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Here's a question whose answer turns out to be very useful: Given two vectors, what is the angle between them? It may not be immediately clear that the question makes sense, but it's not hard to turn it into a question that does. Since vectors have no position, we are as usual free to place vectors wherever we like. If the two vectors are placed tail-to-tail, there is now a reasonable interpretation of the question: we seek the measure of the smallest angle between the two vectors, in the plane in which they lie. Figure 14.3.1 illustrates the situation. Since the angle $\theta$ lies in a triangle, we can compute it using a bit of trigonometry, namely, the law of cosines. The lengths of the sides of the triangle in figure 14.3.1 are $\|{\bf A}\|$, $\|{\bf B}\|$, and $\|{\bf A}-{\bf B}\|$. Let $\ds {\bf A}=\langle a_1,a_2,a_3\rangle$ and $\ds {\bf B}=\langle b_1,b_2,b_3\rangle$; then $$\eqalign{ \|{\bf A}-{\bf B}\|^2&=\|{\bf A}\|^2+\|{\bf B}\|^2-2\|{\bf A}\|\|{\bf B}\|\cos\theta\cr 2\|{\bf A}\|\|{\bf B}\|\cos\theta&=\|{\bf A}\|^2+\|{\bf B}\|^2-\|{\bf A}-{\bf B}\|^2\cr &=a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2-(a_1-b_1)^2-(a_2-b_2)^2-(a_3-b_3)^2\cr &=a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2\cr &\qquad-(a_1^2-2a_1b_1+b_1^2) -(a_2^2-2a_2b_2+b_2^2)-(a_3^2-2a_3b_3+b_3^2)\cr &=2a_1b_1+2a_2b_2+2a_3b_3\cr \|{\bf A}\|\|{\bf B}\|\cos\theta&=a_1b_1+a_2b_2+a_3b_3\cr \cos\theta&=(a_1b_1+a_2b_2+a_3b_3)/(\|{\bf A}\|\|{\bf B}\|)\cr }$$ So a bit of simple arithmetic with the coordinates of $\bf A$ and $\bf B$ allows us to compute the cosine of the angle between them. If necessary we can use the arccosine to get $\theta$, but in many problems $\cos\theta$ turns out to be all we really need. The numerator of the fraction that gives us $\cos\theta$ turns up alot, so we give it a name and more compact notation: we call itthe dot product, and write it as$${\bf A}\cdot{\bf B} = a_1b_1+a_2b_2+a_3b_3.$$This is the same symbol we use for ordinary multiplication, but thereshould never be any confusion; you can tell from context whether weare "multiplying'' vectors or numbers. (We might also use the dot forscalar multiplication: $a\cdot{\bf V}=a{\bf V}$; again, it is clearwhat is meant from context.) Example 14.3.1 Find the angle between the vectors ${\bf A}=\langle 1,2,1\rangle$ and ${\bf B}=\langle 3,1,-5\rangle$. We know that $\cos\theta={\bf A}\cdot{\bf B}/(\|{\bf A}\|\|{\bf B}\|)= (1\cdot3 + 2\cdot1 + 1\cdot(-5))/(\|{\bf A}\|\|{\bf B}\|)=0$, so $\theta=\pi/2$, that is, the vectors are perpendicular. Example 14.3.2 Find the angle between the vectors ${\bf A}=\langle 3,3,0\rangle$ and ${\bf B}=\langle 1,0,0\rangle$. We compute $$\eqalign{ \cos\theta &= (3\cdot1 + 3\cdot0 + 0\cdot0)/(\sqrt{9+9+0}\sqrt{1+0+0})\cr &= 3/\sqrt{18} = 1/\sqrt2\cr}$$ so $\theta=\pi/4$. Example 14.3.3 Some special cases are worth looking at: Find the angles between ${\bf A}$ and ${\bf A}$; ${\bf A}$ and ${\bf -A}$; ${\bf A}$ and ${\bf 0}=\langle 0,0,0\rangle$. $\ds \cos\theta= {\bf A}\cdot{\bf A}/(\|{\bf A}\|\|{\bf A}\|)=(a_1^2+a_2^2+a_3^2)/ (\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{a_1^2+a_2^2+a_3^2})=1$, so the angle between ${\bf A}$ and itself is zero, which of course is correct. $\ds\cos\theta={\bf A}\cdot{\bf -A}/(\|{\bf A}\|\|{\bf -A}\|)=(-a_1^2-a_2^2-a_3^2)/ (\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{a_1^2+a_2^2+a_3^2})=-1$, so the angle is $\pi$, that is, the vectors point in opposite directions, as of course we already knew. $\ds \cos\theta= {\bf A}\cdot{\bf 0}/(\|{\bf A}\|\|{\bf 0}\|)=(0+0+0)/ (\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{0^2+0^2+0^2})$, which is undefined. On the other hand, note that since ${\bf A}\cdot{\bf 0}=0$ it looks at first as if $\cos\theta$ will be zero, which as we have seen means that vectors are perpendicular; only when we notice that the denominator is also zero do we run into trouble. One way to "fix'' this is to adopt the convention that the zero vector ${\bf 0}$ is perpendicular to all vectors; then we can say in general that if ${\bf A}\cdot{\bf B}=0$, $\bf A$ and $\bf B$ are perpendicular. Generalizing the examples, note the following useful facts: 1. If $\bf A$ is parallel or anti-parallel to $\bf B$ then ${\bf A}\cdot{\bf B}/(\|{\bf A}\|\|{\bf B}\|)=\pm1$, and conversely, if ${\bf A}\cdot{\bf B}/(\|{\bf A}\|\|{\bf B}\|)=1$, $\bf A$ and $\bf B$ are parallel, while if ${\bf A}\cdot{\bf B}/(\|{\bf A}\|\|{\bf B}\|)=-1$, $\bf A$ and $\bf B$ are anti-parallel. (Vectors are parallel if they point in the same direction, anti-parallel if they point in opposite directions.) 2. If $\bf A$ is perpendicular to $\bf B$ then ${\bf A}\cdot{\bf B}/(\|{\bf A}\|\|{\bf B}\|)=0$, and conversely if ${\bf A}\cdot{\bf B}/(\|{\bf A}\|\|{\bf B}\|)=0$ then $\bf A$ and $\bf B$ are perpendicular. Given two vectors, it is often useful to find the projection of one vector onto the other, because this turns out to have importantmeaning in many circumstances. More precisely, given ${\bf A}$ and${\bf B}$, we seek a vector parallel to $\bf B$ but with lengthdetermined by $\bf A$ in a natural way, as shown infigure 14.3.2. $\bf V$ is chosen so that thetriangle formed by $\bf A$, $\bf V$, and ${\bf A}-{\bf V}$is a right triangle. Using a little trigonometry, we see that $$ \|{\bf V}\|=\|{\bf A}\|\cos\theta= \|{\bf A}\|{{\bf A}\cdot{\bf B}\over\|{\bf A}\|\|{\bf B}\|}= {{\bf A}\cdot{\bf B}\over\|{\bf B}\|};$$this is sometimes called the scalar projection of $\bf A$ onto $\bf B$. To get $\bf V$itself, we multiply this length by a vector of length one parallel to$\bf B$: $$ {\bf V}= {{\bf A}\cdot{\bf B}\over\|{\bf B}\|}{{\bf B}\over\|{\bf B}\|}= {{\bf A}\cdot{\bf B}\over\|{\bf B}\|^2}{\bf B}.$$Be sure that you understand why ${\bf B}/\|{\bf B}\|$ is a vector oflength one (also called a unit vector) parallel to $\bf B$. The discussion so far implicitly assumed that $0\le\theta\le\pi/2$. If $\pi/2< \theta\le\pi$, the picture is like figure 14.3.3. In this case ${\bf A}\cdot {\bf B}$ is negative, so the vector $${{\bf A}\cdot{\bf B}\over\|{\bf B}\|^2}{\bf B}$$ is anti-parallel to $\bf B$, and its length is $$\left\|{{\bf A}\cdot{\bf B}\over\|{\bf B}\|}\right\|.$$ So in general, the scalar projection of $\bf A$ onto $\bf B$ may be positive or negative. If it is negative, it means that the projection vector is anti-parallel to $\bf B$ and that the length of the projection vector is the absolute value of the scalar projection. Of course, you can also compute the length of the projection vector as usual, by applying the distance formula to the vector. Note that the phrase "projection onto $\bf B$'' is a bit misleading if taken literally; all that $\bf B$ provides is a direction; the length of $\bf B$ has no impact on the final vector. In figure 14.3.4, for example, $\bf B$ is shorter than the projection vector, but this is perfectly acceptable. Example 14.3.4 Physical force is a vector quantity. It is often necessary to compute the "component'' of a force acting in a different direction than the force is being applied. For example, suppose a ten pound weight is resting on an inclined plane—a pitched roof, for example. Gravity exerts a force of ten pounds on the object, directed straight down. It is useful to think of the component of this force directed down and parallel to the roof, and the component down and directly into the roof. These forces are the projections of the force vector onto vectors parallel and perpendicular to the roof. Suppose the roof is tilted at a $\ds 30^\circ$ angle, as in figure 14.3.5. A vector parallel to the roof is $\ds \langle-\sqrt3,-1\rangle$, and a vector perpendicular to the roof is $\ds \langle 1,-\sqrt3\rangle$. The force vector is ${\bf F}=\langle 0,-10\rangle$. The component of the force directed down the roof is then $$\eqalign{ {\bf F}_1&={{\bf F}\cdot \langle-\sqrt3,-1\rangle\over\|\langle-\sqrt3,-1\rangle\|^2} \langle-\sqrt3,-1\rangle ={10\over 2}{\langle-\sqrt3,-1\rangle\over2}= \langle -5\sqrt3/2,-5/2\rangle\cr }$$ with length 5. The component of the force directed into the roof is $$\eqalign{ {\bf F}_2&={{\bf F}\cdot \langle1,-\sqrt3\rangle\over\|\langle1,-\sqrt3\rangle\|^2} \langle1,-\sqrt3\rangle ={10\sqrt3\over 2}{\langle1,-\sqrt3\rangle\over2}= \langle 5\sqrt3/2,-15/2\rangle\cr }$$ with length $\ds 5\sqrt3$. Thus, a force of 5 pounds is pulling the object down the roof, while a force of $\ds 5\sqrt3$ pounds is pulling the object into the roof. The dot product has some familiar-looking properties that will be useful later, so we list them here. These may be proved by writing the vectors in coordinate form and then performing the indicated calculations; subsequently it can be easier to use the properties instead of calculating with coordinates. 1. $\ds {\bf u}\cdot{\bf u} = \|{\bf u}\|^2$ 2. ${\bf u}\cdot{\bf v} = {\bf v}\cdot{\bf u}$ 3. ${\bf u}\cdot({\bf v}+{\bf w}) = {\bf u}\cdot{\bf v}+{\bf u}\cdot{\bf w}$ 4. $(a{\bf u})\cdot{\bf v}=a({\bf u}\cdot{\bf v}) ={\bf u}\cdot(a{\bf v})$ Exercises 14.3 Ex 14.3.1Find $\langle 1,1,1\rangle\cdot\langle 2,-3,4\rangle$.(answer) Ex 14.3.2Find $\langle 1,2,0\rangle\cdot\langle 0,0,57\rangle$.(answer) Ex 14.3.3Find $\langle 3,2,1\rangle\cdot\langle 0,1,0\rangle$.(answer) Ex 14.3.4Find $\langle -1,-2,5\rangle\cdot\langle 1,0,-1 \rangle$.(answer) Ex 14.3.5Find $\langle 3,4,6\rangle\cdot\langle 2,3,4\rangle$.(answer) Ex 14.3.6Find the cosine of the angle between $\langle 1,2,3\rangle$and $\langle 1,1,1\rangle$; use a calculator if necessary to find the angle.(answer) Ex 14.3.7Find the cosine of the angle between $\langle -1, -2,-3\rangle$and $\langle 5,0,2\rangle$; use a calculator if necessary to find the angle.(answer) Ex 14.3.8Find the cosine of the angle between $\langle 47,100,0\rangle$and $\langle 0,0,5\rangle$; use a calculator if necessary to find the angle.(answer) Ex 14.3.9Find the cosine of the angle between $\langle 1,0,1 \rangle$and $\langle 0,1,1\rangle$; use a calculator if necessary to find the angle.(answer) Ex 14.3.10Find the cosine of the angle between $\langle 2,0,0\rangle$and $\langle -1,1,-1\rangle$; use a calculator if necessary to find the angle.(answer) Ex 14.3.11Find the angle between the diagonal of a cube and one of theedges adjacent to the diagonal.(answer) Ex 14.3.12Find the scalar and vector projections of $\langle 1,2,3\rangle$onto $\langle 1,2,0\rangle$.(answer) Ex 14.3.13Find the scalar and vector projections of $\langle 1,1,1\rangle$onto $\langle 3,2,1\rangle$.(answer) Ex 14.3.14A force of 10 pounds is applied to a wagon, directed at anangle of $\ds 30^\circ$. Find the component of this force pulling thewagon straight up, and the component pulling it horizontally alongthe ground.(answer) Ex 14.3.15A force of 15 pounds is applied to a wagon, directed at anangle of $\ds 45^\circ$. Find the component of this force pulling thewagon straight up, and the component pulling it horizontally alongthe ground.(answer) Ex 14.3.16Use the dot product to find a non-zero vector ${\bf w}$perpendicular to both ${\bf u}=\langle 1,2,-3\rangle$ and ${\bf v}=\langle 2,0,1\rangle$.(answer) Ex 14.3.17Let ${\bf x}=\langle 1,1,0 \rangle$ and ${\bf y}=\langle2,4,2 \rangle$. Find a unit vector that is perpendicular to both $\bfx$ and $\bf y$.(answer) Ex 14.3.18Do the three points $(1,2,0)$, $(-2,1,1)$, and $(0,3,-1)$form a right triangle?(answer) Ex 14.3.19Do the three points $(1,1,1)$, $(2,3,2)$, and $(5,0,-1)$form a right triangle?(answer) Ex 14.3.20Show that $\|{\bf A}\cdot{\bf B}\|\le\|{\bf A}\|\|{\bf B}\|$ Ex 14.3.21Let $\bf x$ and $\bf y$ be perpendicular vectors. UseTheorem 14.3.5 to prove that $\ds \|{\bf x}\|^2+\|{\bf y}\|^2=\|{\bf x}+{\bf y}\|^2$. What is this result betterknown as? Ex 14.3.22Prove that the diagonals of a rhombus intersect at right angles. Ex 14.3.23Suppose that ${\bf z}=\|{\bf x}\| {\bf y} + \|{\bf y}\| {\bf x}$where $\bf x$, $\bf y$, and $\bf z$ are all nonzero vectors. Provethat $\bf z$ bisects the angle between $\bf x$ and $\bf y$. Ex 14.3.24Prove Theorem 14.3.5.
Proceedings of the Centre for Mathematics and its Applications Proc. Centre Math. Appl. Workshop/Miniconference on Functional Analysis and Optimization. Simon Fitzpatrick and John Giles, eds. Proceedings of the Centre for Mathematical Analysis, v. 20. (Canberra AUS: Centre for Mathematics and its Applications, Mathematical Sciences Institute, The Australian National University, 1988), 100 - 115 The convergence of entropic estimates for moment problems Abstract We show that if $x_n$ is optimal for the problem \[ sup\left\{\sum_{x_n}{1} log x(s)ds \| \sum_0^1 (x(s)- \hat{x}(s))s^i ds = 0 , i = 0, \cdots,n , 0 \leq x \in L_1[0,1]\right\}, \] then $\frac{1}{x_n} \rightarrow \frac{1}{\hat{x}}$ weakly in $L_1$ (providing $\hat{x}$ is continuous and strictly positive). This result is a special case of a theorem for more general entropic objectives and underlying spaces. Article information Source Workshop/Miniconference on Functional Analysis and Optimization. Simon Fitzpatrick and John Giles, eds. Proceedings of the Centre for Mathematical Analysis, v. 20. (Canberra AUS: Centre for Mathematics and its Applications, Mathematical Sciences Institute, The Australian National University, 1988), 100-115 Dates First available in Project Euclid: 18 November 2014 Permanent link to this document https://projecteuclid.org/ euclid.pcma/1416340107 Mathematical Reviews number (MathSciNet) MR1009598 Zentralblatt MATH identifier 0673.41021 Citation Lewis, A. S. The convergence of entropic estimates for moment problems. Workshop/Miniconference on Functional Analysis and Optimization, 100--115, Centre for Mathematics and its Applications, Mathematical Sciences Institute, The Australian National University, Canberra AUS, 1988. https://projecteuclid.org/euclid.pcma/1416340107
I'm implementing a interpreter for lambda calculus, and now I want to add the equality type. The introduction rule for it is easy, but the elimination rule is rather obscure for me. I found this stackoverflow thread, but it explains the J axiom only in one sentence. How can it be understood intuitively? A complete understanding of what J was actually saying and why has only come fairly recently. This blog post discusses it. While thinking in terms of homotopy and continuous functions provides a lot of intuition and connects to a very rich area of mathematics, I'm going to try to keep the discussion below at the logical level. Let's say you axiomatized equality types directly (these are groupoid operations and laws): $$\frac{\Gamma\vdash x:A}{\Gamma\vdash \mathbf{refl}_x : x=_A x} \qquad \frac{\Gamma, x:A, y:A\vdash p : x=_A y}{\Gamma, x:A, y:A\vdash p^{-1} : y =_A x} \qquad \frac{\Gamma, x:A, y:A\vdash p : x=_A y \quad \Gamma,y:A, z:A\vdash q : y=_A z}{\Gamma, x:A, y:A, z:A\vdash p\cdot q : x =_A z}$$ $$(p^{-1})^{-1} \equiv p \qquad p\cdot p^{-1} \equiv \mathbf{refl} \qquad (p\cdot q)^{-1} \equiv q^{-1}\cdot p^{-1} \\ \mathbf{refl} \cdot p \equiv p \equiv p \cdot \mathbf{refl} \qquad (p \cdot q) \cdot r \equiv p \cdot (q \cdot r)$$ We have substitution which is functorial. $$\frac{\Gamma, z : A\vdash F(z) : \mathbb{U}\qquad\Gamma,x:A,y:A\vdash p : x=_A y}{\Gamma, x:A,y:A\vdash \text{subst}(F,p) : F(x) \to F(y)}$$ $$\text{subst}(F,\mathbf{refl}) = \text{id} \quad \text{subst}(F,p\cdot q) = \text{subst}(F,q)\circ\text{subst}(F,p)\\\text{subst}(\lambda x.c=_A x,p)(q) = q\cdot p$$ Finally, we would have congruence rules for everything, saying everything respects this equality. Here is one of the most important congruences. $$\frac{\Gamma, x:A, y:A\vdash p : x=_Ay \quad \Gamma,z:A\vdash B(z) : \mathbb{U}}{\Gamma, x:A, y:A, b : B(x)\vdash \text{lift}_B(b,p) : \langle x,b\rangle =_{\Sigma x:A.B(x)}\langle y,\text{subst}(B,p)(b)\rangle}$$ Now let's consider a special case of substitution. $$\frac{\Gamma\vdash c:A \quad \Gamma,t:\Sigma x\!:\!A.c=_Ax\vdash C(t):\mathbb{U} \quad \Gamma\vdash b : C(\langle c,\mathbf{refl}_c\rangle)}{\Gamma,y:A,p:c=_Ay\vdash\text{subst}(C,\text{lift}_{\lambda z.c = z}(\mathbf{refl}_c,p))(b) : C(\langle y,p\rangle)}$$ This is J. We can use currying to get the nicer looking form: $$\frac{\Gamma\vdash c:A \quad \Gamma, y:A, p : c=_A y \vdash C(y,p): \mathbb{U} \quad \Gamma\vdash b : C(c,\mathbf{refl}_c)}{\Gamma, y:A,p : c=_A y\vdash J_{A,c}(C,b,y , p) : C(y,p)}$$ Of course, if we start with J we can rederive all the other structure I defined. Now if we have $p : x =_A y$ and $q : y =_A y'$ then $\text{lift}_{\lambda z.x=z}(p, q) : \langle y,p\rangle = \langle y',p\cdot q\rangle$. So if we have $\langle y', p' \rangle$, there's no way to get to it from $\langle y,p\rangle$ via a pre-selected $q$ in general unless $p \cdot q = p'$. (From the homotopy perspective, $p \cdot q = p'$ says we can fill in the triangle with edges $p$, $q$, and $p'$.) To make it more blunt, set $p = \mathbf{refl}_x$ (and $y=x$) and we get $q = p'$ being the required equality which is not true in general (because a value of type $x =_A y'$ can represent an arbitrary equivalence and there's nothing saying two equivalences have to be the same, or, equivalently, because we know groupoids can be non-trivial). The point of this is to demonstrate that things can be equal in more than one way, i.e. one value of $y = y'$ is not as good as another in general. A key thing to understand is that J says the type $\Sigma y\!:\!A.x=_Ay$ is inductively defined, and says little about the type $x=_Ay$ for fixed $x$ and $y$. One way to see this, and why J is the way it is, is to look at what congruence at equality types with matching endpoints looks like. We have the following rule (ignoring the proof term, it's expressible with $\text{subst}$ or J):$$\frac{\Gamma, x:A\vdash p : x=_Ax}{\Gamma, x:A, q : x =_A x\vdash \_ : q =_{x=_Ax} p \cdot q \cdot p^{-1}}$$ With $\text{lift}_{\lambda z.x=z}$ we have the option of doing $\text{lift}_{\lambda z.x=z}(p, p^{-1}\cdot p') : \langle y,p\rangle = \langle y',p'\rangle$ so every point was equivalent to every other point (though not necessarily trivially). With both endpoints matched, we don't have the flexibility of choosing the equalities to first cancel out the input equality and then performing an arbitrary equality. While J carefully respects the non-trivial groupoid structure of equality types, in typical dependently typed languages there's no way to actually define a non-trivial element of an equality type. At this point you hit a fork in the road. One route is to add Axiom K which says that the groupoid is actually trivial which makes many proofs much simpler. The other route is to add axioms that allow you to articulate the non-trivial groupoid structure. The most dramatic instance of this is the Univalence Axiom which leads to Homotopy Type Theory.
Now as for an intuitive explanation concerning Pete Clark's question: I would be interested in, at least, a reference for the fact that K(j(E),h(E[N])) contains the N-ray class field of K for arbitrary orders. Here is my intuitive point of view on it (I put it in another answer because I won't try to be perfectly rigorous), in terms of moduli. Let $E_1$ be a an elliptic curve with maximal CM by $O_K$, and $E_2$ be a curve with CM by $O$, an order $O$ of conductor $F$ in $O_K$. $E_2$ is defined over $K(j(E_2)) \supset K(j(E_1))$, so there is a rational isogeny $E_2 \to E_1$ of degree $F$. If we add the field of definition of the points of $m$-torsion of $E_2$ then it is clear that if $F$ is prime to $m$, these points get transported to the points of $m$-torsion of $E_1$, so we already have the $m$-ray class field. What is more surprising is that it works also when $m$ is not prime to $F$. So let's assume that $m \mid F$ and see why we can still transport the $m$-torsion from $E_2$ to $E_1$. The reason is as follow: when we are in the ring class field of $O$, all isogenies of degree $F$ between $E_1$ and an elliptic curve with endomorphism by $O$ are already rational; this means that the Galois action on $E_1[F]$ is given by a diagonal matrix. (And of course being in the $F$-ray class field means that the Galois action on $E_1[F]$ is the identity.) So in particular the kernel $K_1$ of the isogeny $E_1 \to E_2$ has a rational complement $K_2$. Now if we have the $m$-torsion on $E_2$, pushing it through the dual isogeny $E_2 \to E_1$, we have that at least the points of $m$-torsion of $K_1$ are all rationals.But because the points of $m$-torsion are rationals in $E_2$, the $m$-roots of unity are rationals, and so by looking at the Weil pairing we see that the points of $m$-torsion in $K_2$ are rationals. So all points of $m$-torsion in $E_1$ are rationals.
Yes it would, by a few percent. It may or may not be a goal worth pursuing, but there is more to it.Different reactions would be slowed down to a different extent. Tiny as they are, these discrepancies suffice to disrupt the delicate biochemical machinery of the living cell. No life form more advanced than bacteria is OK with that. Deuterated water in high ... Thermodynamics. The second law of thermodynamics states that entropy always increases in an isolated system. This is taken as a fundamental postulate---we simply accept this statement as a fact regarding how the world works, and our justification is that no experiment has ever shown the second law incorrect. In the framework of macroscopic thermodynamics, ... No.The reason why a gas particle in a large volume has a large entropy is not because it has a lot of space to move around per se. A better explanation is that for a given energy, there are many accessible translational states (these states can be derived from the particle in a box model). If we assume that all of these translational states are equally ... A process is thermodynamically reversible if it is essentially at equilibrium. Specifically, the system and its surroundings stay infinitesimally close to equilibrium with each other throughout a reversible process. Small changes in intensive variables of the system are perfectly balanced by changes in those variables in the surroundings. For example, $T_{\... While it may seem that randomness always increases when a crystal is dissolved into a liquid phase, it does not have to be that way.Concerning sugar, the molecule has a large number of hydroxyl groups and is generally rather large when compared with the water molecule around it — much larger than your average sodium or chloride ions. Every hydroxyl will ... It appears you're looking for an ELI5-style answer, not an elaborate definition.Entropy just happens – as long as the universe isn't frozen solid, things will always be moving around, and that movement tends to introduce randomness more than it tends to introduce order.Consider a deck of cards. Shuffle it. Is it perfectly sorted? No. Why? There are 10^67 ... Without significant mixing, diffusion takes a long time to mix gases. Our understanding of entropy tells us that we will indeed finish with mixed layers, but that doesn't give us a time frame for that mixing, only an outcome.Given a slow but steady production of a dense gas, layers absolutely will form due to density differences. There are plenty of ... Yes, the kinetic isotope effect is the main reason due to the differential lowering of the zero point energy in reactants and transition state, which has the effect of increasing, slightly, the activation energy. However, this effect is small, a few percent in a single reaction. The reason that deuteration has an effect overall is due to the fact that ... For example, suppose you have a solid block of TNT.It explodes and releases much energy. $\Delta H$ is negative.Gaseous products like nitrogen, carbon dioxide and water vapor are formed. The system has become more disordered, so entropy has increased. The most common way of measuring $\Delta S^\circ$ for a chemical reaction is probably by making a van't Hoff plot. You measure the equilibrium constant $K$ at different temperatures and plot $\ln K$ vs $T^{-1}$. The $y$-intercept = $R\Delta S^\circ$ and the slope = $-R\Delta H^\circ$.Another option is to measure $\Delta H^\circ$ by calorimetry and measure ... Briefly, spontaneous processes tend to proceed from states of low probability to states of higher probability. The higher-probability states tend to be those that can be realized in many different ways.Entropy is a measure of the number of different ways a state with a particular energy can be realized. Specifically, $$S=k\ln W$$ where $k$ is Boltzmann's ... The second Law of Thermodynamics states that the entropy of the universe always increases. $$\text{d}S > 0$$In the case of adsorption the entropy of the system; the gas being adsorbed; decreases but the entropy of the surroundings;the rest of the gas and the surface (and everything else in the universe); increases and this outweighs the decrease in ... "Reversible" is not binary. Both the forward and backward reactions always occur and the equilibrium system never has zero reactants or zero products. Thus, irreversible reactions are called this not because they cannot be reversed - they absolutely can - but because reversal is impractical. The equilibrium constant may be so skewed toward product that ... All right, someone bearing the standard of thermodynamics will give you the equations shortly... From a layman to another, here goes my attempt at a simpler explanation.Entropy may be seen as the "disorderlyness" in some settings, but that is not a very useful way of seeing it. The metaphor is often used, but creates the wrong conclusions when looking ... Do not think of entropy as 'disorder' as this is misleading, better is that it is a 'measure of disorder' but this is equally vague. It is better to think of entropy as the number of ways that 'particles' or quanta (say vibrational or rotational quanta in a molecule) can be placed among the various energy levels available.Thus at zero energy all the ... Your definition of entropy is incorrect. The significance of the Clausius inequality is that it shows that the definition of entropy, i.e. $\mathrm{\delta S=\cfrac{\delta q_{rev}}{T}}$ (note that entropy change is defined for a reversible process) is consistent with observed reality: the entropy of an isolated system does not decrease spontaneously.We ... There is a lot of uncertainty regarding the far future of our Universe, but it seems that chemistry as we know it will be gone long before the end. Both free and bound protons (and neutrons) are expected predicted to decay through at least one of several mechanisms, with a half-life somewhere between the range of $10^{35}$-$10^{200}$ yr (far shorter than the ... You can't measure entropy directly, any more than you can measure interatomic distances. You measure other quantities -- for instance often you can measure energy gain/loss and temperature, and then you integrate $dS=dE/T$.How to explain it? One of the best expositions I know is The Second Law by Henry A. Bent. It is full of insightful examples, lays ... NIST is the best place to turn for lots of data. However, more easily parsed, smaller datasets are available in a couple of other locations.The CHNOSz package in R has thermochemical data for a variety of species, mostly inorganic. Their database is referenced back to the chemical literature. See an answer I gave to an old question for an example of how ... You alluded to the answer when you mention activation energy. Kinetically the equilibrium constant is $K_e = k_f/k_b$ where $k_f$ and $k_b$ are the forward are reverse reaction rate constants in the reaction $\ce{A <=> B}$. The reason that there is a finite and not zero back reaction rate constant, is that the activation barrier going B to A is not ... The reason behind this is the hydrophobic effect. Everyone has seen it if they pour a spoonful of vegetable oil into a pot of water, e.g. to cook pasta. As long as nothing is disturbing the vegetable oil, it will collect itself together in one big bubble rather than form many small bubbles.Polar solvents will always be arranged in a way that positively ... I consider watching any video a waste of time, so I'll be judging from your words alone. (Anyway, your question is essentially self-consistent, which is good.)Yes, entropy is a measure of disorder (sort of).No, $dS={\delta Q\over T}$ only for reversible processes.And no, entropy is not the Q-T ratio for two reasons. First, because the above formula deals ... A simple explanation would be that the "o on top" denotes standard state and the one without the "o on top" denotes conditions that are not standard state. However, that may not allow you to "get it" so let's look at the equations.$\Delta G^{\circ} = -RT\ln K$$\Delta G = \Delta G^{\circ} + RT \ln Q$The first one allows us to find the Gibbs free ... The entropy change between two thermodynamic equilibrium states of a system can definitely be directly measured experimentally. To do so, one needs to devise (dream up) a reversible path between the initial and final states. Any convenient reversible path will do, since the integral of dq/T is the same for all reversible paths. So you have to identify a ... Consider two tanks of water. One hot, one cold.You open a valve between the two. The temperature between the two bleeds, and the hot tank loses entropy, and the cold tank gains entropy. However, entropy is also gained just by this process happening, as it would be impossible to force the 'cold' tank to retransfer the heat back to the 'hot' tank.Also, you ... Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$.Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation$$\begin{align}Q_\mathrm{tr,... The definition of entropy in 'Classical Thermodynamics' is $$\Delta S = \int\frac{\delta Q}{T}$$The quantity on the left is the entropy change associated with a physical process. $\delta Q$ is an inexact differential, made exact by an integrating factor $\frac{1}{T}$ This notion was devolped by Kelvin and Cartheodory (I believe). This comes out of the ... If you don't mind me asking, why would you expect the $\ce{Na}$ to have a higher entropy than the $\ce{NaCl}$? The entropy of a state is defined as,$$S=k_b\ln W$$where $W$ is the number of possible microstates which can make up that macrostate.In this case, we are interested in the number of possible ways that we can arrange $\ce{Na}$ into a lattice or the ... If you write $\Delta G = \Delta H -T\Delta S$, you implicitly admit that temperature is constant and is the same for a non-isolated system and its environment.You basically assume that the system and the environment are in thermal equilibrium, at a given temperature $T$. Thermal equilibrium means that when system loses heat, environment absorbs it: this ...
