}$O), methyl $\beta$-D-Glucopyranoside (2-$^{17}$O), methyl $\alpha$-D-Glucopyranoside (4-$^{17}$O), methyl $\alpha$-D-Glucopyranoside (6-$^{17}$O), and $\alpha$-D-Glucopyranosyl ($1$$\rightarrow$$6$) $\alpha$-D-Glucopyranoside (6-$^{17}$O). The $^{17}$O quadrupolar coupling constants and asymmetry parameters measured can be predicted with a model based entirely on the first-coordination sphere around oxygen. For the hydroxyl sites observed in the methyl glucosides the quadrupolar coupling parameters are nearly identical, within 10\%, as predicted, given their nearly identical first coordination sphere structures.
Image Dimensions Describing the fields of the Canvas Properties Dialog The user access the image dimensions in the Canvas Properties Dialog. The 'Others' tab Here some properties can simply be locked (such that they can't be changed) and linked (so that changes in one entry simultaneously change other entries as well). The 'Image' tab Obviously here the image dimensions can be set. There seem to be basically three groups of fields to edit: The on-screen size(?) The fields Widthand Heighttell synfigstudio how many pixels the image shall cover at a zoom level of 100%. The physical size The physical width and height should tell how big the image is on some physical media. That could be when printing out images on paper, or maybe even on transparencies or film. Not all file formats can save this on exporting/rendering images. The mysterious Image Area Given as two points (upper-left and lower-right corner) which also define the image span (Pythagoras: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \scriptstyle\text{span}=\sqrt{\Delta x^2 + \Delta y^2}}). The unit seems to be not pixels but units, which are at 60 pixels each. If the ratio of the image size and image area dimensions are off, for example circles will appear as an ellipse (see image). These settings seem to influence how large one Image Sizepixel is being rendered. This might be useful when one has to deal with non-square output pixels.
Notice: If you happen to see a question you know the answer to, please do chime in and help your fellow community members. We encourage our fourm members to be more involved, jump in and help out your fellow researchers with their questions. GATK forum is a community forum and helping each other with using GATK tools and research is the cornerstone of our success as a genomics research community.We appreciate your help! Test-drive the GATK tools and Best Practices pipelines on Terra Check out this blog post to learn how you can get started with GATK and try out the pipelines in preconfigured workspaces (with a user-friendly interface!) without having to install anything. Genotype Refinement workflow for germline short variants Contents Overview Summary of workflow steps Output annotations Example More information about priors Mathematical details 1. Overview The core GATK Best Practices workflow has historically focused on variant discovery --that is, the existence of genomic variants in one or more samples in a cohorts-- and consistently delivers high quality results when applied appropriately. However, we know that the quality of the individual genotype calls coming out of the variant callers can vary widely based on the quality of the BAM data for each sample. The goal of the Genotype Refinement workflow is to use additional data to improve the accuracy of genotype calls and to filter genotype calls that are not reliable enough for downstream analysis. In this sense it serves as an optional extension of the variant calling workflow, intended for researchers whose work requires high-quality identification of individual genotypes. While every study can benefit from increased data accuracy, this workflow is especially useful for analyses that are concerned with how many copies of each variant an individual has (e.g. in the case of loss of function) or with the transmission (or de novo origin) of a variant in a family. If a “gold standard” dataset for SNPs is available, that can be used as a very powerful set of priors on the genotype likelihoods in your data. For analyses involving families, a pedigree file describing the relatedness of the trios in your study will provide another source of supplemental information. If neither of these applies to your data, the samples in the dataset itself can provide some degree of genotype refinement (see section 5 below for details). After running the Genotype Refinement workflow, several new annotations will be added to the INFO and FORMAT fields of your variants (see below). Note that GQ fields will be updated, and genotype calls may be modified. However, the Phred-scaled genotype likelihoods (PLs) which indicate the original genotype call (the genotype candidate with PL=0) will remain untouched. Any analysis that made use of the PLs will produce the same results as before. 2. Summary of workflow steps Input Begin with recalibrated variants from VQSR at the end of the germline short variants pipeline. The filters applied by VQSR will be carried through the Genotype Refinement workflow. Step 1: Derive posterior probabilities of genotypes Tool used: CalculateGenotypePosteriors Using the Phred-scaled genotype likelihoods (PLs) for each sample, prior probabilities for a sample taking on a HomRef, Het, or HomVar genotype are applied to derive the posterior probabilities of the sample taking on each of those genotypes. A sample’s PLs were calculated by HaplotypeCaller using only the reads for that sample. By introducing additional data like the allele counts from the 1000 Genomes project and the PLs for other individuals in the sample’s pedigree trio, those estimates of genotype likelihood can be improved based on what is known about the variation of other individuals. SNP calls from the 1000 Genomes project capture the vast majority of variation across most human populations and can provide very strong priors in many cases. At sites where most of the 1000 Genomes samples are homozygous variant with respect to the reference genome, the probability of a sample being analyzed of also being homozygous variant is very high. For a sample for which both parent genotypes are available, the child’s genotype can be supported or invalidated by the parents’ genotypes based on Mendel’s laws of allele transmission. Even the confidence of the parents’ genotypes can be recalibrated, such as in cases where the genotypes output by HaplotypeCaller are apparent Mendelian violations. Step 2: Filter low quality genotypes Tool used: VariantFiltration After the posterior probabilities are calculated for each sample at each variant site, genotypes with GQ < 20 based on the posteriors are filtered out. GQ20 is widely accepted as a good threshold for genotype accuracy, indicating that there is a 99% chance that the genotype in question is correct. Tagging those low quality genotypes indicates to researchers that these genotypes may not be suitable for downstream analysis. However, as with the VQSR, a filter tag is applied, but the data is not removed from the VCF. Step 3: Annotate possible de novo mutations Tool used: VariantAnnotator Using the posterior genotype probabilities, possible de novo mutations are tagged. Low confidence de novos have child GQ >= 10 and AC < 4 or AF < 0.1%, whichever is more stringent for the number of samples in the dataset. High confidence de novo sites have all trio sample GQs >= 20 with the same AC/AF criterion. Step 4: Functional annotation of possible biological effects Tool options: Funcotator (experimental) Especially in the case of de novo mutation detection, analysis can benefit from the functional annotation of variants to restrict variants to exons and surrounding regulatory regions. Funcotator is a new tool that is currently still in development. If you would prefer to use a more mature tool, we recommend you look into SnpEff or Oncotator, but note that these are not GATK tools so we do not provide support for them. 3. Output annotations The Genotype Refinement workflow adds several new info- and format-level annotations to each variant. GQ fields will be updated, and genotypes calculated to be highly likely to be incorrect will be changed. The Phred-scaled genotype likelihoods (PLs) carry through the pipeline without being changed. In this way, PLs can be used to derive the original genotypes in cases where sample genotypes were changed. Population Priors New INFO field annotation PG is a vector of the Phred-scaled prior probabilities of a sample at that site being HomRef, Het, and HomVar. These priors are based on the input samples themselves along with data from the supporting samples if the variant in question overlaps another in the supporting dataset. Phred-Scaled Posterior Probability New FORMAT field annotation PP is the Phred-scaled posterior probability of the sample taking on each genotype for the given variant context alleles. The PPs represent a better calibrated estimate of genotype probabilities than the PLs are recommended for use in further analyses instead of the PLs. Genotype Quality Current FORMAT field annotation GQ is updated based on the PPs. The calculation is the same as for GQ based on PLs. Joint Trio Likelihood New FORMAT field annotation JL is the Phred-scaled joint likelihood of the posterior genotypes for the trio being incorrect. This calculation is based on the PLs produced by HaplotypeCaller (before application of priors), but the genotypes used come from the posteriors. The goal of this annotation is to be used in combination with JP to evaluate the improvement in the overall confidence in the trio’s genotypes after applying CalculateGenotypePosteriors. The calculation of the joint likelihood is given as: where the GLs are the genotype likelihoods in [0, 1] probability space. Joint Trio Posterior New FORMAT field annotation JP is the Phred-scaled posterior probability of the output posterior genotypes for the three samples being incorrect. The calculation of the joint posterior is given as: where the GPs are the genotype posteriors in [0, 1] probability space. Low Genotype Quality New FORMAT field filter lowGQ indicates samples with posterior GQ less than 20. Filtered samples tagged with lowGQ are not recommended for use in downstream analyses. High and Low Confidence De Novo New INFO field annotation for sites at which at least one family has a possible de novo mutation. Following the annotation tag is a list of the children with de novo mutations. High and low confidence are output separately. 4. Example Before: 1 1226231 rs13306638 G A 167563.16 PASS AC=2;AF=0.333;AN=6;… GT:AD:DP:GQ:PL 0/0:11,0:11:0:0,0,249 0/0:10,0:10:24:0,24,360 1/1:0,18:18:60:889,60,0 After: 1 1226231 rs13306638 G A 167563.16 PASS AC=3;AF=0.500;AN=6;…PG=0,8,22;… GT:AD:DP:GQ:JL:JP:PL:PP 0/1:11,0:11:49:2:24:0,0,249:49,0,287 0/0:10,0:10:32:2:24:0,24,360:0,32,439 1/1:0,18:18:43:2:24:889,60,0:867,43,0 The original call for the child (first sample) was HomRef with GQ0. However, given that, with high confidence, one parent is HomRef and one is HomVar, we expect the child to be heterozygous at this site. After family priors are applied, the child’s genotype is corrected and its GQ is increased from 0 to 49. Based on the allele frequency from 1000 Genomes for this site, the somewhat weaker population priors favor a HomRef call (PG=0,8,22). The combined effect of family and population priors still favors a Het call for the child. The joint likelihood for this trio at this site is two, indicating that the genotype for one of the samples may have been changed. Specifically a low JL indicates that posterior genotype for at least one of the samples was not the most likely as predicted by the PLs. The joint posterior value for the trio is 24, which indicates that the GQ values based on the posteriors for all of the samples are at least 24. (See above for a more complete description of JL and JP.) 5. More information about priors The Genotype Refinement Pipeline uses Bayes’s Rule to combine independent data with the genotype likelihoods derived from HaplotypeCaller, producing more accurate and confident genotype posterior probabilities. Different sites will have different combinations of priors applied based on the overlap of each site with external, supporting SNP calls and on the availability of genotype calls for the samples in each trio. Input-derived Population Priors If the input VCF contains at least 10 samples, then population priors will be calculated based on the discovered allele count for every called variant. Supporting Population Priors Priors derived from supporting SNP calls can only be applied at sites where the supporting calls overlap with called variants in the input VCF. The values of these priors vary based on the called reference and alternate allele counts in the supporting VCF. Higher allele counts (for ref or alt) yield stronger priors. Family Priors The strongest family priors occur at sites where the called trio genotype configuration is a Mendelian violation. In such a case, each Mendelian violation configuration is penalized by a de novo mutation probability (currently 10-6). Confidence also propagates through a trio. For example, two GQ60 HomRef parents can substantially boost a low GQ HomRef child and a GQ60 HomRef child and parent can improve the GQ of the second parent. Application of family priors requires the child to be called at the site in question. If one parent has a no-call genotype, priors can still be applied, but the potential for confidence improvement is not as great as in the 3-sample case. Caveats Right now family priors can only be applied to biallelic variants and population priors can only be applied to SNPs. Family priors only work for trios. 6. Mathematical details Note that family priors are calculated and applied before population priors. The opposite ordering would result in overly strong population priors because they are applied to the child and parents and then compounded when the trio likelihoods are multiplied together. Review of Bayes’s Rule HaplotypeCaller outputs the likelihoods of observing the read data given that the genotype is actually HomRef, Het, and HomVar. To convert these quantities to the probability of the genotype given the read data, we can use Bayes’s Rule. Bayes’s Rule dictates that the probability of a parameter given observed data is equal to the likelihood of the observations given the parameter multiplied by the prior probability that the parameter takes on the value of interest, normalized by the prior times likelihood for all parameter values: $$ P(\theta\|Obs) = \frac{P(Obs\|\theta)P(\theta)}{\sum_{\theta} P(Obs\|\theta)P(\theta)} $$ In the best practices pipeline, we interpret the genotype likelihoods as probabilities by implicitly converting the genotype likelihoods to genotype probabilities using non-informative or flat priors, for which each genotype has the same prior probability. However, in the Genotype Refinement Pipeline we use independent data such as the genotypes of the other samples in the dataset, the genotypes in a “gold standard” dataset, or the genotypes of the other samples in a family to construct more informative priors and derive better posterior probability estimates. Calculation of Population Priors Given a set of samples in addition to the sample of interest (ideally non-related, but from the same ethnic population), we can derive the prior probability of the genotype of the sample of interest by modeling the sample’s alleles as two independent draws from a pool consisting of the set of all the supplemental samples’ alleles. (This follows rather naturally from the Hardy-Weinberg assumptions.) Specifically, this prior probability will take the form of a multinomial Dirichlet distribution parameterized by the allele counts of each allele in the supplemental population. In the biallelic case the priors can be calculated as follows: $$ P(GT = HomRef) = \dbinom{2}{0} \ln \frac{\Gamma(nSamples)\Gamma(RefCount + 2)}{\Gamma(nSamples + 2)\Gamma(RefCount)} $$ $$ P(GT = Het) = \dbinom{2}{1} \ln \frac{\Gamma(nSamples)\Gamma(RefCount + 1)\Gamma(AltCount + 1)}{\Gamma(nSamples + 2)\Gamma(RefCount)\Gamma(AltCount)} $$ $$ P(GT = HomVar) = \dbinom{2}{2} \ln \frac{\Gamma(nSamples)\Gamma(AltCount + 2)}{\Gamma(nSamples + 2)\Gamma(AltCount)} $$ where Γ is the Gamma function, an extension of the factorial function. The prior genotype probabilities based on this distribution scale intuitively with number of samples. For example, a set of 10 samples, 9 of which are HomRef yield a prior probability of another sample being HomRef with about 90% probability whereas a set of 50 samples, 49 of which are HomRef yield a 97% probability of another sample being HomRef. Calculation of Family Priors Given a genotype configuration for a given mother, father, and child trio, we set the prior probability of that genotype configuration as follows: $$ P(G_M,G_F,G_C) = P(\vec{G}) \cases{ 1-10\mu-2\mu^2 & no MV \cr \mu & 1 MV \cr \mu^2 & 2 MVs} $$ where the 10 configurations with a single Mendelian violation are penalized by the de novo mutation probability μ and the two configurations with two Mendelian violations by μ^2. The remaining configurations are considered valid and are assigned the remaining probability to sum to one. This prior is applied to the joint genotype combination of the three samples in the trio. To find the posterior for any single sample, we marginalize over the remaining two samples as shown in the example below to find the posterior probability of the child having a HomRef genotype: This quantity P(Gc\|D) is calculated for each genotype, then the resulting vector is Phred-scaled and output as the Phred-scaled posterior probabilities (PPs).