When you first encountered the trigonometric functions it was probablyin the context of "triangle trigonometry,'' defining, for example,the sine of an angle as the "side opposite over the hypotenuse.''While this will still be useful in an informal way, we need to use amore expansive definition of the trigonometric functions. First animportant note: while degree measure of angles is sometimes convenientbecause it is so familiar, it turns out to be ill-suited tomathematical calculation, so (almost) everything we do will be interms of radian measure of angles. To define the radian measurement system, we consider the unit circle in the $xy$-plane: An angle, $x$, at the center of the circle isassociated with an arc of the circle which is said to subtend the angle. In the figure, this arc is the portion ofthe circle from point $(1,0)$ to point $A$. The length of this arc isthe radian measure of the angle $x$; the fact that the radian measureis an actual geometric length is largely responsible for theusefulness of radian measure. The circumference of the unit circle is$2\pi r=2\pi(1)=2\pi$, so the radian measure of the full circularangle (that is, of the 360 degree angle) is $2\pi$. While an angle with a particular measure can appear anywhere around the circle, we need a fixed, conventional location so that we can use the coordinate system to define properties of the angle. The standard convention is to place the starting radius for the angle on the positive $x$-axis, and to measure positive angles counterclockwise around the circle. In the figure, $x$ is the standard location of the angle $\pi/6$, that is, the length of the arc from $(1,0)$ to $A$ is $\pi/6$. The angle $y$ in the picture is $-\pi/6$, because the distance from $(1,0)$ to $B$ along the circle is also $\pi/6$, but in a clockwise direction. Now the fundamental trigonometric definitions are: the cosine of $x$ and the sine of $x$ are the first and second coordinates of the point $A$, as indicated in the figure. The angle $x$ shown can be viewed as an angle of a right triangle, meaning the usual triangle definitions of the sine and cosine also make sense. Since the hypotenuse of the triangle is 1, the "side opposite over hypotenuse'' definition of the sine is the second coordinate of point $A$ over 1, which is just the second coordinate; in other words, both methods give the same value for the sine. The simple triangle definitions work only for angles that can "fit'' in a right triangle, namely, angles between 0 and $\pi/2$. The coordinate definitions, on the other hand, apply to any angles, as indicated in this figure: The arc bordering the green area initially has length $7\pi/6$, so the angle it subtends is the angle $7\pi/6$. Both coordinates of the red point in this figure are negative, so the sine and cosine of $7\pi/6$ are both negative. The remaining trigonometric functions can be most easily defined in terms of the sine and cosine, as usual: $$\eqalign{ \tan x &= {\sin x\over \cos x}\cr \cot x &= {\cos x \over \sin x}\cr \sec x &= {1\over \cos x}\cr \csc x &= {1\over \sin x}\cr }$$ and they can also be defined as the corresponding ratios of coordinates. Although the trigonometric functions are defined in terms of the unit circle, the unit circle diagram is not what we normally consider the graph of a trigonometric function. (The unit circle is the graph of, well, the circle.) We can easily get a qualitatively correct idea of the graphs of the trigonometric functions from the unit circle diagram. Consider the sine function, $y=\sin x$. As $x$ increases from 0 in the unit circle diagram, the second coordinate of the point $A$ goes from 0 to a maximum of 1, then back to 0, then to a minimum of $-1$, then back to 0, and then it obviously repeats itself. So the graph of $y=\sin x$ must look something like this: Similarly, as angle $x$ increases from 0 in the unit circle diagram, the first coordinate of the point $A$ goes from 1 to 0 then to $-1$, back to 0 and back to 1, so the graph of $y=\cos x$ must look something like this: Exercises 4.1 Some useful trigonometric identities are in chapter 18. Ex 4.1.1Find all values of $\theta$ such that$\sin(\theta) = -1$; give your answer in radians.(answer) Ex 4.1.2Find all values of $\theta$ such that$\cos(2\theta) = 1/2$; give your answer in radians.(answer) Ex 4.1.3Use an angle sum identity to compute $\cos(\pi/12)$.(answer) Ex 4.1.4Use an angle sum identity to compute $\tan(5\pi/12)$.(answer) Ex 4.1.5Verify the identity $\ds \cos^2(t)/(1-\sin(t)) = 1+\sin(t)$. Ex 4.1.6Verify the identity $2\csc(2\theta)=\sec(\theta)\csc(\theta)$. Ex 4.1.7Verify the identity $\sin(3\theta) - \sin(\theta) = 2\cos(2\theta) \sin(\theta)$. Ex 4.1.8Sketch $y=2\sin(x)$. Ex 4.1.9Sketch $y=\sin(3x)$. Ex 4.1.10Sketch $y=\sin(-x)$. Ex 4.1.11Find all of the solutions of $\ds 2\sin(t) -1 -\sin^2(t) =0$ in the interval $[0,2\pi]$.(answer)
We have already seen that a convenient way to describe a line in three dimensions is to provide a vector that "points to'' every point on the line as a parameter $t$ varies, like $$\langle 1,2,3\rangle+t\langle 1,-2,2\rangle =\langle 1+t,2-2t,3+2t\rangle.$$ Except that this gives a particularly simple geometric object, there is nothing special about the individual functions of $t$ that make up the coordinates of this vector—any vector with a parameter, like $\langle f(t),g(t),h(t)\rangle$, will describe some curve in three dimensions as $t$ varies through all possible values. As $t$ varies, the first two coordinates in all three functions trace out the points on the unit circle, starting with $(1,0)$ when $t=0$ and proceeding counter-clockwise around the circle as $t$ increases. In the first case, the $z$ coordinate is always 0, so this describes precisely the unit circle in the $x$-$y$ plane. In the second case, the $x$ and $y$ coordinates still describe a circle, but now the $z$ coordinate varies, so that the height of the curve matches the value of $t$. When $t=\pi$, for example, the resulting vector is $\langle -1,0,\pi\rangle$. A bit of thought should convince you that the result is a helix. In the third vector, the $z$ coordinate varies twice as fast as the parameter $t$, so we get a stretched out helix. Both are shown in figure 15.1.1. On the left is the first helix, shown for $t$ between 0 and $4\pi$; on the right is the second helix, shown for $t$ between 0 and $2\pi$. Both start and end at the same point, but the first helix takes two full "turns'' to get there, because its $z$ coordinate grows more slowly. A vector expression of the form $\langle f(t),g(t),h(t)\rangle$ is calleda vector function; it is a function fromthe real numbers $\R$ to the set of all three-dimensional vectors.We can alternately think of it as three separate functions, $x=f(t)$, $y=g(t)$, and $z=h(t)$, that describe points in space. Inthis case we usually refer to the set of equations as parametric equations for the curve, justas for a line. While the parameter $t$ in a vector function mightrepresent any one of a number of physical quantities, or be simply a"pure number'', it is often convenient and useful to think of $t$ asrepresenting time. The vector function then tells you where in spacea particular object is at any time. Vector functions can be difficult to understand, that is, difficult to picture. When available, computer software can be very helpful. When working by hand, one useful approach is to consider the "projections'' of the curve onto the three standard coordinate planes. We have already done this in part: in example 15.1.1 we noted that all three curves project to a circle in the $x$-$y$ plane, since $\langle \cos t,\sin t\rangle$ is a two dimensional vector function for the unit circle. Example 15.1.2 Graph the projections of $\langle \cos t,\sin t,2t\rangle$ onto the $x$-$z$ plane and the $y$-$z$ plane. The two dimensional vector function for the projection onto the $x$-$z$ plane is $\langle \cos t, 2t\rangle$, or in parametric form, $x=\cos t$, $z=2t$. By eliminating $t$ we get the equation $x=\cos(z/2)$, the familiar curve shown on the left in figure 15.1.2. For the projection onto the $y$-$z$ plane, we start with the vector function $\langle \sin t, 2t\rangle$, which is the same as $y=\sin t$, $z=2t$. Eliminating $t$ gives $y=\sin(z/2)$, as shown on the right in figure 15.1.2. Exercises 15.1 You can use Sage to graph spacecurves to check your work: Ex 15.1.1Describe the curve ${\bf r}=\langle \sin t,\cos t,\cos8t\rangle$. Ex 15.1.2Describe the curve ${\bf r}=\langle t\cos t,t\sin t,t\rangle$. Ex 15.1.3Describe the curve ${\bf r}=\langle t,t^2,\cos t\rangle$. Ex 15.1.4Describe the curve ${\bf r}=\langle \cos(20t)\sqrt{1-t^2},\sin(20t)\sqrt{1-t^2},t\rangle$ Ex 15.1.5Find a vector function for the curve of intersection of$x^2+y^2=9$ and $y+z=2$.(answer) Ex 15.1.6A bug is crawling outward along the spoke of a wheel that lies alonga radius of the wheel. The bug is crawling at 1 unit per second andthe wheel is rotating at 1 radian per second. Suppose the wheel liesin the $y$-$z$ plane with center at the origin, and at time $t=0$ thespoke lies along the positive $y$ axis and the bug is at the origin.Find a vector function ${\bf r}(t)$for the position of the bug at time $t$.(answer) Ex 15.1.7What is the difference between the parametric curves$f(t)=\langle t, t, t^2 \rangle$, $g(t)=\langle t^2, t^2, t^4\rangle$, and $h(t)=\langle \sin(t), \sin(t), \sin^2(t) \rangle$as $t$runs over all real numbers? Ex 15.1.8Plot each of the curves below in 2 dimensions, projectedonto each of the three standard planes (the $x$-$y$, $x$-$z$, and$y$-$z$ planes). a. $f(t)=\langle t, t^3, t^2 \rangle$, $t$ ranges over all real numbers b. $f(t)=\langle t^2, t-1, t^2+5 \rangle$ for $0\leq t \leq 3$ Ex 15.1.9Given points $A=(a_1, a_2, a_3)$ and $B=(b_1, b_2, b_3)$, giveparametric equations for the line segment connecting $A$ and$B$. Be sure to give appropriate $t$ values. Ex 15.1.10With a parametric plot and a set of $t$ values, we can associatea `direction'. For example, the curve $\langle \cos t, \sin t\rangle$ is the unit circle traced counterclockwise. How can we amenda set of given parametric equations and $t$ values to get the samecurve, only traced backwards?
Since the question about $U^c\times V^c$ is answered I answer just the other ones. The projections $$p_X:X\times Y\to X,~p_X(x,y)=x\text{ and }p_Y:X\times Y\to Y,~p_Y(x,y)=y$$are continuous in the product topology. If $A\subset X$ closed, $U\subset X$ open, $B\subset Y$ closed and $V\subset Y$ open, then by continuoity of $p_X$ and $p_Y$ you get$$X\times V=p_Y^{-1}(V),~U\times Y=p_X^{-1}(U), U\times V=p_X^{-1}(U)\cap p_Y^{-1}(V)$$are open. The same way are$$X\times B=p_Y^{-1}(B),~A\times Y=p_X^{-1}(A),~A\times B=p_X^{-1}(A)\cap p_Y^{-1}(B)$$closed. The product of open sets in $X$ and $Y$ yields an open set in $X\times Y$ and the product of closed sets in $X$ and $Y$ yields a closed set in $X\times Y$. That are obviously not all open and closed sets. You can unite an arbitrary number of open sets to an open set or intersect finite many open set to an open set. The same way you can unite finite many closed sets to a closed set and intersect arbitrary many closed sets to a closed set. Your last question is why $X\times(Y\setminus V)$ is closed while $X\times V$ is open. That holds since$$X\times (Y\setminus V)=(X\times Y)\setminus(X\times V)=(X\times V)^c.$$
Set-theoretic arguments often make use of the fact that a particular property $\varphi$ is local, in the sense that instances of the property can be verified by checking certain facts in only a bounded part of the set-theoretic universe, such as inside some rank-initial segment $V_\theta$ or inside the collection $H_\kappa$ of all sets of hereditary size less than $\kappa$. It turns out that this concept is exactly equivalent to the property being $\Sigma_2$ expressible in the language of set theory. Theorem. For any assertion $\varphi$ in the language of set theory, the following are equivalent: $\varphi$ is ZFC-provably equivalent to a $\Sigma_2$ assertion. $\varphi$ is ZFC-provably equivalent to an assertion of the form “$\exists \theta\, V_\theta\models\psi$,” where $\psi$ is a statement of any complexity. $\varphi$ is ZFC-provably equivalent to an assertion of the form “$\exists \kappa\, H_\kappa\models\psi$,” where $\psi$ is a statement of any complexity. Just to clarify, the $\Sigma_2$ assertions in set theory are those of the form $\exists x\,\forall y\,\varphi_0(x,y)$, where $\varphi_0$ has only bounded quantifiers. The set $V_\theta$ refers to the rank-initial segment of the set-theoretic universe, consisting of all sets of von Neumann rank less than $\theta$. The set $H_\kappa$ consists of all sets of hereditary size less than $\kappa$, that is, whose transitive closure has size less than $\kappa$. Proof. ($3\to 2$) Since $H_\kappa$ is correctly computed inside $V_\theta$ for any $\theta>\kappa$, it follows that to assert that some $H_\kappa$ satisfies $\psi$ is the same as to assert that some $V_\theta$ thinks that there is some cardinal $\kappa$ such that $H_\kappa$ satisfies $\psi$. ($2\to 1$) The statement $\exists \theta\, V_\theta\models\psi$ is equivalent to the assertion $\exists\theta\,\exists x\,(x=V_\theta\wedge x\models\psi)$. The claim that $x\models\psi$ involves only bounded quantifiers, since the quantifiers of $\psi$ become bounded by $x$. The claim that $x=V_\theta$ is $\Pi_1$ in $x$ and $\theta$, since it is equivalent to saying that $x$ is transitive and the ordinals of $x$ are precisely $\theta$ and $x$ thinks every $V_\alpha$ exists, plus a certain minimal set theory (so far this is just $\Delta_0$, since all quantifiers are bounded), plus, finally, the assertion that $x$ contains every subset of each of its elements. So altogether, the assertion that some $V_\theta$ satisfies $\psi$ has complexity $\Sigma_2$ in the language of set theory. ($1\to 3$) This implication is a consequence of the following absoluteness lemma. Lemma. (Levy) If $\kappa$ is any uncountable cardinal, then $H_\kappa\prec_{\Sigma_1} V$. Proof. Suppose that $x\in H_\kappa$ and $V\models\exists y\,\psi(x,y)$, where $\psi$ has only bounded quantifiers. Fix some such witness $y$, which exists inside some $H_\gamma$ for perhaps much larger $\gamma$. By the Löwenheim-Skolem theorem, there is $X\prec H_\gamma$ with $\text{TC}(\{x\})\subset X$, $y\in X$ and $X$ of size less than $\kappa$. Let $\pi:X\cong M$ be the Mostowski collapse of $X$, so that $M$ is transitive, and since it has size less than $\kappa$, it follows that $M\subset H_\kappa$. Since the transitive closure of $\{x\}$ was contained in $X$, it follows that $\pi(x)=x$. Thus, since $X\models\psi(x,y)$ we conclude that $M\models \psi(x,\pi(y))$ and so hence $\pi(y)$ is a witness to $\psi(x,\cdot)$ inside $H_\kappa$, as desired. QED Using the lemma, we now prove the remaining part of the theorem. Consider any $\Sigma_2$ assertion $\exists x\,\forall y\, \varphi_0(x,y)$, where $\varphi_0$ has only bounded quantifiers. This assertion is equivalent to $\exists\kappa\, H_\kappa\models\exists x\,\forall y\,\varphi_0(x,y)$, simply because if there is such a $\kappa$ with $H_\kappa$ having such an $x$, then by the lemma this $x$ works for all $y\in V$ since $H_\kappa\prec_{\Sigma_1}V$; and conversely, if there is an $x$ such that $\forall y\, \varphi_0(x,y)$, then this will remain true inside any $H_\kappa$ with $x\in H_\kappa$. QED In light of the theorem, it makes sense to refer to the $\Sigma_2$ properties as the locally verifiable properties, or perhaps as semi-local properties, since positive instances of $\Sigma_2$ assertions can be verified in some sufficiently large $V_\theta$, without need for unbounded search. A truly local property, therefore, would be one such that positive and negative instances can be verified this way, and these would be precisely the $\Delta_2$ properties, whose positive and negative instances are locally verifiable. Tighter concepts of locality are obtained by insisting that the property is not merely verified in some $V_\theta$, perhaps very large, but rather is verified in a $V_\theta$ where $\theta$ has a certain closeness to the parameters or instance of the property. For example, a cardinal $\kappa$ is measurable just in case there is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, and this is verified inside $V_{\kappa+2}$. Thus, the assertion “$\kappa$ is measurable,” has complexity $\Sigma^2_1$ over $V_\kappa$. One may similarly speak of $\Sigma^n_m$ or $\Sigma^\alpha_m$ properties, to refer to properties that can be verified with $\Sigma_m$ assertions in $V_{\kappa+\alpha}$. Alternatively, for any class function $f$ on the ordinals, one may speak of $f$-local properties, meaning a property that can be checked of $x\in V_\theta$ by checking a property inside $V_{f(\theta)}$. This post was made in response to a question on MathOverflow.
Answer The solution set to this problem is $$\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$ Work Step by Step $$(\csc x+2)(\csc x-\sqrt2)=0$$ over interval $[0,2\pi)$ 1) Solve the equation: $$(\csc x+2)(\csc x-\sqrt2)=0$$ $$\csc x+2=0\hspace{1cm}\text{or}\hspace{1cm}\csc x-\sqrt2=0$$ $$\csc x=-2\hspace{1cm}\text{or}\hspace{1cm}\csc x=\sqrt2$$ 2) Apply the inverse function: - $\csc x=-2$: $\csc\lt0$ means that the angle of $x$ lies either in quadrant III or IV. In quadrant I, $\csc x=2$ refers to angle $\frac{\pi}{6}$; therefore, in quadrant III and IV, $\csc x=-2$ would refer to angle $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ - $\csc x=\sqrt2$: $\csc\gt0$ means that the angle of $x$ lies either in quadrant I or II. In quadrant I, $\csc x=\sqrt2$ refers to angle $\frac{\pi}{4}$; in quadrant II, it refers to angle $\frac{3\pi}{4}$ Therefore, $$x\in\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$ In other words, the solution set to this problem is $$\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$
I have the following delay-differential-algebraic system: \begin{align} c(t) =& \kappa\min (1, \tfrac{\beta}{\alpha} a(t)), \tag{1a}\\ a(t) =& \left\{ \begin{array}{lcl} \frac{\alpha}{\beta} & \text{if} & s(t) > 0,\\ \min(\frac{\alpha}{\beta}, c(t-\Delta)) & \text{if} & s(t) = 0 \end{array}\right. ,\tag{1b}\\ \frac{\text{d} s}{\text{d} t}(t) =& c(t-\Delta) - a(t).\tag{1c} \end{align} which I am trying to numerically estimate by the following code: ClearAll[a, c, t, s, inic, inia]ClearAll["Global'"]kappa = 2;alpha = 1;beta = 2;delta = 1;amax = 10;T = 1.1;inic[t_] := -t;inia[t_] := inic[t - delta]eq1 = c[t] == kappaMin[1, (beta/alpha)*a[t]]eq2 = a[t] == Piecewise[{{Min[alpha/beta, c[t - delta]], s[t] == 0}, {alpha/beta, s[t] > 0}}]eq3 = s'[t] == c[t - delta] - a[t]initialConditions = { a[t /; t <= 0] == inia[t], c[t /; t <= 0] == inic[t], s[t /; t <= 0] == 0}sol = NDSolve[ Union[{eq1, eq2, eq3}, initialConditions ], {a, c, s}, {t, -delta, T}]Plot[Evaluate[{c[t], a[t], s[t]} /. sol], {t, -delta, T}] I've only included a small segment after $t=1$, because all three functions completely explode to $10^{20}$ and upwards at $t=2$. This is inconsistent with the equations, which state that $c(t), a(t)$ can attain at most values of $2$ and $\frac{1}{2}$ respectively, which they attain in the plot between $t=0$ and $t=1$. Moreover the plot doesn't show the values of the functions at interval $t=[-1,0]$, even though these are given as initial conditions. Why does it give this strange result? A piece of the puzzle may be that NDSolve gives the following error: NDSolve::ivcon: The given initial conditions were not consistent with the differential-algebraic equations. NDSolve will attempt to correct the values. >>
I have $$a(x,t+2) - a(x,t) = -\cos{\theta} [a(x+1,t+1) + a(x-1,t+1)]$$ Setting $t' = -t\cos(\theta)$ fetches under the continuum approximation $$2 \frac{\partial a(x,t')}{\partial t'} = a(x-1,t') - a(x+1,t')$$ which resembles Bessel functions's recurrence relations. I used RSolve but it is of no help. In[1]:= RSolve[a[x, t + 2] - a[x, t] - a[x - 1, t + 1] + a[x + 1, t + 1] == 0, a[x, t], {x, t}]Out[1]= RSolve[-a[-1 + x, 1 + t] - a[x, t] + a[x, 2 + t] + a[1 + x, 1 + t] == 0, a[x, t], {x, t}] Also, I tried using partial differential equation version which wasn't helpful either. In[6]:= RSolve[2 D[a[x, t], t] - a[x - 1, t + 1] + a[x + 1, t + 1] == 0, a[x, t], {x, t}]Out[6]= RSolve[-a[-1 + x, 1 + t] + a[1 + x, 1 + t] + 2*Derivative[0, 1][a][x, t] == 0, a[x, t], {x, t}] This is one of the cases I have to verify and I have several such equations for which I don't know the solutions.
Is there an "interesting" class of analytically-integrable, divergence-free vector fields over $\mathbb{R}^2$ and/or $\mathbb{R}^3$? That is, I'm looking for a large class of vector fields given by $V(x;\eta)$ for $x\in\mathbb{R}^k$, $k\in\{2,3\}$, parameterized by a fixed set of values $\eta\in\mathbb{R}^n$ such that: $\nabla_x \cdot V(x;\eta)=0\ \forall x\in\mathbb{R}^k$ There exists an easily-evaluated (closed-form?) function $\Phi(t;x_0,\eta)$ such that the ODE $p'(t)=V(p(t);\eta)$ is solved by taking $p(t)=\Phi(t;p(0),\eta).$ For example, the vector fields $V(x;A)=Ax$ admit $\Phi(t;p_0,\eta)=e^{At}p_0$, but the subset of $A$'s giving divergence-free vector fields is not too rich/interesting.
@JosephWright Well, we still need table notes etc. But just being able to selectably switch off parts of the parsing one does not need... For example, if a user specifies format 2.4, does the parser even need to look for e syntax, or ()'s? @daleif What I am doing to speed things up is to store the data in a dedicated format rather than a property list. The latter makes sense for units (open ended) but not so much for numbers (rigid format). @JosephWright I want to know about either the bibliography environment or \DeclareFieldFormat. From the documentation I see no reason not to treat these commands as usual, though they seem to behave in a slightly different way than I anticipated it. I have an example here which globally sets a box, which is typeset outside of the bibliography environment afterwards. This doesn't seem to typeset anything. :-( So I'm confused about the inner workings of biblatex (even though the source seems.... well, the source seems to reinforce my thought that biblatex simply doesn't do anything fancy). Judging from the source the package just has a lot of options, and that's about the only reason for the large amount of lines in biblatex1.sty... Consider the following MWE to be previewed in the build in PDF previewer in Firefox\documentclass[handout]{beamer}\usepackage{pgfpages}\pgfpagesuselayout{8 on 1}[a4paper,border shrink=4mm]\begin{document}\begin{frame}\[\bigcup_n \sum_n\]\[\underbrace{aaaaaa}_{bbb}\]\end{frame}\end{d... @Paulo Finally there's a good synth/keyboard that knows what organ stops are! youtube.com/watch?v=jv9JLTMsOCE Now I only need to see if I stay here or move elsewhere. If I move, I'll buy this there almost for sure. @JosephWright most likely that I'm for a full str module ... but I need a little more reading and backlog clearing first ... and have my last day at HP tomorrow so need to clean out a lot of stuff today .. and that does have a deadline now @yo' that's not the issue. with the laptop I lose access to the company network and anythign I need from there during the next two months, such as email address of payroll etc etc needs to be 100% collected first @yo' I'm sorry I explain too bad in english :) I mean, if the rule was use \tl_use:N to retrieve the content's of a token list (so it's not optional, which is actually seen in many places). And then we wouldn't have to \noexpand them in such contexts. @JosephWright \foo:V \l_some_tl or \exp_args:NV \foo \l_some_tl isn't that confusing. @Manuel As I say, you'd still have a difference between say \exp_after:wN \foo \dim_use:N \l_my_dim and \exp_after:wN \foo \tl_use:N \l_my_tl: only the first case would work @Manuel I've wondered if one would use registers at all if you were starting today: with \numexpr, etc., you could do everything with macros and avoid any need for \<thing>_new:N (i.e. soft typing). There are then performance questions, termination issues and primitive cases to worry about, but I suspect in principle it's doable. @Manuel Like I say, one can speculate for a long time on these things. @FrankMittelbach and @DavidCarlisle can I am sure tell you lots of other good/interesting ideas that have been explored/mentioned/imagined over time. @Manuel The big issue for me is delivery: we have to make some decisions and go forward even if we therefore cut off interesting other things @Manuel Perhaps I should knock up a set of data structures using just macros, for a bit of fun [and a set that are all protected :-)] @JosephWright I'm just exploring things myself “for fun”. I don't mean as serious suggestions, and as you say you already thought of everything. It's just that I'm getting at those points myself so I ask for opinions :) @Manuel I guess I'd favour (slightly) the current set up even if starting today as it's normally \exp_not:V that applies in an expansion context when using tl data. That would be true whether they are protected or not. Certainly there is no big technical reason either way in my mind: it's primarily historical (expl3 pre-dates LaTeX2e and so e-TeX!) @JosephWright tex being a macro language means macros expand without being prefixed by \tl_use. \protected would affect expansion contexts but not use "in the wild" I don't see any way of having a macro that by default doesn't expand. @JosephWright it has series of footnotes for different types of footnotey thing, quick eye over the code I think by default it has 10 of them but duplicates for minipages as latex footnotes do the mpfoot... ones don't need to be real inserts but it probably simplifies the code if they are. So that's 20 inserts and more if the user declares a new footnote series @JosephWright I was thinking while writing the mail so not tried it yet that given that the new \newinsert takes from the float list I could define \reserveinserts to add that number of "classic" insert registers to the float list where later \newinsert will find them, would need a few checks but should only be a line or two of code. @PauloCereda But what about the for loop from the command line? I guess that's more what I was asking about. Say that I wanted to call arara from inside of a for loop on the command line and pass the index of the for loop to arara as the jobname. Is there a way of doing that?
Let $N$ be a set for which $$\forall a\left(a\in N\iff a=\{\}\lor\exists b\left(b\in N\land a=\left\{b\right\}\right)\right)$$ Prove that $$\forall P\left(\left\{\right\}\in P\land \forall a\left(a\in N\implies\left(a\in P\implies\{a\}\in P\right)\right)\implies\forall a\left(a\in N\implies a\in P\right)\right)$$ I'm having troubles with this problem. The first thing I tried to do is to expand the definition of set $N$. It seems that there is exactly one set $N$ for which the first statement is true. The second statement should be true for all $P$ and $a$ (we need to prove it). My first attempt was to replace $a\in N$ in the second statement with $a=\{\}\lor\exists \left(b\in N\land a=\left\{b\right\}\right)$. The part $\{\}\in P\implies\{\}\in N$ is true, according to the definition of $N$. Now, I'm not sure how to deal with $\forall a\left(a\in N\implies\left(a\in P\implies\{a\}\in P\right)\right)$. Becuase there is an implication after this statement, the first idea which came to my mind is to try contradiction approach. Suppose that $\forall a\left(a\in N\implies\left(a\in P\implies\{a\}\in P\right)\right)$ is not true. So, I need to prove that then exists some $a$ from $N$ which is not in $P$. But, how to do that? I ran out of ideas. One thing I noticed is that both in the definition of $N$ and the second statement, which we need to prove, appear $\{b\}$ (and in the second $\{a\}$). I suppose the trick is to somehow apply this part of the definition of set $N$ (or a set $N$, if there may be multiple sets $N$) such that these two parts cancel out using some logical implications. I'm looking for an axiom-level proof. I mean, applying logical and ZFC axioms to this statement until we obtain an equivalence where on the left side is the statement we need to prove and on the right side is something which is true by definition. Simply, what would be the easiest way to prove this? Thank you in advance if you decide to help :)
Answer Please see the work below. Work Step by Step We know that $H=kA\frac{\Delta T}{\Delta x}$ We plug in the known values to obtain: $H=(1)(8.0\times 12)(\frac{20-10}{0.23})$ $H=4200W$ You can help us out by revising, improving and updating this answer.Update this answer After you claim an answer you’ll have 24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback.