A search for new physics is performed based on all-hadronic events with large missing transverse momentum produced in proton-proton collisions at sqrt(s) = 13 TeV. The data sample, corresponding to an integrated luminosity of 2.3 inverse femtobarns, was collected with the CMS detector at the CERN LHC in 2015. The data are examined in search regions of jet multiplicity, tagged bottom quark jet multiplicity, missing transverse momentum, and the scalar sum of jet transverse momenta. The observed numbers of events in all search regions are found to be consistent with the expectations from standard model processes. Exclusion limits are presented for simplified supersymmetric models of gluino pair production. Depending on the assumed gluino decay mechanism, and for a massless, weakly interacting, lightest neutralino, lower limits on the gluino mass from 1440 to 1600 GeV are obtained, significantly extending previous limits. A search for new phenomena is performed in final states containing one or more jets and an imbalance in transverse momentum in pp collisions at a centre-of-mass energy of 13 $\,\text {TeV}$ . The analysed data sample, recorded with the CMS detector at the CERN LHC, corresponds to an integrated luminosity of 2.3 $\,\text {fb}^{-1}$ . Several kinematic variables are employed to suppress the dominant background, multijet production, as well as to discriminate between other standard model and new physics processes. The search provides sensitivity to a broad range of new-physics models that yield a stable weakly interacting massive particle. The number of observed candidate events is found to agree with the expected contributions from standard model processes, and the result is interpreted in the mass parameter space of fourteen simplified supersymmetric models that assume the pair production of gluinos or squarks and a range of decay modes. For models that assume gluino pair production, masses up to 1575 and 975 $\,\text {GeV}$ are excluded for gluinos and neutralinos, respectively. For models involving the pair production of top squarks and compressed mass spectra, top squark masses up to 400 $\,\text {GeV}$ are excluded. A search for top squark pair production in pp collisions at $ \sqrt{s}=13 $ TeV is performed using events with a single isolated electron or muon, jets, and a large transverse momentum imbalance. The results are based on data collected in 2016 with the CMS detector at the LHC, corresponding to an integrated luminosity of 35.9 fb$^{−1}$. No significant excess of events is observed above the expectation from standard model processes. Exclusion limits are set in the context of supersymmetric models of pair production of top squarks that decay either to a top quark and a neutralino or to a bottom quark and a chargino. Depending on the details of the model, we exclude top squarks with masses as high as 1120 GeV. Detailed information is also provided to facilitate theoretical interpretations in other scenarios of physics beyond the standard model. Results are reported from a search for supersymmetric particles in proton-proton collisions in the final state with a single lepton, multiple jets, including at least one b-tagged jet, and large missing transverse momentum. The search uses a sample of proton-proton collision data at s=13 TeV recorded by the CMS experiment at the LHC, corresponding to an integrated luminosity of 35.9 fb−1. The observed event yields in the signal regions are consistent with those expected from standard model backgrounds. The results are interpreted in the context of simplified models of supersymmetry involving gluino pair production, with gluino decay into either on- or off-mass-shell top squarks. Assuming that the top squarks decay into a top quark plus a stable, weakly interacting neutralino, scenarios with gluino masses up to about 1.9 TeV are excluded at 95% confidence level for neutralino masses up to about 1 TeV. Results are reported from a search for physics beyond the standard model in proton–proton collisions at a center-of-mass energy of s=13TeV . The search uses a signature of a single lepton, large jet and bottom quark jet multiplicities, and high sum of large-radius jet masses, without any requirement on the missing transverse momentum in an event. The data sample corresponds to an integrated luminosity of 35.9 fb −1 recorded by the CMS experiment at the LHC. No significant excess beyond the prediction from standard model processes is observed. The results are interpreted in terms of upper limits on the production cross section for R -parity violating supersymmetric extensions of the standard model using a benchmark model of gluino pair production, in which each gluino decays promptly via g˜→tbs . Gluinos with a mass below 1610 GeV are excluded at 95% confidence level. <p>A search for new long-lived particles decaying to leptons is presented using proton-proton collisions produced by the LHC at <inline-formula><mml:math display="inline"><mml:mrow><mml:msqrt><mml:mrow><mml:mi>s</mml:mi></mml:mrow></mml:msqrt><mml:mo>=</mml:mo><mml:mn>8</mml:mn><mml:mtext> </mml:mtext><mml:mtext> </mml:mtext><mml:mi>TeV</mml:mi></mml:mrow></mml:math></inline-formula>. Data used for the analysis were collected by the CMS detector and correspond to an integrated luminosity of <inline-formula><mml:math display="inline"><mml:mrow><mml:mn>19.7</mml:mn><mml:mtext> </mml:mtext><mml:mtext> </mml:mtext><mml:mrow><mml:msup><mml:mrow><mml:mi>fb</mml:mi></mml:mrow><mml:mrow><mml:mo>-</mml:mo><mml:mn>1</mml:mn></mml:mrow></mml:msup></mml:mrow></mml:mrow></mml:math></inline-formula>. Events are selected with an electron and muon with opposite charges that both have transverse impact parameter values between 0.02 and 2 cm. The search has been designed to be sensitive to a wide range of models with nonprompt <inline-formula><mml:math display="inline"><mml:mrow><mml:mi>e</mml:mi><mml:mtext>-</mml:mtext><mml:mi>μ</mml:mi></mml:mrow></mml:math></inline-formula> final states. Limits are set on the “displaced supersymmetry” model, with pair production of top squarks decaying into an <inline-formula><mml:math display="inline"><mml:mrow><mml:mi>e</mml:mi><mml:mtext>-</mml:mtext><mml:mi>μ</mml:mi></mml:mrow></mml:math></inline-formula> final state via <inline-formula><mml:math display="inline"><mml:mi>R</mml:mi></mml:math></inline-formula>-parity-violating interactions. The results are the most restrictive to date on this model, with the most stringent limit being obtained for a top squark lifetime corresponding to <inline-formula><mml:math display="inline"><mml:mrow><mml:mi>c</mml:mi><mml:mi>τ</mml:mi><mml:mo>=</mml:mo><mml:mn>2</mml:mn><mml:mtext> </mml:mtext><mml:mtext> </mml:mtext><mml:mi>cm</mml:mi></mml:mrow></mml:math></inline-formula>, excluding masses below 790 GeV at 95% confidence level.</p> Results are reported from a search for supersymmetric particles in proton-proton collisions in the final state with a single, high transverse momentum lepton, multiple jets, including at least one b-tagged jet, and large missing transverse momentum. The data sample corresponds to an integrated luminosity of 2.3 fb$^{−1}$ at $ \sqrt{s}=13 $ TeV, recorded by the CMS experiment at the LHC. The search focuses on processes leading to high jet multiplicities, such as gluino pair production with $ \tilde{\mathrm{g}}\to \mathrm{t}\overline{\mathrm{t}}{\tilde{\chi}}_1^0 $ . The quantity M$_{J}$ , defined as the sum of the masses of the large-radius jets in the event, is used in conjunction with other kinematic variables to provide discrimination between signal and background and as a key part of the background estimation method. The observed event yields in the signal regions in data are consistent with those expected for standard model backgrounds, estimated from control regions in data. Exclusion limits are obtained for a simplified model corresponding to gluino pair production with three-body decays into top quarks and neutralinos. Gluinos with a mass below 1600 GeV are excluded at a 95% confidence level for scenarios with low $ {\tilde{\chi}}_1^0 $ mass, and neutralinos with a mass below 800 GeV are excluded for a gluino mass of about 1300 GeV. For models with two-body gluino decays producing on-shell top squarks, the excluded region is only weakly sensitive to the top squark mass. A search for supersymmetry in the context of general gauge-mediated (GGM) breaking with the lightest neutralino as the next-to-lightest supersymmetric particle and the gravitino as the lightest is presented. The data sample corresponds to an integrated luminosity of 36 inverse picobarns recorded by the CMS experiment at the LHC. The search is performed using events containing two or more isolated photons, at least one hadronic jet, and significant missing transverse energy. No excess of events at high missing transverse energy is observed. Upper limits on the signal cross section for GGM supersymmetry between 0.3 and 1.1 pb at the 95% confidence level are determined for a range of squark, gluino, and neutralino masses, excluding supersymmetry parameter space that was inaccessible to previous experiments. A search for physics beyond the standard model in final states with at least one photon, large transverse momentum imbalance, and large total transverse event activity is presented. Such topologies can be produced in gauge-mediated supersymmetry models in which pair-produced gluinos or squarks decay to photons and gravitinos via short-lived neutralinos. The data sample corresponds to an integrated luminosity of 35.9 inverse femtobarns of proton-proton collisions at sqrt(s) = 13 TeV recorded by the CMS experiment at the LHC in 2016. No significant excess of events above the expected standard model background is observed. The data are interpreted in simplified models of gluino and squark pair production, in which gluinos or squarks decay via neutralinos to photons. Gluino masses of up to 1.50-2.00 TeV and squark masses up to 1.30-1.65 TeV are excluded at 95% confidence level, depending on the neutralino mass and branching fraction. A search for supersymmetry is presented based on proton-proton collision events containing identified hadronically decaying top quarks, no leptons, and an imbalance pTmiss in transverse momentum. The data were collected with the CMS detector at the CERN LHC at a center-of-mass energy of 13 TeV, and correspond to an integrated luminosity of 35.9 fb−1. Search regions are defined in terms of the multiplicity of bottom quark jet and top quark candidates, the pTmiss, the scalar sum of jet transverse momenta, and the mT2 mass variable. No statistically significant excess of events is observed relative to the expectation from the standard model. Lower limits on the masses of supersymmetric particles are determined at 95% confidence level in the context of simplified models with top quark production. For a model with direct top squark pair production followed by the decay of each top squark to a top quark and a neutralino, top squark masses up to 1020 GeV and neutralino masses up to 430 GeV are excluded. For a model with pair production of gluinos followed by the decay of each gluino to a top quark-antiquark pair and a neutralino, gluino masses up to 2040 GeV and neutralino masses up to 1150 GeV are excluded. These limits extend previous results. A search for supersymmetry is presented based on events with at least one photon, jets, and large missing transverse momentum produced in proton–proton collisions at a center-of-mass energy of 13 $\,\text {Te}\text {V}$ . The data correspond to an integrated luminosity of 35.9 $\,\text {fb}^{-1}$ and were recorded at the LHC with the CMS detector in 2016. The analysis characterizes signal-like events by categorizing the data into various signal regions based on the number of jets, the number of $\mathrm {b}$ -tagged jets, and the missing transverse momentum. No significant excess of events is observed with respect to the expectations from standard model processes. Limits are placed on the gluino and top squark pair production cross sections using several simplified models of supersymmetric particle production with gauge-mediated supersymmetry breaking. Depending on the model and the mass of the next-to-lightest supersymmetric particle, the production of gluinos with masses as large as 2120 $\,\text {Ge}\text {V}$ and the production of top squarks with masses as large as 1230 $\,\text {Ge}\text {V}$ are excluded at 95% confidence level. A search for exotic decays of the Higgs boson to a pair of light pseudoscalar particles a$_1$ is performed under the hypothesis that one of the pseudoscalars decays to a pair of opposite sign muons and the other decays to b$\overline{\mathrm{b}}$. Such signatures are predicted in a number of extensions of the standard model (SM), including next-to-minimal supersymmetry and two-Higgs-doublet models with an additional scalar singlet. The results are based on a data set of proton-proton collisions corresponding to an integrated luminosity of 35.9 fb$^{-1}$, accumulated with the CMS experiment at the CERN LHC in 2016 at a centre-of-mass energy of 13 TeV. No statistically significant excess is observed with respect to the SM backgrounds in the search region for pseudoscalar masses from 20 GeV to half of the Higgs boson mass. Upper limits at 95% confidence level are set on the product of the production cross section and branching fraction, $\sigma_{\mathrm{h}}\mathcal{B}$(h $\to$ a$_1$ a$_1$ $\to$ $\mu^+\mu^-\mathrm{b}\bar{\mathrm{b}}$), ranging from 5 to 33 fb, depending on the pseudoscalar mass. Corresponding limits on the branching fraction, assuming the SM prediction for $\sigma_{\mathrm{h}}$, are (1$-$7)$\times$ 10$^{-4}$.
Current browse context: cond-mat.str-el Change to browse by: References & Citations Bookmark(what is this?) Condensed Matter > Strongly Correlated Electrons Title: Textured electronic states of the triangular lattice Hubbard model and Na$_x$CoO$_2$ (Submitted on 2 Sep 2013 (v1), last revised 3 Nov 2014 (this version, v4)) Abstract: We show that geometric frustration and strong correlation in the triangular lattice Hubbard model lead a rich and novel phase structure of $\sqrt{3}\times\sqrt{3}$ spin-charge textured electronic states over a wide region of electron doping $0\le x \le 0.40$. In addition to the 120$^\circ$ N\'eel ordered insulator at half-filling, we found a novel spin-charge ordered insulator at $x=1/3$ with collinear antiferromagnetic (AF) order on the underlying unfrustrated honeycomb lattice. Separating the two insulating phases is a Lifshitz transition between a noncollinear AF ordered metal and one with coexisting charge order. We obtain the phase diagram and the evolution of the Fermi surface (FS). Remarkably, the correlated ground states near $x=1/3$ emerges as doping the "1/3 AF insulator" by excess carriers, leading to electron and hole FS pockets with important implications for the cobaltate superconducting state. Submission historyFrom: Kun Jiang [view email] [v1]Mon, 2 Sep 2013 20:01:18 GMT (261kb) [v2]Wed, 4 Jun 2014 19:43:32 GMT (279kb) [v3]Fri, 6 Jun 2014 21:27:46 GMT (279kb) [v4]Mon, 3 Nov 2014 15:40:00 GMT (279kb)
Translate Word Phrases to Math Notation Earlier in this section, we translated math notation for division into words. Now we’ll translate word phrases into math notation. Some of the words that indicate division are given in Table $\PageIndex{2}$. Operation Word Phrase Example Expression Division divided by 12 divided by 4 12 ÷ 4 $\frac{12}{4}$ quotient of the quotient of 12 and 4 12/4 divided into 4 divided into 12 $4 \overline{\smash{)}12}$ Translate and simplify: the quotient of $51$ and $17$. Solution The word quotient tells us to divide. Translate. 51 ÷ 17 Divide. 3 We could just as correctly have translated the quotient of $51$ and $17$ using the notation $17 \overline{\smash{)}51}$ or $\frac{51}{17}$. exercise $\PageIndex{23}$ Translate and simplify: the quotient of $91$ and $13$. Answer $91 \div 13; 7$ exercise $\PageIndex{24}$ Translate and simplify: the quotient of $52$ and $13$. Answer $52 \div 13; 4$ Divide Whole Numbers in Applications We will use the same strategy we used in previous sections to solve applications. First, we determine what we are looking for. Then we write a phrase that gives the information to find it. We then translate the phrase into math notation and simplify it to get the answer. Finally, we write a sentence to answer the question. Example $\PageIndex{13}$: translate and simplify Cecelia bought a $160$-ounce box of oatmeal at the big box store. She wants to divide the $160$ ounces of oatmeal into $8$-ounce servings. She will put each serving into a plastic bag so she can take one bag to work each day. How many servings will she get from the big box? Solution We are asked to find the how many servings she will get from the big box. Write a phrase. 160 ounces divided by 8 ounces Translate to math notation. 160 ÷ 8 Simplify by dividing. 20 Write a sentence to answer the question. Cecelia will get 20 servings from the big box. exercise $\PageIndex{25}$ Marcus is setting out animal crackers for snacks at the preschool. He wants to put $9$ crackers in each cup. One box of animal crackers contains $135$ crackers. How many cups can he fill from one box of crackers? Answer Marcus can fill $15$ cups. exercise $\PageIndex{26}$ Andrea is making bows for the girls in her dance class to wear at the recital. Each bow takes $4$ feet of ribbon, and $36$ feet of ribbon are on one spool. How many bows can Andrea make from one spool of ribbon? Answer Andrea can make $9$ bows. Access Additional Online Resources Key Concepts Operation Notation Expression Read as Result $\begin{align} \div &\\ \frac{a}{b} &\\ b\overline{\smash{)} a} &\\ a/b & \end{align}$ $\begin{align} 12\div 4 &\\ \frac{12}{4} &\\ 4\overline{\smash{)}12} &\\ 12/4 & \end{align}$ Division Properties of One Any number (except $0$) divided by itself is one. Any number divided by one is the same number. Division Properties of Zero Zero divided by any number is $0$. $ Dividing a number by zero is undefined. Divide whole numbers. Divide the first digit of the dividend by the divisor. If the divisor is larger than the first digit of the dividend, divide the first two digits of the dividend by the divisor, and so on. Write the quotient above the dividend. Multiply the quotient by the divisor and write the product under the dividend. Subtract that product from the dividend. Bring down the next digit of the dividend. Repeat from Step 1 until there are no more digits in the dividend to bring down. Check by multiplying the quotient times the divisor. Glossary dividend When dividing two numbers, the dividend is the number being divided. divisor When dividing two numbers, the divisor is the number dividing the dividend. quotient The quotient is the result of dividing two numbers. Practice Makes Perfect Use Division Notation In the following exercises, translate from math notation to words. 54 ÷ 9 \(\frac{56}{7}$ $\frac{32}{8}$ $6 \overline{\smash{)}42}$ 48 ÷ 6 $\frac{63}{9}$ $7 \overline{\smash{)}63}$ 72 ÷ 8 Model Division of Whole Numbers In the following exercises, model the division. 15 ÷ 5 10 ÷ 5 $\frac{14}{7}$ $\frac{18}{6}$ $4 \overline{\smash{)}20}$ $3 \overline{\smash{)}15}$ 24 ÷ 6 16 ÷ 4 Divide Whole Numbers In the following exercises, divide. Then check by multiplying. 18 ÷ 2 14 ÷ 2 $\frac{27}{3}$ $\frac{30}{3}$ $4 \overline{\smash{)}28}$ $4 \overline{\smash{)}36}$ $\frac{45}{5}$ $\frac{35}{5}$ 72 / 8 $8 \overline{\smash{)}64}$ $\frac{35}{7}$ 42 ÷ 7 $15 \overline{\smash{)}15}$ $12 \overline{\smash{)}12}$ 43 ÷ 43 37 ÷ 37 $\frac{23}{1}$ $\frac{29}{1}$ 19 ÷ 1 17 ÷ 1 0 ÷ 4 0 ÷ 8 $\frac{5}{0}$ $\frac{9}{0}$ $\frac{26}{0}$ $\frac{32}{0}$ $12 \overline{\smash{)}0}$ $16 \overline{\smash{)}0}$ 72 ÷ 3 57 ÷ 3 $\frac{96}{8}$ $\frac{78}{6}$ $5\overline{\smash{)}465}$ $4\overline{\smash{)}528}$ 924 ÷ 7 861 ÷ 7 $\frac{5,226}{6}$ $\frac{3,776}{8}$ $4\overline{\smash{)}31,324}$ $5\overline{\smash{)}46,855}$ 7,209 ÷ 3 4,806 ÷ 3 5,406 ÷ 6 3,208 ÷ 4 $4\overline{\smash{)}2,816}$ $6 \overline{\smash{)}3624}$ $\frac{91,881}{9}$ $\frac{83,256}{8}$ 2,470 ÷ 7 3,741 ÷ 7 $8\overline{\smash{)}55,305}$ $9\overline{\smash{)}51,492}$ $\frac{431,174}{5}$ $\frac{297,277}{4}$ 130,016 ÷ 3 105,609 ÷ 2 $15\overline{\smash{)}5,735}$ $\frac{4,933}{21}$ 56,883 ÷ 67 43,725 / 75 $\frac{30,144}{314}$ 26,145 ÷ 415 $273\overline{\smash{)}542,195}$ 816,243 ÷ 462 Mixed Practice In the following exercises, simplify. 15(204) 74 • 391 256 − 184 305 − 262 719 + 341 647 + 528 $25\overline{\smash{)}875}$ 1104 ÷ 23 Translate Word Phrases to Algebraic Expressions In the following exercises, translate and simplify. the quotient of 45 and 15 the quotient of 64 and 16 the quotient of 288 and 24 the quotient of 256 and 32 Divide Whole Numbers in Applications In the following exercises, solve. Trail mixRic bought 64 ounces of trail mix. He wants to divide it into small bags, with 2 ounces of trail mix in each bag. How many bags can Ric fill? CrackersEvie bought a 42 ounce box of crackers. She wants to divide it into bags with 3 ounces of crackers in each bag. How many bags can Evie fill? Astronomy classThere are 125 students in an astronomy class. The professor assigns them into groups of 5. How many groups of students are there? Flower shopMelissa’s flower shop got a shipment of 152 roses. She wants to make bouquets of 8 roses each. How many bouquets can Melissa make? BakingOne roll of plastic wrap is 48 feet long. Marta uses 3 feet of plastic wrap to wrap each cake she bakes. How many cakes can she wrap from one roll? Dental flossOne package of dental floss is 54 feet long. Brian uses 2 feet of dental floss every day. How many days will one package of dental floss last Brian? Mixed Practice In the following exercises, solve. Miles per gallonSusana’s hybrid car gets 45 miles per gallon. Her son’s truck gets 17 miles per gallon. What is the difference in miles per gallon between Susana’s car and her son’s truck? DistanceMayra lives 53 miles from her mother’s house and 71 miles from her motherin-law’s house. How much farther is Mayra from her mother-in-law’s house than from her mother’s house? Field tripThe 45 students in a Geology class will go on a field trip, using the college’s vans. Each van can hold 9 students. How many vans will they need for the field trip? Potting soilAki bought a 128 ounce bag of potting soil. How many 4 ounce pots can he fill from the bag? HikingBill hiked 8 miles on the first day of his backpacking trip, 14 miles the second day, 11 miles the third day, and 17 miles the fourth day. What is the total number of miles Bill hiked? ReadingLast night Emily read 6 pages in her Business textbook, 26 pages in her History text, 15 pages in her Psychology text, and 9 pages in her math text. What is the total number of pages Emily read? PatientsLaVonne treats 12 patients each day in her dental office. Last week she worked 4 days. How many patients did she treat last week? ScoutsThere are 14 boys in Dave’s scout troop. At summer camp, each boy earned 5 merit badges. What was the total number of merit badges earned by Dave’s scout troop at summer camp? Writing Exercises Explain how you use the multiplication facts to help with division. Oswaldo divided 300 by 8 and said his answer was 37 with a remainder of 4. How can you check to make sure he is correct? Everyday Math Contact lensesJenna puts in a new pair of contact lenses every 14 days. How many pairs of contact lenses does she need for 365 days? Cat foodOne bag of cat food feeds Lara’s cat for 25 days. How many bags of cat food does Lara need for 365 days? Self Check (a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. (b) Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not? Contributors Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (formerly of Santa Ana College). This content produced by OpenStax and is licensed under a Creative Commons Attribution License 4.0 license.