This relation can be obtained if we realize that Total Cost can be written as a function of output (which in turn is a function of input factors), but also, that Total Cost equals total payments to factors of production. In real terms, $$TC=TC(Q) = (w/p)N + (r/p)K$$with the payment rates exogenous. But $Q = Q(N,K)$. Differentiate both sides by $N$: $$\frac {\partial TC(Q(N,K))}{\partial N} = w/p $$ $$\Rightarrow \frac {\partial TC(Q(N,K))}{\partial Q}\cdot \frac {\partial Q(N,K)}{\partial N} = w/p$$ $$\Rightarrow MC \cdot MPN = w/p \Rightarrow MC = \frac {w/p}{MPN}$$ One can perform the same exercise to obtain the analogous relation with respect to capital: $$MC = \frac {r/p}{MPK}$$
Strongly compact cardinal The strongly compact cardinals have their origins in the generalization of the compactness theorem of first order logic to infinitary languages, for anuncountable cardinal $\kappa$ is strongly compact if the infinitary logic $L_{\kappa,\kappa}$ exhibits the $\kappa$-compactness property. It turns out that this model-theoretic concept admits fruitful embedding characterizations, which as with so many large cardinal notions, has become the focus of study. Strong compactness rarefies into a hierarchy, and a cardinal $\kappa$ is strongly compact if and only if it is $\theta$-strongly compact for every ordinal $\theta\geq\kappa$. The strongly compact embedding characterizations are closely related to that of supercompact cardinals, which are characterized by elementary embeddings with a high degree of closure: $\kappa$ is $\theta$-supercompact if and only if there is an embedding $j:V\to M$ with critical point $\kappa$ such that $\theta<j(\kappa)$ and every subset of $M$ of size $\theta$ is an element of $M$. By weakening this closure requirement to insist only that $M$ contains a small cover for any subset of size $\theta$, or even just a small cover of the set $j''\theta$ itself, we arrive at the $\theta$-strongly compact cardinals. It follows that every $\theta$-supercompact cardinal is $\theta$-strongly compact and so every supercompact cardinal is strongly compact. Furthermore, since every ultrapower embedding $j:V\to M$ with critical point $\kappa$ has $M^\kappa\subset M$, for $\theta$-strong compactness we may restrict our attention to the case when $\kappa\leq\theta$. Contents 1 Diverse characterizations 1.1 Strong compactness characterization 1.2 Strong compactness embedding characterization 1.3 Cover property characterization 1.4 Fine measure characterization 1.5 Filter extension characterization 1.6 Discontinuous ultrapower characterization 1.7 Discontinuous embedding characterization 1.8 Ketonen characterization 1.9 Regular ultrafilter characterization 2 Strongly compact cardinals and forcing 3 Relation to other large cardinal notions 4 Topological Relevance 5 References Diverse characterizations There are diverse equivalent characterizations of the strongly compact cardinals. Strong compactness characterization An uncountable cardinal $\kappa$ is strongly compact if every $\kappa$-satisfiable theory in the infinitary logic $L_{\kappa,\kappa}$ is satisfiable. The signature of an $L_{\kappa,\kappa}$ language consists, just as in the first order context, of a set of finitary function, relation and constant symbols. The $L_{\kappa,\kappa}$ formulas, however, are built up in an infinitary process, by closing under infinitary conjunctions $\wedge_{\alpha<\delta}\varphi_\alpha$ and disjunctions $\vee_{\alpha<\delta}\varphi_\alpha$ of any size $\delta<\kappa$, as well as infinitary quantification $\exists\vec x$ and $\forall\vec x$ over blocks of variables $\vec x=\langle x_\alpha\mid\alpha<\delta\rangle$ of size less than $\kappa$. A theory in such a language is satisfiable if it has a model under the natural semantics. A theory is $\kappa$-satisfiable if every subtheory consisting of fewer than $\kappa$ many sentences of it is satisfiable. First order logic is precisely $L_{\omega,\omega}$, and the classical compactness theorem asserts that every $\omega$-satisfiable $L_{\omega,\omega}$ theory is satisfiable. Similarly, an uncountable cardinal $\kappa$ is defined to be strongly compact if every $\kappa$-satisfiable $L_{\kappa,\kappa}$ theory is satisfiable (and we call this the $\kappa$-compactness property}). The cardinal $\kappa$ is weakly compact, in contrast, if every $\kappa$-satisfiable $L_{\kappa,\kappa}$ theory, in a language having at most $\kappa$ many constant, function and relation symbols, is satisfiable. Strong compactness embedding characterization A cardinal $\kappa$ is $\theta$-strongly compact if and only if there is an elementary embedding $j:V\to M$ of the set-theoretic universe $V$ into a transitive class $M$ with critical point $\kappa$, such that $j''\theta\subset s\in M$ for some set $s\in M$ with $\|s\|^M\lt j(\kappa)$. [1] Cover property characterization A cardinal $\kappa$ is $\theta$-strongly compact if and only if there is an ultrapower embedding $j:V\to M$, with critical point $\kappa$, that exhibits the $\theta$-strong compactness cover property, meaning that for every $t\subset M$ of size $\theta$ there is $s\in M$ with $t\subset s$ and $\|s\|^M<j(\kappa)$. Fine measure characterization An uncountable cardinal $\kappa$ is $\theta$-strongly compact if and only if there is a fine measure on $\mathcal{P}_\kappa(\theta)$. The notation $\mathcal{P}_\kappa(\theta)$ means $\{\sigma\subset\theta\mid \|\sigma\|<\kappa\}$. [1] Filter extension characterization An uncountable cardinal $\kappa$ is $\theta$-strongly compact if and only if every $\kappa$-complete filter of size at most $\theta$ on a set extends to a $\kappa$-complete ultrafilter on that set. [1] Discontinuous ultrapower characterization A cardinal $\kappa$ is $\theta$-strongly compact if and only if there is an ultrapower embedding $j:V\to M$ with critical point $\kappa$, such that $\sup j''\lambda<j(\lambda)$ for every regular $\lambda$ with $\kappa\leq\lambda\leq\theta^{\lt\kappa}$. In other words, the embedding is discontinuous at all such $\lambda$. Discontinuous embedding characterization A cardinal $\kappa$ is $\theta$-strongly compact if and only if for every regular $\lambda$ with $\kappa\leq\lambda\leq\theta^{\lt\kappa}$, there is an embedding $j:V\to M$ with critical point $\kappa$ and $\sup j''\lambda<j(\lambda)$. Ketonen characterization An uncountable regular cardinal $\kappa$ is $\theta$-strongly compact if and only if there is a $\kappa$-complete uniform ultrafilter on every regular $\lambda$ with $\kappa\leq\lambda\leq\theta^{\lt\kappa}$. An ultrafilter $\mu$ on a cardinal $\lambda$ is uniform if all final segments $[\beta,\lambda)= \{\alpha<\lambda\mid \beta\leq\alpha\}$ are in $\mu$. When $\lambda$ is regular, this is equivalent to requiring that all elements of $\mu$ have the same cardinality. Regular ultrafilter characterization An uncountable cardinal $\kappa$ is $\theta$-strongly compact if and only if there is a $(\kappa,\theta)$-regular ultrafilter on some set. An ultrafilter $\mu$ is $(\kappa,\theta)$-regular if it is $\kappa$-complete and there is a family $\{X_\alpha\mid\alpha<\theta\}\subset \mu$ such that $\bigcap_{\alpha\in I}X_\alpha=\emptyset$ for any $I$ with $\|I\|=\kappa$. Strongly compact cardinals and forcing If there is proper class-many strongly compact cardinals, then there is a generic model of $\text{ZF}$ + "all uncountable cardinals are singular". If each strongly compact cardinal is a limit of measurable cardinals, and if the limit of any sequence of strongly compact cardinals is singular, then there is a forcing extension V[G] that is a symmetric model of $\text{ZF}$ + "all uncountable cardinals are singular" + "every uncountable cardinal is both almost Ramsey and a Rowbottom cardinal carrying a Rowbottom filter". This also directly follows from the existence of a proper class of supercompact cardinals, as every supercomact cardinal is simultaneously strongly compact and a limit of measurable cardinals. Relation to other large cardinal notions Strongly compact cardinals are measurable. The least strongly compact cardinal can be equal to the least measurable cardinal, or to the least supercompact cardinal, by results of Magidor. [2] (It cannot be equal to both at once because the least measurable cardinal cannot be supercompact.) Even though strongly compact cardinals imply the consistency of the negation of the singular cardinal hypothesis, SCH, for any singular strong limit cardinal $\kappa$ above the least strongly compact cardinal, $2^\kappa=\kappa^+$ (also known as "SCH holds above strong compactness"). [2] If there is a strongly compact cardinal $\kappa$ then for all $\lambda\geq\kappa$ and $A\subseteq\lambda$, $\lambda^+$ is ineffable in $L[A]$. It is not currently known whether the existence of a strongly compact cardinal is equiconsistent with the existence of a supercompact cardinal. The ultrapower axiom gives a positive answer to this, but itself isn't known to be consistent with the existence of a supercompact in the first place. Every strongly compact cardinal is strongly tall, although the existence of a strongly compact cardinal is equiconsistent with "the least measurable cardinal is the least strongly compact cardinal, and therefore the least strongly tall cardinal", so it could be the case that the least of the measurable, tall, strongly tall, and strongly compact cardinals all line up. Topological Relevance Strongly compact cardinals are related to the topological notion of compactness, interestingly enough. Intuition A topological space $X$ is called $\kappa$-compact when every open cover has a subcover of size below $\kappa$. More intuitively, it "looks" as though it has size below $\kappa$. For example, the $\aleph_0$-compact subspaces of the real number line are just the subspaces which are bounded. For example, a shape with finite area could be considered $\aleph_0$-compact, even though the amount of points is not only infinite but continuum-sized. The product of a collection of spaces is a little difficult to describe intuitively. However, it notably increases the amount of "dimensions" to a space. For example, the product of $n$-copies of the real number line is just the $n$-dimensional euclidean space (the line, the plane, etc.). Also, the general intuition is that it doesn't make spaces any bigger than the biggest one in the collection, so the product of a bunch of small spaces and a big space should be no 'bigger' than the big space. The idea is that the product of $\kappa$-compact spaces should itself be $\kappa$-compact, since the product doesn't make spaces any "bigger." However, there are examples of two $\aleph_1$-compact spaces (they "look countably infinite") which combine to make a space which isn't $\aleph_1$-compact ("looks uncountable"). However, if $\kappa$ is strongly compact, then this intuition holds; the product of any $\kappa$-compact spaces is strongly compact. One could maybe see why strongly compact cardinals are so big then; they imply that combining a bunch of small-relative-to-$\kappa$ spaces together by adding arbitrarily many dimensions keeps the space looking small relative to $\kappa$.' Tychonoff's theorem is precisely the statement that the product of $\aleph_0$-compact spaces is $\aleph_0$-compact; that is, if you combine a bunch of finite-looking spaces together and keep adding more and more dimensions, you get a space which is finite-looking. (Sources to be added) References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Jech, Thomas J. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www bibtex Set Theory.
I am trying to solve this diffusion equation : $\dfrac{\partial D\dfrac{\partial f}{\partial x}}{\partial x}+S = \dfrac{\partial f}{\partial t}$ ($D$ is not constant and varies according to $x$) with the following BC: $f(x,0)=1 , f(0,t)=0, \dfrac{\partial f}{\partial x}(1,t)=0$ I am using a central finite difference scheme (2nd order) for space and Euler explicit for time (1st order). The discretized $[0..1]$ domain contains the $x_{i}$ points, $i \in [0..N]$ ($x_{0}=0 , x_{N}=1$). The implementation of Dirichlet condition at $x=0$ is simple. For the Neumann boundary condition, I saw a lot of references that treated simple cases, adequate for constant diffusion coefficients and stationnary cases, like the ghost point, that I can't consider in my case(Diffusion coefficient can't be evaluated out of the domain, ghost point envolves evaluation the equation on a point that is not inside the interior domain... ). So I am trying something at my own and I want confirmation or correction if I omitted an important aspect in my work: The discretized equation evaluated at $x_{N-1}$ involves the value of $f_{N}$. If we had Diricihlet condition at $x_{N}$, the problem would have been solved directly. But in our case, $f_{N}$ is unknown. What I did is that I treated the Neumann BC with a backward scheme (at the 2nd order to preserve the general order of the whole scheme) in order to have an expression of $f^{k}_{N}$: $\dfrac{\partial f}{\partial x}(x_{N},k\Delta t)=\dfrac{-3f^{k}_{N}+4f^{k}_{N-1}-f^{k}_{N-2}}{2\Delta x}=0$ $f^{k}_{N}=4/3 f^{k}_{N-1}-1/3 f^{k}_{N-2}$ then I injected the new expression of $f^{k}_{N}$ in the discretized equation at $x_{N-1}$. Meaning that I did not add an equation to the problem as I would do for a stationnary problem but I injected the expression of $f^{k}_{N}$ inside the last equation. So the unknown vector $f^{k}$ in a case of matrix writing would only contain $f^{k}_{i}, i\in [1..N-1]$ (like in a full Dirichlet situation) Is it correct ? For information, the matrix form of the discretized problem is : $f^{k+1}=Af^{k}+S+B$, with $B$ the vector containing the B.C , and $S$ the 2nd member vector
Worldly cardinal Every inaccessible cardinal is worldly. Nevertheless, the least worldly cardinal is singular and hence not inaccessible. The least worldly cardinal has cofinality $\omega$. Indeed, the next worldly cardinal above any ordinal, if any exist, has cofinality $\omega$. Any worldly cardinal $\kappa$ of uncountable cofinality is a limit of $\kappa$ many worldly cardinals. Degrees of worldliness A cardinal $\kappa$ is $1$-worldly if it is worldly and a limit of worldly cardinals. More generally, $\kappa$ is $\alpha$-worldly if it is worldly and for every $\beta\lt\alpha$, the $\beta$-worldly cardinals are unbounded in $\kappa$. The cardinal $\kappa$ is hyper-worldly if it is $\kappa$-worldly. One may proceed to define notions of $\alpha$-hyper-worldly and $\alpha$-hyper${}^\beta$-worldly in analogy with the hyper-inaccessible cardinals. Every inaccessible cardinal $\kappa$ is hyper${}^\kappa$-worldly, and a limit of such kinds of cardinals. The consistency strength of a $1$-worldly cardinal is stronger than that of a worldly cardinal, the consistency strength of a $2$-worldly cardinal is stronger than that of a $1$-worldly cardinal, etc. The worldly cardinal terminology was introduced in lectures of J. D. Hamkins at the CUNY Graduate Center and at NYU. Replacement Characterization As long as $\kappa$ is an uncountable cardinal, $V_\kappa$ already satisfies $\text{ZF}^-$ ($\text{ZF}$ without the axiom schema of replacement). So, $\kappa$ is worldly if and only if $\kappa$ is uncountable and $V_\kappa$ satisfies the axiom schema of replacement. More analytically, $\kappa$ is worldly if and only if $\kappa$ is uncountable and for any function $f:A\rightarrow V_\kappa$ definable from parameters in $V_\kappa$ for some $A\in V_\kappa$, $f"A\in V_\kappa$ also.
I'm going through the paper Weight Uncertainty in Neural Networks by Google Deepmind. In the final line of the proof of proposition 1, the integral and the derivative are swapped. Then the derivative is taken. But this somehow yields 2 derivatives of $f$ with respect to $\theta$. I thought that this was the result of a product rule applied to $q(\epsilon)$ and $f(w,\theta)$ and then the chain rule. But that does not yield the same outcome as $\frac{\partial q(\epsilon)}{\partial \theta} = 0 $. My question is: does anyone understand how the equation in the last line comes about? I'm going through the paper we start with $\frac{\partial}{\partial \theta} \mathbb E_{q(\mathbf w\mid\theta)}[f(\mathbf w, \theta)]$ using definition of expectation for continuous case: $\mathbb E[X] = \int xp(x) dx$ for the first equation we get: $\frac{\partial}{\partial \theta} \int f(\mathbf w, \theta)q(\mathbf w \mid \theta) d\mathbf w$ we swap $q(\mathbf w \mid \theta) d\mathbf w$ for $q(\epsilon)d\epsilon$ and we get: $\frac{\partial}{\partial \theta} \int f(\mathbf w, \theta)q(\epsilon)d\epsilon$ using definition of expectation again we have: $\frac{\partial}{\partial \theta} \mathbb E_{q(\epsilon)}[f(\mathbf w,\theta)]$ i believe dominant convergence theorem let's us interchange expectation and derivative and we have: $\mathbb E_{q(\epsilon)}[\frac{\partial}{\partial \theta}f(\mathbf w,\theta)]$ we have a function $f$ that depends on two variables $\mathbf w$ and $\theta$. In order to get a full derivative of such function we need to take derivative of $f$ with respect to both variables. We have: $\mathbb E_{q(\epsilon)}[\frac{\partial f(\mathbf w,\theta)}{\partial \theta} + \frac{\partial f(\mathbf w,\theta)}{\partial \mathbf w} \frac{\partial \mathbf w}{\partial \theta}]$ First term inside expectation is derivative of function $f$ with respect to $\theta$ directly, and in the second term $f$ is derivated with respect to $\mathbf w$, but we need to apply chain rule to get full derivative with respect to $\theta$
It looks exactly like an integrator to me. Since $$y[k] = y[k-1]+x[k] = y[k-2] + x[k-1] +x[k] = \sum{x}$$ Applying the Z-transform gives \begin{align} Y(z) &= Y(z)\cdot z^{-1} + X(z)\\ \Rightarrow\frac{Y(z)}{X(z)} &= \frac{1}{1-z^{-1}} \end{align} When I convert $\frac{1}{1-z^{-1}}$ into Lapace transform in Matlab using d2c, it does not return $\frac{k}{s}$. Instead, it returns $\frac{1+s}{s}$. It seems there is a proportional term inside, when thinking in PID control point of view. Could anyone explain on that?
hw4 discussion (ECE438BoutinFall09) Put your content here . . . Shouldn't one of the bounds in the first part of question 1 be non-inclusive since if you plug in $ \frac{\pi}{2} $ for the top part, you get $ \frac{2}{\pi} - 1 $ instead of 0 ? - pclay The bounds are correct. The expression over the interval $ \| \omega \| \leq \pi / 2 $ is the following: $ 1 - \frac{\|\omega\|}{\pi / 2} $ It appears that you have read it as $ \frac{1 - \|\omega \|}{\pi / 2} $ -crtaylor hmmm, yes. yes, it appears I have. lol --pclay
How do I express $\frac{1}{3^n}$, or $3^{-n}$ as a typical geometric series of the form: $ar^{n-1}$ ? This is a small part of a larger question, in which I'm trying to find the convergence of $\sum_{n=0}^\infty \frac{1+2^n}{3^n}$. I broke the series into two separate series', $\sum_{n=0}^\infty \frac{1}{3^n} + \sum_{n=0}^\infty (\frac23)^n $. However I am stuck trying to represent the first term as a geometric series. I know that it is a geometric series only because Wolfram Alpha told me so here.
I get so frustrated with modular arithmetic. It seems like every example I look at leaves steps out. I am trying to solve this problem: Solve the linear congruence equations for x: $x \equiv 2 \mod 7$ $x \equiv 1 \mod 3$ Ok, so I start We know that 1st equation has a solution when $7 \mid (x-2)$. So there exists an integer k where $x = 2 + 7k$. Ok, great. So I substitute into the 2nd equation: $ 2+7k \equiv 1 \mod 3 \implies \\ 7k \equiv -1 \mod 3 \implies \\ 7k \equiv 2 \mod 3 $ Now I need to find an inverse of this last congruence. How do I do that? I know there is one solution because gcd(7,3) = 1. This is the step I'm having problems on. If I can get the solution to $7k \equiv 2 \mod 3$ into the form $k = a + bj$ where $a,b \in \mathbb{N}$ then I know how to solve it. Thank you.
Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues? Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson... Hmm, it seems we cannot just superimpose gravitational waves to create standing waves The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line [The Cube] Regarding The Cube, I am thinking about an energy level diagram like this where the infinitely degenerate level is the lowest energy level when the environment is also taken account of The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings @Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer). Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it? Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks. I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh... @0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P) Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio... the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = \|\vec{A}\|\|\vec{B}\|\cos \theta$$$$\vec{A}\cdot\... @ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there. @CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer
Contents Practice Question on Computing the Output of an LTI system by Convolution The unit impulse response h(t) of a DT LTI system is $ h(t)= e^{-3t }u(t). \ $ Use convolution to compute the system's response to the input $ x(t)= u(t). \ $ You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too! Answer 1 $ y(t)=x(t)*h(t)=\int_{-\infty}^\infty u(\tau)e^{-3(t-\tau)}u(t-\tau)d\tau=e^{-3t}\int_0^\infty e^{3\tau}u(t-\tau)d\tau=\Bigg(e^{-3t}\int_0^t e^{3\tau}d\tau\Bigg)u(t)=\Bigg(\frac{1}{3}e^{-3t}\bigg(e^{3t}-1\bigg)\Bigg)u(t) $ $ y(t)=\Bigg(\frac{1}{3}-\frac{e^{-3t}}{3}\Bigg)u(t) $ --Cmcmican 21:00, 4 February 2011 (UTC) Instructor's comments: yes, that it! You really got the hang of it. That's exactly how you should answer a question like this on the test. -pm Answer 2 Write it here. Answer 3 Write it here.
Buchholz's ψ functions Buchholz's functions are a hierarchy of single-argument ordinal functions $ (\psi _{\nu }:On\rightarrow On)_{\nu\le\omega}$ introduced by German mathematician Wilfried Buchholz in 1981. Contents Basic Notions Small Greek letters always denote ordinals. Each ordinal $\alpha$ is identified with the set of its predecessors $\alpha=\{\beta\|\beta<\alpha\}$. $On$ denotes the class of all ordinals. We define $\Omega_0=1$ and $\Omega_{\nu}=\aleph_{\nu}$ for $\nu>0$. An ordinal $\alpha$ is an additive principal number if $\alpha>0$ and $\xi+\eta<\alpha$ for all $\xi,\eta<\alpha$. Let $P$ denote the set of all additive principal numbers i.e. $P=\{\alpha\in On\|0<\alpha\wedge\forall\xi,\eta<\alpha(\xi+\eta\in\alpha)\}=\{\omega^\beta\|\beta\in On\}$ For every $\alpha\notin P$ there exist unique set $P(\alpha)=\{\alpha_1, \alpha_2, ... ,\alpha_n\}$ such that $\alpha=\alpha_1+\alpha_2+ \cdots+\alpha_n$ and $\alpha>\alpha_1\geq\alpha_2\geq \cdots\geq\alpha_n$ and $\alpha_1, \alpha_2, ... ,\alpha_n\in P$ $\alpha=_{NF}\alpha_1+\alpha_2+\cdots+\alpha_n$ iff $\alpha=\alpha_1+\alpha_2+\cdots+\alpha_n$ and $\alpha>\alpha_1\geq\alpha_2\geq\cdots\geq\alpha_n$ and $\alpha_1,\alpha_2,...,\alpha_n\in P$ Definition Buchholz's functions are defined as follows: $C_\nu^0(\alpha) = \Omega_\nu$, $C_\nu^{n+1}(\alpha) = C_\nu^n(\alpha) \cup \{\gamma \| P(\gamma) \subseteq C_\nu^n(\alpha)\} \cup \{\psi_\mu(\xi) \| \xi \in \alpha \cap C_\nu^n(\alpha) \wedge \xi \in C_\mu(\xi) \wedge \mu \leq \omega\}$, $C_\nu(\alpha) = \bigcup_{n < \omega} C_\nu^n (\alpha)$, $\psi_\nu(\alpha) = \min\{\gamma \| \gamma \not\in C_\nu(\alpha)\}$. In other words $\psi_\nu(\alpha)$ is the least ordinal number which cannot be generated from ordinals less than $\Omega_\nu$ by applying of addition and the functions $\psi_{\mu}(\eta)$ with $\eta < \alpha$ and $\mu \le \omega$. We define $\alpha=_{NF}\psi_\nu(\beta)$ iff $\alpha=\psi_\nu(\beta)$ and $\beta\in C_\nu(\beta)$ Properties Buchholz showed the following properties of those functions: $\psi_\nu(0)=\Omega_\nu$, $\psi_\nu(\alpha)\in P$, $\psi_\nu(\alpha+1)=\text{min}\{\gamma\in P\| \psi_\nu(\alpha)<\gamma\}\text{ if }\alpha\in C_\nu(\alpha)$, $\Omega_\nu\le\psi_\nu(\alpha)<\Omega_{\nu+1}$, $\alpha\le\beta\Rightarrow\psi_\nu(\alpha)\le\psi_\nu(\beta)$, $\psi_0(\alpha)=\omega^\alpha \text{ if }\alpha<\varepsilon_0$, $\psi_\nu(\alpha)=\omega^{\Omega_\nu+\alpha} \text{ if }\alpha<\varepsilon_{\Omega_\nu+1} \text{ and } \nu\neq 0$. Fundamental sequences The fundamental sequence for an ordinal number $\alpha$ with cofinality $\text{cof}(\alpha)=\beta$ is a strictly increasing sequence $(\alpha[\eta])_{\eta<\beta}$ with length $\beta$ and with limit $\alpha$, where $\alpha[\eta]$ is the $\eta$-th element of this sequence. We define the set $T$ consisting of zero and all ordinals expressible using Buchholz's functions and the operation of addition $0 \in T$ if $\alpha=_{NF}\alpha_1+\alpha_2+\cdots+\alpha_n$ and $\alpha_1,\alpha_2,...,\alpha_n\in T$ then $\alpha \in T$ if $\alpha=_{NF}\psi_\nu(\beta)$ and $\nu,\beta\in T$ and $\nu\le\omega$ then $\alpha \in T$ For nonzero ordinals $\alpha\in T$ we define the fundamental sequences as follows: if $\alpha=\alpha_1+\alpha_2+\cdots+\alpha_n$ then $\text{cof} (\alpha)= \text{cof} (\alpha_n)$ and $\alpha[\eta]=\alpha_1+\alpha_2+\cdots+(\alpha_n[\eta])$ if $\alpha=\psi_0(0)$ or $\alpha=\psi_{\nu+1}(0)$ then $\operatorname{cof}(\alpha)=\alpha$ and $\alpha[\eta]=\eta$ if $\alpha=\psi_\omega(0)$ then $\operatorname{cof}(\alpha)=\omega$ and $\alpha[\eta]=\psi_\eta(0)$ if $\alpha=\psi_{\nu}(\beta+1)$ then $\operatorname{cof}(\alpha)=\omega$ and $\alpha[\eta]=\psi_{\nu}(\beta)\cdot \eta$ if $\alpha=\psi_{\nu}(\beta)$ and $\operatorname{cof}(\beta)\in\{\omega\}\cup\{\Omega_{\mu+1}\mid\mu<\nu\}$ then $\operatorname{cof}(\alpha)=\operatorname{cof}(\beta)$ and $\alpha[\eta]=\psi_{\nu}(\beta[\eta])$ if $\alpha=\psi_{\nu}(\beta)$ and $\operatorname{cof}(\beta)\in\{\Omega_{\mu+1}\mid\mu\geq\nu\}$ then $\operatorname{cof}(\alpha)=\omega$ and $\alpha[\eta]=\psi_\nu(\beta[\gamma[\eta]])$ with $\gamma[0]=\Omega_\mu$ and $\gamma[\eta+1]=\psi_\mu(\beta[\gamma[\eta]])$ Takeuti-Feferman-Buchholz ordinal The Takeuti-Feferman-Buchholz ordinal is equal to $\psi_0(\varepsilon_{\Omega_\omega+1})$ using Buchholz $\psi$-notaion and also it is equal to $\theta_{\varepsilon_{\Omega_\omega+1}}(0)$ using Feferman $\theta$-notation. This ordinal is the limit of both notations. The name of the ordinal was proposed by David Madore. See also Other ordinal collapsing functions: References 1. W.Buchholz. A New System of Proof-Theoretic Ordinal Functions. Annals of Pure and Applied Logic (1986),32
I realize that there has already been an answer to this problem. But I want to know if my proof was correct. Thank you for your time. Problem Suppose A,B, and C are sets and $f: A \rightarrow B$ Suppose that C has at least two elements, and for all functions $g$ and $h$ from B to C, if $g \circ h = h \circ f$ then $g = h$. Prove that $f$ is onto. Proof Suppose f is not onto. Then there is $b_1$ such that for all $a \in A$, $f(a) \neq b_1$. Suppose $(b_1, c_1) \in g$ and $(b_1, c_2) \in h$. For the assumption that $g \circ h = h \circ f$ then $g = h$ to be true, $(b_1, c_1) = (b_1, c_2)$ whenever $g \circ f = h \circ f $. Assume $g \circ f = h \circ f$. Since $b_1 \notin Ran(f)$, $(a,c_1) \notin g \circ f$ and $(a,c_2) \notin h \circ f$. This means that as long as there are at least two elements in C, it is possible that $(b_1, c_2) \neq (b_1, c_2)$ while $g \circ f = h \circ f$. (Even if $c_1 \neq c_2$, it can still be true that $g \circ f = h \circ f$ since $c_1$ and $c_2$ are not in the range of $g \circ f$ and $h \circ f$.) This is a contradiction, hence $f$ is onto.