Difference between revisions of "Applied/ACMS/absS15" (→ACMS Abstracts: Spring 2015) (→Murad Banaji (Portsmouth)) Line 49: Line 49: === Murad Banaji (Portsmouth) === === Murad Banaji (Portsmouth) === − Nonexpansivity in chemical reaction networks + Nonexpansivity in chemical reaction networks This work is motivated by the observation that quite often systems of differential equations describing chemical reaction networks (CRNs) display simple global behaviour such as convergence of all orbits to a unique equilibrium under only weak and physically reasonable assumptions on the reaction rates (kinetics). We are led to wonder if the structure of a CRN may sometimes force some distance between solutions to decrease (or at least not increase) with time. If so, how can we find this nonincreasing quantity? We explore different ways in which CRNs can define nonexpansive semiflows (recall that a semiflow <math>(\phi_t)_{t \geq 0}</math> on some Banach space <math>(X, \|\cdot\|)</math> is nonexpansive if <math>\|\phi_t(x)-\phi_t(y)\| \leq \|x-y\|</math> for all <math>x,y \in X</math> and all <math>t \geq 0</math>). It turns out that in CRNs the natural evolution of chemical concentrations may be nonexpansive; or a nonexpansive semiflow may be obtained from the evolution of the so-called "extents" of reactions. In both cases we may be able to draw global conclusions about convergence of chemical concentrations. In each case the challenge is to find the correct norm to get nonexpansivity for arbitrary kinetics. To construct such norms and show nonexpansivity we appeal to the theory of monotone dynamical systems. Families of CRNs which can be analysed in this way are presented; however characterising fully the class of CRNs to which this theory applies remains an open - and undoubtedly difficult - task. This work is motivated by the observation that quite often systems of differential equations describing chemical reaction networks (CRNs) display simple global behaviour such as convergence of all orbits to a unique equilibrium under only weak and physically reasonable assumptions on the reaction rates (kinetics). We are led to wonder if the structure of a CRN may sometimes force some distance between solutions to decrease (or at least not increase) with time. If so, how can we find this nonincreasing quantity? We explore different ways in which CRNs can define nonexpansive semiflows (recall that a semiflow <math>(\phi_t)_{t \geq 0}</math> on some Banach space <math>(X, \|\cdot\|)</math> is nonexpansive if <math>\|\phi_t(x)-\phi_t(y)\| \leq \|x-y\|</math> for all <math>x,y \in X</math> and all <math>t \geq 0</math>). It turns out that in CRNs the natural evolution of chemical concentrations may be nonexpansive; or a nonexpansive semiflow may be obtained from the evolution of the so-called "extents" of reactions. In both cases we may be able to draw global conclusions about convergence of chemical concentrations. In each case the challenge is to find the correct norm to get nonexpansivity for arbitrary kinetics. To construct such norms and show nonexpansivity we appeal to the theory of monotone dynamical systems. Families of CRNs which can be analysed in this way are presented; however characterising fully the class of CRNs to which this theory applies remains an open - and undoubtedly difficult - task. This is joint work with Bas Lemmens (University of Kent) and Pete Donnell (University of Portsmouth). This is joint work with Bas Lemmens (University of Kent) and Pete Donnell (University of Portsmouth). Revision as of 12:58, 5 March 2015 Contents ACMS Abstracts: Spring 2015 Irene Kyza (U Dundee) Adaptivity and blowup detection for semilinear evolution convection-diffusion equations based on a posteriori error control We discuss recent results on the a posteriori error control and adaptivity for an evolution semilinear convection-diffusion model problem with possible blowup in finite time. This belongs to the broad class of partial differential equations describing e.g., tumor growth,chemotaxis and cell modelling. In particular, we derive a posteriori error estimates that are conditional (estimates which are valid under conditions of a posteriori type) for an interior penalty discontinuous Galerkin (dG) implicit-explicit (IMEX) method using a continuation argument. Compared to a previous work, the obtained conditions are more localised and allow the efficient error control near the blowup time. Utilising the conditional a posteriori estimator we are able to propose an adaptive algorithm that appears to perform satisfactorily. In particular, it leads to good approximation of the blowup time and of the exact solution close to the blowup. Numerical experiments illustrate and complement our theoretical results. This is joint work with A. Cangiani, E.H. Georgoulis, and S. Metcalfe from the University of Leicester. Daniel Vimont (UW) Linear Inverse Modeling of Central and East Pacific El Niño / Southern Oscillation (ENSO) Events Research on the structure and evolution of individual El Niño / Southern Oscillation (ENSO) events has identified two categories of ENSO event characteristics that can be defined by maximum equatorial SST anomalies centered in the Central Pacific (around the dateline to 150 deg. W; CP events) or in the Eastern Pacific (east of about 150 deg. W; EP events). The distinction between these two events is not just academic: both types of event evolve differently, implying different predictability; the events tend to have different maximum amplitude; and the global teleconnection differs between each type of event. In this presentation I will (i) describe the Linear Inverse Modeling (LIM) technique, (ii) apply LIM to determine an empirical dynamical operator that governs the evolution of tropical Pacific climate variability, (iii) define norms under which initial conditions can be derived that optimally lead to growth of CP or EP ENSO events, and (iv) identify patterns of stochastic forcing that are responsible for exciting each type of event. Saverio Spagnolie (UW) Sedimentation in viscous fluids: flexible filaments and boundary effects The deformation and transport of elastic filaments in viscous fluids play central roles in many biological and technological processes. Compared with the well-studied case of sedimenting rigid rods, the introduction of filament compliance may cause a significant alteration in the long-time sedimentation orientation and filament geometry. In the weakly flexible regime, a multiple-scale asymptotic expansion is used to obtain expressions for filament translations, rotations and shapes which match excellently with full numerical simulations. In the highly flexible regime we show that a filament sedimenting along its long axis is susceptible to a buckling instability. Embedding the analytical results for a single filament into a mean-field theory, we show how flexibility affects a well established concentration instability in a sedimenting suspension. Another problem of classical interest in fluid mechanics involves the sedimentation of a rigid particle near a wall, but most studies have been numerical or experimental in nature. We have derived ordinary differential equations describing the sedimentation of arbitrarily oriented prolate and oblate spheroids near a vertical or inclined plane wall which may be solved analytically for many important special cases. Full trajectories are predicted which compare favorably with complete numerical simulations performed using a novel double layer boundary integral formulation, a Method of Stresslet Images. Several trajectory-types emerge, termed tumbling, glancing, reversing, and sliding, along with their fully three-dimensional analogues. Jonathan Freund (UIUC) Adjoint-based optimization for understanding and reducing flow noise Advanced simulation tools, particularly large-eddy simulation techniques, are becoming capable of making quality predictions of jet noise for realistic nozzle geometries and at engineering relevant flow conditions. Increasing computer resources will be a key factor in improving these predictions still further. Quality prediction, however, is only a necessary condition for the use of such simulations in design optimization. Predictions do not of themselves lead to quieter designs. They must be interpreted or harnessed in some way that leads to design improvements. As yet, such simulations have not yielded any simplifying principals that offer general design guidance. The turbulence mechanisms leading to jet noise remain poorly described in their complexity. In this light, we have implemented and demonstrated an aeroacoustic adjoint-based optimization technique that automatically calculates gradients that point the direction in which to adjust controls in order to improve designs. This is done with only a single flow solutions and a solution of an adjoint system, which is solved at computational cost comparable to that for the flow. Optimization requires iterations, but having the gradient information provided via the adjoint accelerates convergence in a manner that is insensitive to the number of parameters to be optimized. The talk will review the formulation of the adjoint of the compressible flow equations for optimizing noise-reducing controls and present examples of its use. We will particularly focus on some mechanisms of flow noise that have been revealed via this approach. Markos Katsoulakis (U Mass Amherst) Information Theory methods for parameter sensitivity and coarse-graining of high-dimensional stochastic dynamics In this talk we discuss path-space information theory-based sensitivity analysis and parameter identification methods for complex high-dimensional dynamics, as well as information-theoretic tools for parameterized coarse-graining of non-equilibrium extended systems. Furthermore, we establish their connections with goal-oriented methods in terms of new, sharp, uncertainty quantification inequalities. The combination of proposed methodologies is capable to (a) handle molecular-level models with a very large number of parameters, (b) address and mitigate the high-variance in statistical estimators, e.g. for sensitivity analysis, in spatially distributed Kinetic Monte Carlo (KMC), (c) tackle non-equilibrium processes, typically associated with coupled physicochemical mechanisms, boundary conditions, etc. (such as reaction-diffusion systems), and where even steady states are unknown altogether, e.g. do not have a Gibbs structure. Finally, the path-wise information theory tools, (d) yield a surprisingly simple, tractable and easy-to-implement approach to quantify and rank parameter sensitivities, as well as (e) provide reliable molecular model parameterizations for coarse-grained molecular systems and their dynamics, based on fine-scale data and rational model selection methods through suitable path-space (dynamics-based) information criteria. The proposed methods are tested against a wide range of high-dimensional stochastic processes, ranging from complex biochemical reaction networks with hundreds of parameters, to spatially extended Kinetic Monte Carlo models in catalysis and Langevin dynamics of interacting molecules with internal degrees of freedom. Tao Zhou (Chinese Academy of Sciences) The Christoffel function weighted least-squares for stochastic collocation approximations: applications to Uncertainty Quantification We shall consider the multivariate stochastic collocation methods on unstructured grids. The motivation for such a study is the applications in parametric Uncertainty Quantification (UQ). We will first give a general framework of stochastic collocation methods, which include approaches such as compressed sensing, least-squares, and interpolation. Particular attention will be then given to the least-squares approach, and we will review recent progresses in this topic. Elaine Spiller (Marquette) TBA Murad Banaji (Portsmouth) "Nonexpansivity in chemical reaction networks" This work is motivated by the observation that quite often systems of differential equations describing chemical reaction networks (CRNs) display simple global behaviour such as convergence of all orbits to a unique equilibrium under only weak and physically reasonable assumptions on the reaction rates (kinetics). We are led to wonder if the structure of a CRN may sometimes force some distance between solutions to decrease (or at least not increase) with time. If so, how can we find this nonincreasing quantity? We explore different ways in which CRNs can define nonexpansive semiflows (recall that a semiflow [math](\phi_t)_{t \geq 0}[/math] on some Banach space [math](X, \|\cdot\|)[/math] is nonexpansive if [math]\|\phi_t(x)-\phi_t(y)\| \leq \|x-y\|[/math] for all [math]x,y \in X[/math] and all [math]t \geq 0[/math]). It turns out that in CRNs the natural evolution of chemical concentrations may be nonexpansive; or a nonexpansive semiflow may be obtained from the evolution of the so-called "extents" of reactions. In both cases we may be able to draw global conclusions about convergence of chemical concentrations. In each case the challenge is to find the correct norm to get nonexpansivity for arbitrary kinetics. To construct such norms and show nonexpansivity we appeal to the theory of monotone dynamical systems. Families of CRNs which can be analysed in this way are presented; however characterising fully the class of CRNs to which this theory applies remains an open - and undoubtedly difficult - task. This is joint work with Bas Lemmens (University of Kent) and Pete Donnell (University of Portsmouth).