(One version of) the Whitehead theorem states that a homology equivalence between simply connected CW complexes is a homotopy equivalence. Does the following generalisation hold true? Suppose $X,Y$ are two connected CW complexes and $f:X\to Y$ is a continuous map that induces isomorphisms of the fundamental groups and on homology. Then $f$ is a weak equivalence. It would be enough to show that the map $\tilde{f}$ obtained by lifting $f$ to the universal covers (for some arbitrary choice of base points) is a homology equivalence, but I've never studied the homology spectral sequence, so I don't know the relationship between the homology of the universal cover $\widetilde{X}$ and that of $X$, and I don't know how to prove that $\tilde{f}$ is a homology equivalence. Also, are there any good, reasonably self contained and short accounts of the homology of covering spaces, preferably online? EDIT 1 I've seen some lecture notes (second paragraph on page 5) where this theorem is used (without further comment). They used this proposition to establish that a certain type of homotopy colimit is well defined up to weak equivalence. EDIT 2 Theorem 6.71 from Kirk and Davis (page 179 of the document or 164 internally) is probably what I need, but it involves isomorphisms with local coefficient systems for homology for the base spaces. EDIT 3 Thank you @studiosus for the reference. I don't quite understand yet, could you help out? If I get it (using the theorem of the source), there are finite connected two dimensional complexes $X,Y$ and a homotopy equivalence $X\vee S^2\simeq Y\vee S^2$, yet $X$ and $Y$ aren't homotopy equivalent. From the assumption it follows immediately that $$X\hookrightarrow X\vee S^2\to Y\vee S^2\twoheadrightarrow Y$$ gives an isomorphism of the fundamental groups. From $$\tilde{H}(X)\oplus\tilde{H}(S^2)\simeq\tilde{H}(X\vee S^2)\simeq\tilde{H}(Y\vee S^2)\simeq\tilde{H}(X)\oplus\tilde{H}(S^2)\;,$$ $\tilde{H}(S^2)\simeq \Bbb Z[2]$ and the fundamental theorem of finitely generated abelian groups, it follows at once that $\tilde{H}(X)\simeq\tilde{H}(Y)$ abstractly. The only problem I have is that it doesn't seem to me that we get a map $X\to Y$ that induces isomorphisms on $\pi_1$ and is a homology equivalence, actually, I don't see why there should be a homology equivalence $X\to Y$ at all.
In «New Introduction to Multiple Time Series», page 90, we have the following formulas for the ML estimators of a stable Gaussian VAR$(p)$ process: where $\tilde \alpha = vec(\tilde A_1,...,\tilde A_p)$. And here is exactly where my doubt resides. If for $\tilde \alpha$ estimator, I need $\tilde \mu$ estimator, but then for $\tilde \mu$ I'm going to need also $\tilde \alpha$. So, how is this dependence broken, or is this solved by an iterative algorithm that will reach a fixed point? I've also posted this question in CV, but I didn't get any answer. Any help would be appreciated.
While reading an introduction on elliptic curves, I stumbled upon something called reduction modulo $p$. The definition states that we want to create a group homomorphism that maps an elliptic curve on the rational field to the integers modulo a prime $p$, i.e.: $$E(\mathbb{Q}) \rightarrow \tilde{E}(\mathbb{F}_p), \;\;\;\; P \mapsto \tilde{P}$$ Where $P = (x,y) \in E(\mathbb{Q}),$ and $\tilde{P} = (\tilde{x},\tilde{y}) \in \tilde{E}(\mathbb{F}_p)$. Such a homomorphism is worked out trivially when $x$ and $y$ are in $\mathbb{Z}$, while in the rational case the reader has to write $x = \frac{a}{b}$. Then, if $p \nmid b$, then $\tilde{b}$ has an inverse in $\mathbb{F}_p$, hence $\tilde{x} = \tilde{a}\tilde{b}^{-1}$ in $\mathbb{F}_p$. The last step is quite obscure to me. The paper provides some examples: Let $E: y^2 = x^3 + 2x + 4.$ Then, $P = (2,4)$ and $Q = \left(\frac{1}{4},\frac{17}{8}\right)$ are points in $E(\mathbb{Q})$. Then the reduction modulo 11 map $E(\mathbb{Q}) \rightarrow \tilde{E}(\mathbb{F}_{11})$ gives: $\tilde{P} = (2,4)$ and $\tilde{Q} = (3,9)$. I feel like I'm missing something trivial, but I really have no idea of how the author found the value of $\tilde{Q}$. Can someone help?
Page 552: Problem 20 $ \int_{0}^{\frac{1}{\sqrt{2}}}2x\sin^{-1}(x^2)dx $ Using 2x as u gets rid of it eventually, but the inverse sin just gets worse and worse. Actually the integral of the inverse sin is just the inverse sin minus some radical. So it just cycles through over and over. So using the inverse sin looks better... It comes up with $ [x^2\sin^{-1}(x^2)]_0^{\frac{1}{\sqrt{2}}} - \int_{0}^{\frac{1}{\sqrt{2}}}\frac{x^2}{\sqrt{1-x^4}}dx $ This looks better... but then I can't figure out how to solve that integral. Anyone? I've tried using the bottom as dv and going back to the inverse sin but that just comes up with A = ##### + A and the A's cancel out, so that doesn't work.. Thanks. Idryg 16:10, 13 October 2008 (UTC) I attempted to do the same thing at first with this problem as well. I couldn't get anywhere either. So I tried something totally different. I did a substitution before I integrated by parts. let $ p=x^2,dp=2xdx $ Try that and see what you get from there. Oh yeah, I had to a second substitution after the integration by parts in the second integral. if you were going to do it the first way, I would think it should be ...$ -\int_{0}^{\frac{1}{\sqrt{2}}}\frac{2x^3}{\sqrt{1-x^4}}dx $, because you forgot to take the du for sin^-1. ----Gary Brizendine II 15:14, 14 October 2008 (UTC) Good point Gary, I think that should get you the same answer in the end, and a lot cleaner too. I obviously forgot to take the proper derivative of inverse sine as well. His Awesomeness, Josh Hunsberger
Variable Importance in Random Forests can suffer from severe overfitting Predictive vs. interpretational overfitting There appears to be broad consenus that random forests rarely suffer from “overfitting” which plagues many other models. (We define overfitting as choosing a model flexibility which is too high for the data generating process at hand resulting in non-optimal performance on an independent test set.) By averaging many (hundreds) of separately grown deep trees -each of which inevitably overfits the data – one often achieves a favorable balance in the bias variance tradeoff. For similar reasons, the need for careful parameter tuning also seems less essential than in other models. This post does not attempt to contribute to this long standing discussion (see e.g. https://stats.stackexchange.com/questions/66543/random-forest-is-overfitting) but points out that random forests’ immunity to overfitting is restricted to the predictions only and not to the default variable importance measure! We assume the reader is familiar with the basic construction of random forests which are averages of large numbers of individually grown regression/classification trees. The random nature stems from both “row and column subsampling’’: each tree is based on a random subset of the observations, and each split is based on a random subset of candidate variables. The tuning parameter – which for popular software implementations has the default $\lfloor p/3 \rfloor$ for regression and $\sqrt{p}$ for classification trees – can have profound effects on prediction quality as well as the variable importance measures outlined below. At the heart of the random forest library is the CART algorithm which chooses the split for each node such that maximum reduction in overall node impurity is achieved. Due to the CART bootstrap row sampling, $36.8\%$ of the observations are (on average) not used for an individual tree; those “out of bag” (OOB) samples can serve as a validation set to estimate the test error, e.g.: \[\begin{equation} E\left( Y – \hat{Y}\right)^2 \approx OOB_{MSE} = \frac{1}{n} \sum_{i=1}^n{\left( y_i – \overline{\hat{y}}_{i, OOB}\right)^2} \end{equation}\] where $\overline{\hat{y}}_{i, OOB}$ is the average prediction for the $i$th observation from those trees for which this observation was OOB. Variable Importance The default method to compute variable importance is the mean decrease in impurity (or gini importance) mechanism: At each split in each tree, the improvement in the split-criterion is the importance measure attributed to the splitting variable, and is accumulated over all the trees in the forest separately for each variable. Note that this measure is quite like the $R^2$ in regression on the training set. The widely used alternative as a measure of variable importance or short permutation importance is defined as follows: \[\begin{equation} \label{eq:VI} \mbox{VI} = OOB_{MSE, perm} – OOB_{MSE} \end{equation}\] Gini importance can be highly misleading We use the well known titanic data set to illustrate the perils of putting too much faith into the Gini importance which is based entirely on training data – not on OOB samples – and makes no attempt to discount impurity decreases in deep trees that are pretty much frivolous and will not survive in a validation set. In the following model we include passengerID as a feature along with the more reasonable Age, Sex and Pclass: randomForest(Survived ~ Age + Sex + Pclass + PassengerId, data=titanic_train[!naRows,], ntree=200,importance=TRUE,mtry=2) The figure below shows both measures of variable importance and surprisingly passengerID turns out to be ranked number 2 for the Gini importance (right panel). This unexpected result is robust to random shuffling of the ID. The permutation based importance (left panel) is not fooled by the irrelevant ID feature. This is maybe not unexpected as the IDs shold bear no predictive power for the out-of-bag samples. Noise Feature Let us go one step further and add a Gaussian noise feature, which we call PassengerWeight: titanic_train$PassengerWeight = rnorm(nrow(titanic_train),70,20) rf4 =randomForest(Survived ~ Age + Sex + Pclass + PassengerId + PassengerWeight, data=titanic_train[!naRows,], ntree=200,importance=TRUE,mtry=2) Again, the blatant “overfitting” of the Gini variable importance is troubling whereas the permutation based importance (left panel) is not fooled by the irrelevant features. (Encouragingly, the importance measures for ID and weight are even negative!) In the remainder we investigate if other libraries suffer from similar spurious variable importance measures. h2o library Unfortunately, the h2o random forest implementation does not offer permutation importance: Coding passenger ID as integer is bad enough: Coding passenger ID as factor makes matters worse: Let’s look at a single tree from the forest: If we scramble ID, does it hold up? partykit conditional inference trees are not being fooled by ID: And the variable importance in cforest is indeed unbiased python’s sklearn Unfortunately, like h2o the python random forest implementation offers only Gini importance, but this insightful post offers a solution: Gradient Boosting Boosting is highly robust against frivolous columns: mdlGBM = gbm(Survived ~ Age + Sex + Pclass + PassengerId +PassengerWeight, data= titanic_train, n.trees = 300, shrinkage = 0.01, distribution = "gaussian") Conclusion Sadly, this post is 12 years behind: It has been known for while now that the Gini importance tends to inflate the importance of continuous or high-cardinality categorical variables: the variable importance measures of Breiman’s original Random Forest method … are not reliable in situations where potential predictor variables vary in their scale of measurement or their number of categories. Single Trees I am still struggling with the extent of the overfitting. It is hard to believe that passenger ID could be chosen as a split point early in the tree building process given the other informative variables! Let us inspect a single tree ## rowname left daughter right daughter split var split point status ## 1 1 2 3 Pclass 2.5 1 ## 2 2 4 5 Pclass 1.5 1 ## 3 3 6 7 PassengerId 10.0 1 ## 4 4 8 9 Sex 1.5 1 ## 5 5 10 11 Sex 1.5 1 ## 6 6 12 13 PassengerId 2.5 1 ## prediction ## 1 <NA> ## 2 <NA> ## 3 <NA> ## 4 <NA> ## 5 <NA> ## 6 <NA> This tree splits on passenger ID at the second level !! Let us dig deeper: The help page states For numerical predictors, data with values of the variable less than or equal to the splitting point go to the left daughter node. So we have the 3rd class passengers on the right branch. Compare subsequent splits on (i) sex, (ii) Pclass and (iii) passengerID: Starting with a parent node Gini impurity of 0.184 Splitting on sex yields a Gini impurity of 0.159 1 2 0 72 303 1 71 50 Splitting on passengerID yields a Gini impurity of 0.183 FALSE TRUE 0 2 373 1 3 118 And how could passenger ID accrue more importance than sex ? // add bootstrap table styles to pandoc tables function bootstrapStylePandocTables() { $('tr.header').parent('thead').parent('table').addClass('table table-condensed'); } $(document).ready(function () { bootstrapStylePandocTables(); });
given: $$m_1+m_2+...+m_k=m$$ How can I find $\Theta(log(m_1)+...+log(m_k))$ as related to $m$? I know that i can doing that: $O(log(m_1)+...log(m_k))=O(log(m)+...+log(m))=O(k \cdot log(m))$ , but can I find something without $k$ ? Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It only takes a minute to sign up.Sign up to join this community The $\log$ function is concave so, by Jensen's inequality, $$\frac{1}{k}\sum_{i=1}^k\log m_i \leq \log\Big(\frac1k\sum_{i=1}^k m_i\Big)$$ i.e., $$\sum_{i=1}^k \log m_i \leq k\log\frac{m}{k} = k\,(\log m - \log k)\,,$$ which is basically what you got by bounding $\log m_i\leq \log m$. We can't do any better than this because we could have $m_1 = \dots = k_m = \tfrac{m}{k}$, in which case $\sum_i\log m_i = k\log\tfrac{m}{k}$. However, if $k$ is a fixed constant, then $k\log\tfrac{m}{k}=O(\log m)$.
Question of the week came up in my schools logic club but there is not much information to it. Here is the question: Show that $$ \exists x\,[R(x)\wedge \lnot Q(x)],\ \forall x\,[P(x)\to Q(x)],\, \forall x[R(x)\to(P(x)\vee S(x))] \models \exists x\,[R(x)\wedge S(x)] $$ Describe in details all your work Now I know that it does not want us to use formal deduction, so what method do they want us to use then? Formal Deduction is the only way I am familiar with at the moment but apparently that has been ruled out as a possible method.
For probably twenty years, category theorists have known of some objects in the Platonic universe called "(weak) $\infty$-categories", in which there are $k$-morphisms for all $k\in \mathbb N$, with some weak forms of composition and associativity and .... It was recognized a bit later that it is worth recording two numbers for each category: an $(m,n)$-category for $m,n \in \mathbb N\cup\lbrace \infty\rbrace$ with $n\leq m$ has morphisms up to dimension $m$ (and above that only equalities), but everything above dimension $n$ is invertible (in some weak sense). Thus an $(\infty,0)$-category is a "higher groupoid", which Grothendieck's homotopy hypothesis says should be the same as a homotopy type. (Many people take the homotopy hypothesis to be a definition; others prefer to take it as a check, so that a definition is good if the homotopy hypothesis is a theorem.) Just to confuse the language, an $(\infty,1)$-category is what many people now call by the name "$\infty$-category". In any case, for quite a while these objects were expected to exist but definitions were not available (or perhaps too available, but without enough known to decide which was the correct one). More than a decade ago, Leinster provided A survey of definitions of n-category (Theory and Applications of Categories, Vol. 10, 2002, No. 1, pp 1-70.). Much more recently, Bergner and Rezk have begun a Comparison of models for (∞,n)-categories (to appear in Geometry and Topology). These works tend to focus on $(\infty,n)$-categories with $n<\infty$ (although some of Leinster's definitions do include the $n=\infty$ case). My impression, almost certainly very biased by the people that I happen to hang out with, is that by now it is known that "$(\infty,0)$-category" might as well mean "Kan simplicial set" and "$(\infty,1)$-category" might as well mean "simplicial set in which all inner horns are fillable". By "might as well mean", I mean that there are many reasonable definitions, but all are known to be Quillen-equivalent. For $n< \infty$, "$(\infty,n)$-category" can mean "iterated complete Segal space", for example. It is not clear how to take $n=\infty$ in this approach. A very tempting definition of "$(\infty,n)$-category", which I believe is discussed in Lurie's (to my knowledge not yet fully rigorous) paper On the Classication of Topological Field Theories, is that an $(\infty,n)$-category should be an $(\infty,1)$-category enriched in $(\infty,n-1)$-categories. This is based on the observation that the coherence axioms of being an enriched category only require invertible morphisms above dimension $1$. Thus as soon as you've settled on a notion of "$(\infty,1)$-category" (this seems to have been settled) and a notion of "enriched" (settled?), you get a notion of "$(\infty,n)$-category". This approach is developed in Bergner–Rezk, for example. In any case, it is less clear to me how to use these inductive definitions to get to $(\infty,\infty)$. Let $\mathrm{Cat}_n$ denote the $(\infty,n+1)$-category of $(\infty,n)$-categories. Then there are clearly inclusion functors $\mathrm{Cat}_n \to \mathrm{Cat}_{n+1}$, but the colimit of this sequence does not deserve to be called $\mathrm{Cat}_\infty$, since any particular object in it has morphisms only up to some dimension. I have seen it proposed that one should instead take the (inverse) limit of the "truncation" functors $\mathrm{Cat}_{n+1} \to \mathrm{Cat}_n$; I am unsure exactly how to define these, and even less sure why this deserves the name $\mathrm{Cat}_\infty$. Thus my question: Is there by now an accepted / consensus notion of $(\infty,\infty)$-category? For instance, if I have decided on my favorite meaning of $(\infty,1)$-category, is there some agreed-upon procedure that produces a notion of $(\infty,\infty)$-category? Or are there known comparison results that can push all the way to $n=\infty$? A closely related MathOverflow question was asked a little less than a year ago (or via Wayback Machine here), but received no answers. Also, I am certainly unaware of many papers in the literature, and likely I have misrepresented even those papers I mentioned above. My apologies to all parties for doing so. It is precisely because of these gaps in my knowledge of the literature that I pose this question.
So far we have been investigating functions of the form $y=f(x)$, with one independent and one dependent variable. Such functions can be represented in two dimensions, using two numerical axes that allow us to identify every point in the plane with two numbers. We now want to talk about three-dimensional space; to identify every point in three dimensions we require three numerical values. The obvious way to make this association is to add one new axis, perpendicular to the $x$ and $y$ axes we already understand. We could, for example, add a third axis, the $z$ axis, with the positive $z$ axis coming straight out of the page, and the negative $z$ axis going out the back of the page. This is difficult to work with on a printed page, so more often we draw a view of the three axes from an angle: You must then imagine that the $z$ axis is perpendicular to the other two. Just as we have investigated functions of the form $y=f(x)$ in two dimensions, we will investigate three dimensions largely by considering functions; now the functions will (typically) have the form $z=f(x,y)$. Because we are used to having the result of a function graphed in the vertical direction, it is somewhat easier to maintain that convention in three dimensions. To accomplish this, we normally rotate the axes so that $z$ points up; the result is then: Note that if you imagine looking down from above, along the $z$ axis, the positive $z$ axis will come straight toward you, the positive $y$ axis will point up, and the positive $x$ axis will point to your right, as usual. Any point in space is identified by providing the three coordinates of the point, as shown; naturally, we list the coordinates in the order $(x,y,z)$. One useful way to think of this is to use the $x$ and $y$ coordinates to identify a point in the $x$-$y$ plane, then move straight up (or down) a distance given by the $z$ coordinate. It is now fairly simple to understand some "shapes'' in three dimensions that correspond to simple conditions on the coordinates. In two dimensions the equation $x=1$ describes the vertical line through $(1,0)$. In three dimensions, it still describes all points with $x$-coordinate 1, but this is now a plane, as in figure 14.1.1. Recall the very useful distance formula in two dimensions: the distance between points $\ds (x_1,y_1)$ and $\ds (x_2,y_2)$ is $\ds \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$; this comes directly from the Pythagorean theorem. What is the distance between two points $\ds (x_1,y_1,z_1)$ and $\ds (x_2,y_2,z_2)$ in three dimensions? Geometrically, we want the length of the long diagonal labeled $c$ in the "box'' in figure 14.1.2. Since $a$, $b$, $c$ form a right triangle, $\ds a^2+b^2=c^2$. $b$ is the vertical distance between $\ds (x_1,y_1,z_1)$ and $\ds (x_2,y_2,z_2)$, so $\ds b=\|z_1-z_2\|$. The length $a$ runs parallel to the $x$-$y$ plane, so it is simply the distance between $\ds (x_1,y_1)$ and $\ds (x_2,y_2)$, that is, $\ds a^2=(x_1-x_2)^2+(y_1-y_2)^2$. Now we see that $\ds c^2=(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2$ and $\ds c=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$. It is sometimes useful to give names to points, for example we might let $P_1=(x_1,y_1,z_1)$, or more concisely we might refer to the point $P_1(x_1,y_1,z_1)$, and subsequently use just $P_1$. Distance between two points in either two or three dimensions is sometimes denoted by $d$, so for example the formula for the distance between $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ might be expressed as $$d(P_1,P_2)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}.$$ In two dimensions, the distance formula immediately gives us the equation of a circle: the circle of radius $r$ and center at $(h,k)$ consists of all points $(x,y)$ at distance $r$ from $(h,k)$, so the equation is $\ds r=\sqrt{(x-h)^2+(y-k)^2}$ or $\ds r^2=(x-h)^2+(y-k)^2$. Now we can get the similar equation $\ds r^2=(x-h)^2+(y-k)^2+(z-l)^2$, which describes all points $(x,y,z)$ at distance $r$ from $(h,k,l)$, namely, the sphere with radius $r$ and center $(h,k,l)$. Exercises 14.1 Ex 14.1.1Sketch the location of the points $(1,1,0)$, $(2,3,-1)$,and $(-1,2,3)$ on a single set of axes. Ex 14.1.2Describe geometrically the set of points $(x,y,z)$ thatsatisfy $z=4$. Ex 14.1.3Describe geometrically the set of points $(x,y,z)$ thatsatisfy $y=-3$. Ex 14.1.4Describe geometrically the set of points $(x,y,z)$ thatsatisfy $x+y=2$. Ex 14.1.5The equation $x+y+z=1$ describes some collection of pointsin $\ds \R^3$. Describe and sketch the points that satisfy $x+y+z=1$ andare in the $x$-$y$ plane, in the $x$-$z$ plane, and in the $y$-$z$ plane. Ex 14.1.6Find the lengths of the sides of the triangle with vertices $(1,0,1)$, $(2,2,-1)$, and $(-3,2,-2)$.(answer) Ex 14.1.7Find the lengths of the sides of the triangle with vertices $(2,2,3)$, $(8,6,5)$, and $(-1,0,2)$. Why do the results tellyou that this isn't really a triangle?(answer) Ex 14.1.8Find an equation of the sphere with center at $(1,1,1)$ andradius 2.(answer) Ex 14.1.9Find an equation of the sphere with center at $(2,-1,3)$ andradius 5.(answer) Ex 14.1.10Find an equation of the sphere with center $(3, -2, 1)$ andthat goes through the point $(4, 2, 5)$.(answer) Ex 14.1.11Find an equation of the sphere with center at $(2,1,-1)$ andradius 4. Find an equation for the intersection of this sphere withthe $y$-$z$ plane; describe this intersection geometrically.(answer) Ex 14.1.12Consider the sphere of radius 5 centered at $(2,3,4)$. What isthe intersection of this sphere with each of the coordinate planes? Ex 14.1.13Show that for all values of $\theta$ and $\phi$, the point$(a\sin\phi\cos\theta,a\sin\phi\sin\theta,a\cos\phi)$ lies on thesphere given by $\ds x^2+y^2+z^2=a^2$. Ex 14.1.14Prove that the midpoint of the line segment connecting$\ds (x_1,y_1,z_1)$ to $\ds (x_2,y_2,z_2)$ is at $\ds\left({x_1+x_2\over 2},{y_1+y_2\over 2},{z_1+z_2\over 2}\right)$. Ex 14.1.15Any three points $P_1(x_1,y_1,z_1)$, $P_2(x_2,y_2,z_2)$,$P_3(x_3,y_3,z_3)$, lie in a plane and form a triangle. The triangle inequality says that $\dsd(P_1,P_3)\le d(P_1,P_2)+d(P_2,P_3)$. Prove the triangle inequalityusing either algebra (messy) or the law of cosines (less messy). Ex 14.1.16Is it possible for a plane to intersect a sphere in exactly twopoints? Exactly one point? Explain.
The displacement of charge in response to the force exerted by an electric field constitutes a reduction in the potential energy of the system (Section 5.8). If the charge is part of a steady current, there must be an associated loss of energy that occurs at a steady rate. Since power is energy per unit time, the loss of energy associated with current is expressible as power dissipation. In this section, we address two questions: How much power is dissipated in this manner, and What happens to the lost energy? First, recall that work is force times distance traversed in response to that force (Section 5.8). Stated mathematically: \[\Delta W = +{\bf F}\cdot \Delta {\bf l}\] where the vector ${\bf F}$ is the force (units of N) exerted by the electric field, the vector $\Delta {\bf l}$ is the direction and distance (units of m) traversed, and $\Delta W$ is the work done (units of J) as a result. Note that a “$+$” has been explicitly indicated; this is to emphasize the distinction from the work being considered in Section 5.8. In that section, the work “$\Delta W$” represented energy from an external source that was being used to increase the potential energy of the system by moving charge “upstream” relative to the electric field. Now, $\Delta W$ represents this internal energy as it is escaping from the system in the form of kinetic energy; therefore, positive $\Delta W$ now means a reduction in potential energy, hence the sign change (note: it is bad form to use the same variable to represent both tallies; nevertheless, it is common practice and so we simply remind the reader that it is important to be aware of the definitions of variables each time they are (re)introduced). The associated power $\Delta P$ (units of W) is $\Delta W$ divided by the time $\Delta t$ (units of s) required for the distance $\Delta{\bf l}$ to be traversed: \[\Delta P = \frac{\Delta W}{\Delta t} = {\bf F}\cdot \frac{\Delta {\bf l}}{\Delta t}\] Now we’d like to express force in terms of the electric field exerting this force. Recall that the force exerted by an electric field intensity ${\bf E}$ (units of V/m) on a particle bearing charge $q$ (units of C) is $q{\bf E}$ (Section 2.2). However, we’d like to express this force in terms of a current, as opposed to a charge. An expression in terms of current can be constructed as follows. First, note that the total charge in a small volume “cell” is the volume charge density $\rho_v$ (units of C/m$^3$) times the volume $\Delta v$ of the cell; i.e., $q = \rho_v \Delta v$ (Section 5.3). Therefore: \[{\bf F} = q{\bf E} = \rho_v~\Delta v~{\bf E}\] and subsequently \[\Delta P = \rho_v~\Delta v~{\bf E} \cdot \frac{\Delta {\bf l}}{\Delta t} = {\bf E} \cdot \left( \rho_v \frac{\Delta {\bf l}}{\Delta t} \right) \Delta v\] The quantity in parentheses has units of C/m$^3$ $\cdot$ m $\cdot$ s$^{-1}$, which is A/m$^2$. Apparently this quantity is the volume current density ${\bf J}$, so we have \[\Delta P = {\bf E} \cdot {\bf J}~\Delta v\] In the limit as $\Delta v\to 0$ we have \[dP = {\bf E} \cdot {\bf J}~dv\] and integrating over the volume ${\mathcal V}$ of interest we obtain \[P = \int_{\mathcal V} dP = \int_{\mathcal V} {\bf E} \cdot {\bf J}~dv \label{m0106_eJLx}\] The above expression is commonly known as Joule’s Law. In our situation, it is convenient to use Ohm’s Law for Electromagnetics (${\bf J}=\sigma{\bf E}$; Section 6.3) to get everything in terms of materials properties ($\sigma$), geometry (${\mathcal V}$), and the electric field: \[P = \int_{\mathcal V} {\bf E} \cdot \left(\sigma{\bf E}\right)~dv\] which is simply \[\boxed{ P = \int_{\mathcal V} \sigma \left\|{\bf E}\right\|^2~dv } \label{m0106_eJL}\] Thus: The power dissipation associated with current is given by Equation \ref{m0106_eJL}. This power is proportional to conductivity and proportional to the electric field magnitude squared. This result facilitates the analysis of power dissipation in materials exhibiting loss; i.e., having finite conductivity. But what is the power dissipation in a perfectly conducting material? For such a material, $\sigma\to\infty$ and ${\bf E}\to 0$ no matter how much current is applied (Section 6.3). In this case, Equation \ref{m0106_eJL} is not very helpful. However, as we just noted, being a perfect conductor means ${\bf E}\to 0$ no matter how much current is applied, so from Equation \ref{m0106_eJLx} we have found that: No power is dissipated in a perfect conductor. When conductivity is finite, Equation \ref{m0106_eJL} serves as a more-general version of a concept from elementary circuit theory, as we shall now demonstrate. Let ${\bf E}=\hat{\bf z}E_z$, so $\left\|{\bf E}\right\|^2=E_z^2$. Then Equation \ref{m0106_eJL} becomes: \[P = \int_{\mathcal V} \sigma E_z^2~dv = \sigma E_z^2~\int_{\mathcal V}~dv \label{m0106_ePintegrals}\] The second integral in Equation \ref{m0106_ePintegrals} is a calculation of volume. Let’s assume ${\mathcal V}$ is a cylinder aligned along the $z$ axis. The volume of this cylinder is the cross-sectional area $A$ times the length $l$. Then the above equation becomes: \[P = \sigma E_z^2~A~l\] For reasons that will become apparent very shortly, let’s reorganize the above expression as follows: \[P = \left(\sigma E_z A\right) \left(E_z l\right)\] Note that $\sigma E_z$ is the current density in A/m$^2$, which when multiplied by $A$ gives the total current. Therefore, the quantity in the first set of parentheses is simply the current $I$. Also note that $E_z l$ is the potential difference over the length $l$, which is simply the node-to-node voltage $V$ (Section 5.8). Therefore, we have found: \[P = IV\] as expected from elementary circuit theory. Now, what happens to the energy associated with this dissipation of power? The displacement of charge carriers in the material is limited by the conductivity, which itself is finite because, simply put, other constituents of the material get in the way. If charge is being displaced as described in this section, then energy is being used to displace the charge-bearing particles. The motion of constituent particles is observed as heat – in fact, this is essentially the definition of heat. Therefore: The power dissipation associated with the flow of current in any material that is not a perfect conductor manifests as heat. This phenomenon is known as joule heating, ohmic heating, and by other names. This conversion of electrical energy to heat is the method of operation for toasters, electric space heaters, and many other devices that generate heat. It is of course also the reason that all practical electronic devices generate heat. Contributors Ellingson, Steven W. (2018) Electromagnetics, Vol. 1. Blacksburg, VA: VT Publishing. https://doi.org/10.21061/electromagnetics-vol-1 Licensed with CC BY-SA 4.0 https://creativecommons.org/licenses/by-sa/4.0. Report adoption of this book here. 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Aleksander from Gdynia Bilingual HighSchool No 3, Poland, David from Guilford County (is that UK, US orelsewhere?) Andrei from Romania and Chris from CSN used differentmethods to solve this problem, all of them leading to solving aquadratic equation. Chris used the observation that, removing threeresistors leaves the resistance of the remaining network unchanged,and quickly arrived at the quadratic equation which gives thesolution. Aleksander visualised the network in a different, butequivalent, configuration and used continued fractions and David'smethod was similar. Andrei arrived at the same quadratic equationby first proving that the sequence of resistances $R_0, R_1, R_2,...$ from A to B, corresponding to different number of resistance'squares', is a steadily decreasing sequence and, as $R_n > 0$,the sequence tends to a limit.This is Chris's solution: Following the hint, keep in mind that the equations for seriesresistances and parallel resistances are: $R = R_1 + R_2$ forseries and $R = (R_1 \times R_2)/(R_1 + R_2)$ for parallel. The resistance $R$ between A and B is the equivalent of aresistance of 1 ohm in parallel with two resistances of 1 ohm and aresistance R in series and we can write the following equation: $$R= {1 \times (2 + R)\over 1 + (2 + R)} = {(2 + R)\over (3 + R)}.$$Reworking the above equation we get: $R(3 + R) = (2 + R)$ or $$R^2+ 2R - 2 = 0.$$ Solving the quadratic equation we get twosolutions, one of which we cannot use because it is negative, so $R= - 2.73205$ is not valid because a resistance cannot have anegative value. The correct solution is: $$R =\sqrt 3 -1 \approx0.732051.$$ This is Aleksander'ssolution: Generally it is easier to calculate the resistance of acircuit when you present it in a way that one end of circuit (hereA) is on the left and the other is on the right. Here we have anillustration of our ladder circuit in such a way making it mucheasier to justify whether two resistors are in parallel or inseries.: From the picture we know that resistors labeled 1, 2 and thegroup of resistors 4,5,6,7,8,9 (all 1 ohm) are connected in series.We can also observe that the resistor 3 is connected in parallel tothe resistors mentioned before. Therefore, where $R$ is theresistance of the whole circuit, and $R_n$ is the resistance of the$n$th resistor, we can write: $$R = {1\over {1\over R_3} + {1\overR_1 + R_2 +...}}.$$ The three dots refer to the resistances of theresistors not mentioned in the formula, but occurring in thefurther part of the circuit. We can observe that the pattern for the first three resistorsrepeats for the rest so $$R = {1\over {1\over R_3} + {1\over R_1+R_2 + {1\over {1\over R_6} + {1\over R_4 +R_5+...}}}}$$ and thefraction grows like that to infinity. As all the resistances are 1ohm this gives the continued fraction: $$R = {1\over 1 + {1\over 2+ {1\over 1 + {1\over 2 +...}}}} = {1\over 1 + {1\over 2 + R}}$$After a few simplifications we obtain the quadratic equation : $R^2+ 2R - 2 = 0.$ As $R$ must be positive the solution is $R = \sqrt 3- 1$ ohms. This is Andrei'ssolution: First I shall prove that the resistance from A to B tends to alimit as more 'squares' are added to the network. I shall denotethe value of each resistance by $R$ and the values of theequivalent resistances between A and B, corresponding to differentnumber of resistance 'squares', by $R_0, R_1, R_2, ... $ etc.Evidently $R_0 = R$ and $R_1 = 3R/4$. From the diagram I observe that: $${1\over R_n} = {1\over R} +{1\over 2R +R_{n-1}}.$$ where $R = 1$. I observe that $R_1< R_0$, and I shall prove by induction that $R_n< R_{n-1}$. This,by the recurrence relation, is equivalent to: $$R_n< R_{n-1}\Leftrightarrow {1\over R_n}> {1\over R_{n-1}}\Leftrightarrow {1\over R} + {1\over 2R + R_{n-1}} > {1\over R}+ {1\over 2R + R_{n-2}}\Leftrightarrow R_{n-1} < R_{n-2}.$$ So,the sequence $(Rn)$ is strictly decreasing. Evidently $R_n > 0$so the sequence $(R_n)$ is convergent. By considering the limit inthe recurrence relation (putting $R=1$), $$\lim _{n\to \infty} R_n= {1\over {1+ {1\over 2 + \lim_{n\to \infty}R_{n-1}}}}$$ where$$\lim _{n\to \infty} R_n = \lim _{n\to \infty} R_{n-1} = R_{L}.$$This gives $R_{L}= {2+ R_{L}\over 3 + R_{L}}$ and hence thequadratic equation $R_{L}^2 + 2R_{L} - 2= 0$. The positive solutionof this equation gives the resistance of the network: $\sqrt 3 - 1$ohms.
What is the difference between linearly and affinely independent vectors? Why does affine independence not imply linear independence necessarily? Can someone explain using an example? To augment Lord Shark's answer, I just wanted to talk a little about the intuition behind it. Intuitively, a set of vectors is linearly dependent if there are more vectors than necessary to generate their span, i.e. the smallest subspace containing them. On the other hand, a set of vectors is affinely dependent if there are more vectors than necessary to generate their affine hull, i.e. the smallest flat (translate of a linear space) containing them. A single vector $v$ in a vector space generates an affine hull of $\lbrace v \rbrace$, which is just the trivial subspace $\lbrace 0 \rbrace$ translated by $v$. But, if $v \neq 0$, the span is the entire line between $0$ and $v$, as $0$ must be part of any subspace. To generate that line as an affine hull, you could look at the list $v, 0$. So, $v, 0$ are linearly dependent (e.g. $0 = 0 \cdot v + 5 \cdot 0$) as $0$ is not necessary to generate the span (just $v$ would have done fine), but both are necessary to generate the line as the affine hull, so they are affinely independent. To prove this, suppose $\lambda_1 + \lambda_2 = 0$ and, $$\lambda_1 \cdot v + \lambda_2 \cdot 0 = 0.$$ Then $\lambda_1 \cdot v = 0$, which implies $\lambda_1 = 0$, since $v \neq 0$. Since $\lambda_1 + \lambda_2 = 0$, we therefore also have $\lambda_2 = 0$. This proves affine independence. Vectors $v_1,\ldots,v_n$ are linearly dependent if there are$\lambda_1,\ldots,\lambda_n$, not all zero, with $\lambda_1 v_1+\cdots+\lambda_n v_n=0$. Vectors $v_1,\ldots,v_n$ are affinely dependent if there are$\lambda_1,\ldots,\lambda_n$, not all zero, with $\lambda_1 v_1+\cdots+\lambda_n v_n=0$ and $\lambda_1+\cdots+\lambda_n=0$. In $\Bbb R^1$ any two distinct vectors are linearly dependent but affinely independent. The set $\{ v_1,...,v_n\}$ is linearly independent if for $c_1, c_2,...,c_n$, not all zero, such that $$\sum_{k=1}^n c_kv_k =0 \Leftrightarrow c_j=0, \: \forall j=1,2,...,n.$$ The set is called affinely linearly independent if they are linearly independent and also $$ \sum_{k=1}^n c_k=1.$$
Answer The current when the resistance is $10$ ohms is $11$ Amperes. Work Step by Step We have to write the relation between $ I $ and $ R $ as, $\begin{align} & I\propto \frac{1}{R} \\ & I=\frac{k}{R} \end{align}$ Where $ k $ is the constant of proportionality. Put the value of $ R=22$ and $ I=5$ in the equation to get the value of $ k $ as: $\begin{align} & 5=\frac{k}{22} \\ & k=5\cdot 22 \\ & k=110 \end{align}$ So, the equation becomes $ I=\frac{110}{R}$. Then, one has to find the value of $ I $ when $ R $ is $10$ as: $\begin{align} & I=\frac{110}{10} \\ & =11 \end{align}$ Thus, the current when resistance is $10$ ohms is $11$ Amperes.
You'll probably find a partial answer in this post. The reversible/irreversible tag refers to what is happening to the system, not the surroundings. Events in the surroundings are as a rule treated as reversible. You should also keep in mind that heat is a signed quantity, and that there can be different conventions with regard to what is a positive/negative heat transfer. Usually chemists say q>0 when the system receives energy as heat, and when the surroundings loses energy as heat. Since heat is regarded as the sole source of change in the entropy of the surroundings (assumed to be an inifinitely large reservoir at constant T$^\ast$), we write $$dS_{surroundings} = -\frac{dq}{T}$$ for an infinitesimal heat transfer and $$\Delta S_{surroundings} = -\frac{q}{T}$$ for a finite heat transfer. $^\ast$ An infinite reservoir can provide or accept energy in the form of heat without temperature change (or any other change in its state).
Skills to Develop Explain all the quantities involved in the ideal gas law. Evaluate the gas constant $R$ from experimental results. Calculate $T$, $V$, $P$, or $n$ of the ideal gas law: $P V = n R T$. Describe the ideal gas law using graphics. The Ideal Gas Law The volume ( V) occupied by n moles of any gas has a pressure ( P) at temperature ( T) in Kelvin. The relationship for these variables, \[P V = n R T\] where R is known as the gas constant, is called the ideal gas law or equation of state. Properties of the gaseous state predicted by the ideal gas law are within 5% for gases under ordinary conditions. In other words, given a set of conditions, we can predict or calculate the properties of a gas to be within 5% by applying the ideal gas law. How to apply such a law for a given set of conditions is the focus of general chemistry. At a temperature much higher than the critical temperature and at low pressures, however, the ideal gas law is a very good model for gas behavior. When dealing with gases at low temperature and at high pressure, correction has to be made in order to calculate the properties of a gas in industrial and technological applications. One of the common corrections made to the ideal gas law is the van der Waal's equation, but there are also other methods dealing with the deviation of gas from ideality. The Gas Constant R Repeated experiments show that at standard temperature (273 K) and pressure (1 atm or 101325 N/m 2), one mole ( n = 1) of gas occupies 22.4 L volume. Using this experimental value, you can evaluate the gas constant R, $\begin{align} R &= \dfrac{P V}{n T} = \mathrm{\dfrac{1\: atm\:\: 22.4\: L}{1\: mol\:\: 273\: K}}\\ \\ &= \mathrm{0.08205\: \dfrac{L\: atm}{mol\cdot K}} \end{align}$ When SI units are desirable, P = 101325 N/m 2 (Pa for pascal) instead of 1 atm. The volume is 0.0224 m 3. The numerical value and units for R are $\begin{align} R &= \mathrm{\dfrac{101325\: \dfrac{N}{m^2}\: 0.0224\: m^3}{1\: mol\: 273\: K}}\\ \\ &= \mathrm{8.314\: \dfrac{J}{mol\cdot K}} \end{align}$ Note that $\mathrm{1\: L\: atm = 0.001\: m^3 \times 101325\: \dfrac{N}{m^2} = 101.325\: J\: (or\: N\: m)}$. Since energy can be expressed in many units, other numerical values and units for R are frequently in use. For your information, the gas constant can be expressed in the following values and units. $\begin{align} R &= \mathrm{0.08205\: \dfrac{L\: atm}{mol \cdot K}} &&\textrm{Notes:} \\ &= \mathrm{8.3145\: \dfrac{L\: kPa}{mol\cdot K}} &&\mathrm{1\: atm = 101.32\: kPa} \\ &= \mathrm{8.3145\: \dfrac{J}{mol\cdot K}} &&\mathrm{1\: J = 1\: L\: kPa} \\ &= \mathrm{1.987\: \dfrac{cal}{mol\cdot K}} &&\mathrm{1\: cal = 4.182\: J} \\ &= \mathrm{62.364\: \dfrac{L\: torr}{mol\cdot K}} &&\mathrm{1\: atm = 760\: torr}\\ \end{align}$ The gas constant R is such a universal constant for all gases that its values are usually listed in the "Physical Constants" of textbooks and handbooks. It is also listed in Constants of our HandbookMenu at the left bottom. Although we try to use SI units all the time, the use of atm for pressure is still common. Thus, we often use R = 8.314 J / (mol·K) or 8.3145 J / mol·K. The volume occupied by one mole, n = 1, of substance is called the molar volume, $V_{\textrm{molar}} = \dfrac{V}{n}$. Using the molar volume notation, the ideal gas law is: $P V_{\textrm{molar}} = R T$ Applications of the Ideal Gas Law The ideal gas law has four parameters and a constant, R, $P V = n R T$, and it can be rearranged to give an expression for each of P, V, n or T. For example, $P = \dfrac{n R T}{V}$ (Boyle's law) $P = \left(\dfrac{n R}{V}\right) T$ (Charles's law) These equations are Boyle's law and Charles's law respectively. Similar expressions can be derived for V, n and T in terms of other variables. Thus, there are many applications. However, you must make sure that you use the proper numerical value for the gas constant R according to the units you have for the parameters. Furthermore, $\dfrac{n}{V}$ is number of moles per unit volume, and this quantity has the same units as the concentration ( C). Thus, the concentration is a function of pressure and temperature, $C = \dfrac{P}{R T}$ At 1.0 atm pressure and room temperature of 298 K, the concentration of an ideal gas is 0.041 mol/L. Avogadro's law can be further applied to correlate gas density $\rho$ (weight per unit volume or n M / V) and molecular mass M of a gas. The following equation is easily derived from the ideal gas law: \[P M =\dfrac{n M}{V}R T\] Thus, we have $\begin{align} P M &= \dfrac{d R T}{M}\\ \rho &= \dfrac{n M}{V} \leftarrow \textrm{definition, and}\\ \rho &= \dfrac{P M}{R T}\\ M &= \dfrac{d R T}{P} \end{align}$ Example 1 An air sample containing only nitrogen and oxygen gases has a density of 1.3393 g / L at STP. Find the weight and mole percentages of nitrogen and oxygen in the sample. SOLUTION From the density $\rho$, we can evaluate an average molecular weight (also called molar mass). $\begin{align} P M &= d R T\\ M &= 22.4 \times d\\ &= \mathrm{22.4\: L/mol \times 1.3393\: g/L}\\ &= \mathrm{30.0\: g / mol} \end{align}$ Assume that we have 1.0 mol of gas, and x mol of which is nitrogen, then (1 - x) is the amount of oxygen. The average molar mass is the mole weighted average, and thus, $\mathrm{28.0\, x + 32.0 (1 - x) = 30.0}$ $\mathrm{- 4\, x = - 2}$ $\mathrm{x = 0.50\: mol\: of\: N_2,\: and\: 1.0 - 0.50 = 0.50\: mol\: O_2}$ Now, to find the weight percentage, find the amounts of nitrogen and oxygen in 1.0 mol (30.0 g) of the mixture. $\mathrm{Mass\: of\: 0.5\: mol\: nitrogen = 0.5 \times 28.0 = 14.0\: g}$ $\mathrm{Mass\: of\: 0.5\: mol\: oxygen = 0.5 \times 32.0 = 16.0\: g}$ $\mathrm{Percentage\: of\: nitrogen = 100 \times \dfrac{14.0}{30.0} = 46.7 \% }$ $\mathrm{Percentage\: of\: oxygen = 100 \times \dfrac{16.0}{30.0} = 100 - 46.7 = 53.3 \%}$ DISCUSSION We can find the density of pure nitrogen and oxygen first and evaluate the fraction from the density. $\rho \mathrm{\: of\: N_2 = \dfrac{28.0}{22.4} = 1.2500\: g/L}$ $\rho \mathrm{\: of\: O_2 = \dfrac{32.0}{22.4} = 1.4286\: g/L}$ $\mathrm{1.2500\, x + 1.4286 (1 - x) = 1.3393}$ Solving for x gives $\mathrm{x = 0.50}$ (same result as above) Exercise 1 Now, repeat the calculations for a mixture whose density is 1.400 g/L. Example 2 What is the density of acetone ($\ce{C3H6O}$) vapor at 1.0 atm and 400 K? SOLUTION The molar mass of acetone = 312.0 + 61.0 + 16.0 = 58.0. Thus, $\begin{align} \rho &= \dfrac{P M}{R T}\\ \\ &= \mathrm{\dfrac{1.0 \times 58.0\: atm\: \dfrac{g}{mol}} {0.08205\: \dfrac{L\: atm}{mol\: K} \times 400\: K}}\\ \\ &= \mathrm{1.767\: g / L} \end{align}$ Exercise 2 The density of acetone is 1.767 g/L; calculate its molar mass. Confidence Building Questions A closed system means no energy or mass flow into or out of a system. In a closed system, how many independent variables are there amongNote: an independent variable can be of any arbitrary values. n, T, Vand Pfor a gas? Hint: one Contributors Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)
The difference between the "stabilising" and "destabilising" examples that you have chosen are that the "stabilising" examples are static molecules, whereas the "destabilising" examples are chemical reactions, in which a lot is happening. We'll look at the "stabilising" cases first, as these are easier to understand. As with any question involving "stability", it is important to consider what we are measuring stabilisation with respect to. In the example of the anomeric effect (e.g. in esters), we are comparing the true electronic state of the ester against a hypothetical scenario, where the lone pair and σ* do not overlap, but everything else is kept the same. In such cases, it is nearly always true that allowing overlap, or delocalisation, leads to a lower energy. This is captured quite well in the last picture in your question body. (See also: Physical reason for why delocalisation leads to stability?) In a chemical reaction, as we move along the reaction coordinate (RC), we are creating overlap between two different orbitals, just as in the previous case. Naively we might think that this necessarily leads to stabilisation. But this is not physically sensible: if the reaction proceeds via a transition state (TS), this TS must be higher in energy than the reactants (at infinite separation). [1] The problem is that when we bring two molecules together in space, it is not just overlap that is affected: the energies of every orbital will change as nuclear positions are moved, electron-electron repulsions are different, etc. This means that all the orbital energies change in a way which depends not only on overlap with other orbitals (as was the case before), but also a number of other factors. Instead of going from reactants at infinite separation to the TS directly along the RC, as would be physically realistic, it may be instructive to consider hypothetically doing it in two stages. In the first step, we bring the reactants together from infinite separation to the TS geometry while forbidding any orbital overlap; secondly, we stay at the TS geometry, but turn on the overlap. In real life, you cannot separate these two things: changing the geometry comes with formation of overlap. But thinking about it this way allows us to link it to the "stabilising" cases talked about before. Let's use the addition of a nucleophile to a carbonyl group as an example to make this more concrete. In the reactants, there are three key frontier orbitals we need to consider: the nucleophile lone pair ($n$), and the π-type orbitals of the carbonyl group ($\pi$ and $\pi^$). In general, as we bring the nucleophile and carbonyl together, mixing of all three orbitals will occur continuously. As the reaction proceeds, we will therefore have three MOs $\{\psi^{(i)}\}$ $(i = 1,2,3)$ which are linear combinations of the three frontier orbitals listed earlier: $$\psi^{(i)} = c^{(i)}_n\phi_n + c^{(i)}_\pi \phi_\pi + c^{(i)}_{\pi^}\phi_{\pi^}$$ For well-separated reactants, the relevant MOs will (for the most part) be the original three frontier orbitals. So, for example, we could have $\psi^{(1)} = \phi_\pi$, $\psi^{(2)} = \phi_n$, $\psi^{(3)} = \phi_{\pi^}$. But at the end of the reaction, the MOs of the product will be a complicated mixture of the three orbitals. For example, the $\ce{C-Nu}$ $\sigma$ orbital might be made up of something that resembles this: $$\psi^{(1,\text{new})} \approx \frac{1}{\sqrt{3}}(\phi_n + \phi_\pi + \phi_{\pi^})$$ The reason why these linear combinations (or mixtures) can form is because there is non-zero overlap between all of these orbitals as the reaction proceeds. If there is no overlap between two orbitals, then they cannot mix. So, the way to forbid any overlap from occurring in the first step is to simply not change the coefficients of the MOs as we bring the reactants closer to each other (towards the TS geometry). Of course, this is not physically possible, which is why it is purely a hypothetical scenario. Even though the MOs themselves do not change in this first step, their energies do change, because the Hamiltonian is changing (by virtue of the nuclear positions changing). So, the energies of the MOs will change. To throw an additional wrench into the works, it is not just the electronic energy that we must consider, but also the nuclear energy (when we bring nuclei closer to each other, they will repel each other, and this is not captured in MO diagrams which only illustrate electronic energy). As you can imagine, it is very hard to make any real analysis as to what happens in this step. In any case, once we reach the TS geometry, the energy of the system will definitely have changed in some way. We can now carry out the second step by "turning on" the overlap, i.e. finding the appropriate linear combinations of ${n, \pi, \pi^}$. Since nothing else here is changing (in particular, no nuclear movement), except for us "allowing" overlap to occur, it is exactly analogous to the "stabilising" case which was discussed earlier. So the second step always leads to a decrease in energy. The problem is that, from the infinitely-separated starting materials, we have carried out two steps to get to the TS. In the second step, the energy has decreased – a "stabilisation" of sorts. But in the first step, we don't know what has happened. If a TS is to be higher in energy than the reactants, then it follows that there must have been an increase in energy in the first step. [2] Simple LCAO-MO diagrams don't quite capture the complexity involved in the first step – they can only really properly describe the second step. A better way of drawing it might be to do something like this: I haven't drawn the orbitals in the TS, but you can imagine that they are starting to morph into what they will be in the products. So $\psi^{(1)}$ will be somewhat intermediate between $\pi_\ce{C=O}$ and $\sigma_\ce{C-Nu}$. Of course, this is purely hypothetical - especially in the middle, the orbitals and their energies have no physical meaning at all. In a real reaction you would go straight from SM to TS, and not via an unphysical intermediate stage as is depicted here. So, therein lies the difference between the two cases you have described. In the "stabilising" case, the only thing that is happening is a magical "switching on" of overlap. In the "destabilising" case of a chemical reaction, in addition to the overlap (which is stabilising), you also have a glut of other factors which can lower or (more frequently) raise the energy of the system as it move towards the TS. The problem is that we cannot separate the two from each other. Creating overlap comes with all the other strings attached, and so overall as we move along the RC towards the TS, there is (typically) an increase in energy. Finally, how do we relate this back to the description of nucleophilic addition in Clayden? The wording is definitely somewhat loose (organic chemists frequently do this for ease of use), but it does have a grounding in quantum theory. I'll reproduce the image here and try to relate each phrase: Electrons in HOMO begin to interact with LUMO This refers to the developing overlap that we looked at in step 2, i.e. the formation of MOs which are linear combinations of the HOMO ($n$) and LUMO ($\pi^$). On its own this seems to be a stabilising effect, but recall that this does not come on its own: we also have to consider the different nuclear positions and its implications on the energy. Filling of $\pi^$ I think the way to interpret it is that, at the start of the reaction the two filled orbitals $\psi^{(1)}$ and $\psi^{(2)}$ are simply $n$ and $\pi$, so the coefficients $c_{\pi^}^{(1)}$ and $c_{\pi^}^{(2)}$ are zero. As the reaction progresses, these coefficients steadily increase, which in a sense corresponds to a population of the $\pi^$-orbital, even though the $\pi^$-orbital ceases to exist in its own right once we bring the starting materials together. causes $\pi$ bond to break In a sense we could define a "$\pi$ bond order" to be some kind of difference between the coefficients of the $\pi$ orbital in the filled MOs, and the coefficients of the $\pi^$ orbital in the filled MOs. At the start these are $1$ and $0$ respectively, for a bond order of $1$. As the reaction proceeds and the orbitals mix, we find that $c_\pi$ decreases in the filled MOs, $c_{\pi^}$ increases, and at the end of the reaction we have a bond order of $0$. So loosely speaking, it is true that "population of the $\pi^$ orbital" (i.e. increasing $c_{\pi^}$ in the filled MOs) leads to the cleavage of the $\pi$-bond. Electrons from $\pi$ bond end up as negative charge on oxygen This is somewhat true again since the filled MOs morph from $\pi_\ce{C=O}$ plus a lone pair to $\sigma_\ce{C-Nu}$ and a lone pair, but it's important to remember that it is not a one-to-one correlation, i.e. the $\pi_\ce{C=O}$ MO does not become the oxygen lone pair all by itself. [3] More properly, the oxygen lone pair is a linear combination of multiple MOs in the starting material. It's also important to not ascribe too much importance to exactly which electrons go where, as this violates quantum indistinguishability. Notes [1] There are barrierless reactions which do not proceed via a transition state, such as $\ce{H + H -> H2}$, and in such cases it is quite true that we always get stabilisation as we move along the RC towards products. However, these reactions don't fully capture the complexity involved in other reactions with TSes and it's better to think of these as being special cases of the other reactions. [2] Barrierless reactions don't obey this line of thinking, of course. In fact, they don't even have a TS, so there isn't a well-defined geometry for us to distort towards in the first step. You can pick any geometry along the reaction coordinate between reactants and products, and it would probably work. [3] Under some special circumstances it is possible to have a one-to-one relationship between reactant MOs and product MOs. See e.g. the first correlation diagram in Theoretical basis behind orbital correlation diagram for pericyclic reactions. But note that here there is only one orbital of each symmetry species, so there is no mixing going on and no linear combinations being formed. See also the last section in the answer on "Real pericyclic reactions".
In addition to the moment of inertia, the product of inertia is commonly used. Here only the product of the area is defined and discussed. The product of inertia defined as \[I_{x_{i}x_{j}} = \int_{A} x_{i} x_{j} dA \tag{47}\] For example, the product of inertia for $x$ and $y$ axes is \[I_{xy} = \int_{A} x y dA \tag{48}\] Product of inertia can be positive or negative value as oppose the moment of inertia. The calculation of the product of inertia isn't different much for the calculation of the moment of inertia. The units of the product of inertia are the same as for moment of inertia. Transfer of Axis Theorem Same as for moment of inertia there is also similar theorem. \[I_{x'y'} = \int_{A} x'y'dA = \int_{A}\left(x + \Delta x\right) \left(y + \Delta y\right) dA \tag{49}\] expanding equation 49 results in \[I_{x'y'} = \int_{A} xydA + \int_{A} x\Delta ydA + \int_{A} \Delta xydA + \int_{A} \Delta x \Delta ydA\tag{50}\] The final form is \[I_{x'y'} = I_{xy} + \Delta x \Delta y dA \tag{51}\] There are several relationships should be mentioned \[I_{xy} = I_{yx}\tag{52}\] Symmetrical area has zero product of inertia because integration of odd function (asymmmertial function) left part cancel the right part. Fig. 3.13. Product of inertia for triangle. Example 3.8 Calculate the product of inertia of straight edge triangle. Solution 3.8 The equation of the line is \[y = \dfrac{a}{b} x + a \tag{53}\] The product of inertia at the center is zero. The total product of inertia is \[I_{x^{'}y^{'}} = 0 + \overbrace{\dfrac{a}{3}}^{\Delta x} \overbrace{\dfrac{b}{3}}^{\Delta y} \overbrace{\left(\dfrac{a\,b}{2}\right)}^{A} = \dfrac{a^2\,b^2}{18} \tag{54} \] Contributors Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.
I dropped the $n\rightarrow\infty$ in the title as it was exceeding the character limit. In the book I'm currently reading, the author claims that $\liminf\limits_{n\rightarrow\infty} E_{n_0+n} = \liminf\limits_{n\rightarrow\infty} E_{n}$ and $\limsup\limits_{n\rightarrow\infty} E_{n_0+n} = \limsup\limits_{n\rightarrow\infty} E_{n}$ where $\mu(E_{n_{0}}) < \infty$ and $E_n$ is a decreasing sequence? ($\mu$ is an arbitrary measure!) My attempt at a proof is as follows: Let $x \in \liminf\limits_{n\rightarrow\infty} E_{n_0+n} = \bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n_0+n}$ Then $x \in \bigcap_{k\geq n}E_{n_0+n_{0}'}$ for some $n_{0}' \in \mathbb{N}$ But $\bigcap_{k\geq n}E_{n_0+n_{0}'} \subset \bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n} = \liminf\limits_{n\rightarrow\infty} E_{n}$ and so $\liminf\limits_{n\rightarrow\infty} E_{n_0+n} \subset \liminf\limits_{n\rightarrow\infty} E_{n}$. I however couldn't prove the reverse inclusion in my attempt below: Let $x \in \liminf\limits_{n\rightarrow\infty} E_{n} = \bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n}$ Then $x \in \bigcap_{k\geq n}E_{n_{0}''}$ for some $n_{0}'' \in \mathbb{N}$ I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $\bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n_0+n}$ from $\bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}E_{n}$. I suspect the proof for $\limsup\limits_{n\rightarrow\infty} E_{n_0+n} = \limsup\limits_{n\rightarrow\infty} E_{n}$ is similar and would also fall apart at the same point as my proof for $\liminf\limits_{n\rightarrow\infty} E_{n_0+n} = \liminf\limits_{n\rightarrow\infty} E_{n}$ ($n_{0} \neq n_{0}''$) and so I didn't attempt it. Perhaps I'm missing something about how $n_0$ was chosen such that $\mu(E_{n_{0}}) <\infty$... Original Text: The main part of my question comes from here: Theorem 1.26: And finally, Lemma 1.7: Hopefully someone can shed some light!