A long while ago I promised to take you from the action by the modular group $\Gamma=PSL_2(\mathbb{Z})$ on the lattices at hyperdistance $n$ from the standard orthogonal laatice $L_1$ to the corresponding ‘monstrous’ Grothendieck dessin d’enfant. Speaking of dessins d’enfant, let me point you to the latest intriguing paper by Yuri I. Manin and Matilde Marcolli, ArXived a few days ago Quantum Statistical Mechanics of the Absolute Galois Group, on how to build a quantum system for the absolute Galois group from dessins d’enfant (more on this, I promise, later). Where were we? We’ve seen natural one-to-one correspondences between (a) points on the projective line over $\mathbb{Z}/n\mathbb{Z}$, (b) lattices at hyperdistance $n$ from $L_1$, and (c) coset classes of the congruence subgroup $\Gamma_0(n)$ in $\Gamma$. How to get from there to a dessin d’enfant? The short answer is: it’s all in Ravi S. Kulkarni’s paper, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math 113 (1991) 1053-1135. It is a complete mystery to me why Tatitscheff, He and McKay don’t mention Kulkarni’s paper in “Cusps, congruence groups and monstrous dessins”. Because all they do (and much more) is in Kulkarni. I’ve blogged about Kulkarni’s paper years ago: – In the Dedekind tessalation it was all about assigning special polygons to subgroups of finite index of $\Gamma$. – In Modular quilts and cuboid tree diagram it did go on assigning (multiple) cuboid trees to a (conjugacy class) of such finite index subgroup. – In Hyperbolic Mathieu polygons the story continued on a finite-to-one connection between special hyperbolic polygons and cuboid trees. – In Farey codes it was shown how to encode such polygons by a Farey-sequence. – In Generators of modular subgroups it was shown how to get generators of the finite index subgroups from this Farey sequence. The modular group is a free product \[ \Gamma = C_2 \ast C_2 = \langle s,u~\|~s^2=1=u^3 \rangle \] with lifts of $s$ and $u$ to $SL_2(\mathbb{Z})$ given by the matrices \[ S=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},~\qquad U= \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \] As a result, any permutation representation of $\Gamma$ on a set $E$ can be represented by a $2$-coloured graph (with black and white vertices) and edges corresponding to the elements of the set $E$. Each white vertex has two (or one) edges connected to it and every black vertex has three (or one). These edges are the elements of $E$ permuted by $s$ (for white vertices) and $u$ (for black ones), the order of the 3-cycle determined by going counterclockwise round the vertex. Clearly, if there’s just one edge connected to a vertex, it gives a fixed point (or 1-cycle) in the corresponding permutation. The ‘monstrous dessin’ for the congruence subgroup $\Gamma_0(n)$ is the picture one gets from the permutation $\Gamma$-action on the points of $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$, or equivalently, on the coset classes or on the lattices at hyperdistance $n$. Kulkarni’s paper (or the blogposts above) tell you how to get at this picture starting from a fundamental domain of $\Gamma_0(n)$ acting on teh upper half-plane by Moebius transformations. Sage gives a nice image of this fundamental domain via the command FareySymbol(Gamma0(n)).fundamental_domain() Here’s the image for $n=6$: The boundary points (on the halflines through $0$ and $1$ and the $4$ half-circles need to be identified which is indicaed by matching colours. So the 2 halflines are identified as are the two blue (and green) half-circles (in opposite direction). To get the dessin from this, let’s first look at the interior points. A white vertex is a point in the interior where two black and two white tiles meet, a black vertex corresponds to an interior points where three black and three white tiles meet. Points on the boundary where tiles meet are coloured red, and after identification two of these reds give one white or black vertex. Here’s the intermediate picture The two top red points are identified giving a white vertex as do the two reds on the blue half-circles and the two reds on the green half-circles, because after identification two black and two white tiles meet there. This then gives us the ‘monstrous’ modular dessin for $n=6$ of the Tatitscheff, He and McKay paper: Let’s try a more difficult example: $n=12$. Sage gives us as fundamental domain giving us the intermediate picture and spotting the correct identifications, this gives us the ‘monstrous’ dessin for $\Gamma_0(12)$ from the THM-paper: In general there are several of these 2-coloured graphs giving the same permutation representation, so the obtained ‘monstrous dessin’ depends on the choice of fundamental domain. You’ll have noticed that the domain for $\Gamma_0(6)$ was symmetric, whereas the one Sage provides for $\Gamma_0(12)$ is not. This is caused by Sage using the Farey-code \[ \xymatrix{ 0 \ar@{-}[r]_1 & \frac{1}{6} \ar@{-}[r]_1 & \frac{1}{5} \ar@{-}[r]_2 & \frac{1}{4} \ar@{-}[r]_3 & \frac{1}{3} \ar@{-}[r]_4 & \frac{1}{2} \ar@{-}[r]_4 & \frac{2}{3} \ar@{-}[r]_3 & \frac{3}{4} \ar@{-}[r]_2 & 1} \] One of the nice results from Kulkarni’s paper is that for any $n$ there is a symmetric Farey-code, giving a perfectly symmetric fundamental domain for $\Gamma_0(n)$. For $n=12$ this symmetric code is \[ \xymatrix{ 0 \ar@{-}[r]_1 & \frac{1}{6} \ar@{-}[r]_2 & \frac{1}{4} \ar@{-}[r]_3 & \frac{1}{3} \ar@{-}[r]_4 & \frac{1}{2} \ar@{-}[r]_4 & \frac{2}{3} \ar@{-}[r]_3 & \frac{3}{4} \ar@{-}[r]_2 & \frac{5}{6} \ar@{-}[r]_1 & 1} \] It would be nice to see whether using these symmetric Farey-codes gives other ‘monstrous dessins’ than in the THM-paper. Remains to identify the edges in the dessin with the lattices at hyperdistance $n$ from $L_1$. Using the tricks from the previous post it is quite easy to check that for any $n$ the monstrous dessin for $\Gamma_0(n)$ starts off with the lattices $L_{M,\frac{g}{h}} = M,\frac{g}{h}$ as below Let’s do a sample computation showing that the action of $s$ on $L_n$ gives $L_{\frac{1}{n}}$: \[ L_n.s = \begin{bmatrix} n & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -n \\ 1 & 0 \end{bmatrix} \] and then, as last time, to determine the class of the lattice spanned by the rows of this matrix we have to compute \[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -n \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -n \end{bmatrix} \] which is class $L_{\frac{1}{n}}$. And similarly for the other edges.2 Comments
Abstract We prove an upper bound for characters of the symmetric groups. In particular, we show that there exists a constant $a>0$ with a property that for every Young diagram $\lambda$ with $n$ boxes, $r(\lambda)$ rows and $c(\lambda)$ columns $$ \left\| \frac{\mathrm{Tr}\, \rho^{\lambda}(\pi)}{\mathrm{Tr}\, \rho^{\lambda}(e)} \right\| \leq \left[a \max\left(\frac{r(\lambda)}{n},\frac{c(\lambda)}{n},\frac{\|\pi\|}{n} \right)\right]^{\|\pi\|}, $$ where $\|\pi\|$ is the minimal number of factors needed to write $\pi\in S_n$ as a product of transpositions. We also give uniform estimates for the error term in the Vershik-Kerov’s and Biane’s character formulas and give a new formula for free cumulants of the transition measure. [Biane1997] P. 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Kerov, "Characters and factor representations of the infinite symmetric group," Dokl. Akad. Nauk SSSR, vol. 257, iss. 5, pp. 1037-1040, 1981. @article {VershikKerov1981, MRKEY = {0614033}, AUTHOR = {Vershik, A. M. and Kerov, S. V.}, TITLE = {Characters and factor representations of the infinite symmetric group}, JOURNAL = {Dokl. Akad. Nauk SSSR}, FJOURNAL = {Doklady Akademii Nauk SSSR}, VOLUME = {257}, YEAR = {1981}, NUMBER = {5}, PAGES = {1037--1040}, ISSN = {0002-3264}, MRCLASS = {20C32}, MRNUMBER = {0614033}, ZBLNUMBER = {0534.20008}, } [VoiculescuDykemaNica1992] D. V. Voiculescu, K. J. Dykema, and A. Nica, Free Random Variables, Providence, RI: Amer. Math. Soc., 1992, vol. 1. @book {VoiculescuDykemaNica1992, MRKEY = {1217253}, AUTHOR = {Voiculescu, D. V. and Dykema, K. J. and Nica, A.}, TITLE = {Free Random Variables}, SERIES = {{\rm CRM} Monogr. Series}, VOLUME = {1}, NOTE = {A noncommutative probability approach to free products with applications to random matrices, operator algebras and harmonic analysis on free groups}, PUBLISHER = {Amer. Math. Soc.}, ADDRESS = {Providence, RI}, YEAR = {1992}, PAGES = {vi+70}, ISBN = {0-8218-6999-X}, MRCLASS = {46L50 (46L10 47B80 60H99)}, MRNUMBER = {1217253}, MRREVIEWER = {Roland Speicher}, ZBLNUMBER = {0795.46049}, } [Zvonkin1997] A. Zvonkin, "Matrix integrals and map enumeration: an accessible introduction," Math. Comput. Modelling, vol. 26, iss. 8-10, pp. 281-304, 1997. @article {Zvonkin1997, MRKEY = {1492512}, AUTHOR = {Zvonkin, A.}, TITLE = {Matrix integrals and map enumeration: an accessible introduction}, NOTE = {Combinatorics and physics (Marseilles, 1995)}, JOURNAL = {Math. Comput. Modelling}, FJOURNAL = {Mathematical and Computer Modelling}, VOLUME = {26}, YEAR = {1997}, NUMBER = {8-10}, PAGES = {281--304}, ISSN = {0895-7177}, CODEN = {MCMOEG}, MRCLASS = {81Q30 (05C30 81T18)}, MRNUMBER = {1492512}, MRREVIEWER = {R{\u{a}}zvan Gelca}, DOI = {10.1016/S0895-7177(97)00210-0}, ZBLNUMBER = {1185.81083}, }
The potentials $\phi$ and $\vec{A}$ in electromagnetism are the canonical examples of quantities which aren't gauge-invariant. For instance, we are free to shift the electric potential $\phi$ by a constant without changing the physics; this shift is called a gauge transformation. It is a change in our description of the system, but not a change in any physical properties of the system. All physical quantities must be gauge-invariant, therefore. Examples include $E$ or $E \cdot B$. Usually when people use the word 'scalar' in this context, they mean 'number which is invariant under Lorentz transformations'. So there is no such thing as a scalar which is not Lorentz invariant. If by 'scalar' one means 'number which is invariant under rotations', then $E\cdot E$ does the trick for your first question. This quantity changes under Lorentz transformations. Finding numbers which don't change under Lorentz transformations is easiest to do if we reformulate electromagnetism in relativistic terms. The canonical examples are $E \cdot B$ and $E^2-B^2$.
Difference between revisions of "Fujimura's problem" (→n=5) (→General n) Line 64: Line 64: since deleting the bottom row of a equilateral-triangle-free-set gives another equilateral-triangle-free-set. since deleting the bottom row of a equilateral-triangle-free-set gives another equilateral-triangle-free-set. + + Revision as of 04:45, 13 February 2009 Let [math]\overline{c}^\mu_n[/math] the largest subset of the triangular grid [math]\Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c=n \}[/math] which contains no equilateral triangles. Fujimura's problem is to compute [math]\overline{c}^\mu_n[/math]. This quantity is relevant to a certain "hyper-optimistic" version of DHJ(3). n=0 It is clear that [math]\overline{c}^\mu_0 = 1[/math]. n=1 It is clear that [math]\overline{c}^\mu_1 = 2[/math]. n=2 It is clear that [math]\overline{c}^\mu_2 = 4[/math] (e.g. remove (0,2,0) and (1,0,1) from [math]\Delta_2[/math]). n=3 Deleting (0,3,0), (0,2,1), (2,1,0), (1,0,2) from [math]\Delta_3[/math] shows that [math]\overline{c}^\mu_3 \geq 6[/math]. In fact [math]\overline{c}^\mu_3 = 6[/math]: just note (3,0,0) or something symmetrical has to be removed, leaving 3 triangles which do not intersect, so 3 more removals are required. n=4 [math]\overline{c}^\mu_4=9:[/math] The set of all [math](a,b,c)[/math] in [math]\Delta_4[/math] with exactly one of a,b,c =0, has 9 elements and doesn’t contain any equilateral triangles. Let [math]S\subset \Delta_4[/math] be a set without equilateral triangles. If [math](0,0,4)\in S[/math], there can only be one of [math](0,x,4-x)[/math] and [math](x,0,4-x)[/math] in S for [math]x=1,2,3,4[/math]. Thus there can only be 5 elements in S with [math]a=0[/math] or [math]b=0[/math]. The set of elements with [math]a,b\gt0[/math] is isomorphic to [math]\Delta_2[/math], so S can at most have 4 elements in this set. So [math]\|S\|\leq 4+5=9[/math]. Similar if S contain (0,4,0) or (4,0,0). So if [math]\|S\|\gt9[/math] S doesn’t contain any of these. Also, S can’t contain all of [math](0,1,3), (0,3,1), (2,1,1)[/math]. Similar for [math](3,0,1), (1,0,3),(1,2,1)[/math] and [math](1,3,0), (3,1,0), (1,1,2)[/math]. So now we have found 6 elements not in S, but [math]\|\Delta_4\|=15[/math], so [math]S\leq 15-6=9[/math]. n=5 [math]\overline{c}^\mu_5=12[/math] The set of all (a,b,c) in [math]\Delta_5[/math] with exactly one of a,b,c=0 has 12 elements and doesn’t contain any equilateral triangles. Let [math]S\subset \Delta_5[/math] be a set without equilateral triangles. If [math](0,0,5)\in S[/math], there can only be one of (0,x,5-x) and (x,0,5-x) in S for x=1,2,3,4,5. Thus there can only be 6 elements in S with a=0 or b=0. The set of element with a,b>0 is isomorphic to [math]\Delta_3[/math], so S can at most have 6 elements in this set. So [math]\|S\|\leq 6+6=12[/math]. Similar if S contain (0,5,0) or (5,0,0). So if \|S\| >12 S doesn’t contain any of these. S can only contain 2 point in each of the following equilateral triangles: (3,1,1),(0,4,1),(0,1,4) (4,1,0),(1,4,0),(1,1,3) (4,0,1),(1,3,1),(1,0,4) (1,2,2),(0,3,2),(0,2,3) (3,2,0),(2,3,0),(2,2,1) (3,0,2),(2,1,2),(2,0,3) So now we have found 9 elements not in S, but [math]\|\Delta_5\|=21[/math], so [math]S\leq 21-9=12[/math]. General n [Cleanup required here] A lower bound for [math]\overline{c}^\mu_n[/math] is 3(n-1), made of all points in [math]\Delta_n[/math] with exactly one coordinate equal to zero. A trivial upper bound is [math]\overline{c}^\mu_{n+1} \leq \overline{c}^\mu_n + n+2[/math] since deleting the bottom row of a equilateral-triangle-free-set gives another equilateral-triangle-free-set. Another upper bound comes from counting the triangles. There are [math]\binom{n+2}{3}[/math] triangles, and each point belongs to n of them. So you must remove at least (n+2)(n+1)/6 points to remove all triangles, leaving (n+2)(n+1)/3 points as an upper bound for [math]\overline{c}^\mu_n[/math].
What is the smallest possible size of a set of points in $\mathbb{F}_q^3$ which intersects (blocks) every line? Clearly the union of three affine hyperplanes that intersect in a singleton, say $x = 0, y = 0$ and $z = 0$, forms such a set. If we denote by $s(q)$ the smallest possible size, then this gives us $$s(q) \leq 3q^2 - 3q + 1$$ for all $q$. Since any two points of $\mathbb{F}_2^3$ form a line, we get $s(2) = 2^3 - 1 = 7 = 12 - 6 + 1$. For $q = 3$ we can do better. In $\mathbb{F}_3^3$ the complement of such a set is a cap, i.e., a set of points no three of which are collinear. We know that the largest size of a cap in $\mathbb{F}_3^3$ is $9$ (corresponding to a quadric), and hence $s(3) = 18 < 27 - 9 + 1$. (side note: the problem of finding such a blocking set in $\mathbb{F}_3^n$ is thus equivalent to the famous cap set problem. See the survey article by Bierbrauer and Edel, large caps in projective Galois spaces and the paper by Bateman and Katz, new bounds on cap sets) Question 1: Can we improve the upper bound in general? We can also give a lower bound on $s(q)$. Jamison/Brouwer-Schrijver proved using the polynomial method that the smallest possible size of a blocking set in $\mathbb{F}_q^2$ is $2q - 1$. See this, this, this and this for various proofs of their result. Now take any $q$ parallel affine planes in $\mathbb{F}_q^3$, then the intersection of a blocking set with these hyperplanes must have size at least $2q - 1$, and hence $$2q^2 - q \leq s(q).$$ Question 2: Can we improve this lower bound in general? The Jamison/Brouwer-Schrijver result gives us another way of constructing a blocking set of size $3q^2 - 3q + 1$. Again take $q$ parallel hyperplanes $H_1, \dots, H_q$. Let $B_2, \dots, B_{q}$ be blocking sets of size $2q - 1$ in $H_2, \dots, H_{q}$. Then $B = H_1 \cup B_2 \cup \dots \cup B_{q}$ is a blocking set of size $(q-1)(2q-1) + q^2 = 3q^2 - 3q + 1$. Note that the problem of determining $s(q)$ is trivial for projective spaces. It's a classical result that a line blocking set in $PG(3,q)$ has size at least $1 + q + q^2$ with equality if and only if it is a hyperplane. See Chapter 3 of current research topics in Galois geometry for a recent survey on projective blocking sets. Edit 1: After Douglas Zare's answer below we have $s(q) \geq 2q^2 - 1$ for all $q$ and $s(q) \leq 3q^2 - 3q$ for $q \geq 3$. Can this be improved further? I have also found two references that prove this lower bound of $2q^2 - 1$, Proposition 4.1 in Nuclei of pointsets in $PG(n,q)$ (1997) and Theorem 3.1 in On Nuclei and Blocking Sets in Desarguesian Spaces (1999). In fact, Sziklai has mentioned the same argument as Douglas Zare after his proof of Proposition 4.1. Their proofs are generalisations of the polynomial technique introduced by Blokhuis in On nuclei and affine blocking sets (1994).
I am currently taking a course on communication complexity with Alexander Sherstov. Much of communication complexity involves matrix analysis, so yesterday we did a brief review of results from linear algebra. In the review, Sherstov gave the following definition for the rank of a matrix $M \in \mathbf{F}^{n \times m}$: \[ \mathrm{rank}(M) = \min{k \| M = A B, A \in \mathbf{F}^{n \times k}, B \in \mathbf{F}^{k \times m}} \] Since this “factor” definition of rank appears very different from the standard definition given in a linear algebra course, I thought I would prove that the two definitions are equivalent. The “standard” definition of rank is \[ \mathrm{rank}(M) = \mathrm{dim}\ \mathrm{span} {\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m} \] where $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m$ are the columns of $M$. Equivalently, $\mathrm{rank}(M)$ is the number of linearly independent columns of $M$. An easy consequence of the standard definition of rank is that the rank of a product of matrices is at most the rank of the first matrix: $\mathrm{rank}(A B) \leq \mathrm{rank}(A)$. Indeed, the columns of $AB$ are linear combinations of the columns of $A$, hence the span of the columns of $AB$ is a subspace of the span of the columns of $A$. Thus, if we can write $M = A B$ with $A \in \mathbf{F}^{n \times k}$ and $B \in \mathbf{F}^{k \times m}$, we must have $\mathrm{rank}(M) \leq k$. It remains to show that if $\mathrm{rank}(M) = k$ then we can factor $M = A B$ with $A \in \mathbf{F}^{n \times k}$ and $B \in \mathbf{F}^{k \times m}$. To this end, assume without loss of generality that the first $k$ columns of $M$ are linearly independent. Write these columns as vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$. Denote the remaining columns by $\mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_{m-k}$. Since these vectors lie in the span of the first $k$ columns of $M$, we can write the $\mathbf{w}_j$ as linear combinations of the $\mathbf{v}_i$: \[ \mathbf{w}_j = \sum_{i = 1}^k b_{i, j} \mathbf{v}_i \quad\text{for}\quad j = 1, 2, \ldots, m – k. \] Now define the matrices \[ A = (\mathbf{v}_1\ \mathbf{v}_2\ \cdots\ \mathbf{v}_k) \] and \[ B = (\mathbf{e}_1\ \mathbf{e}_2\ \cdots\ \mathbf{e}_k\ \mathbf{b}_1\ \mathbf{b}_2\ \cdots\ \mathbf{b}_{m – k}) \] where $\mathbf{e}_j$ is the $j$th standard basis vector in $\mathbf{F}^k$ and \[ \mathbf{b}_j = \begin{pmatrix} b_{1, j}\ b_{2, j}\ \vdots\ b_{k, j} \end{pmatrix} \quad\text{for}\quad j = 1, 2, \ldots, m – k. \] It is straightforward to verify that we have $M = A B$, which proves the equivalence of the two definitions of rank.
Let $S \subset \mathbf{R}^n$. We call $x \in \mathbf{R}^n$ a boundary point of $S$ if for every $\varepsilon > 0$ the ball $B(x, \varepsilon)$ centered at $x$ of radius $\varepsilon$ contains at least one point $y \in S$ and at least one point $z \notin S$. The set of all boundary points of $S$ is called the boundary of $S$ denoted $\partial S$. We say that $S$ is closed if $\partial S \subset S$. Now let ${S_\alpha}$ be a collection of closed sets indexed by $A$ (that is $\alpha$ ranges over all possible values in $A$). We would like to show that the set \[ S = \bigcap_{\alpha \in A} S_\alpha \] is closed. (Recall that $x \in \bigcap_{\alpha \in A} S_\alpha$ if and only if $x \in S_\alpha$ for all $\alpha \in A$.) To this end, we must show that every $x \in \partial S$ is also in $S$. To see this is the case, let $x \in \partial S$. Since $x \in \partial S$, for every $\varepsilon > 0$, there exist $y \in S$ and $z \notin S$ such that $y, z \in B(x, \varepsilon)$. Since $y \in S$, from the definition of $S$, we must have $y \in S_\alpha$ for all $\alpha$. Therefore, $x \in S_\alpha \cup \partial S_\alpha$ for each $\alpha$, hence $x \in S_\alpha$ for each $\alpha$ as $\partial S_\alpha \subset S_\alpha$. Thus we may conclude that \[ x \in \bigcap_{\alpha \in A} S_\alpha = S \] which is what we needed to show. Now we move on to consider unions of closed sets. First let $S = \bigcup_{i = 1}^k S_i$ be a finite union of closed sets. Suppose $x \in \partial S$. Again, for every $\varepsilon > 0$ there exists $y \in S$ and $z \notin S$ with $y, z \in B(x, \varepsilon)$. We would like to show that $x \in S$, hence $S$ is closed. Consider a sequence $y_1, y_2, \ldots \in S$ with \(\|x – y_i\|
I try to plot the following graph using Graphics and Table. The graph consists of infinitely many isosceles triangles (without their bases). The plot is within $[0,1]\times[0,1]$. The first isosceles triangle is of base $\frac{1}{2}$ and height $1$, the 2nd is of base $\frac{1}{4}$ and height $\frac{1}{2}$, etc. In general the $k$ th is of base length $2^{-k}$ and height $\frac{1}{k}$. There are infinitely many such triangles and the total base length is hence $\sum\limits_{k = 1}^\infty {{2^{ - k}}} = 1$. The code I tried is linelist = Table[ {{Boole[Mod[i, 2] == 0](1 - 2^(-i/2)) + Boole[Mod[i, 2] == 1](1 - 2^(-(i + 1)/2) - 2^(-(i + 1)/2 - 1)), Boole[Mod[i, 2] == 0]0 + Boole[Mod[i, 2] == 1]2/(i + 1)}}, {i, 1, 100} ];Graphics[lineList, Axes -> True]; but something wrong occurs and the output is empty. I don't know what is wrong with the code. Thank you! The graph I want to plot is sketched as the following:
Search Now showing items 1-10 of 26 Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV (Elsevier, 2017-12-21) We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ... Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV (American Physical Society, 2017-09-08) The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ... Online data compression in the ALICE O$^2$ facility (IOP, 2017) The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ... Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (American Physical Society, 2017-09-08) In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ... J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (American Physical Society, 2017-12-15) We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($\|y\| < 0.9$) ... Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions (Nature Publishing Group, 2017) At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ... K$^{}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV (American Physical Society, 2017-06) The production of K$^{}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ... Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Springer, 2017-06) The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ... Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC (Springer, 2017-01) The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ... Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC (Springer, 2017-06) We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ...
That was an excellent post and qualifies as a treasure to be found on this site! wtf wrote: When infinities arise in physics equations, it doesn't mean there's a physical infinity. It means that our physics has broken down. Our equations don't apply. I totally get that . In fact even our friend Max gets that.http://blogs.discovermagazine.com/crux/ ... g-physics/ Thanks for the link and I would have showcased it all on its own had I seen it first The point I am making is something different. I am pointing out that: All of our modern theories of physics rely ultimately on highly abstract infinitary mathematics That doesn't mean that they necessarily do; only that so far, that's how the history has worked out. I see what you mean, but as Max pointed out when describing air as seeming continuous while actually being discrete, it's easier to model a continuum than a bazillion molecules, each with functional probabilistic movements of their own. Essentially, it's taking an average and it turns out that it's pretty accurate. But what I was saying previously is that we work with the presumed ramifications of infinity, "as if" this or that were infinite, without actually ever using infinity itself. For instance, y = 1/x as x approaches infinity, then y approaches 0, but we don't actually USE infinity in any calculations, but we extrapolate. There is at the moment no credible alternative. There are attempts to build physics on constructive foundations (there are infinite objects but they can be constructed by algorithms). But not finitary principles, because to do physics you need the real numbers; and to construct the real numbers we need infinite sets. Hilbert pointed out there is a difference between boundless and infinite. For instance space is boundless as far as we can tell, but it isn't infinite in size and never will be until eternity arrives. Why can't we use the boundless assumption instead of full-blown infinity? 1) The rigorization of Newton's calculus culminated with infinitary set theory. Newton discovered his theory of gravity using calculus, which he invented for that purpose. I didn't know he developed calculus specifically to investigate gravity. Cool! It does make sense now that you mention it. However, it's well-known that Newton's formulation of calculus made no logical sense at all. If $\Delta y$ and $\Delta x$ are nonzero, then $\frac{\Delta y}{\Delta x}$ isn't the derivative. And if they're both zero, then the expression makes no mathematical sense! But if we pretend that it does, then we can write down a simple law that explains apples falling to earth and the planets endlessly falling around the sun. I'm going to need some help with this one. If dx = 0, then it contains no information about the change in x, so how can anything result from it? I've always taken dx to mean a differential that is smaller than can be discerned, but still able to convey information. It seems to me that calculus couldn't work if it were based on division by zero, and that if it works, it must not be. What is it I am failing to see? I mean, it's not an issue of 0/0 making no mathematical sense, it's a philosophical issue of the nonexistence of significance because there is nothing in zero to be significant. 2) Einstein's gneral relativity uses Riemann's differential geometry. In the 1840's Bernhard Riemann developed a general theory of surfaces that could be Euclidean or very far from Euclidean. As long as they were "locally" Euclidean. Like spheres, and torii, and far weirder non-visualizable shapes. Riemann showed how to do calculus on those surfaces. 60 years later, Einstein had these crazy ideas about the nature of the universe, and the mathematician Minkowski saw that Einstein's ideas made the most mathematical sense in Riemann's framework. This is all abstract infinitary mathematics. Isn't this the same problem as previous? dx=0? 3) Fourier series link the physics of heat to the physics of the Internet; via infinite trigonometric series. In 1807 Joseph Fourier analyzed the mathematics of the distribution of heat through an iron bar. He discovered that any continuous function can be expressed as an infinite trigonometric series, which looks like this: $$f(x) = \sum_{n=0}^\infty a_n \cos(nx) + \sum_{n=1}^\infty b_n \sin(nx)$$ I only posted that because if you managed to survive high school trigonometry, it's not that hard to unpack. You're composing any motion into a sum of periodic sine and cosine waves, one wave for each whole number frequency. And this is an infinite series of real numbers, which we cannot make sense of without using infinitary math. I can't make sense of it WITH infinitary math lol! What's the cosine of infinity? What's the infnite-th 'a'? 4) Quantum theory is functional analysis . If you took linear algebra, then functional analysis can be thought of as infinite-dimensional linear algebra combined with calculus. Functional analysis studies spaces whose points are actually functions; so you can apply geometric ideas like length and angle to wild collections of functions. In that sense functional analysis actually generalizes Fourier series. Quantum mechanics is expressed in the mathematical framework of functional analysis. QM takes place in an infinite-dimensional Hilbert space. To explain Hilbert space requires a deep dive into modern infinitary math. In particular, Hilbert space is complete , meaning that it has no holes in it. It's like the real numbers and not like the rational numbers. QM rests on the mathematics of uncountable sets, in an essential way. Well, thanks to Hilbert, I've already conceded that the boundless is not the same as the infinite and if it were true that QM required infinity, then no machine nor human mind could model it. It simply must be true that open-ended finites are actually employed and underpin QM rather than true infinite spaces. Like Max said, "Not only do we lack evidence for the infinite but we don’t need the infinite to do physics. Our best computer simulations, accurately describing everything from the formation of galaxies to tomorrow’s weather to the masses of elementary particles, use only finite computer resources by treating everything as finite. So if we can do without infinity to figure out what happens next, surely nature can, too—in a way that’s more deep and elegant than the hacks we use for our computer simulations." We can claim physics is based on infinity, but I think it's more accurate to say pretend or fool ourselves into thinking such. Max continued with, "Our challenge as physicists is to discover this elegant way and the infinity-free equations describing it—the true laws of physics. To start this search in earnest, we need to question infinity. I’m betting that we also need to let go of it." He said, "let go of it" like we're clinging to it for some reason external to what is true. I think the reason is to be rid of god, but that's my personal opinion. Because if we can't have infinite time, then there must be a creator and yada yada. So if we cling to infinity, then we don't need the creator. Hence why Craig quotes Hilbert because his first order of business is to dispel infinity and substitute god. I applaud your effort, I really do, and I've learned a lot of history because of it, but I still cannot concede that infinity underpins anything and I'd be lying if I said I could see it. I'm not being stubborn and feel like I'm walking on eggshells being as amicable and conciliatory as possible in trying not to offend and I'm certainly ready to say "Ooooohhh... I see now", but I just don't see it. ps -- There's our buddy Hilbert again. He did many great things. William Lane Craig misuses and abuses Hilbert's popularized example of the infinite hotel to make disingenuous points about theology and in particular to argue for the existence of God. That's what I've got against Craig. Craig is no friend of mine and I was simply listening to a debate on youtube (I often let youtube autoplay like a radio) when I heard him quote Hilbert, so I dug into it and posted what I found. I'm not endorsing Craig lol 5) Cantor was led to set theory from Fourier series. In every online overview of Georg Cantor's magnificent creation of set theory, nobody ever mentions how he came upon his ideas. It's as if he woke up one day and decided to revolutionize the foundations of math and piss off his teacher and mentor Kronecker. Nothing could be further from the truth. Cantor was in fact studing Fourier's trigonometric series! One of the questions of that era was whether a given function could have more than one distinct Fourier series. To investigate this problem, Cantor had to consider the various types of sets of points on which two series could agree; or equivalently, the various sets of points on which a trigonometric series could be zero. He was thereby led to the problem of classifying various infinite sets of real numbers; and that led him to the discovery of transfinite ordinal and cardinal numbers. (Ordinals are about order in the same way that cardinals are about quantity). I still can't understand how one infinity can be bigger than another since, to be so, the smaller infinity would need to have limits which would then make it not infinity. In other words, and this is a fact that you probably will not find stated as clearly as I'm stating it here: If you begin by studying the flow of heat through an iron rod; you will inexorably discover transfinite set theory. Right, because of what Max said about the continuum model vs the actual discrete. Heat flow is actually IR light flow which is radiation from one molecule to another: a charged particle vibrates and vibrations include accelerations which cause EM radiation that emanates out in all directions; then the EM wave encounters another charged particle which causes vibration and the cycle continues until all the energy is radiated out. It's a discrete process from molecule to molecule, but is modeled as continuous for simplicity's sake. I've long taken issue with the 3 modes of heat transmission (conduction, convention, radiation) because there is only radiation. Atoms do not touch, so they can't conduct, but the van der waals force simply transfers the vibrations more quickly when atoms are sufficiently close. Convection is simply vibrating atoms in linear motion that are radiating IR light. I have many issues with physics and have often described it as more of an art than a science (hence why it's so difficult). I mean, there are pages and pages on the internet devoted to simply trying to define heat.https://www.quora.com/What-is-heat-1https://www.quora.com/What-is-meant-by-heathttps://www.quora.com/What-is-heat-in-physicshttps://www.quora.com/What-is-the-definition-of-heathttps://www.quora.com/What-distinguishes-work-and-heat Physics is a mess. What gamma rays are, depends who you ask. They could be high-frequency light or any radiation of any frequency that originated from a nucleus. But I'm digressing.... I do not know what that means in the ultimate scheme of things. But I submit that even the most ardent finitist must at least give consideration to this historical reality. It just means we're using averages rather than discrete actualities and it's close enough. I hope I've been able to explain why I completely agree with your point that infinities in physical equations don't imply the actual existence of infinities. Yet at the same time, I am pointing out that our best THEORIES of physics are invariably founded on highly infinitary math. As to what that means ... for my own part, I can't help but feel that mathematical infinity is telling us something about the world. We just don't know yet what that is. I think it means there are really no separate things and when an aspect of the universe attempts to inspect itself in order to find its fundamentals or universal truths, it will find infinity like a camera looking at its own monitor. Infinity is evidence of the continuity of the singular universe rather than an existing truly boundless thing. Infinity simply means you're looking at yourself. Anyway, great post! Please don't be mad. Everyone here values your presence and are intimidated by your obvious mathematical prowess Don't take my pushback too seriously I'd prefer if we could collaborate as colleagues rather than competing.
In JIMS 4, p.78, Question 359 was asked by Ramanujan. (See The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society, p. 9, by Bruce Berndt, et al.) If, $$\sin(x+y) = 2\sin\big(\tfrac{1}{2}(x-y)\big)\tag1$$ $$\sin(y+z) = 2\sin\big(\tfrac{1}{2}(y-z)\big)\tag2$$ prove that, $$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}\tag3$$ The example by Ramanujan was, $$\begin{aligned} x &= \frac{\pi-\arcsin\big((\sqrt{5}-2)^3(4+\sqrt{15})^2\big)}{2}=1.094\dots\\ y &=\frac{\arcsin\big(\sqrt{5}-2\big)}{2}=0.119\dots\\ z &=\frac{\arcsin\big((\sqrt{5}-2)^3(4-\sqrt{15})^2\big)}{2}=0.0001\dots \end{aligned}$$ Ten years later, a 3-page proof was given in JIMS 15, p.114-117. I got an email asking if there was a shorter proof. Considering the problem, I observed the following. Given the quartic, $$a^3w^4+(1-3a^2)w^3+3a(1-a^2)w^2+a^2(3-a^2)w-a=0\tag4$$with $a=\tan(\color{blue}y/4)$. Define,$$x=4\tan^{-1} u\\z=4\tan^{-1} v$$ where $u,v$ are two appropriate roots of the quartic, then we get the bizarre relation as Ramanujan, same $$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2\color{blue}y)^{1/12}\tag5$$ Equivalently, those two roots $u,v$ obey, $$\left(\frac{1}{2}\,\frac{4(u-u^3)}{(u^2+1)^2}\frac{v^4-6v^2+1}{(v^2+1)^2} \right)^{1/4}+\left(\frac{1}{2}\,\frac{4(v-v^3)}{(v^2+1)^2}\frac{u^4-6u^2+1}{(u^2+1)^2} \right)^{1/4}=\big(\sin 2\color{blue}y\big)^{1/12}\tag6$$ For example, let $y=1$, so $a=\tan(1/4)$, then $u,v$ are the two real roots of (4). : Anyone knows a short proof for (3) and (6)? Question
TIPS FOR SOLVING QUESTIONS RELATED TO SIMPLE INTEREST: 1. Principal: The money borrowed or lent out for a certain period is called the Principal. 2. Interest: Interest is the sum which is paid for the use of other's money. 3. Simple Interest:Simple Interest is the interest on a sum borrowed for a certain period reckoned uniformly i.e. on initial principal only. 4. Formula for calculating Simple Interest: If Principal = P, Rate = R% per annum, and Time = T years, then \begin{aligned} \text{Simple Interest (S.I.)} = \left( \frac{P\times R\times T}{100}\right) \\ \end{aligned} \begin{aligned} \text{Principal (P) } = \left( \frac{100\times S.I.}{R\times T}\right) \\ \text{Rate (R)} = \left( \frac{100\times S.I.}{P\times T}\right) \\ \text{Time (T)} = \left( \frac{100\times S.I.}{P\times R}\right) \\ \end{aligned} 5. If the rate percent is given half yearly or quarterly, then to find the rate percent per annum multiply by 2 or 4 respectively. 6. One ordinary year (1 year) = 365 days. 7. One leap year = 366 days.