The enthalpy of combustion $\Delta_\mathrm cH$ of a substance can be calculated from the enthalpy of formation $\Delta_\mathrm fH$. For a compound containing only carbon, hydrogen, and oxygen, the general combustion reaction is $$\ce{C_$a$H_$b$O_$c$ + ($a$ + 1/4 $b$ – 1/2 $c$) O2 -> $a$ CO2(g) + 1/2 $b$ H2O(l)}$$ The corresponding standard enthalpy of combustion is $$\Delta_\mathrm cH^\circ=-a\Delta_\mathrm fH^\circ(\ce{CO2,g})-\frac12b\Delta_\mathrm fH^\circ(\ce{H2O,l})+\Delta_\mathrm fH^\circ(\ce{C_$a$H_$b$O_$c$})$$ This equation applies if the reactants start in their standard states and the products return to the same conditions. The standard state pressure is $p^\circ=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ (note that most data published before 1982 used a standard pressure of one ‘standard atmosphere’, i.e. $p=1\ \mathrm{atm}=101325\ \mathrm{Pa}$). The definition of standard state makes no reference to fixed temperature; thus, the standard enthalpy of formation and the standard enthalpy of combustion remain functions of temperature. The most widely used reference temperature is $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$. The following values for the standard molar enthalpy of formation at $p=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ and $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ are taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL. $$\begin{align}\Delta_\mathrm fH^\circ(\text{propan-1-ol},\ \mathrm l)&=-302.6\ \mathrm{kJ\ mol^{-1}}\\\Delta_\mathrm fH^\circ(\text{propan-2-ol},\ \mathrm l)&=-318.1\ \mathrm{kJ\ mol^{-1}}\\\Delta_\mathrm fH^\circ(\ce{H2O, l})&=-285.8\ \mathrm{kJ\ mol^{-1}}\\\Delta_\mathrm fH^\circ(\ce{CO2, g})&=-393.5\ \mathrm{kJ\ mol^{-1}}\end{align}$$ During your experiment, the products certainly did not completely return to the initial temperature. Anyway, for the difference in enthalpy of combustion of propan-1-ol and propan-2-ol, the exact state of the products is not relevant (provided that the conditions are reproduced) since the produced amounts of $\ce{CO2}$ and $\ce{H2O}$ are identical. The remaining difference is caused by the difference in enthalpy of formation of propan-1-ol and propan-2-ol.
Given two smooth projective schemes $X$ and $Y$ over some algebraically closed field $k$, we have $X\times Y$ with the projections $p$ to $X$ and $q$ to $Y$. Furthermore we have a "nice" sheaf of algebras $R$ on $X$, i.e. locally free and of global dimension at most dim($X$), e.g. Azumaya or something similar. Given two $p^{}R$-modules $M$ and $N$ on $X\times Y$, which are coherent and torsion free. Like in the commutative case, i define the i-th relative $\mathcal{E}xt$-sheaf on $Y$ to be: $\mathcal{E}xt^i_{p^{}R,q}(M,N):=(R^i(q_{}\mathcal{H}om_{p^{}R}(M,-))(N)$ Can I expect them to have the same properties as in the commutative case? For example: (1) Do we have $\mathcal{E}xt^i_{p^{}R,q}(M,N)=0$ for $i>dim(X)$? (2) Given $y\in Y$ is there a map $\mathcal{E}xt^i_{p^{}R,q}(M,N)\otimes k(y) \rightarrow Ext_R^i(M_y,N_y)$ (3) If $Ext_R^i(M_y,N_y)=0$ for all $y\in Y$ does this imply $\mathcal{E}xt^i_{p^{}R,q}(M,N)=0$? (4) Is there a kind of base change theorem for the $\mathcal{E}xt^i_{p^{}R,q}(M,N)$? Or do I need more conditions for $M$ and $N$ to have the desired properties? I'm especially interested in the case, where $M=p^{*}P$ for some $R$-module $P$ on $X$.
Let $M$ be a compact symplectic manifold equipped with a line bundle $\mathcal L$ with curvature $\omega$. Denote by $H^0(M,\mathcal L)$ the space of smooth global sections of $\mathcal L$ (this space is infinite-dimensional). Let $N$ parameterize complex structures on $M$ compatible with $\omega$. For $J\in N$, we may consider the space $\mathcal H^0_J(M,\mathcal L)\subseteq H^0(M,\mathcal L)$, the finite-dimensional subspace of $J$-holomorphic sections of $\mathcal L$. The Hitchin connection is a natural flat connection on $\mathcal H^0_J(M,\mathcal L)$ considered as a bundle over $N$. The same construction of course applies to $\mathcal L^{\otimes k}$ for any $k\geq 1$. Suppose we have (over some small open set in $N$) two flat sections of $\mathcal H^0_J(M,\mathcal L^{\otimes a})$ and $\mathcal H^0_J(M,\mathcal L^{\otimes b})$. Clearly their tensor product is a section of $\mathcal H^0_J(M,\mathcal L^{\otimes(a+b)})$. Is it flat? I am mainly interested in the case where $M$ is the character variety of a compact Riemann surface.
As we briefly discussed in section 1.1, the most famous graph coloring problem is certainly the map coloring problem, proposed in the nineteenth century and finally solved in 1976. Definition 5.8.1 A proper coloring of a graphis an assignment of colors to the vertices of the graph so that no twoadjacent vertices have the same color. Usually we drop the word "proper'' unless other types of coloring are also under discussion. Of course, the "colors'' don't have to be actual colors; they can be any distinct labels—integers, for example. If a graph is not connected, each connected component can be colored independently; except where otherwise noted, we assume graphs are connected. We also assume graphs are simple in this section. Graph coloring has many applications in addition to its intrinsic interest. Example 5.8.2 If the vertices of a graph represent academic classes, and two vertices are adjacent if the corresponding classes have people in common, then a coloring of the vertices can be used to schedule class meetings. Here the colors would be schedule times, such as 8MWF, 9MWF, 11TTh, etc. Example 5.8.3 If the vertices of a graph represent radio stations, and two vertices are adjacent if the stations are close enough to interfere with each other, a coloring can be used to assign non-interfering frequencies to the stations. Example 5.8.4 If the vertices of a graph represent traffic signals at an intersection, and two vertices are adjacent if the corresponding signals cannot be green at the same time, a coloring can be used to designate sets of signals than can be green at the same time. Graph coloring is closely related to the concept of an independent set. Definition 5.8.5 A set $S$ of vertices in a graph is independent if no two vertices of $S$ are adjacent. If a graph is properly colored, the vertices that are assigned a particular color form an independent set. Given a graph $G$ it is easy to find a proper coloring: give every vertex a different color. Clearly the interesting quantity is the minimum number of colors required for a coloring. It is also easy to find independent sets: just pick vertices that are mutually non-adjacent. A single vertex set, for example, is independent, and usually finding larger independent sets is easy. The interesting quantity is the maximum size of an independent set. Definition 5.8.6 The chromatic number of agraph $G$ is the minimum number of colors required in a proper coloring;it is denoted $\chi(G)$. The independence number of $G$ isthe maximum size of an independent set; it is denoted $\alpha(G)$. The natural first question about these graphical parametersis: how small or large can they be in a graph $G$ with $n$vertices. It is easy to see that$$\eqalign{1&\le \chi(G)\le n\cr1&\le \alpha(G)\le n\cr}$$and that the limits are all attainable: A graph with no edges haschromatic number 1 and independence number $n$, while a complete graphhas chromatic number $n$ and independence number 1. These inequalitiesare thus not very interesting. We will see some that are moreinteresting. Another natural question: What is the relation between the chromatic number of a graph $G$ and chromatic number of a subgraph of $G$? This too is simple, but quite useful at times. Theorem 5.8.7 If $H$ is a subgraph of $G$, $\chi(H)\le \chi(G)$. Proof. Any coloring of $G$ provides a proper coloring of $H$, simply by assigning the same colors to vertices of $H$ that they have in $G$. This means that $H$ can be colored with $\chi(G)$ colors, perhaps even fewer, which is exactly what we want. Often this fact is interesting "in reverse''. For example, if $G$ hasa subgraph $H$ that is a complete graph $K_m$, then $\chi(H)=m$ and so$\chi(G)\ge m$. A subgraph of $G$ that is a complete graph is called a clique, and there is an associated graphicalparameter. Definition 5.8.8 The clique number of a graph $G$ isthe largest $m$ such that $K_m$ is a subgraph of $G$. It is tempting to speculate that the only way a graph $G$could require $m$ colors is by having such a subgraph. This is false;graphs can have high chromatic number while having low clique number;see figure 5.8.1. It is easy to see that thisgraph has $\chi\ge 3$, because there are many 3-cliques in thegraph. In general it can be difficult to show that a graph cannot becolored with a given number of colors, but in this case it is easy tosee that the graph cannot in fact be colored with three colors,because so much is "forced''. Suppose the graph can be colored with 3colors. Starting at the leftif vertex $v_1$gets color 1, then $v_2$ and $v_3$ must be colored 2 and 3, and vertex$v_4$ must be color 1. Continuing, $v_{10}$ must be color 1, but this isnot allowed, so $\chi>3$. On the other hand, since $v_{10}$ can becolored 4, we see $\chi=4$. Paul Erdős showed in 1959 that there are graphs with arbitrarilylarge chromatic number and arbitrarily large girth (the girth is the size of the smallest cycle ina graph). This is much stronger than the existence of graphs with highchromatic number and low clique number. Bipartite graphs with at least one edge have chromatic number 2, since the two parts are each independent sets and can be colored with a single color. Conversely, if a graph can be 2-colored, it is bipartite, since all edges connect vertices of different colors. This means it is easy to identify bipartite graphs: Color any vertex with color 1; color its neighbors color 2; continuing in this way will or will not successfully color the whole graph with 2 colors. If it fails, the graph cannot be 2-colored, since all choices for vertex colors are forced. If a graph is properly colored, then each color class (a color class is the set of all vertices of asingle color) is an independent set. Theorem 5.8.9 In any graph $G$ on $n$ vertices, $\ds {n\over \alpha}\le\chi$. Proof. Suppose $G$ is colored with $\chi$ colors. Since each color class is independent, the size of any color class is at most $\alpha$. Let the color classes be $V_1,V_2,\ldots,V_\chi$. Then $$ n=\sum_{i=1}^\chi \|V_i\| \le \chi\alpha, $$ as desired. We can improve the upper bound on $\chi(G)$ as well. In any graph $G$, $\Delta(G)$ is the maximum degree of any vertex. Proof. We show that we can always color $G$ with $\Delta+1$ colors by a simple greedy algorithm: Pick a vertex $v_n$, and list thevertices of $G$ as $v_1,v_2,\ldots,v_n$ so that if $i< j$, $\d(v_i,v_n) \ge\d(v_j,v_n)$, that is, we list the vertices farthestfrom $v_n$ first. We use integers $1,2,\ldots, \Delta+1$ as colors.Color $v_1$ with 1. Then for each $v_i$ inorder, color $v_i$ with the smallest integer that does not violate theproper coloring requirement, that is, which is different than thecolors already assigned to the neighbors of $v_i$. For $i< n$, we claimthat $v_i$ is colored with one of $1,2,\ldots,\Delta$. This is certainly true for $v_1$. For $1< i< n$, $v_i$ has at least one neighbor that is not yet colored, namely, a vertex closer to $v_n$ on a shortest path from $v_n$ to $v_i$. Thus, the neighbors of $v_i$ use at most $\Delta-1$ colors from the colors $1,2,\ldots,\Delta$, leaving at least one color from this list available for $v_i$. Once $v_1,\ldots,v_{n-1}$ have been colored, all neighbors of $v_n$ have been colored using the colors $1,2,\ldots,\Delta$, so color $\Delta+1$ may be used to color $v_n$. Note that if $\d(v_n)< \Delta$, even $v_n$ may be colored with one of the colors $1,2,\ldots,\Delta$. Since the choice of $v_n$ was arbitrary, we may choose $v_n$ so that $\d(v_n)< \Delta$, unless all vertices have degree $\Delta$, that is, if $G$ is regular. Thus, we have proved somewhat more than stated, namely, that any graph $G$ that is not regular has $\chi\le\Delta$. (If instead of choosing the particular order of $v_1,\ldots,v_n$ that we used we were to list them in arbitrary order, even vertices other than $v_n$ might require use of color $\Delta+1$. This gives a slightly simpler proof of the stated theorem.) We state this as a corollary. There are graphs for which $\chi=\Delta+1$: any cycle of odd length has $\Delta=2$ and $\chi=3$, and $K_n$ has $\Delta=n-1$ and $\chi=n$. Of course, these are regular graphs. It turns out that these are the only examples, that is, if $G$ is not an odd cycle or a complete graph, then $\chi(G)\le\Delta(G)$. The greedy algorithm will not always color a graph with the smallest possible number of colors. Figure 5.8.2 shows a graph with chromatic number 3, but the greedy algorithm uses 4 colors if the vertices are ordered as shown. In general, it is difficult to compute $\chi(G)$, that is, it takes a large amount of computation, but there is a simple algorithm for graph coloring that is not fast. Suppose that $v$ and $w$ are non-adjacent vertices in $G$. Denote by $G+\{v,w\}=G+e$ the graph formed by adding edge $e=\{v,w\}$ to $G$. Denote by $G/e$ the graph in which $v$ and $w$ are "identified'', that is, $v$ and $w$ are replaced by a single vertex $x$ adjacent to all neighbors of $v$ and $w$. (But note that we do not introduce multiple edges: if $u$ is adjacent to both $v$ and $w$ in $G$, there will be a single edge from $x$ to $u$ in $G/e$.) Consider a proper coloring of $G$ in which $v$ and $w$ are different colors; then this is a proper coloring of $G+e$ as well. Also, any proper coloring of $G+e$ is a proper coloring of $G$ in which $v$ and $w$ have different colors. So a coloring of $G+e$ with the smallest possible number of colors is a best coloring of $G$ in which $v$ and $w$ have different colors, that is, $\chi(G+e)$ is the smallest number of colors needed to color $G$ so that $v$ and $w$ have different colors. If $G$ is properly colored and $v$ and $w$ have the same color, then this gives a proper coloring of $G/e$, by coloring $x$ in $G/e$ with the same color used for $v$ and $w$ in $G$. Also, if $G/e$ is properly colored, this gives a proper coloring of $G$ in which $v$ and $w$ have the same color, namely, the color of $x$ in $G/e$. Thus, $\chi(G/e)$ is the smallest number of colors needed to properly color $G$ so that $v$ and $w$ are the same color. The upshot of these observations is that $\ds\chi(G)=\min(\chi(G+e),\chi(G/e))$. This algorithm can be applied recursively, that is, if $G_1=G+e$ and $G_2=G/e$ then $\ds\chi(G_1)=\min(\chi(G_1+e),\chi(G_1/e))$ and $\ds\chi(G_2)=\min(\chi(G_2+e),\chi(G_2/e))$, where of course the edge $e$ is different in each graph. Continuing in this way, we can eventually compute $\chi(G)$, provided that eventually we end up with graphs that are "simple'' to color. Roughly speaking, because $G/e$ has fewer vertices, and $G+e$ has more edges, we must eventually end up with a complete graph along all branches of the computation. Whenever we encounter a complete graph $K_m$ it has chromatic number $m$, so no further computation is required along the corresponding branch. Let's make this more precise. Theorem 5.8.13 The algorithm above correctly computes the chromatic number in a finite amount of time. Proof. Suppose that a graph $G$ has $n$ vertices and $m$ edges. The number of pairs of non-adjacent vertices is $\na(G)={n\choose 2}-m$. The proof is by induction on $\na$. If $\na(G)=0$ then $G$ is a complete graph and the algorithm terminates immediately. Now we note that $\na(G+e)< \na(G)$ and $\na(G/e)< \na(G)$: $$ \na(G+e)={n\choose 2}-(m+1)=\na(G)-1 $$ and $$ \na(G/e)={n-1\choose 2}-(m-c), $$ where $c$ is the number of neighbors that $v$ and $w$ have in common. Then $$\eqalign{ \na(G/e)&={n-1\choose 2}-m+c\cr &\le {n-1\choose 2}-m+n-2\cr &={(n-1)(n-2)\over 2}-m+n-2\cr &={n(n-1)\over 2}-{2(n-1)\over 2}-m+n-2\cr &={n\choose 2} -m -1\cr &=\na(G)-1.\cr }$$ Now if $\na(G)>0$, $G$ is not a complete graph, so there are non-adjacent vertices $v$ and $w$. By the induction hypothesis the algorithm computes $\chi(G+e)$ and $\chi(G/e)$ correctly, and finally it computes $\chi(G)$ from these in one additional step. While this algorithm is very inefficient, it is sufficiently fast to be used on small graphs with the aid of a computer. Example 5.8.14 We illustrate with a very simple graph: The chromatic number of the graph at the top is $\min(3,4)=3$. (Of course, this is quite easy to see directly.) Exercises 5.8 Ex 5.8.1Suppose $G$ has $n$ vertices and chromatic number$k$. Prove that $G$ has at least $k\choose2$ edges. Ex 5.8.2Find the chromatic number of the graph below by usingthe algorithm in this section. Draw all of the graphs $G+e$ and $G/e$generated by the alorithm in a "tree structure'' with the completegraphs at the bottom, label each complete graph with its chromaticnumber, then propogate the values up to the original graph. Ex 5.8.3Show that $\chi(G-v)$ is either $\chi(G)$ or$\chi(G)-1$. Ex 5.8.4Prove theorem 5.8.10 without assuming any particularproperties of the order $v_1,\ldots,v_n$. If $G$ is not 2-connected, show that the blocks of $G$ may colored with $\Delta(G)$ colors, and then the colorings may be altered slightly so that they combine to give a proper coloring of $G$. If $G$ is 2-connected, show that there are vertices $u$, $v$, $w$ such that $u$ is adjacent to both $v$ and $w$, $v$ and $w$ are not adjacent, and $G-v-w$ is connected. Given such vertices, color $v$ and $w$ with color 1, then color the remaining vertices by a greedy algorithm similar to that in theorem \xrefnexternal{thm:almost brooks}{cgt.pdf}, with $u$ playing the role of $v_n$. To show the existence of $u$, $v$, $w$ as required, let $x$ be a vertex not adjacent to all other vertices. If $G-x$ is 2-connected, let $v=x$, let $w$ be at distance 2 from $v$ (justify this), and let a path of length 2 be $v,u,w$. Use theorem 5.7.4 to show that $u$, $v$, $w$ have the required properties. If $G-x$ is not 2-connected, let $u=x$ and let $v$ and $w$ be (carefully chosen) vertices in two different endblocks of $G-x$. Show that $u$, $v$, $w$ have the required properties. Brooks proved the theorem in 1941; this simpler proof is due to Lovász, 1975.
Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$. Show that $$ \frac{x^2+y^2+1}{xy}= 3 \;.$$ I have been solving this for a week and I do not know how to prove the statement. I saw this in a book and I am greatly challenged. Can anyone give me a hint on how to attack the problem? thanks Use the very tricky technique called Vieta Jumping. The idea is considering a polynomial $f(x,y)$ that is quadratic in both $x$ and $y$, with integer coefficients and symmetrical (that is $f(x,y)=f(y,x)$. We have that if $f(x,y)$ has some property when $x,y$ are integers and we want to prove something regarding $x$ and $y$. Suppose that some pair $x_1,y_1$ of integers satisfies the property, since $f$ is symmetrical, we can suppose WLOG that $x_1>y_1$ (the case $x_1=y_1$ is usually easy). Recall the vieta formulas: If $z_1$ and $z_2$ are the roots of $x^2+bx+c=0$, then $z_1z_2=c$ and $z_1+z_2=-b$. Those formulas are very useful, particularly the last one, since it is a simple sum. Now since $f(x,y)$ is quadratic in $x$, we apply the vieta formulas in $x_1$ and we find some integer $x_0$ with $x_0<y_1$ that satisfy the same property, Now we do the same with $y_1$ and find another integer $y_0$ with $y_0<x_0$ that also satisfy the property. Continuing this way we get a pair $(a,b)$ of integers that satisfy the property with $a$ and $b$ really small (like $a=1$). It's easy to prove what we want when the integers are small. Now since all these pairs were satisfying the same property, what we proved about $(a,b)$, also applies to the initial $(x_1,y_1)$. Well, that was kinda long. I hope i have explained the main point. Try to use this on the problem and then back and post your results :) Suppose $xy\mid x^2+y^2+1$ and let $t=\displaystyle\frac{x^2+y^2+1}{xy}$ such that $t\in\mathbb{N}$. Construct the set, $$S=\left\{(x,y)\in\mathbb{N}\times\mathbb{N} : \frac{x^2+y^2+1}{xy}=t\in\mathbb{N}\right\}$$ We deduce that $\displaystyle\frac{x^2+y^2+1}{xy} \ge 3$ because $\displaystyle\frac{x^2+y^2+1}{xy}<3$ implies $x^2+y^2+1 \le 2xy \le x^2+y^2$ which is clearly a contradiction. Now fix $t$ and suppose that $t>3$. Since $S\neq \varnothing$, we can choose $(a,b)\in S$ such that $a+b$ is minimal and $t>3$. WLOG assume $a\ge b>0$. Let us consider the quadratic $$p(w)=w^2-tbw+b^2+1=0$$ It follows that $a$ is a solution since $(a,b)\in S$ and hence satisfies the quadratic equation, that is $a$ is a root. By applying Vieta's formulas we obtain the other root $c$. Hence $a+c=tb$ and $ac=b^2+1$. Since $c=tb-a$ we have $c\in\mathbb{Z}$. Now it remains to prove that $(a,c)\in S$. To this end suppose $c<0$. It follows that $$\displaystyle 0<a^2+ac+1-3ab=a^2+ac-\frac{3c}{t}(a+c)<0$$ which is clearly a contradiction. It also immediately follows that $c\neq 0$ as this implies $b^2+1=0$. Therefore $c\in\mathbb{N}$ and $(a,c)\in S$. Now we show that this $c$ contradicts the minimality of $a$, that is $c<a$. Suppose $c>a$ so it follows that $a+1\le c$. But from Vieta's equations we obtain $\displaystyle a+1\le c=\frac{b^2+1}{a}\le a+\frac{1}{a}$ which is impossible since this inequality holds in $\mathbb{N}$ if and only if $a=1$ and hence $a=b=1$ implying $t=3$ which contradicts our assumption that $t>3$. Therefore $c\le a$. But if $c=a$ then this implies that $\displaystyle a^2=b^2+1>\frac{9}{4}b^2$ which again is a contradiction. Hence we conclude that $c<a$ and as a result $c+b$ contradicts the minimality of $a+b$. Hence $t=3$ I found this on Wolfram under the Fibonacci Number page: Catalan's Identity: $F(n)^2 - F(n-r)F(n+r) = (-1)^{(n-r)}F(r)^2$ For $r = 1$, n is even $F(n)^2 - F(n-1)F(n+1) = -1$ Replace F(n) using $F(n) = F(n+1) - F(n-1)$ and you get $(F(n+1) - F(n-1))^2 - F(n-1)F(n+1) = -1$ or $F(n+1)^2 + F(n-1)^2 + 1 = 3F(n-1)F(n+1)$ is of the form: $x^2 + y^2 + 1 = 3xy$ I do not believe it shows that only Fibonacci numbers are solutions. The relationship that I was looking for on my other solution uses Catalan's Identity, with r = 2. My try(wrong, post factum) $$\frac{x^2+y^2+1}{xy}=k$$ Or: $$\frac{x}{y}+\frac{y}{x}+\frac{1}{xy}=k$$ Now, we try to bound that integer: Edit: $$?>\frac{x}{y}+\frac{y}{x}+\frac{1}{xy}\geq\frac{3}{\sqrt[3]{xy}} \ \ \ \ \ (1)$$ We should deduce that $k=3$ $(1)$-GM $x$ divides $x^2 + y^2 + 1$ implies $y^2 = ax - 1$. Then $y$ divides $x(x+a)$ Case 1 - $y$ divides $x$, so $x = by$. $$1/b + b + 1/(by^2) = k$$. $$b=x=y=1$$ $$k=3$$ Or, $b=x=2$, $y=1$, $k=3$ Case 2 - $y$ divides $x+a$, so $y = -a \text{ mod } x$, $y^2 = 1 \text{ mod } x$. $$x^2 + y^2 +1 = 2 \text{ mod } x$$ $x$ is $1$ or $2$. (Otherwise, $x$ does not divide the equation) If $x = 1$, $x^2 + y^2 +1 = 2 \text{ mod } y$, $y$ is $1$ or $2$ If $x = 2$, $y$ is $1$ or $5$. Solutions: $(1,1),(2,1),(5,2)$, $k$ is always $3$ (Tough to type on the phone)
Express $\sin e^i$ in $a + ib$ form. $$\sin e^i = \frac{e^{ie^i}-e^{-ie^i}}{2i}$$ I feel like I can express $ie^i$ in a simpler way, but I'm not sure how. Thanks for your attention. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community Express $\sin e^i$ in $a + ib$ form. $$\sin e^i = \frac{e^{ie^i}-e^{-ie^i}}{2i}$$ I feel like I can express $ie^i$ in a simpler way, but I'm not sure how. Thanks for your attention. $$\displaylines{ e^i=\cos1+i\sin1\cr ie^i=-\sin1+i\cos1\cr e^{ie^i}=e^{-\sin1}(\cos(\cos1)+i\sin(\cos1))\cr e^{-ie^i}=e^{\sin1}(\cos(\cos1)-i\sin(\cos1))\cr e^{ie^i}-e^{-ie^i}=-2\sinh(\sin1)\cos(\cos1)+2i\cosh(\sin1)\sin(\cos1)\cr \sin(e^i)=\cosh(\sin1)\sin(\cos1)+i\sinh(\sin1)\cos(\cos1)\cr }$$ Use the most beautiful equation in mathematics: $e^{i \pi} = -1$ So $e^i = (-1)^{1/\pi} = (i^2)^{1/\pi} = i^{2/\pi}$. Multiply by $i$ to get: $i e^i = i^{(\pi + 2)/\pi} = -0.841471 + 0.540302 i$. as one "simple" solution (among others). "Simple"? You be the judge.
How to describe a $\operatorname{LL(1)}$ parsing algorithm for strings generated by a given grammar? I have to design a parser for a specific grammar. Let $G$ be the grammar described as: $$S \rightarrow aBC $$ $$B \rightarrow bB \mid cC $$ $$C \rightarrow c \mid S $$ My approach 1. Test if the grammar is $\operatorname{LL(1)}$ First I have to determine if the grammar is $\operatorname{LL(1)}$ or not. It seems that G is $\operatorname{LL(1)}$ because: $\operatorname{FIRST}(aBC) \cap \operatorname{FIRST}(bB \mid cC) \cap \operatorname{FIRST}(c \mid S) = \{a\} \cap \{b,c\} \cap \{c,a\} = \emptyset $ 2. Write a string generated by G $s = abccaccc$ 3. Design the algorithm for parsing strings I'm getting stuck here 4. Test the algorithm for a given string Lets test the algorithm for a given string $s = abccaccc$
Ordinal numbers An elegant formulation of the ordinal concept in ZFC was provided by von Neumann: an ordinal is simply a transitive set well-ordered by the set membership relation $\in$. Equivalently, an ordinal is a hereditarily transitive set, meaning that it is transitive, and all of its elements are transitive. The ordinals are ordered by the relation $\alpha\lt\beta$ just in case $\alpha\in\beta$, and one can show that this is a total order, indeed, a well-order. The collection of all ordinals is a transitive proper class. Successor ordinals If $\alpha$ is an ordinal, then so is the set $\alpha\cup\{\alpha\}$, and it is easy to prove that $\alpha\cup\{\alpha\}$ is the successor ordinal to $\alpha$, the smallest ordinal above $\alpha$, and is accordingly denoted $\alpha+1$. Limit ordinals A limit ordinal is a nonzero ordinal with no immediate predecessor. Every ordinal is either $0$, a successor ordinal or a limit ordinal. Transfinite induction Transinite induction is a method of proving that a statement $\varphi(\alpha)$ holds of all ordinals $\alpha$. Since the ordinals are well-ordered by $\in$, it follows that every nonempty set or class $X$ of ordinals contains a smallest ordinal. Consequently, one can prove that a statement $\varphi(\alpha)$ holds for all ordinals $\alpha$ by proving that it admits of no least counterexample; in other words, one need only prove that whenever $\varphi(\beta)$ holds for all $\beta\lt\alpha$, then $\varphi(\alpha)$ holds. It follows that it holds for all ordinals, since there can be no least failure. It is sometimes convenient to break the transfinite inductive argument into cases, by proving that $\varphi(0)$ holds, that $\varphi(\alpha)\to\varphi(\alpha+1)$ and that $[\forall\beta\lt\lambda\ \varphi(\beta)]\to \varphi(\lambda)$, when $\lambda$ is a limit ordinal. Transfinite recursion Transinite recursion is a method of constructing a well-ordered sequence of objects $a_\alpha$, by specifying how $a_\alpha$ is constructed, assuming one has already constructed $a_\beta$ for $\beta\lt\alpha$. This article is a stub. Please help us to improve Cantor's Attic by adding information.