I'm working on a certain problem of slow, non-Newtonian, thin-film flow. This problem can be modelled with the incompressible Stokes equations: $\nabla\cdot 2\mu(\dot\varepsilon)\dot\varepsilon - \nabla p + \rho g = 0, \qquad \nabla\cdot u = 0$ where $u$ is the velocity, $\mu$ is the (nonlinear) viscosity, $\dot\varepsilon$ is the strain rate tensor, $p$ the pressure, $\rho$ the fluid density and $g$ the gravitational acceleration vector. In my field, there's a good approximation for when the fluid doesn't slip at all along the lower boundary, and a different approximation for when the fluid is in free slip along the lower boundary. The Stokes equations are costly to solve in the whole domain, so some folks have tried to come up with approximations to it which work in both the no-slip and free-slip regimes. My idea to approach this is to use semi-discretization in the $z$-direction, but with spectral methods. From looking at experimental evidence, I have a strong hunch that most of the vertical variation could be captured with the Legendre polynomials up to degree 4. The velocity field would be approximated as $u(x, y, z) = \sum_lp_l(z)u_l(x, y)$ to derive a coupled system of PDE for the fields $u_l$. This coupled system for $\{u_l\}$ can then be solved using, e.g. Galerkin finite element methods, finite difference methods, etc. I'm looking for any references or case studies where this approach has been used. I googled and found this paper on solving the Navier-Stokes equations in domains with only one direction of periodicity. There's also this paper which is more from a model reduction point of view. If there's a buzzword for this kind of approach I'd like to know what it is. Finally, any software implementations would be helpful. I can write this from the ground up using deal.II myself. I'm less familiar with FEniCS, Firedrake, or LifeV; do any of them have a way to create Frankenstein finite element spaces like this?
In this note we prove a general large deviation principle (LDP) for the longest success run in a sequence of independent Bernoulli trails. This study not only recovers several recently derived LDPs, but also gives new LDPs for the longest success run. The method is based on the Bryc’s inverse Varadhan lemma, which can be intuitively generalized to the longest success run in a two-state (success and failure) Markov chain. Bernstein processes over a finite time interval are simultaneously forward and backward Markov processes with arbitrarily fixed initial and terminal probability distributions. In this paper, a large deviation principle is proved for a family of Bernstein processes (depending on a small parameter ħ which is called the Planck constant) arising naturally in Euclidean quantum physics. The method consists in nontrivial Girsanov transformations of integral forms, suitable equivalence forms for large deviations and the (local and global) estimates on the parabolic kernel of the Schrödinger operator. Funding agencies: NTU MOE Tier 2 Grant [ARC3/13]; FCT [PTDC/MAT-STA/0975/2014] In this paper, we obtain new estimates on upper tail probabilities of suitable random series involving a distribution having an exponential tail. These estimates are exact, and the distribution is not heavy-tailed. In this paper we consider a family of generalized Brownian bridges with a small noise, which was used by Brennan and Schwartz [3] to model the arbitrage profit in stock index futures in the absence of transaction costs. More precisely, we study the large deviation principle of these generalized Brownian bridges as the noise becomes infinitesimal. (C) 2015 Elsevier Inc. All rights reserved. We obtain new upper tail probabilities of m-times integrated Brownian motions under the uniform norm and the L (p) norm. For the uniform norm, Talagrands approach is used, while for the L (p) norm, Zolotares approach together with suitable metric entropy and the associated small ball probabilities are used. This proposed method leads to an interesting and concrete connection between small ball probabilities and upper tail probabilities (large ball probabilities) for general Gaussian random variables in Banach spaces. As applications, explicit bounds are given for the largest eigenvalue of the covariance operator, and appropriate limiting behaviors of the Laplace transforms of m-times integrated Brownian motions are presented as well. Funding Agencies\|Simons Foundation [246211] Let $\{X_n\}_{n\geq1}$ be a sequence of i.i.d. standard Gaussian random variables, let $S_n=\sum_{i=1}^nX_i$ be the Gaussian random walk, and let $T_n=\sum_{i=1}^nS_i$ be the integrated (or iterated) Gaussian random walk. In this paper we derive the following upper and lower bounds for the conditional persistence:\begin{align}\mathbb{P}\left\{\max_{1\leq k \leq n}T_{k} \leq 0\,\,\Big\|\,\,T_n=0,S_n=0\right\}&\lesssim n^{-1/2},\\\mathbb{P}\left\{\max_{1\leq k \leq 2n}T_{k} \leq 0\,\,\Big\|\,\,T_{2n}=0,S_{2n}=0\right\}&\gtrsim\frac{n^{-1/2}}{\log n},\end{align}for $n\rightarrow\infty,$ which partially proves a conjecture by Caravenna and Deuschel (2008). We prove a version of the Feynman-Kac formula for Levy processes andintegro-differential operators, with application to the momentum representationof suitable quantum (Euclidean) systems whose Hamiltonians involve L´evytypepotentials. Large deviation techniques are used to obtain the limitingbehavior of the systems as the Planck constant approaches zero. It turns outthat the limiting behavior coincides with fresh aspects of the semiclassical limitof (Euclidean) quantum mechanics. Non-trivial examples of Levy processes areconsidered as illustrations and precise asymptotics are given for the terms inboth configuration and momentum representations. In this paper, we consider a family of Markov bridges with jumps constructed from truncated stable processes. These Markov bridges depend on a small parameter h greater than 0, and have fixed initial and terminal positions. We propose a new method to prove a large deviation principle for this family of bridges based on compact level sets, change of measures, duality and various global and local estimates of transition densities for truncated stable processes. In this note, we consider the non-negative least-square method with a random matrix. This problem has connections with the probability that the origin is not in the convex hull of many random points. As related problems, suitable estimates are obtained as well on the probability that a small ball does not hit the convex hull.
Physics307L:People/Josey/Rough Draft Contents Measuring and Predicting Background Radiation Using Poisson Statistics SJK 12:41, 28 November 2010 (EST) Author: Brian P. Josey Experimentalists: Brian P Josey and Kirstin G G Harriger Junior Lab, Department of Physics & Astronomy, University of New Mexico Albuquerque, NM 87131 bjosey@unm.edu Abstract SJK 12:53, 28 November 2010 (EST) In this paper, the results of the measurements of the background radiation were measured with differing time intervals over which the measurements took place. The fact that this data followed the Poisson distribution was demonstrated. After establishing the Poisson distribution, the results of the data were then used to predict the average rate of radiation per second. This estimated value was found to be 30 ± 4 radiations per second. A new set of data was generated from a run representing the rate for radiation per second. This data indicated that the average rate in the laboratory to be 30 ± 5 radiations per second, indicating that the technique is useful to model and measure background radiation. Introduction SJK 13:00, 28 November 2010 (EST) The Poisson distribution, first described by Siméon Denis Poisson in 1838 1, is used to describe events that occur in discrete numbers at random and independent intervals, but at a definitive average rate. Examples of this are radiation from nuclei over a range of time, number of births in a maternity ward over time, or divisions of cells in a culture 2. Mathematically, the Poisson distribution is given as: [math] P_{\mu} (\nu) = e^{-\mu} \frac {\mu ^ {\nu}} {\nu !} [/math] where, P-the probability of μ(ν) νcounts in a definitive interval, μ- mean number of counts in the time interval, This distribution has the unique property that the square root of the average is also the standard deviation 2. Here the Poisson distribution was demonstrated by measuring the background radiation in a standard physics teaching laboratory at 5000 ft elevation. Several sets of data were created by varying the size of the time interval of the radiation, referred to as the "window size." After taking several different sets of data with the window size ranging from 10 ms to 800 ms, the average rate per second was calculated, and then compared to data taken from a set of data of 1 second window size. Methods and Materials SJK 13:12, 28 November 2010 (EST) For this experiment, a combined scintillator-photomultiplier tube (scintillator-PMT), Figure 1 was used to collect data 3. To do this, every time the scintillator detected radiation, it would fire a beam of ultraviolet light to the PMT. The PMT would then convert this light signal into a single voltage SJK 13:06, 28 November 2010 (EST) . This voltage would be carried to an internal MCS card in a computer, where it would be analyzed by a UCS 30 software. This software counts each signal voltage and create a set of data containing the size of the window over which the data was collected, the time and the number of radiation events to occur in that window. In order for the scintillator to detect the radiation, it had to have a potential gradient that would pick up ions created in the radiation event. This potential was supplied by a Spectech Universal Computer Spectrometer power supply, Figure 2, and set to 1200 V throughout the course of the experiment. The collection of the data was carried out by using the UCS 30 software. It would create a consecutive series of windows of set interval of time, and count the number of signal currents, which represent the number of radiation events, that occurred within each window. This data was then saved into data file that could be manipulated and processed using MATLAB v. 2009a. To demonstrate the Poisson distribution, the scintillator-PMT, power supply, and computer were all turned on. The UCS 30 software was then uploaded, and the window size was set to various lengths of time. Data was collected over a series of sets of windows, the window sizes were set at 10, 20, 40, 80, 100, 200, 400 and then 800 ms. Each run contained 2064 windows of the given size. The data was then analyzed, see results and discussion below, to demonstrate that background radiation did follow a Poisson distribution. This data was then used to predict the behavior of a similar set of data that occurred for a 1 second window size. After predicting its behavior, the system was ran again, using a 2064 windows of 1 second length. This data was then compared to the predicted results form the initial data. SJK 13:09, 28 November 2010 (EST) Results SJK 13:51, 28 November 2010 (EST) From the scintillator-PMT and the UCS 30 software, the data was processed using Google Docs to determine the range of the number of radiation events per window, and the number of windows with the given number of radiation events. A fractional distribution of the data was also determined. This data is summarized in Table 1 below: {{#widget:Google Spreadsheet \|key=0AjJAt7upwcA4dERfb3NIc0xhbmt1dnExMWhUdG1QS2c \|width=600 \|height=325 }} Table 1This table summarizes the raw data from each of the initial experimental runs. For each window size, the whole range of number of radiation events per window is given in the first column. In the second column, the number of windows that had this given number of radiation events is given. For example, for the 10 ms run, only a single window had four radiation events in it. The third and final column gives the fractional probability, out of 1, of such event happening in the given run. SJK 13:53, 28 November 2010 (EST) SJK 13:56, 28 November 2010 (EST) From this data, a graph was generated to illustrate the characteristics of the data. This graph is given in Figure 3. For the sake of simplicity only four of the nine data runs are graphed, however, even from this small sample of data, the trends across the whole of the sample are still clear. While the exact implications of this data is discussed in the discussion section below, the data makes it clear that as the window sizes increases, the most probable number of radiations per window increases, and the distribution of the probabilities spreads. This spreading in the data is a result of a greater standard error. Together these two trends are qualitative reasons to believe that the data follows a Poisson distribution. This is further discussed bellow in the discussion section 4. A more potent argument for the Poisson distribution is to determine the averages and standard deviations for each set of data. This data is summarized in Table 2 below: {{#widget:Google Spreadsheet \|key=0AjJAt7upwcA4dF8yb09KUzVVYTg5LUcxaG43bTlvdHc \|width=600 \|height=325 }} Table 2This table summarizes the raw data collected from the runs with varying window sizes, from 10 ms to 800 ms. The first column represents the average radiation event per window as calculated directly from the data. The second and third rows represent the standard deviation of the data as the square root of the average, and as directly calculated from the data. The fourth row represents the percent error of the standard deviation from the data from the square root standard deviation. Because of the very low difference, it is clear that the data follows a Poisson distribution. The last two rows are the average and standard deviation converted from window size to rate per second. As this table illustrates, the average radiation event per window sizes grows as the window sizes increases, and the standard deviation in proportion to the average also increases. There are two standard deviations in the table. The first is the square root of the average, while the second is directly calculated from the data. The reason for this, discussed in greater detail below, is that the standard deviation of a Poisson distribution is identical to the square root of the average. As the percent error between the two values shows, the data does follow this trend very closely, and the small differences, that never exceed 1.5 %, indicate that this is true. The averages and standard deviations from the data were then divided by the window size to give the rate and error as number of radiations per second. These values are then graphed in Figure 4. Clearly the standard deviation decreases as the window size approaches 1 second, indicating the greater accuracy in the data. From this data, the average rate and standard deviation for a windows size of 1 second was calculated. This value, x wav, was calculated using weighted averages 2: [math] x_{wav} = \frac {\sum {w_i x_i}} {\sum {w_i}} [/math] where x- is the best estimate for the average number of radiations per second, wav w- is the weight of each average, this value is just the inverse of the square of the standard error for a given run, i x- is the average value from each run i The standard deviation is then given as: [math] \sigma_{wav} = \frac {1} {\sqrt {\sum {w_i}}} [/math] where again, w i is the weight of each average, and σis the best estimate for the standard deviation in the final average. These calculations then give a predicted value of 30 ± 4 radiations per second. Another experiment was ran, this time the window size was raised to 1 second. The data from this experiment is summarized in wav Table 3below. It is important to note that the predicted value for the average number of radiations per second is only off by the measured value by 0.725 %. SJK 14:07, 28 November 2010 (EST) {{#widget:Google Spreadsheet \|key=0AjJAt7upwcA4dC04WnpvSmM4Zzd2Rld1elo2QWo2YkE \|width=600 \|height=325 }} Table 3This table summarizes the data taken for a window size of 1 second. On the left is the average predicted by the calculations, its square root, and predicted standard deviation. Below this is the actual average, its square root, standard deviation and the percent difference between the prediction and the measured amount. On the left is the range of radiations per minute, the raw number of windows with a given number and their fractional frequency. Discussion SJK 14:10, 28 November 2010 (EST) Qualitatively, the Poisson distribution has a specific trend in its behavior, namely, as the window size over which the counts were taken increases, the average number of events per window increases, and the distribution spreads out over time 4. This results in probability graphs moving to the right and flattening in a standard x-y graph as Figure 3 demonstrates. This behavior in the data indicates that a Poisson distribution could be the mostly likely way to model and represent the data. However, there is a much stronger argument for this conclusion. The qualitative behavior in the Poisson distribution is actually the result of the change in the standard deviation from the mean 4. As Table 2 illustrates, the square root of the mean is close enough to the experimental deviation. In fact, using the theory of large numbers, it can be shown that the standard deviation of a Poisson distribution and the square root of the mean are the same 2. The data for the first several experimental runs illustrates that the standard deviation and square root of the mean never differ by more than 1.3 %, indicating a very close relationship. This relationship is so close that it is clear that the background radiation in the laboratory can be modeled by a Poisson distribution. Also, this means that as the mean number of radiations per window increases, the standard deviation increases proportional to the square root of the mean, giving the spreading behavior in Figure 3. Armed with this knowledge of the behavior of the background radiation, the means and standard deviations for the experimental sets between 10 ms and 800 ms were used to determine the average rate of radiation per second. As discussed in the results section, this value was found by using the weighted average method to account for varying degrees of certainty in the measurements. This generated a value of 30 ± 4 radiations per second, and the experimental test for the one second time window gave a final result of 30 ± 6 radiations per minute. As summarized in Table 3 the difference between the two values was only 0.725 %. Because of this close relationship between the two values, it is clear that the method of applying the Poisson distribution to calculate the average radiation rate is successful. This can then be used for any other window size, where determining the final result is only a function of scale. Conclusions SJK 14:12, 28 November 2010 (EST) For this paper, the radiation rates were measured for varying time intervals to determine its behavior and find a mathematical representation of it. Because the behavior in the probability distributions followed a pattern where the mean grew as the interval increased and the experimental standard deviation grew as the square root of the mean, it became clear that the Poisson distribution is the best way to model the behavior of the background radiation. The difference between an ideal Poisson distribution and the data was found to never vary more than 1.3 % for all the data. Prompted by this, the average rate of radiation per second was calculated from the data. This gave an average rate of 30 ± 4 radiations per second. These values were checked experimentally, giving a value of 30 ± 6 radiations per second, showing that the model was acceptable and could create accurate predictions. The utility of this models is found for any process that, like background radiation, occurs at random independent times, but a definitive average rate. The methods presented here can be used to predict the behavior of these processes to gain very accurate results. Acknowledgments SJK 14:14, 28 November 2010 (EST) As always, I would like to thank the professor for this course, Dr. Steve Koch, the TA, Katie Richardson and my fellow experimenter, Kirstin Harriger. All of whom were very helpful in conducting the lab. I would like to also thank Dr. Michael Gold, the original professor of the course for his manual and work in initially producing this experiment and procedure. References SJK 12:39, 28 November 2010 (EST) Poisson, Siméon Denis Recherches sur la probabilité des jugements en matière criminelle et en matière civile(1838) Link here (in French). Taylor, John R. An Introduction to Error Analysis: The Study of Uncertainties in Physical Measurements2e (1997) pg. 174-5 for weighted average 245- 260 for Poisson, Amazon link Gold, Michael Physics 307L: Junior Lab Manual(2006) link here Wikipedia Poisson Distributionarticle form the web, external link Koch Comments Steve Koch 12:39, 28 November 2010 (EST): This was a page-turner, even on first read-through! Excellent first draft, and great approach to the open-ended lab. My main criticisms were: (1) incomplete methods, (2) I question the weighted mean method/pretty sure I'm right, and (3) lack of peer-reviewed citations. All very easy to fix. So, what to do for the "extra data/analysis" week? Well, first, try the correct to the weighted mean and make the new graph. Other than that, it's up to you. Can you determine whether you're really looking at mostly background radiation? Is it time dependent (time of day/longer term)? What do the energies of your pulses look like. If you make the MCS too sensitive, do you start triggering on electrical noise, and does it become non-Poissonian? Another idea would be to write Matlab code to simulate poisson data. I am think I may lecture about this tomorrow so you'd know how to do it. Then you could analyze the simulated data with your same methods and see how well it works for known poisson data. Finally, you could implement further methods, such as in the paper that I sent you. For that, you would need to know the arrival times for all the events, and I don't think the software can do that. But you could use simulated data. OK, thanks again for the excellent first draft!
User:Jjtorrens/BancoDePruebas $\newcommand{\abs}[1]{\lvert #1\rvert} \newcommand{\norm}[1]{\lVert #1\rVert} \newcommand{\bfl}{\mathbf{l}}$ A space $W^l_p(\Omega)$ of functions $f=f(x)=f(x_1,\ldots,x_n)$ on a set $\Omega\subset\RR^n$ (usually open) such that the $p$-th power of the absolute value of $f$ and of its generalized derivatives (cf. Generalized derivative) up to and including order $l$ are integrable ($1\leq p\leq \infty$). The norm of a function $f\in W^l_p(\Omega)$ is given by \begin{equation}\label{eq:1} \norm{f}_{W^l_p(\Omega)}=\sum_{\abs{k}\leq l} \norm{f^{(k)}}_{L_p(\Omega)}. \end{equation} Here \begin{equation} f^{(k)}=\frac{\partial^{\lvert k\rvert}f}{\partial x_1^{k_1}\cdots \partial x_n^{k_n}},\qquad f^{(0)}=f, \end{equation} is the generalized partial derivative of $f$ of order $\abs{k}=\sum_{j=1}^n k_j$, and \begin{equation} \norm{\psi}_{L_p(\Omega)} =\left( \int_\Omega \abs{\psi(x)}^p\,dx \right)^{1/p} \qquad (1\leq p\leq \infty). \end{equation} When $p=\infty$, this norm is equal to the essential supremum: \begin{equation} \norm{\psi}_{L_\infty(\Omega)} =\operatorname{ess sup}_{x\in\Omega}\abs{\psi(x)} \qquad (p=\infty), \end{equation} that is, to the greatest lower bound of the set of all $A$ for which $A<\abs{\psi(x)}$ on a set of measure zero. Since its definition involves generalized derivatives rather than ordinary ones, it is complete, that is, it is a Banach space. $W^l_p(\Omega)$ is considered in conjunction with the linear subspace $W^l_{pc}(\Omega)$ consisting of functions having partial derivatives of order $l$ that are uniformly continuous on $\Omega$. $W^l_{pc}(\Omega)$ has advantages over $W^l_p(\Omega)$, although it is not closed in the metric of $W^l_p(\Omega)$ and is not a complete space. However, for a wide class of domains (those with a Lipschitz boundary, see below) the space $W^l_{pc}(\Omega)$ is dense in $W^l_p(\Omega)$ for all $p$, $1\leq p<\infty$, that is, for such domains the space $W^l_p(\Omega)$ acquires a new property in addition to completeness, in that every function belonging to it can be arbitrarily well approximated in the metric of $W^l_p(\Omega)$ by functions from $W^l_{pc}(\Omega)$. It is sometimes convenient to replace the expression \eqref{eq:1} for the norm of $f\in W^l_p(\Omega)$ by the following: \begin{equation}\label{eq:2} \norm{f}^\prime_{W^l_p(\Omega)}=\left( \int_\Omega \sum_{\abs{k}\leq l} \abs{f^{(k)}(x)}^p \,dx \right)^{1/p} \qquad (1\leq p<\infty). \end{equation} The norm \eqref{eq:2} is equivalent to the norm \eqref{eq:1}, i.e. $c_1 \norm{f}\leq\norm{f}^\prime\leq c_2\norm{f}$, where $c_1, c_2>0$ do not depend on $f$. When $p=2$, \eqref{eq:2} is a Hilbert norm, and this fact is widely used in applications. The boundary $\Gamma$ of a bounded domain $\Omega$ is said to be Lipschitz if for any $x^0\in\Gamma$ there is a rectangular coordinate system $\xi=(\xi_1,\ldots,\xi_n)$ with origin $x^0$ so that the box \begin{equation} \Delta=\{ \xi : \abs{\xi_j}<\delta,\ j=1,\ldots,n \} \end{equation} is such that the intersection $\Gamma\cap\Delta$ is described by a function $\xi_n=\psi(\xi')$, with \begin{equation} \xi'=(\xi_1,\ldots,\xi_n)\in\Delta'=\{\abs{\xi_j}<\delta,\ j=1,\ldots,n-1\}, \end{equation} which satisfies on $\Delta'$ (the projection of $\Delta$ onto the plane $\xi_n=0$) the Lipschitz condition \begin{equation} \abs{\psi(\xi'_1)-\psi(\xi'_2)}\leq M \abs{\xi'_1-\xi'_2},\quad \xi'_1,\xi'_2\in\Delta', \end{equation} where the constant $M$ does not depend on the points $\xi'_1,\xi'_2$, and $\abs{\xi}^2=\sum_{j=1}^{n-1}\xi_j^2$. All smooth and many piecewise-smooth boundaries are Lipschitz boundaries. For a domain with a Lipschitz boundary, \eqref{eq:1} is equivalent to the following: \begin{equation} \norm{f}_{W^l_p(\Omega)}=\norm{f}_{L_p(\Omega)}+\norm{f}'_{w^l_p(\Omega)}, \end{equation} where \begin{equation} \norm{f}'_{w^l_p(\Omega)}=\sum_{\abs{k}=l}\norm{f^{(k)}}_{L_p(\Omega)}. \end{equation} One can consider more general anisotropic spaces (classes) $W^\bfl_p(\Omega)$, where $\bfl=(l_1,\ldots,l_n)$ is a positive vector (see Imbedding theorems). For every such vector $\bfl$ one can define, effectively and to a known extent exhaustively, a class of domains $\mathfrak{M}^{(\bfl)}$ with the property that if $\Omega\in\mathfrak{M}^{(\bfl)}$, then any function $f\in W^\bfl_p(\Omega)$ can be extended to $\R^n$ within the same class. More precisely, it is possible to define a function $\overline{f}$ on $\R^n$ with the properties \begin{equation} \overline{f}(x)=f(x),\quad x\in\Omega, \quad \norm{\overline{f}}_{W^\bfl_p(\R^n)}\leq c \norm{f}_{W^\bfl_p(\R^n)}, \end{equation} where $c$ does not depend on $f$ (see [3]). In virtue of this property, inequalities of the type found in imbedding theorems for functions $f\in W^\bfl_p(\R^n)$ automatically carry over to functions $f\in W^\bfl_p(\Omega)$, $\Omega\in\mathfrak{M}^{(\bfl)}$. For vectors $\bfl=(l,\ldots,l)$, the domains $\Omega\in\mathfrak{M}^{(\bfl)}$ have Lipschitz boundaries, and $W^\bfl_p(\Omega)=W^l_p(\Omega)$. The investigation of the spaces (classes) $W^\bfl_p(\Omega)$ ($\Omega\in\mathfrak{M}^{(\bfl)}$) is based on special integral representations for functions belonging to these classes. The first such representation was obtained (see [1], [2]) for an isotropic space $W^\bfl_p(\Omega)$ of a domain $\Omega$, star-shaped with respect to some sphere. For the further development of this method see, for example, [3]. The classes $W^\bfl_p$ and $W^l_p$ can be generalized to the case of fractional $l$, or vectors $\bfl=(l_1,\ldots,l_n)$ with fractional components $l_j$. The space $W^l_p(\Omega)$ can also be defined for negative integers $l$. Its elements are usually generalized functions, that is, linear functionals $(f,\phi)$ on infinitely-differentiable functions $\phi$ with compact support in $\Omega$. By definition, a generalized function $f$ belongs to the class $W^{-l}_p(\Omega)$ ($l=1,2,\ldots$) if \begin{equation} \norm{f}_{W^{-l}_p(\Omega)}=\sup(f,\phi) \end{equation*} is finite, where the supremum is taken over all functions $\phi\in W^l_q(\Omega)$ with norm at most one ($1/p+1/q=1$). The functions $f\in W^{-l}_p(\Omega)$ form the space adjoint to the Banach space $W^l_q(\Omega)$. References [1] S.L. Sobolev, "On a theorem of functional analysis" Transl. Amer. Math. Soc. (2) , 34 (1963) pp. 39–68 Mat. Sb. , 4 (1938) pp. 471–497 [2] S.L. Sobolev, "Some applications of functional analysis in mathematical physics" , Amer. Math. Soc. (1963) (Translated from Russian) [3] O.V. Besov, V.P. Il'in, S.M. Nikol'skii, "Integral representations of functions and imbedding theorems" , 1–2 , Wiley (1978) (Translated from Russian) [4] S.M. Nikol'skii, "Approximation of functions of several variables and imbedding theorems" , Springer (1975) (Translated from Russian) Comments References [a1] V.G. Maz'ja, "Sobolev spaces" , Springer (1985) [a2] F. Trèves, "Basic linear partial differential equations" , Acad. Press (1975) pp. Sects. 24–26 [a3] R.A. Adams, "Sobolev spaces" , Acad. Press (1975) How to Cite This Entry: Jjtorrens/BancoDePruebas. Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Jjtorrens/BancoDePruebas&oldid=25987
The following exercise is inspired by an old exam question (note that this is not the question): Let us define a single-letter alphabet $\Sigma = \{a\}$ and the language $L = \{ w \in \Sigma^*: \|w\| = 2^n, n \in \mathbb Z_+\}$. (In other words, $L = \{a^2, a^4, a^8, \ldots \}$.) What is an unrestricted grammar for $L$? I thought of a solution using binary number representation. Consider the grammar: $S \rightarrow 10TaX$ $T \rightarrow 0T \| \epsilon$ $0a \rightarrow aa0$ $0X \rightarrow X$ $1a \rightarrow a1$ $1X \rightarrow \epsilon$ $\epsilon$ signifies the empty string. Using the first two production rules, we can produce a string of the type $10^naX$, for example $100aX$. One route here to get rid of the non-terminal symbols is $100aX \Rightarrow 10aa0X \Rightarrow 1aa0a0X \Rightarrow 1aaaa00X \Rightarrow 1aaaa0X$ $\Rightarrow 1aaaaX \Rightarrow \ldots \Rightarrow aaaa1X \Rightarrow aaaa = a^4$. Assuming that this solution is correct, one interesting thing about it is that by modifying the third production rule to $0a \rightarrow a^k0$ one can get a restriction-free grammar for strings of length $k^n$ instead. But maybe there is a cleaner solution, so I am interested in seeing other examples of unrestricted grammars for the language. However, here comes the actual question: the existence of an unrestricted grammar for $L$ only proves that it is recursively enumerable. Therefore I would like to ask for an unrestricted grammar of $\overline{L}$ (then we have one proof that $L$ is a Turing-decidable language). Edit: In case the question here is unclear, please read the discussion in the comments.
Composition of Functions Example $\PageIndex{1}$ Sociologists in Holland determine that the number of people $y$ waiting in a water ride at an amusement park is given by \[y = \dfrac{1}{50}C^2 + C + 2 \nonumber \] where $C$ is the temperature in degrees $C$. The formula to convert Fahrenheit to Celsius $C$ is given by \[C = \dfrac{5}{9}F + \dfrac{160}{9}. \nonumber \] To get a function of $F$ we compose the two function: \[y(C(F)) = \left(\dfrac{1}{50}\right)\left[\dfrac{5}{9}F + \dfrac{160}{9}\right]^2 + \left[\dfrac{5}{9}F + \dfrac{160}{9}\right] + 2 \nonumber \] Exercise $\PageIndex{1}$ If $f(x) = 3x + 2$ $g(x) = 2x^2 + 1$ $h(x) = \sqrt{x-2}$ $c(x) = 4$ Find $f(g(x))$ $f(h(x))$ $f(f(x))$ $h(c(x))$ $c(f(g(h(x))))$ 1-1 Functions Definition: 1-1 (one-to-one) A function $f(x)$ is 1-1 if \[f(a) = f(b)\] implies that \[a = b.\] Example $\PageIndex{2}$ If \[f(x) = 3x + 1 \nonumber\] then \[3a + 1 = 3b + 1 \nonumber\] implies that \[3a = 3b \nonumber\] hence \[a = b \nonumber\] therefore $f(x)$ is 1-1. Example $\PageIndex{3}$ If \[f(x) = x^2 \nonumber\] then \[a^2 = b^2 \nonumber\] implies that \[a^2-b^2 = 0 \nonumber\] or that \[(a - b)(a + b) = 0 \nonumber\] hence \[a = b \text{ or } a = -b \nonumber\] For example \[f (2) = f (-2) = 4 \nonumber\] Hence $f(x)$ is not 1-1. Horizontal Line Test If every horizontal line passes through $f(x)$ at most once then $f(x)$ is 1-1. Inverse Functions Definition: Inverse function A function $g(x)$ is an inverse of $f(x)$ if \[f(g(x)) = g(f(x)) = x.\] Example $\PageIndex{4}$ The volume of a lake is modeled by the equation \[V(t) = \dfrac{1}{125}h^3. \nonumber\] Show that the inverse is \[h(N) = 5V^{\frac{1}{3}}. \nonumber\] Solution: We have \[h(V(h)) = 5(\dfrac{1}{125}h^3)^{\frac{1}{3}} = \dfrac{5}{5}h = h \nonumber\] and \[v(h(V)) = \dfrac{1}{125}(5V^{\frac{1}{3}})^3 = \dfrac{1}{125}(125V) = V. \nonumber\] Step by Step Process for Finding the Inverse Interchange the variables Solve for $y$ Write in terms of $f^{-1}(x)$ Example$\PageIndex{5}$ Find the inverse of \[f (x) = y = 3x^3 - 5 \nonumber\] Solution: \[\begin{align} x &= 3y^3 - 5 \\ x + 5 &= 3y^3 \\ \dfrac{(x + 5)}{3} &= y^3 ,\\ \left[\dfrac{(x + 5)}{3}\right]^{\frac{1}{3}}&=y \end{align}\] \[f^{-1}(x) = \left[\dfrac{(x + 5)}{ 3 }\right]^{\frac{1}{3}}. \nonumber\] Graphing To graph an inverse we draw the $y = x$ line and reflect the graph across this line. To interactively view the graph of an inverse click here: http://mathcsjava.emporia.edu/~greenlar/Inverse/inverse.html Contributors Larry Green (Lake Tahoe Community College) Integrated by Justin Marshall.