I am learning finite element method(galerkin method) for solving ode/pde. when searching this topic,I often see examples using the hat function UnitTriangle[x] as the basis function of the galerkin approximation. I understand that the Galekin method is a method of expressing the objective function by the sum of a basis function and a coefficient, and solving the algebraic equation that is the result of integrating the residual with each basis function. something like, f is approximate function(solution) u is true(target) function approximated by f R is residual for example,$R[f(x)]=(u(x)-f(x))^2$ then $f=a_1 \phi_1+a_2 \phi_2+a_3 \phi_3+....+a_n \phi_n$ where $\int_{R}\phi_i \phi_j =0$ if $i\neq j$ and for all j=1,2,...,n $\int_{R}R[f(x)]\phi_j=0$ however,this is contrary to my intuition because linear combination of hat function doesn't provide practical approximation for the target function. that can be used for galerkin method the below is example with 10 nodes which can't be used for galerkin method. (Hat function)kernel[j_] := UnitTriangle[x - j](candicate solution)f = Total@Table[c[j]kernel[j], {j, -10, 10, 1}];(L2 norm between target function and candicate solution)L2norm = Total[Power[Table[Sin[j] - (f /. x -> j), {j, -10, 10, 1}], 2 ] ];sol = Last@NMinimize[L2norm, Table[c[j], {j, -10, 10, 1}]];Plot[{Sin[x], f /. sol}, {x, -10, 10}] in the above,inner product of basis function doesn't 0. Table[ (inner product of basis function) NIntegrate[kernel[j]kernel[j + 1], {x, -10, 10} ], {j, -10, 9} ] {0.166667, 0.166667, 0.166667, 0.166667, 0.166667, 0.166667, \ 0.166667, 0.166667, 0.166667, 0.166667, 0.166667, 0.166667, 0.166667, \ 0.166667, 0.166667, 0.166667, 0.166667, 0.166667, 0.166667, 0.166667} next is example which can be used for galerkin method where inner product of the basis functions are always 0. (Hat function)kernel[j_] := UnitTriangle[x - j](candicate solution)f = Total@Table[c[j]kernel[j], {j, -10, 10, 2}];(L2 norm between target function and candicate solution)L2norm = Total[Power[Table[Sin[j] - (f /. x -> j), {j, -10, 10, 1}], 2 ] ];sol = Last@NMinimize[L2norm, Table[c[j], {j, -10, 10, 2}]];Plot[{Sin[x], f /. sol}, {x, -10, 10}]Table[ (inner product of basis function) NIntegrate[kernel[j]kernel[j + 2], {x, -10, 10} ], {j, -10, 8} ] {0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \ 0., 0.} it's clear something wrong...
In Caramello, Theories, Sites, Toposes, I read that Every Grothendieck topos $\mathcal E$ admits a unique (up to isomorphism) geoemtric morphism $\gamma_{\mathcal E}:\mathcal E\to \bf Set$. The direct image of $\gamma_{\mathcal E}$ is the global section functor $Hom_{\mathcal E}:\mathcal E\to \bf Set$, while the inverse image functor $\gamma^*_{\mathcal E}$ is given by $S\to\coprod_{s\in S}1_\mathcal E$. Now I have two questions: 1) if the topos is not cocomplete, does this definition make sense? I do not think that every sheaf topos over a small site is cocomplete. 2) Why is this unique? My guess is the following: by a well-known fact about topoi, it suffices to prove that the inverse image part is unique. But since it commutes with finite limits, then it should take the terminal object to the terminal object; hence, since it commutes with colimits and $\bf Set$ is generated under colimits by its terminal object, it follows that it must be as above. Am I right? Thank you in advance.
I would like to perform the numerical integration of an integral of the form $$ \int_{-\infty}^\infty e^{i \omega 0+} G(i \omega, \mathbf{v}) d \omega ,$$ or, using the symmetry $G(i\omega)^* = G(-i \omega)$, $$ \int_0^\infty \Re(e^{i \omega 0+} G(i \omega, \mathbf{v}) ) d\omega ,$$ where $\mathbf{v}$ is a vector of additional parameters. $G(i \omega)$ is a rational function and behaves for large $\omega$ as $\omega^5/\omega^6 = 1/\omega$ and it can not be integrated analytically. I split the integral, i.e. $\int_0^\infty = \int_0^a + \int_a^\infty$, where $a$ is a large positive constant, such that the approximation $G(i \omega) \approx 1/i \omega$ holds for $\omega \geq a$. The infinite integral gives an additive constant. The convergence factor in $\int_0^a$ can now be dropped because it is a finite integral. Now there is this problem: I use the quad() function from SciPy to evaluate the finite integral for set of different $\mathbf{v}$. If I set the upper limit in quad() as $a=10^8, 10^9$ or $a = \infty$ then it gives consistent and correct results. But for $a = 10^{10} - 10^{14}$, the results are rubbish. For $a = \infty$ quad() gives a warning message that the integrand may be slowly convergent and that there is a roundoff error detected in the extrapolation table. So obviously there is a numerical problem and I don't know what quad() exactly does, when one enters $\infty$ as a limit. From the source code it seems that quad() maps the big finite interval onto $[0,1]$, rescales the integrand and uses a Gauss-Kronrod method for integration, but it is not clear (to me) how big this scaling respectively the finite interval is. Are there other reliable methods to perform a numerical integration over large intervals? Is a method where one divides the big interval into a number of subintervals with fixed width and subsequent integration (e.g. with Simpson's rule) recommended?
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Published papers including conference proceedings Year Title Reference Link 2019 Constraints on mediator-based dark matter and scalar dark energy models using 13 TeV pp collision data collected by the ATLAS detector JHEP 05 (2019) 142 2018 Search for four-top-quark production in the single-lepton and opposite-sign dilepton final states in pp collisions at 13 TeV with the ATLAS detector PRD 99 052009 (2019) 2018 FATALIC: A novel CMOS front-end readout ASIC for the ATLAS Tile Calorimeter JINST 13 P12013 2018 Search for new phenomena in events with same-charge leptons and b-jets in pp collisions at 13 TeV with the ATLAS detector JHEP 12 (2018) 039 2018 FATALIC: a fully integrated electronics for the ATLAS tile calorimeter at the HL-LHC NIMA 936 (2019) 316 2016 Searches for new physics with lepton flavors and multi-lepton final states in ATLAS EPJ Web of Conf 126, 04028 (2016) 2016 Performance of a remote High Voltage power supply for the Phase II upgrade of the ATLAS Tile Calorimeter JINST 11 C02050 2015 Evidence for the Higgs-boson Yukawa coupling to $\tau$ leptons with the ATLAS detector JHEP 04 117 2013 Search for Fermionic Higgs Boson Decays in $pp$ Collisions at the ATLAS and CMS Experiments Int. J. Mod. Phys. Conf. Ser. 31, 1460287 2012 Search for the standard model Higgs boson in tau lepton final states at the DØ experiment Phys. Lett. B 714, 237 2011 Identification of $\tau$ leptons at the DØ experiment Nucl. Phys. Proc. Suppl. 218, 291-295 Public notes Year Title Reference Link 2018 Technical Design Report for the Phase-II Upgrade of the ATLAS Tile Calorimeter CERN-LHCC-2017-019 2016 Search for new phenomena using events with b-jets and a pair of same-charge leptons in 3.2 fb −1 of pp collisions at 13 TeV with the ATLAS detector ATLAS-CONF-2016-032 2014 Performance of the Fast ATLAS Tracking Simulation and Fast Calorimeter Simulation with single particles ATL-SOFT-PUB-2014-001 2013 Evidence for Higgs Boson Decays to the $\tau\tau$ Final State with the ATLAS Detector ATLAS-CONF-2013-108 2012 Search for Higgs boson in $H\to\tau\tau$ decays in proton-proton collisions with the ATLAS detector ATLAS-CONF-2012-160 2011 Search for the Standard Model Higgs boson in final state with 7.3 fb −1 of collisions at 1.96 TeV Conf DØ note 6179 Internal notes Year Title Reference Link 2018 Combined search for four-top quarks production with in 36.1 -1 of pp collisions ATL-COM-PHYS-2018-269 2018 Constraint on a top-coupled dark matter model combining searches in t+EmissT and tt final state ATL-COM-PHYS-2017-1710 2018 Simulation of FATALIC Frond-end Electronics in the ATLAS Tile Calorimeter Software TL-TILECAL-INT-2018-004 2017 Performance of FATALIC v5 using including test beam data ATL-TILECAL-INT-2018-003 2017 Simulation of FATALIC for the full signal reconstruction ATL-TILECAL-INT-2018-002 2017 Signal Reconstruction with FATALIC ATL-TILECAL-INT-2017-002 2016 Search for new phenomena using events with b-jets and a pair of same charge leptons in 36.1 fb -1 of pp collisions ATL-COM-PHYS-2016-1474 2015 Search for new phenomena using events with b-jets and a pair of same-charge leptons in 3.2 fb −1 of pp collisions ATL-COM-PHYS-2015-389 2015 Performances of a Remote High Voltage Power Supply for the Phase II Upgrade of the ATLAS Tile Calorimeter ATL-TILECAL-INT-2015-005 2013 Search for the Higgs boson $H\to\tau\tau$ decays with the ATLAS detector Detector in 8 TeV Proton-Proton Collisions ATL-COM-PHYS-2013-722 2013 Search for Standard Model $H\to\tau_{\text{lep}}\tau_{\text{had}}$ with the ATLAS Detector in 8 TeV ProtonProton Collisions (20.3 fb −1) ATL-COM-PHYS-2013-1494 2012 Search for Standard Model $H\to\tau_{\text{lep}}\tau_{\text{had}}$ with the ATLAS Detector in 8 TeV ProtonProton Collisions (13.0 fb −1) ATL-COM-PHYS-2012-1201 2012 Re-optimized Search for Standard Model $H\to\tau_{\text{lep}}\tau_{\text{had}}$ with ATLAS in 7 TeV Proton-Proton Collisions ATL-COM-PHYS-2012-1087 2011 Search for the Standard Model Higgs boson in final state with 7.3 fb −1 of collisions at 1.96 TeV Int DØ note 6135 2011 Model of W+jets background in $\mu+\tau$ final state and measurement in data Int DØ note 6134 2010 Tau lepton identification using MVA optimizations, preshower detector and bID tools Int DØ note 6061 Other Documents Year Title Reference Link 2015 Dark Matter Benchmark Models for Early LHC Run-2 Searches: Report of the ATLAS/CMS Dark Matter Forum. arXiv:1507.00966 2011 Search for the Higgs boson in the $\mu+\tau_{\text{had}}$ final state and identification of $\tau$ leptons at DØ. DØ Thesis
This page looks at the relationship between orders of reaction and mechanisms in some simple cases. It explores what a mechanism is, and the idea of a rate determining step. It also explains the difference between the sometimes confusing terms "order of reaction" and "molecularity of reaction". Reaction mechanisms In any chemical change, some bonds are broken and new ones are made. Quite often, these changes are too complicated to happen in one simple stage. Instead, the reaction may involve a series of small changes one after the other. A reaction mechanism describes the one or more steps involved in the reaction in a way which makes it clear exactly how the various bonds are broken and made. The following example comes from organic chemistry. Example 1: Organic Substitution Reaction This is a reaction between 2-bromo-2-methylpropane and the hydroxide ions from sodium hydroxide solution: \[(CH_3)_3CBr + OH^- \rightarrow (CH_3)_3COH +Br^-\] The overall reaction replaces the bromine atom in the organic compound by an $OH$ group. The first thing that happens is that the carbon-bromine bond in a small proportion of the organic compound breaks to give ions: Carbon-bromine bonds are reasonably strong, so this is a slow change. If the ions hit each other again, the covalent bond will reform. The curly arrow in the equation shows the movement of a pair of electrons. If there is a high concentration of hydroxide ions present, the positive ion stands a high chance of hitting one of those. This step of the overall reaction will be very fast. A new covalent bond is made between the carbon and the oxygen, using one of the lone pairs on the oxygen atom. Because carbon-oxygen bonds are strong, once the OH group has attached to the carbon atom, it tends to stay attached. The mechanism shows that the reaction takes place in two steps and describes exactly how those steps happen in terms of bonds being broken or made. It also shows that the steps have different rates of reaction - one slow and one fast. The rate determining step The overall rate of a reaction (the one which you would measure if you did some experiments) is controlled by the rate of the slowest step. In the example above, the hydroxide ion cannot combine with the positive ion until that positive ion has been formed. The second step is in a sense waiting around for the first slow step to happen. The slow step of a reaction is known as the rate determining step. As long as there is a lot of difference between the rates of the various steps, when you measure the rate of a reaction, you are actually measuring the rate of the rate determining step. Reaction mechanisms and orders of reaction The examples we use at this level are the very simple ones where the orders of reaction with respect to the various substances taking part are 0, 1 or 2. These tend to have the slow step of the reaction happening before any fast step(s). Example Here is the mechanism we have already looked at. How do we know that it works like this? By doing rate of reaction experiments, you find this rate equation: \[ \text{rate} = k[(CH_3)_3C-Br]\] The reaction is first order with respect to the organic compound, and zero order with respect to the hydroxide ions. The concentration of the hydroxide ions is not affecting the overall rate of the reaction. If the hydroxide ions were taking part in the slow step of the reaction, increasing their concentration would speed the reaction up. Since their concentration does not seem to matter, they must be taking part in a later fast step. Increasing the concentration of the hydroxide ions will speed up the fast step, but that won't have a noticeable effect on the overall rate of the reaction. That is governed by the speed of the slow step. In a simple case like this, where the slow step of the reaction is the first step, the rate equation tells you what is taking part in that slow step. In this case, the reaction is first order with respect to the organic molecule - and that's all. This gives you a starting point for working out a possible mechanism. Having come up with a mechanism, you would need to find more evidence to confirm it. For example, in this case you might try to detect the presence of the positive ion that is formed in the first step. Example 2 At first sight this reaction seems identical with the last one. A bromine atom is being replaced by an OH group in an organic compound. However, the rate equation for this apparently similar reaction turns out to be quite different. That means that the mechanism must be different. \[ \text{rate} = k[(CH_3CH_2-Br][OH^-\] The reaction this time is first order with respect to both the organic compound and the hydroxide ions. Both of these must be taking part in the slow step of the reaction. The reaction must happen by a straightforward collision between them. The carbon atom which is hit by the hydroxide ion has a slight positive charge on it and the bromine a slight negative one because of the difference in their electronegativities. As the hydroxide ion approaches, the bromine is pushed off in one smooth action. Order of reaction The important thing to realize is that this is something which can only be found by doing experiments. It gives you information about which concentrations affect the rate of the reaction. You cannot look at an equation for a reaction and deduce what the order of the reaction is going to be - you have to do some practical work! Having found the order of the reaction experimentally, you may be able to make suggestions about the mechanism for the reaction - at least in simple cases. Molecularity of a reaction This starts at the other end! If you know the mechanism for a reaction, you can write down equations for a series of steps which make it up. Each of those steps has a molecularity. The molecularity of a step simply counts the number of species (molecules, ions, atoms or free radicals) taking part in that step. For example, going back to the mechanisms we've been looking at: This step involves a single molecule breaking into ions. Because only one species is involved in the reaction, it has a molecularity of 1. It could be described as unimolecular. The second step of this mechanism, involves two ions reacting together. This step has a molecularity of 2 - a bimolecular reaction. The other reaction we looked at happened in a single step: Because of the two species involved (one molecule and one ion), this reaction is also bimolecular. Unless an overall reaction happens in one step (like this last one), you cannot assign it a molecularity. You have to know the mechanism, and then each individual step has its own molecularity. There's nothing the least bit complicated about the term molecularity. The only confusion is that you may sometimes find it used as if it meant the same as order. It doesn't! More about reaction mechanisms and orders of reaction Relating orders of reaction to mechanisms is relatively easy where the slow step is the first step of the reaction mechanism. It isn't so easy when it is one of the later steps. I want to look at this in a bit more detail in case your syllabus requires it. We will revisit the simple case first where the slow step is the first step of the mechanism. Cases where the slow step is the first step in the mechanism Suppose you had a reaction between A and B, and it turned out (from doing some experiments) to be first order with respect to both A and B. So the rate equation is: Rate = k[A][B] Which of these two mechanisms is consistent with this experimental finding? Mechanism 1 Mechanism 2 Remember that in simple cases, where the slow step is the first step of the mechanism, the orders tell you what is taking part in the slow step. In this case, the reaction is first order with respect to both A and B, so one molecule of each must be taking part in the slow step. That means that mechanism 2 is possible. However, mechanism 1 must be wrong. One molecule of A is taking part in the slow step, but no B. The rate equation for that would be: \[ \text{Rate} = k[A]\] Cases where the slow step is not the first step in the mechanism This is much more difficult to do and explain. I'm going to start with as simple example as possible Example Suppose the mechanism for a reaction between A and B looks like this: This time the slow step is the second step. Notice that the first (fast) step is reversible. I need to assume that the fast step is much faster than the slow step - for reasons that I'm not going to explain. The rate of the reaction will be governed by the slow step, and so the rate equation might look like this: \[ \text{Rate} = k[A][X]\] . . except, of course, that X isn't one of the things you are starting with! At an introductory level, the flawed discussion often goes something like this: \[ \text{Rate} = k[A][A][B] = k[A]^2[B] \] As it happens, that gives the right answer in this case, but this simplistic view doesn't work in all cases - as I will show below. So let's start again and do it better! The rate of the reaction will be governed by the slow step, and so the rate equation might look like this: \[ \text{Rate} = k[A][X] \] . . . except, of course, that $X$ is not one of the things you are starting with! You need to be able to express the concentration of $X$ in terms of $[A]$ and $[B$], and you can do that because the first step is an equilibrium. The equilibrium constant for the first reaction is: \[K_c = \dfrac{[X]}{[A][B]}\] You can rearrange this to give an expression for $[X]$: \[ [X]=K_c[A][B]\] . . . and then substitute this value into the rate expression we started with: \[ \text{rate}= k[A] K_c[A][B]\] If you sort this out, and combine the two different constants into a new rate constant ($k_1=kK_c$), you get the rate expression: \[ \text{rate}= k_1[A]^2[B]\] So the reaction is second order with respect to $A$ and first order with respect to $B$. Example 2 I am going to take another similar-looking example now, chosen to be deliberately awkward. This is to try to show that you can't reliably work these problems out just by looking at them. This relates to the following reaction: \[ A + B + C \longrightarrow Y + Z\] The mechanism for this is: From the slow step of the reaction, the rate equation would like like this: \[ \text{Rate} = k[C][X]\] Up to now, this looks just the same as the previous example - but it is not! Let's do it properly, and think about the equilibrium expression for the first step in order to find a value for [X] that we can substitute into the rate equation. \[K_c = \dfrac{[X][Y]}{[A][B]}\] The problem is that we have now got an extra variable in this equation that was not there before - we have got the concentration of $Y$. What do we do about that? Well, for every mole of $X$ formed, a mole of $Y$ will be formed as well. Provided that there was no $X$ or $Y$ in the mixture to start with, that means that the concentration of $Y$ is equal to the concentration of X, and so we can substitute a value of [X] in place of the [Y], giving: \[K_c = \dfrac{[X][Y]}{[A][B]}\] \[ = \dfrac{[X]^2}{[A][B]}\] Rearranging that to get an expression for [X] gives: \[ [X] = \sqrt{ K_c[A][B]}\] \[ = K_c^{\frac{1}{2}} [A]^{\frac{1}{2}} [B]^{\frac{1}{2}}\] And now you can substitute this into the rate equation: \[ \text{rate}=k[C] K_c^{\frac{1}{2}} [A]^{\frac{1}{2}} [B]^{\frac{1}{2}}\] . . . and then tidy it up by combining the two constants into a single new rate constant ($k_1=k K_c^{\frac{1}{2}}$): \[ \text{rate}=k_1[A]^{\frac{1}{2}} [B]^{\frac{1}{2}} [C]\] The order is 0.5 with respect to both $A$ and $B$, and 1 with respect to $C$. There is no way you could have come up with that answer if you had just looked at the equations and assumed that the concentration of $X$ is proportional to the concentrations of $A$ and $B$. The presence of the extra $Y$ makes a serious difference to the result - but just looking at the equations, that is not at all obvious. Contributors Jim Clark (Chemguide.co.uk)
Answer The distance to the boat is approximately 845 feet. Work Step by Step We can express the angle in degrees: $\theta = 2^{\circ}10' = (2 + \frac{10}{60})^{\circ} = 2.17^{\circ}$ We can convert the angle to radians: $\theta = (2.17^{\circ})(\frac{\pi~rad}{180^{\circ}}) = 0.03787~rad$ Let $h$ be the height of the mast. We can approximate the distance $d$ to the boat: $h \approx \theta ~d$ $d \approx \frac{h}{\theta}$ $d \approx \frac{32~ft}{0.03787~rad}$ $d \approx 845~ft$ The distance to the boat is approximately 845 feet.
25 2 Homework Statement Converting between wavenumber and wavelength Homework Equations None By dimensional analysis, we have that the wavenumber: $$k = \frac{\text{radians}}{\text{distance}}$$ And the wavelength: $$\lambda = \frac{\text{distance}}{1 \text{wave}}$$ Then: $$\lambda k = \frac{\text{radians}}{\text{distance}}\frac{\text{distance}}{1 \text{wave}} = \frac{\text{radians}}{1 \text{wave}}$$ Now: $$2\pi \text{radians} = 1 \text{wave} $$ $$\frac{\text{radians}}{1 \text{wave}} = \frac{1}{2\pi} $$ So $$\lambda k = \frac{1}{2\pi} $$ But this contradicts the "well known" $$\lambda k = 2\pi $$ So where did I go wrong? And the wavelength: $$\lambda = \frac{\text{distance}}{1 \text{wave}}$$ Then: $$\lambda k = \frac{\text{radians}}{\text{distance}}\frac{\text{distance}}{1 \text{wave}} = \frac{\text{radians}}{1 \text{wave}}$$ Now: $$2\pi \text{radians} = 1 \text{wave} $$ $$\frac{\text{radians}}{1 \text{wave}} = \frac{1}{2\pi} $$ So $$\lambda k = \frac{1}{2\pi} $$ But this contradicts the "well known" $$\lambda k = 2\pi $$ So where did I go wrong?
Suppose that $y$ is a function of $x$, say $y = f(x)$. It is often necessary to know how sensitive the value of $y$ is to small changes in $x$. Example 2.1.1 Take, for example, $\ds y=f(x)=\sqrt{625-x^2}$ (the upper semicircle of radius 25 centered at the origin). When $x=7$, we find that $\ds y=\sqrt{625-49}=24$. Suppose we want to know how much $y$ changes when $x$ increases a little, say to 7.1 or 7.01. In the case of a straight line $y=mx+b$, the slope $m=\Delta y/\Delta x$ measures the change in $y$ per unit change in $x$. This can be interpreted as a measure of "sensitivity''; for example, if $y=100x+5$, a small change in $x$ corresponds to a change one hundred times as large in $y$, so $y$ is quite sensitive to changes in $x$. Let us look at the same ratio $\Delta y/\Delta x$ for our function $\ds y=f(x)=\sqrt{625-x^2}$ when $x$ changes from 7 to $7.1$. Here $\Delta x=7.1-7=0.1$ is the change in $x$, and $$\eqalign{ \Delta y =f(x+\Delta x)-f(x)&=f(7.1)-f(7)\cr &=\sqrt{625-7.1^2}-\sqrt{625-7^2}\approx 23.9706-24=-0.0294.\cr }$$ Thus, $\Delta y/\Delta x\approx -0.0294/0.1=-0.294$. This means that $y$ changes by less than one third the change in $x$, so apparently $y$ is not very sensitive to changes in $x$ at $x=7$. We say "apparently'' here because we don't really know what happens between 7 and $7.1$. Perhaps $y$ changes dramatically as $x$ runs through the values from 7 to $7.1$, but at $7.1$ $y$ just happens to be close to its value at $7$. This is not in fact the case for this particular function, but we don't yet know why. One way to interpret the above calculation is by reference to a line.We have computed the slope of the line through $(7,24)$ and$(7.1,23.9706)$, called a chord of the circle.In general, if we draw the chord from the point $(7,24)$ to a nearbypoint on the semicircle $(7+\Delta x,\,f(7+\Delta x))$, the slope of thischord is the so-called difference quotient$$\hbox{slope of chord}={f(7+\Delta x)-f(7)\over \Delta x}={\sqrt{625-(7+\Delta x)^2}-24\over \Delta x}.$$ For example, if $x$ changes only from 7 to 7.01, then thedifference quotient (slope of the chord) is approximately equal to$(23.997081-24)/0.01=-0.2919$. This is slightly less steep than thechord from $(7,24)$ to $(7.1,23.9706)$. As the second value $7+\Delta x$ moves in towards 7, the chord joining$(7,f(7))$ to $(7+\Delta x,f(7+\Delta x))$ shifts slightly. Asindicated in figure 2.1.1, as $\Delta x$ gets smaller andsmaller, the chord joining $(7,24)$ to $(7+\Delta x,f(7+\Delta x))$gets closer and closer to the tangent line to the circle at the point $(7,24)$. (Recall that thetangent line is the line that just grazes the circle at that point,i.e., it doesn't meet the circle at any second point.) Thus, as$\Delta x$ gets smaller and smaller, the slope $\Delta y/\Delta x$ ofthe chord gets closer and closer to the slope of the tangent line.This is actually quite difficult to see when $\Delta x$ is small,because of the scale of the graph. The values of $\Delta x$ used forthe figure are $1$, $5$, $10$ and $15$, not really very small values.The tangent line is the one that is uppermost at the right handendpoint. So far we have found the slopes of two chords that should be close to the slope of the tangent line, but what is the slope of the tangent line exactly? Since the tangent line touches the circle at just one point, we will never be able to calculate its slope directly, using two "known'' points on the line. What we need is a way to capture what happens to the slopes of the chords as they get "closer and closer'' to the tangent line. Instead of looking at more particular values of $\Delta x$, let's see what happens if we do some algebra with the difference quotient using just $\Delta x$. The slope of a chord from $(7,24)$ to a nearby point is given by $$ \eqalign{ {\sqrt{625-(7+\Delta x)^2} - 24\over \Delta x}&= {\sqrt{625-(7+\Delta x)^2} - 24\over \Delta x}{\sqrt{625-(7+\Delta x)^2}+24\over \sqrt{625-(7+\Delta x)^2}+24}\cr &={625-(7+\Delta x)^2-24^2\over \Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\cr &={49-49-14\Delta x-\Delta x^2\over \Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\cr &={\Delta x(-14-\Delta x)\over \Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\cr &= {-14-\Delta x\over\sqrt{625-(7+\Delta x)^2}+24}\cr } $$ Now, can we tell by looking at this last formula what happens when $\Delta x$ gets very close to zero? The numerator clearly gets very close to $-14$ while the denominator gets very close to $\ds \sqrt{625-7^2}+24=48$. Is the fraction therefore very close to $-14/48 = -7/24 \cong -0.29167$? It certainly seems reasonable, and in fact it is true: as $\Delta x$ gets closer and closer to zero, the difference quotient does in fact get closer and closer to $-7/24$, and so the slope of the tangent line is exactly $-7/24$. What about the slope of the tangent line at $x=12$? Well, 12 can't be all that different from 7; we just have to redo the calculation with 12 instead of 7. This won't be hard, but it will be a bit tedious. What if we try to do all the algebra without using a specific value for $x$? Let's copy from above, replacing 7 by $x$. We'll have to do a bit more than that—for example, the "24'' in the calculation came from $\ds \sqrt{625-7^2}$, so we'll need to fix that too. $$ \eqalign{ &{\sqrt{625-(x+\Delta x)^2} - \sqrt{625-x^2}\over \Delta x}=\cr \qquad&={\sqrt{625-(x+\Delta x)^2} - \sqrt{625-x^2}\over \Delta x}{\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}\over \sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}}\cr &={625-(x+\Delta x)^2-625+x^2\over \Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\cr &={625-x^2-2x\Delta x-\Delta x^2-625+x^2\over \Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\cr &={\Delta x(-2x-\Delta x)\over \Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\cr &= {-2x-\Delta x\over\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}}\cr } $$ Now what happens when $\Delta x$ is very close to zero? Again it seems apparent that the quotient will be very close to $${-2x\over \sqrt{625-x^2}+\sqrt{625-x^2}} ={-2x\over 2\sqrt{625-x^2}}={-x\over \sqrt{625-x^2}}. $$ Replacing $x$ by 7 gives $-7/24$, as before, and now we can easily do the computation for 12 or any other value of $x$ between $-25$ and 25. So now we have a single, simple formula, $\ds {-x/ \sqrt{625-x^2}}$, that tells us the slope of the tangent line for any value of $x$. This slope, in turn, tells us how sensitive the value of $y$ is to changes in the value of $x$. What do we call such a formula? That is, a formula with one variable,so that substituting an "input'' value for the variable produces anew "output'' value? This is a function. Starting with one function,$\ds \sqrt{625-x^2}$, we have derived, by means of some slightly nastyalgebra, a new function, $\ds {-x/ \sqrt{625-x^2}}$, that gives usimportant information about the original function. This new functionin fact is called the derivative of theoriginal function. If the original is referred to as $f$ or $y$ thenthe derivative is often written $f'$ or $y'$ and pronounced "fprime'' or "y prime'', so in this case we might write $\ds f'(x)=-x/\sqrt{625-x^2}$. At a particular point, say $x=7$, we say that$f'(7)=-7/24$ or "$f$ prime of 7 is $-7/24$'' or "the derivative of$f$ at 7 is $-7/24$.'' To summarize, we compute the derivative of $f(x)$ by forming the difference quotient $$\eqalignno{ &{f(x+\Delta x)-f(x)\over \Delta x},& (2.1.1)\cr }$$ which is the slope of a line, then we figure out what happens when $\Delta x$ gets very close to 0. We should note that in the particular case of a circle, there's a simple way to find thederivative. Since the tangent to a circle at a point is perpendicular tothe radius drawn to the point of contact, its slope is the negativereciprocal of the slope of the radius. The radius joining $(0,0)$ to$(7,24)$ has slope 24/7. Hence, the tangent line has slope$-7/24$. In general, a radius to the point $\ds (x,\sqrt{625-x^2})$ hasslope $\ds \sqrt{625-x^2}/x$, so the slope of the tangent line is$\ds {-x/ \sqrt{625-x^2}}$, as before. It is NOT always true that atangent line is perpendicular to a line from the origin—don't usethis shortcut in any other circumstance. As above, and as you might expect, for different values of $x$ we generally get different values of the derivative $f'(x)$. Could it be that the derivative always has the same value? This would mean that the slope of $f$, or the slope of its tangent line, is the same everywhere. One curve that always has the same slope is a line; it seems odd to talk about the tangent line to a line, but if it makes sense at all the tangent line must be the line itself. It is not hard to see that the derivative of $f(x)=mx+b$ is $f'(x)=m$; see exercise 6. Exercises 2.1 Ex 2.1.1Draw the graph of the function $\ds y=f(x)=\sqrt{169-x^2}$ between $x=0$and $x=13$. Find the slope $\Delta y/\Delta x$ of the chord between thepoints of the circle lying over (a) $x=12$ and $x=13$, (b) $x=12$ and$x=12.1$, (c) $x=12$ and $x=12.01$, (d) $x=12$ and $x=12.001$. Now usethe geometry of tangent lines on a circle to find (e) the exact value of thederivative $f'(12)$. Your answers to (a)–(d) should be getting closer andcloser to your answer to (e).(answer) Ex 2.1.2Use geometry to find the derivative $f'(x)$ of the function$\ds f(x)=\sqrt{625-x^2}$ in the text for each of the following $x$: (a) 20,(b) 24, (c) $-7$, (d) $-15$. Draw a graph of the upper semicircle, anddraw the tangent line at each of these four points.(answer) Ex 2.1.3Draw the graph of the function $y=f(x)=1/x$ between $x=1/2$ and $x=4$.Find the slope of the chord between (a) $x=3$ and $x=3.1$, (b) $x=3$ and$x=3.01$, (c) $x=3$ and $x=3.001$. Now use algebra to find a simpleformula for the slope of the chord between $(3,f(3))$ and $(3+\Deltax,f(3+\Delta x))$. Determine what happens when $\Delta x$ approaches 0.In your graph of $y=1/x$, draw the straight line through the point$(3,1/3)$ whose slope is this limiting value of the difference quotient as$\Delta x$ approaches 0.(answer) Ex 2.1.4Find an algebraic expression for the difference quotient $\ds \bigl(f(1+\Deltax)-f(1)\bigr)/\Delta x$ when $\ds f(x)=x^2-(1/x)$. Simplify the expression asmuch as possible. Then determine what happens as $\Delta x$ approaches 0.That value is $f'(1)$.(answer) Ex 2.1.5Draw the graph of $\ds y=f(x)=x^3$ between $x=0$ and $x=1.5$. Find the slopeof the chord between (a) $x=1$ and $x=1.1$, (b) $x=1$ and $x=1.001$, (c)$x=1$ and $x=1.00001$. Then use algebra to find a simple formula for theslope of the chord between $1$ and $1+\Delta x$. (Use the expansion$\ds (A+B)^3=A^3+3A^2B+3AB^2+B^3$.) Determine what happens as $\Delta x$approaches 0, and in your graph of $\ds y=x^3$ draw the straight line throughthe point $(1,1)$ whose slope is equal to the value you just found.(answer) Ex 2.1.6Find an algebraic expression for the difference quotient $(f(x+\Deltax)-f(x))/\Delta x$ when $f(x)=mx+b$. Simplify the expression asmuch as possible. Then determine what happens as $\Delta x$ approaches 0.That value is $f'(x)$.(answer) Ex 2.1.7Sketch the unit circle. Discuss the behavior of the slopeof the tangent line at various angles around the circle. Whichtrigonometric function gives the slope of the tangent line at an angle$\theta$? Why? Hint: think in terms of ratios of sides of triangles. Ex 2.1.8Sketch the parabola $\ds y=x^2$. For what values of $x$ onthe parabola is the slope of the tangent line positive? Negative?What do you notice about the graph at the point(s) where the sign ofthe slope changes from positive to negative and vice versa?