Directional Derivatives Suppose you are given a topographical map and want to see how steep it is from a point that is neither due West or due North. Recall that the slopes due north and due west are the two partial derivatives. The slopes in other directions will be called the directional derivatives. Formally, we define Definition: Directional Derivatives Let $f(x,y)$ be a differentiable function and let u be a unit vector then the directional derivative of $f$ in the direction of u is \[ D_u f(x,y) =\lim_{t \rightarrow 0} \dfrac{f(x+tu_1,y+tu_2) - f(x,y) }{t}. \] Note that if u is $\hat{\textbf{i}}$ then the directional derivative is just $f_x$ and if u is $ \hat{\textbf{i}} $ the it is $f_y$. Just as there is a difficult and an easy way to compute partial derivatives, there is a difficult way and an easy way to compute directional derivatives. Let $f(x,y)$ be a differentiable function, and u be a unit vector with direction $\hat{\textbf{q}} $, then \[ D_u f(x,y) = \left \langle f_x,f_y \right \rangle \cdot \left \langle \cos \theta, \sin \theta \right \rangle .\] Example $\PageIndex{1}$ Let \[ f(x,y) = 2x + 3y^2 - xy \] and \[ \textbf{v} = \left \langle 3,2 \right \rangle .\] Find \[ D_v f(x,y) .\] Solution We have \[ f_x = 2 - y \;\;\; \text{and} \;\;\; f_y = 6y - x \] and \[\|\| \textbf{v} \|\| = \sqrt{9+4} = \sqrt{13} . \] Hence \[ D_v f(x,y) = \left \langle 2 - y, 6y - x \right \rangle \cdot \left \langle \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right \rangle \] \[ = \dfrac{2}{\sqrt{13}} (2-y) + \dfrac{3}{\sqrt{13}} (6y-x). \] Let \[f(x,y)= e^{xy^2} \;\;\; and \;\;\; v=\langle2,-5\rangle . \] Find $D_v \; f(x,y) $. The Gradient We define \[\nabla f = \langle f_x,f_y\rangle . \] Notice that \[ D_u f(x,y) = (\nabla f) \cdot u .\] The gradient has a special place among directional derivatives. The theorem below states this relationship. Theorem If $\nabla f(x,y) = 0$ then for all u, $D_u f(x,y) = 0$. The direction of $\nabla f(x,y)$ is the direction with maximal directional derivative. The direction of $\nabla f(x,y)$ is the direction with the minimal directional derivative. Proof 1. If \[\nabla f(x,y) = 0\] then \[D_u f(x,y) = \nabla f \cdot \textbf{u} = 0 \cdot \textbf{u} = 0. \] 2. \[D_u f(x,y) = \nabla f \cdot \textbf{u} = \|\| \nabla f \|\| \cos q. \] This is a maximum when $q = 0$ and a minimum when $q = p$. If $q = 0$ then $ \nabla f$ and u point in the same direction. If $q = p$ then u and $ \nabla f$ point in opposite directions. This proves 2 and 3. Example $\PageIndex{2}$ Suppose that a hill has altitude \[ w(x,y) = x^2 - y .\] Find the direction that is the steepest uphill and the steepest downhill at the point $(2,3)$. Solution We find \[\nabla w = \langle 2x, -y\rangle = \langle 4, -3\rangle .\] Hence the steepest uphill is in the direction \[\langle 4,-3\rangle \] while the steepest downhill is in the direction \[-\langle 4,-3\rangle = \langle -4,3\rangle .\] The Gradient and Level Curves If $f$ is differentiable at $(a,b)$ and $ \nabla f$ is nonzero at $(a,b)$ then $ \nabla $ is perpendicular to the level curve through $(a,b)$. Contributors Larry Green (Lake Tahoe Community College) Integrated by Justin Marshall.
I suspect there is in general not much difference between GMRES and CG for an SPD matrix. Let's say we are solving $ Ax = b $ with $ A $ symmetric positive definite and the starting guess $ x_0 = 0 $ and generating iterates with CG and GMRES, call them $ x_k^c $ and $ x_k^g $. Both iterative methods will be building $ x_k $ from the same Krylov space $ K_k = \{ b, Ab, A^2b, \ldots \} $. They will do so in slightly different ways. CG is characterized by minimizing the error $ e_k^c = x - x_k^c $ in the energy norm induced by $ A $, so that\begin{equation} (A e_k^c, e_k^c) = (A (x - x_k^c), x - x_k^c) = \min_{y \in K} (A (x-y), x-y).\end{equation} GMRES minimizes instead the residual $ r_k = b - A x^g_k $, and does so in the discrete $ \ell^2 $ norm, so that\begin{equation} (r_k, r_k) = (b - A x_k^g, b - A x_k^g) = \min_{y \in K} (b - Ay, b - Ay).\end{equation}Now using the error equation $ A e_k = r_k $ we can also write GMRES as minimizing\begin{equation} (r_k, r_k) = (A e_k^g, A e_k^g) = (A^2 e_k^g, e_k^g)\end{equation}where I want to emphasize that this only holds for an SPD matrix $ A $. Then we have CG minimizing the error with respect to the $ A $ norm and GMRES minimizing the error with respect to the $ A^2 $ norm. If we want them to behave very differently, intuitively we would need an $ A $ such that these two norms are very different. But for SPD $ A $ these norms will behave quite similarly. To get even more specific, in the first iteration with the Krylov space $ K_1 = \{ b \} $, both CG and GMRES will construct an approximation of the form $ x_1 = \alpha b $. CG will choose\begin{equation} \alpha = \frac{ (b,b) }{ (Ab,b) }\end{equation}and GMRES will choose\begin{equation} \alpha = \frac{ (Ab,b) }{ (A^2b,b) }.\end{equation}If $ A $ is diagonal with entries $ (\epsilon,1,1,1,\ldots) $ and $ b = (1,1,0,0,0,\ldots) $ then as $ \epsilon \rightarrow 0 $ the first CG step becomes twice as large as the first GMRES step. Probably you can construct $ A $ and $ b $ so that this factor of two difference continues throughout the iteration, but I doubt it gets any worse than that.
Here is one explanation of why countable choice is not problematic in constructive mathematics. For this discussion it is useful to formulate the axiom of choice as follows: $(\forall x \in X . \exists y \in Y . R(x,y)) \implies \exists f \in Y^X . \forall x \in X . R(x,f(x))$ This says that a total relation $R \subseteq X \times Y$ contains a function. The usual formulation of the axiom of choice is equivalent to the above one. Indeed, if $(S_i)_{i \in I}$ is a family of non-empty sets we take $X = I$, $Y = \bigcup_i S_i$ and $R(i,x) \iff x \in S_i$ to obtain a choice function $f : I \to \bigcup_i S_i$. Conversely, given a total relation $R \subseteq X \times Y$, consider the the family $(S_x)_{x \in X}$ where $S_x = \lbrace y \in Y \mid R(x,y)\rbrace$ and apply the usual axiom of choice. One way of viewing sets in constructive mathematics is to imagine that they are collections together with given equality, i.e., some sort of "presets" equipped with equivalence relations. This actually makes sense if you think about how we implement abstract sets in computers: each element of the abstract set is represented by a finite sequence of bits, where each element may have many valid representations (and this is unavoidable in general). Let me give two specific examples: a natural number $n \in \mathbb{N}$ is represented in the usual binary system, and let us allow leadings zeroes, so that $42$ is represented by $101010$ as well as $0101010$, $00101010$, etc. a (computable) real $x \in \mathbb{R}$ is represented by machine code (a binary string) that computes arbitrarily good approximations of $x$. Specifically, a piece of code $p$ represents $x$ when $p(n)$ outputs a rational number that differs from $x$ by at most $2^{-n}$. Of course we only represent computable reals this way, and every computable real has many different representations. Let me write $\mathbb{R}$ for the set of computable reals, because those are the only reals relevant to this discussion. An essential difference between the first and the second example is that there is a computable canonical choice of representatives for elements of $\mathbb{N}$ (chop off the leading zeroes), whereas there is no such canonical choice for $\mathbb{R}$, for if we had it we could decide equality of computable reals and consequently solve the Halting problem. According to the constructive interpretation of logic, a statement of the form $\forall x \in X . \exists y \in Y . R(x, y)$ holds if there is a program $p$ which takes as input a representative for $x \in X$ and produces a representative for $y \in Y$, together with a witness for $R(x,y)$. Crucially, $p$ need not respect equality of $X$. For example, $\forall x \in \mathbb{R} . \exists n \in \mathbb{N} . x < n$ is accepted because we can write a program which takes as input a representative of $x$, namely a program $p$ as described above, and outputs a natural number larger than $x$, for example $round(p(0)) + 1000$. However, the number $n$ will necessarily depend on $p$, and there is no way too make it depend only on $x$ (computably). Let us have a look at the axiom of choice again: $(\forall x \in X . \exists y \in Y . R(x,y)) \implies \exists f \in Y^X . \forall x \in X . R(x,f(x))$ We accept this if there is a program which takes as input a $p$ witnessing totality of $R$ and outputs a representative of a choice function $f$, as well as a witness that $\forall x \in X. R(x, f(x))$ holds. This is probematic because $p$ need not respect equality of $X$, whereas a representative for $f$ must respect equality. It is not clear where we could get it from, and in specific examples we can show that there isn't one. Already the following fails: $(\forall x \in \mathbb{R} . \exists n \in \mathbb{N} . x < n) \implies \exists f \in \mathbb{N}^\mathbb{R} . \forall x \in \mathbb{R} . x < f(x)$. Indeed, every computable map $f : \mathbb{R} \to \mathbb{N}$ is constant (because a non-constant one would allow us to write a Halting oracle). However, if we specialize to countable choice $(\forall n \in \mathbb{N} . \exists y \in Y . R(n,y)) \implies \exists f \in Y^\mathbb{N} . \forall n \in \mathbb{N} . R(n,f(n))$ then we can produce the desired program. Given $p$ that witnesses totality of $R$, define the following program $q$ that represents a choice function: $q$ takes as input a binary representation of a natural number $n$, possibly with leading zeroes, chops of the leading zeroes, and applies $p$. Now, even if $p$ did not respect equality of natural numbers, $q$ does because it applies $p$ to canonically chosen representatives. In general, we will accept choice for those sets $X$ that have computable canonical representatives for their elements. Ok, this was a bit quick, but I hope I got the idea accross. Let me finish with a general comment. Most working mathematicians cannot imagine alternative mathematical universes because they were thoroughly trained to think about only one mathemtical universe, namely classical set theory. As a result their mathematical intuition has fallen a victim to classical set theory. The first step towards understanding why someone might call into question a mathematical principle which seems obviously true to them, is to broaden their horizon by studying other mathematical universes. On a smaller scale this is quite obvious: one cannot make sense of non-Euclidean geometry by interpreting points and lines as those of the Euclidean plane. Similarly, you cannot understand in what way the axiom of choice could fail by interpreting it in classical set theory. You must switch to a different universe, even though you think there isn't one... Of course, this takes some effort, but it's a real eye-opener.
Illinois Journal of Mathematics Illinois J. Math. Volume 43, Issue 4 (1999), 733-751. Hardy's inequality and embeddings in holomorphic Triebel-Lizorkin spaces Abstract In this work we study some properties of the holomorphic TriebeI-Lizorkin spaces $H F^{pq}_{s}$, $0 \lt p$, $q \leq \infty$, $s \in \mathbb{R}$, in the unit ball $B$ of $\mathbb{C}^{n}$, motivated by some well-known properties of the Hardy-Sobolev spaces $H^{p}_{s} = H F^{p^{2}}_{s}$, $0 \lt p \lt \infty$. We show that $\sum_{n \geq 0}\|a_{n}\|/(n + 1) \lesssim \|\|\sum_{n \geq 0}a_{n}z^{n}\|\|_{H F^{1 \infty}_{0}}$, which improves the classical Hardy's inequality for holomorphic functions in the Hardy space $H^{1}$ in the disc. Moreover, we give a characterization of the dual of $HF^{1q}_{s}$, which includes the classical result $(H^{1})^{\ast} = \mathrm{BMOA}$. Finally, we prove some embeddings between holomorphic Triebel-Lizorkin and Besov spaces, and we apply them to obtain some trace theorems. Article information Source Illinois J. Math., Volume 43, Issue 4 (1999), 733-751. Dates First available in Project Euclid: 20 October 2009 Permanent link to this document https://projecteuclid.org/euclid.ijm/1256060689 Digital Object Identifier doi:10.1215/ijm/1256060689 Mathematical Reviews number (MathSciNet) MR1712520 Zentralblatt MATH identifier 0936.32004 Subjects Primary: 32A37: Other spaces of holomorphic functions (e.g. bounded mean oscillation (BMOA), vanishing mean oscillation (VMOA)) [See also 46Exx] Secondary: 46E15: Banach spaces of continuous, differentiable or analytic functions Citation Ortega, Joaquín M.; Fàbrega, Joan. Hardy's inequality and embeddings in holomorphic Triebel-Lizorkin spaces. Illinois J. Math. 43 (1999), no. 4, 733--751. doi:10.1215/ijm/1256060689. https://projecteuclid.org/euclid.ijm/1256060689
I have been given this recently in PDE class involving the solutions to the Bessel fucntion in Sturm-Liouville form, asking for Eigenvalues and Eigenfunctions: $ (xy')'+\lambda x y = 0 \space \space \space \space 0<a<x<b $ $ y(a)=y(b)=0 $ Here is where my problems start: If I expand the equation I get: $ xy''+y'+\lambda x y = 0 \to x^2y''+xy'+\lambda x^2 y = 0 $ now depending on whether or not the eigenvalues are positive or negative (checking zero is not one is fairly simple) we obtain the general solution to the Bessel equation or modified Bessel equation of order zero: $ y(x) = AJ_0(\sqrt{\lambda}x) + BY_0(\sqrt{\lambda}x) $ or $ y(x) = AI_0(\sqrt{\lambda}x) + BK_0(\sqrt{\lambda}x) $ So there is not the matter of boundedness at zero resulting in eliminating the singular solution from the linear combination so my problem here is basically applying the boundary conditions like this: $ AJ_0(\sqrt{\lambda}a) + BY_0(\sqrt{\lambda}a) = AJ_0(\sqrt{\lambda}b) + BY_0(\sqrt{\lambda}b) = 0 $ so this equation and deriving $ \lambda $ and the actual eigenfunctions from this is beyond my ability, and that's where I am stuck and need the help. Thanks all.
It is consistent that the answer is no. If we start with $L$ as our ground model then whenever $T$ is a Suslin tree, the forcing $\mathbb{P}_T$ which shoots a branch through $T$ will always introduce new Suslin trees. Thus in $L$, $\mathbb{P}_T$ is an example of a forcing which has the ccc, adds Suslin trees and does not add a Cohen real. To see that $\mathbb{P}_T$ adds a new Suslin tree, note first that whenever $G$ is $L$-generic for $\mathbb{P}_T$, $\diamondsuit$ still holds in the extension $L[G]$: if $(a_{\alpha} \, : \, \alpha < \omega_1)$ is $\diamondsuit$-sequence in $L$ then $(a_{\alpha}^G \, : \, \alpha < \omega_1)$ will be a $\diamondsuit$-sequence in $L[G]$ as can be seen using the ccc of $\mathbb{P}_T$ and the fact that $\mathbb{P}_T$ has size $\aleph_1$. Next observe that $\mathbb{P}_T$ will add a new subset of $\omega_1$ and hence a new stationary subset of $\omega_1$ by a theorem of Solovay, saying that each stationary subset of $\omega_1$ can be partitioned into $\omega_1$-many stationary sets. Finally fix a new stationary subset $B \subset (\omega_1 \cap Lim)$ and, working in $L[G]$, use the restricted diamond sequence $\diamondsuit_B := (a_{\alpha}^G \, : \, \alpha \in B)$ to define a new Suslin tree $S$:Build the tree via induction on its height. If $\alpha$ is a limit ordinal then let $S_{\alpha}:= \bigcup_{\beta< \alpha} S_{\beta}$. If $\alpha$ is a double successor ordinal let $S_{\alpha+1}$ be the extension of $S_{\alpha}$ which adjoins to every top level $x \in S_{\alpha}$ infinitely many immediate successors. If $\alpha$ is limit we distinguish two cases in the definition of $S_{\alpha+1}$. If $\alpha \notin B$ let $S_{\alpha+1}$ be the $<_L$-least normal extension of $S_{\alpha}$ (note here that we can use the $L$-wellorder as $\mathbb{P}_T$ is $\omega$-distributive, hence no new countable sets are added by $\mathbb{P}_T$). If $\alpha \in B$ and $a_{\alpha}^G$ is a maximal antichain in $S_{\alpha}$ let $S_{\alpha+1}$ be the $<_L$-least normal extension of $S_{\alpha}$ such that $a_{\alpha}^G$ remains maximal in $S_{\alpha+1}$. And finally if $\alpha \in B$ but $a_{\alpha}^G$ is not a maximal antichain in $S_{\alpha}$, let $S_{\alpha+1}$ be a normal extension of $S_{\alpha}$ which is not the $<_L$-least. The standard argument shows that $S:= \bigcup_{\alpha < \omega_1} S_{\alpha}$ is a Suslin tree.However the way we defined $S$ ensures that $S$ can not be in $L$.Indeed if we assume that $S \in L$ then one could define $B \subset \omega_1$ via just looking at the stages of $S$ where $S_{\alpha+1}$ is not the $<_L$-least normal extension of $S_{\alpha}$. Thus if $S \in L$, $B$ would be in $L$ as well which is a contradiction.

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