How can I write the Taylor series for $e^z\sin(z)$ at $z=0$ without making the procedure too complicated? Isn't there an easier way than to compute it's derivatives and find a pattern? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community Yes, there are simpler methods based on the power series expansion of $\exp$. I indicate two equivalent methods, the first one being more elementary than the second one. Method 1: Use Euler's formula $\sin(z) = \dfrac{e^{iz} - e^{-iz}}{2i}$. Method 2: For all $x \in \Bbb R$, $$ e^{x}\sin(x) = \Im(e^{(1+i)x}) = \sum_{n=0}^\infty\frac{\Im((1+i)^n)}{n!}x^n= \sum_{n=0}^\infty \frac{2^{n/2}\sin(\frac{n\pi}{4})}{n!}x^n. $$ By analytic continuation, the identity still holds for $z \in \Bbb C$. to find the $nth$ derivative of $$e^{az}\sin (bz+c)$$ differentiating this expression once $$D^1=e^{az}(a\sin (bz+c)+b\cos (bz+c))$$ let $a=r\cos \theta$ and $b=r\sin \theta$. you get $$D^1= e^{az}r\sin (bz+c+\theta)$$ if you observe you see $$D^n= e^{az}r^n\sin (bz+c+n\theta)$$ where $r=\sqrt{a^2+b^2}$ and $\theta=\arctan(\frac{b}{a})$ now substitute for $a,b,c$. I think the easiest procedure is the multiplication of power series. Let \begin{align} g(z)= & a_0+b_1z+a_2z^2+a_3z^3+\ldots+a_nz^n+\ldots\\ f(z)= & b_0+b_1z+b_2z^2+b_3z^3+\ldots+b_nz^n+\ldots \end{align} Then \begin{align} g(z)f(z) = & (a_0b_0) \\ + & (a_1b_0+a_0b_1)z \\ + & (a_0b_2+a_1b_1+a_2b_0)z^2 \\ + & (a_0b_3+a_1b_2+a_2b_1+a_3b_0)z^3 \\ + & (a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0)z^4 \\ \vdots & \\ \vdots & \\ + & \sum_{i+j=k}a_ib_jz^k \end{align} implies $$ g(z)f(z)=\sum_{k=1}^\infty\color{red}{\sum_{i+j=k}a_ib_j}z^k. $$ In your case, \begin{align} e^z\sin(z) = & \left(\sum_{\alpha=0}^{\infty}\frac{1}{\alpha!}z^\alpha \right)\left(\sum_{\beta=0}^{\infty}\frac{(-1)^{\beta}}{(2\beta+1)!}z^{2\beta+1} \right) \\ = & \sum_{k=0}^{\infty}\;\; \color{red}{\sum_{\alpha+(2\beta+1)=k} \left[\frac{1}{\alpha!}\cdot\frac{(-1)^{\beta}}{(2\beta+1)!} \right]} z^{k} \end{align}
LaTeX is a great tool for printable professional-looking documents, but can be also used to generate PDF files with excellent navigation tools. This article describes how to create hyperlinks in your document, and how to set up LaTeX documents to be viewed with a PDF-reader. Contents Let's start with a minimal working example, by simply importing the hyperref package all cross-referenced elements become hyperlinked. \documentclass{book} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{hyperref} \begin{document} \frontmatter \tableofcontents ... \end{document} The lines in the table of contents become links to the corresponding pages in the document by simply adding in the preamble of the document the line \usepackage{hyperref} One must be careful when importing hyperref. Usually, it has to be the last package to be imported, but there might be some exceptions to this rule. The default formatting for links can be changed so the information in your documents is more clearly presented. Below you can see an example: \documentclass{book} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{hyperref} \hypersetup{ colorlinks=true, linkcolor=blue, filecolor=magenta, urlcolor=cyan, } \urlstyle{same} \begin{document} \tableofcontents \chapter{First Chapter} This will be an empty chapter and I will put some text here \begin{equation} \label{eq:1} \sum_{i=0}^{\infty} a_i x^i \end{equation} The equation \ref{eq:1} shows a sum that is divergent. This formula will later be used in the page \pageref{second}. For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com} or open the next file \href{run:./file.txt}{File.txt} It's also possible to link directly any word or \hyperlink{thesentence}{any sentence} in your document. \end{document} This is a complete example, it will be fully explained in the rest of the article. Below is a description of the commands related to the colour and styling of the links. \hypersetup{ ... } \colorlinks=true \linkcolor=blue \filecolor=magenta \urlcolor=cyan \urlstyle{same} Links to a web address or email can added to a LaTeX file using the \url command to display the actual link or \href to use a hidden link and show a word/sentence instead. For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com} There are two commands in the example that generate a link in the final document: \href{http://www.sharelatex.com}{Something Linky} \url{http://www.sharelatex.com} The commands \href and \url presented in the previous section can be used to open local files For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com} or open the next file \href{run:./file.txt}{File.txt} The command \href{run:./file.txt}{File.txt} prints the text "File.txt" that links to a local file called "file.txt" located in the current working directory. Notice the text "run:" before the path to the file. The file path follows the conventions of UNIX systems, using . to refer the current directory and .. for the previous directory. The command \url{} can also be used, with the same syntax described for the path, but it's reported to have some problems. It was mentioned before that all cross-referenced elements become links once hyperref is imported, thus we can use \label anywhere in the document and refer later those labels to create links. This is not the only manner to insert hyperlinks manually. It's also possible to link directly any word or \hyperlink{thesentence}{any sentence} in you document. If you read this text, you will get no information. Really? Is there no information? For instance \hypertarget{thesentence}{this sentence}. There are two commands to create user-defined links. \hypertarget{thesentence}{this sentence} \hyperlink{thesentence}{any sentence} Links in a document are created having in mind a document that will be read in PDF format. The PDF file can be further personalized to add additional information and change the way the PDF viewer displays it. Below an example: \hypersetup{ colorlinks=true, linkcolor=blue, filecolor=magenta, urlcolor=cyan, pdftitle={Sharelatex Example}, bookmarks=true, pdfpagemode=FullScreen, } Using the command \hypersetup, described in the section styles and colours, accepts extra parameters to set up the final PDF file. pdftitle={Sharelatex Example} bookmarks=true pdfpagemode=FullScreen See the reference guide for a full list of options that can be passed to \hypersetup. Linking style options Option Default value Description hyperindex true Makes the page numbers of index entries into hyperlinks linktocpage false Makes the page numbers instead of the text to be link in the Table of contents. breaklinks false Allows links to be broken into multiple lines. colorlinks false Colours the text for links and anchors, these colours will appear in the printed version linkcolor red Colour for normal internal links anchorcolor black Colour for anchor (target) text citecolor green Colour for bibliographical citations filecolor cyan Colour for links that open local files urlcolor magenta Colour for linked URLs frenchlinks false Use small caps instead of colours for links PDF-specific options Option Default value Description bookmarks true Acrobat bookmarks are written, similar to the table of contents. bookmarksopen false Bookmarks are shown with all sub-trees expanded. citebordercolor 0 1 0 Colour of the box around citations in RGB format. filebordercolor 0 .5 .5 Colour of the box around links to files in RGB format. linkbordercolor 1 0 0 Colour of the box around normal links in RGB format. menubordercolor 1 0 0 Colour of the box around menu links in RGB format. urlbordercolor 0 1 1 Colour of the box around links to URLs in RGB format. pdfpagemode empty Determines how the file is opened. Possibilities are UseThumbs (Thumbnails), UseOutlines (Bookmarks) and FullScreen. pdftitle Sets the document title. pdfauthor Sets the document Author. pdfstartpage 1 Determines on which page the PDF file is opened. For more information see
The Hypergeometric Distribution is usually explained via an urn analogy and formulated as the ratio of “favorable outcomes” to all possible outcomes: \[ \displaystyle \boxed{P(x=a;N,A,n,a) = \frac{{A \choose a} \cdot {N-A \choose n-a} }{{N \choose n}}} \] where $N$ is the total number of balls in, $A$ the number of “red” balls in the urn, and $n$ the sample size. The question answered by the above expression is the probability of finding $x=a$ red balls in the sample. However, a different – equally worthy – viewpoint is that of a tree with conditional probabilities. Here is an example for $N=10, A=4, n=3$ So there are 3 leafs with exactly one red ball in the sample. Following the tree we multiply the conditional probabilities of the tree egdes to get to the “and” probability of the leafs. It should be clear that all three probabilites are identical – but for the order of multiplication: \[ P (\mbox{one red}) = 3 \cdot P_{leaf} = 3 \cdot \frac{4}{10} \cdot \frac{6}{9} \cdot \frac{5}{8} = 3 \cdot \frac{4 \cdot 6 \cdot 5}{10 \cdot 9 \cdot 8} \] Alternative formula How many leafs contain exactly one red ball? Exactly ${n \choose a} = {3 \choose 1} = 3$. The probability for the precise event “a red balls followed by n-a blue balls”is: \[ \frac{A}{N} \cdot \frac{A-1}{N-1} \cdots \frac{A-a+1}{N-a+1} \cdot \frac{N-A}{N-a} \cdot \frac{N-A-1}{N-a-1} \cdots \frac{N-A-(n-a-1)}{N-a-(n-a-1)} \] The last denominator is simply $N-n+1$, i.e. the full denominator is $_NV_n=N!/(N-n)!$ The left numerator is simply $_AV_n = A!/(A-a)!$ in analogy the right numerator $_{N-A}V_{n-a} = (N-A)!/(N-A-(n-a))!$ So, all in all \[ {n \choose a} \cdot \frac{_AV_n \cdot _{N-A}V_{n-a}}{_NV_n} = {n \choose a} \cdot \frac{A! \cdot (N-A)! \cdot (N-n)!}{(A-a)! \cdot (N-A-(n-a))! \cdot N!} \] \[ = {n \choose a} \cdot \frac{\frac{A!}{(A-a)!} \cdot \frac{(N-A)!}{(N-A-(n-a))!} }{\frac{N!}{(N-n)!}} = \frac{{A \choose a} \cdot {N-A \choose n-a} }{{N \choose n}} \] which leaves us with \[ \displaystyle \boxed{P(x=a;N,A,n,a) = {n \choose a} \cdot \frac{_AV_n \cdot _{N-A}V_{n-a}}{_NV_n} } \]
If $K(s)$ is the Kolmogorov complexity of the string $s \in \{0,1\}^*$, Can we prove (or disprove) the following statement: "Every string $s$ is a prefix of an incompressible string; i.e. for every string $s$ there exists a string $r$ such that $K(sr) \geq \|sr\|$" ? In a very informal (and perhaps not too meaningful) way: we know that $K(r) \leq \|r\| + O(1)$; if we pick a large enough incompressible string $r$, can we "use" the $O(1)$ to "mask" the compressibility of the given string $s$ ? A similar (but different) result is that for any $c$, we can find $s$ and $r$ such that: $K(sr) > K(s) + K(r) + c$
Let's say we are evaluating two algorithms. Algorithm 1 gets p1 = 80% conversion Algorithm 2 gets p2 = 85% conversion The two sample sizes are n1 and n2 respectively. I'm trying to test the following hypothesis $H_0:$ p1 = p2 $H_1:$ p1 $\neq$ p2 I though when the sample is large $p1-p2$ ~ $N(p1-p2,\ \frac{p1(1-p1)}{n1} + \frac{p2(1-p2)}{n2})$ so the test statistic is $z = \frac{p1-p2}{\sqrt{\frac{p1(1-p1)}{n1} + \frac{p2(1-p2)}{n2}}}$ But then I read in multiple places [http://mlwiki.org/index.php/Binomial_Proportion_Tests] that the two-sample proportions actually has a test statistic of $z = \frac{p1-p2}{\sqrt{p(1-p)[\frac{1}{n1} + \frac{1}{n2}]}}$ where $p$ is the pooled proportion $p = \frac{x1+x2}{n1+n2}$ where $x1,\ x2$ are the number of conversions. What is the difference between the two? And when should I use each of them?
"Abstract: We prove, once and for all, that people who don't use superspace are really out of it. This includes QCDers, who always either wave their hands or gamble with lettuce (Monte Zuma calculations). Besides, all nonsupersymmetric theories have divergences which lead to problems with things like renormalons, instantons, anomalons, and other phenomenons. Also, they can't hide from gravity forever." Can a gravitational field possess momentum? A gravitational wave can certainly possess momentum just like a light wave has momentum, but we generally think of a gravitational field as a static object, like an electrostatic field. "You have to understand that compared to other professions such as programming or engineering, ethical standards in academia are in the gutter. I have worked with many different kinds of people in my life, in the U.S. and in Japan. I have only encountered one group more corrupt than academic scientists: the mafia members who ran Las Vegas hotels where I used to install computer equipment. " so Ive got a small bottle that I filled up with salt. I put it on the scale and it's mass is 83g. I've also got a jup of water that has 500g of water. I put the bottle in the jug and it sank to the bottom. I have to figure out how much salt to take out of the bottle such that the weight force of the bottle equals the buoyancy force. For the buoyancy do I: density of water * volume of water displaced * gravity acceleration? so: mass of bottle * gravity = volume of water displaced * density of water * gravity? @EmilioPisanty The measurement operators than I suggested in the comments of the post are fine but I additionally would like to control the width of the Poisson Distribution (much like we can do for the normal distribution using variance). Do you know that this can be achieved while still maintaining the completeness condition $$\int A^{\dagger}_{C}A_CdC = 1$$? As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room.An altern... You're always welcome to ask. One of the reasons I hang around in the chat room is because I'm happy to answer this sort of question. Obviously I'm sometimes busy doing other stuff, but if I have the spare time I'm always happy to answer. Though as it happens I have to go now - lunch time! :-) @JohnRennie It's possible to do it using the energy method. Just we need to carefully write down the potential function which is $U(r)=\frac{1}{2}\frac{mg}{R}r^2$ with zero point at the center of the earth. Anonymous Also I don't particularly like this SHM problem because it causes a lot of misconceptions. The motion is SHM only under particular conditions :P I see with concern the close queue has not shrunk considerably in the last week and is still at 73 items. This may be an effect of increased traffic but not increased reviewing or something else, I'm not sure Not sure about that, but the converse is certainly false :P Derrida has received a lot of criticism from the experts on the fields he tried to comment on I personally do not know much about postmodernist philosophy, so I shall not comment on it myself I do have strong affirmative opinions on textual interpretation, made disjoint from authoritorial intent, however, which is a central part of Deconstruction theory. But I think that dates back to Heidegger. I can see why a man of that generation would be leaned towards that idea. I do too.
Recap of Lecture 13 Last lecture continued the discussion of symmetry (and direct product tables for odd/even functions). We shows the quantum Harmonic Oscillator wavefunctions alternate between even and odd because of the Hermite polynomial component. This maps into the transition moment integral to predict that only transition in the IR between adjacent wavefunctions will be allowed (i.e., no harmonics will be observed). We demonstrated that this is not what is experimentally observed since overtones can be weakly observed, which means the harmonic oscillator is an approximation of a true bond. We discussed the Taylor expansion of an arbitray potential results in the HO as the first meaningful term and higher order terms are call anharmonic terms. We introduced the Morse oscillator which is a functional form that far better approximates the true potential of vibration. The second part of the lecture introduced rotations, first by motivating classical rotation, then 2D quantum (one quantum number) and finally 3D rotation (two quantum numbers). We continue with that discussion here. Rotation Redux: 2D Quantum Rigid Rotor (rotation in a single plane) The angular momentum of a rigid rotor is \[L_z = pr\] The Hamiltonian is for a rigid rotor is then \[ - \dfrac{\hbar^2}{2I} \dfrac{\partial ^2 }{\partial \phi^2} \label{RR Hamiltonian} \] The corresponding wavefunctions are: \[ \color{red} \|\Phi_m \rangle = \sqrt{\dfrac{1}{2\pi}} e ^{im\phi} \label{eigenstates of a 2D RR}\] with the constraint that: \[ m =\pm \dfrac{\sqrt{2IE}}{\hbar} \label{eigenenergies of a 2D RR}\] The angular momentum is quantized \[\color{red} L_z = m\hbar\] and the energy is also quantized: \[ \color{red} E = \dfrac{L_z^2}{2I} = \dfrac{m^2\hbar^2}{2I} \label{Energies are quantized}\] we find that the quantum numbers for 2D rotation are \[ \color{red} m = 0, \pm 1,\pm, \pm 2, ... \label{quantum number for 2D RR}\] Properties and Notes about 2D rigid Rotor Solutions of the 2D Rigid Rotor rotational Hamiltonian are sine and cosine functions just like the particle in a box. Here the boundary condition is imposed by the circle and the fact that the wavefunction must not interfere with itself. This model is similar to condition in the Bohr model of the atom The energy and wavefunctions are dependent on the quantum number $m$ that varies from $0$ to $\infty$. There is zero point energy for rotation; this is similar to translation of a free particle. Zero point energies are encountered when a particle if bound like in a particle in a box or a harmonic oscillator). 3D Quantum Rigid Rotor Read The Energy Levels of a Rigid Rotator for more complete derivation! For the rigid rotator, we are switching from Cartesian coordinates $x,y,z$ to Spherical coordinates $r,\theta, \phi$, and the variables $r$ and $\ell$ or $r_o$ or $r_{eq}$ can be used interchangeably, as they both refer to the bond distance. Spherical coordinates (r, θ, φ) as often used in mathematics: radial distance r, azimuthal angle θ, and polar angle φ. The meanings of θ and φ have been swapped compared to the physics convention. Image used with permission of Wikipedia. All bond length changes are put in the Harmonic Oscillator problem. This allows for the separation of variables between $\theta$ and $\phi$, as $r$ is part of the Harmonic Oscillator. In Cartesian Coordinates, \[\hat{T} = \hat{H}= \dfrac {-\hbar^2}{2\mu}\triangledown^2 \label{4}\] In Spherical coordinates, with the condition that $r$ is a constant, \[\hat{T} = \hat{H} = \dfrac {-\hbar^2}{2I} \left[\dfrac {1}{\sin(\theta)} \dfrac {d}{d \theta} (\sin(\theta) \dfrac{d}{d \theta}) + \dfrac {1}{\sin^2 ({\theta})} \dfrac {d^2}{d \phi^2})\right]\label{5}\] Using the formula \[T = \dfrac {L^2}{2I}\] we can solve for the angular momentum, \[\hat{L^2} = -\hbar^2 \left [\dfrac {1}{\sin(\theta)} \dfrac {d}{d \theta} \left(\sin(\theta) \dfrac{d}{d \theta}\right) + \dfrac {1}{\sin^2 ({\theta})} \dfrac {d^2}{d \phi^2}) \right ]\label{6}\] with $L$ is angular momentum, $I$ is the moment of inertia, $\mu$ is the reduced mass, and $r_0$ is average bond length. Vibration and Rotation? If we added a vibrational term (e.g., the Harmonic Oscillator) into the $\hat{H}$, the we can can describe a molecule the rotates and vibrates. However, we are considering only a rigid rotor (is that a feasible model?). For the Rigid Rotator, $\hat{H} = \hat {T}_{\theta \phi}$ and $Y(\theta, \phi) = \psi_{RR}$, then the Schrodinger equation is: \[T_{\theta \phi} \| Y_{\ell}^{m} (\theta, \phi) \rangle = E_{RR} \| Y_{J}^{m} (\theta, \phi) \rangle \label{7}\] The energy associated with the Rigid Rotor is quantized in $J$ where $J = 0, 1, 2 ... \infty$. Each $J$ defines an angular momentum state of the molecule \[ \color{red} E_{RR} = E_{J} = \dfrac {\hbar^2}{2I} J(J+1) = BJ(J+1) \label{8}\] where \[B = \dfrac{\hbar^2}{2I} = \dfrac {h^2}{8\pi^2 \mu r_0^2}\] or in terms of wavenumbers \[\tilde{B} = \dfrac{B}{hc} = \dfrac {h}{8\pi^2 c \mu r_0^2}\] and \[ \color{red} J \ge \|m_J\| \label {5.8.29}\] $J$ can be 0 or any positive integer greater than or equal to $m_J$. Each pair of values for the quantum numbers, $J$ and $m_J$, identifies a rotational state with a wavefunction and energy. Equation $\ref{8}$ means that $J$ controls the allowed values of $m_J$. 3D rigid rotors have two quantum numbers: $J$ and $m_J$. Each pair of values for the quantum numbers identifies a specific rotational state and hence a specific wavefunction with accompanies energy. Confusing Quantum Numbers The quantum number for angular momentum and its orientation in space changes depending on the system and if a chemist or physists is doing the problem. For example in the 2D Rigid Rotor, we used $m$ that ranges from $m = 0, \pm 1,\pm, \pm 2$. For the 3D rigid rotor, we used $J$ for total angular moment quantum number where $J = 0, 1, 2 ... \infty$ and $m_J$ for its orientation where $J \ge \|m_J\| $. When we discuss atoms, we switch again to $\ell$ and $m_{\ell}$ (remember from general chemistry) just to make things confusing. So, be careful when comparing equations from different systems. Degeneracy of Eigenstates The solutions form a set of $2J + 1$ eigenstates at each energy \[E(J) = \dfrac{\hbar^2J(J+1)}{2I}.\] A set of levels that are equal in energy is called a degenerate set as discussed within the context of 2D and 3D particle in boxes. $J=0$: The lowest energy state has $J = 0$ and $m_J = 0$. This state has an energy $E_0 = 0$. There is only one state with this energy, i.e. one set of quantum numbers, one wavefunction, and one set of properties for the molecule. $J=1$: The next energy level is $J = 1$ with energy \[E(J=1)=\dfrac {2\hbar ^2}{2I}.\] There are three stateswith this energy because $m_J$ can equal +1, 0, or ‑1. These different states correspond to different orientations of the rotating molecule in space. States with the same energy are said to be degenerate. The degeneracy of an energy level is the number of states with that energy. The degeneracy of the $J = 1$ energy level is 3 because there are three states with the energy $\dfrac {2\hbar ^2}{2I}$. $J=2$: The next energy level is for $J = 2$. The energy is \[E(J=2)=\dfrac {6\hbar ^2}{2I},\] and there are five states with this energy corresponding to $m_J = +2, \,+1,\, 0,\, ‑1,\, ‑2$. The energy level degeneracy is five. Note that the spacing between energy levels increases as J increases. Also note that the degeneracy increases. The degeneracy is always $2J+1$ because $m_J$ ranges from $+J$ to $‑J$ in integer steps, including 0. Exercise Calculate the possible angles a $J = 1$ angular momentum vector can have with respect to the z-axis. Exercise What is the rotational energy and angular momentum of a molecule in the state with $J = 0$? Describe the rotation of a molecule in this state. Uncertainty in Rotations (Vectoral Picture) The classical definition of the orbital angular momentum of such a particle about the origin is ${\bf L} = {\bf x}\times{\bf p}$, giving (i.e., via the cross product): \[ \hat{L}_x = y\, p_z - z\, p_y, \label{6.3.1a}\] \[ \hat{L}_y = z\, p_x - x\, p_z \label{6.3.1b}\] \[ \hat{L}_z = x\,p_y - y \,p_x \label{6.3.1c}\] In Cartesian coordinates, the three components of orbital angular momentum can be written \[\hat{L}_x = -{\rm i}\,\hbar\left(y\,\dfrac{\partial}{\partial z} - z\,\dfrac{\partial} {\partial y}\right) \label{6.3.2a}\] \[ \hat{L}_y = -{\rm i}\,\hbar\left(z\,\dfrac{\partial}{\partial x} - x\,\dfrac{\partial} {\partial z}\right) \label{6.3.2b}\] \[ \hat{L}_z = -{\rm i}\,\hbar\left(x\,\dfrac{\partial}{\partial y} - y\,\dfrac{\partial} {\partial x}\right) \label{6.3.2c}\] using the Schrödinger representation. $L_x$, $L_y$, and $L_z$ can, in principle, be measured. However, to determine if they can be measured simultaneously with infinite precision, these operators must commute. Heisenberg's Uncertainty Principle for linear momentum and position Remember that the fundamental commutation relations satisfied by the position and linear momentum operators are: \[[x_i, x_j] =0 \label{6.3.12}\] \[[p_i, p_j] =0 \label{6.3.13}\] \[[x_i, p_j] = i\,\hbar \,\delta_{ij} \label{6.3.14}\] where $i$ and $j$ stand for either $x$, $y$, or $z$. Consider the commutator of the operators $L_x$ and $L_z$ : \[ \begin{align} [L_x, L_y] & = [(y\,p_z-z\,p_y), (z\,p_x-x \,p_z)] \\ &= y\,[p_z, z]\,p_x + x\,p_y\,[z, p_z] \end{align} \label{6.3.15}\] \[ = {\rm i}\,\hbar\,(-y \,p_x+ x\,p_y) = {\rm i}\,\hbar\, L_z \label{6.3.16}\] This is one example of several cyclic permutations of the fundamental commutation relations satisfied by the components of an orbital angular momentum: \[[L_x, L_y] = {\rm i}\,\hbar\, L_z \label{6.3.17a}\] \[[L_y, L_z] = {\rm i}\,\hbar\, L_x \label{6.3.17b}\] \[[L_z, L_x] = {\rm i}\,\hbar\, L_y \label{6.3.17c}\] Therefore, two orthogonal components of angular momentum (for example $L_x$ and $L_y$) are complementary and cannot be simultaneously known or measured, except in special cases such as $\displaystyle L_{x}=L_{y}=L_{z}=0$. We can introduce a new operator $\hat{L}^2$: \[\hat{L}^2 = L_x^{\,2}+L_y^{\,2}+L_z^{\,2} \label{6.3.5}\] That is the magnitude of the Angular momentum squared. It is possible to simultaneously measure or specify $L^2$ and any one component of $L$; for example, $L^2$ and $L_z$. This is often useful, and the values are characterized by ($J$) and ($m_J$). In this case the quantum state of the system is a simultaneous eigenstate of the operators $L^2$ and $L_z$, but not of $L_x$ or $L_y$. Illustration of the vector model of orbital angular momentum. Image used with permission (Public domain; Maschen). Since the angular momenta are quantum operators, they cannot be drawn as vectors like in classical mechanics. Nevertheless, it is common to depict them heuristically in this way. Depicted above is a set of states with quantum numbers ${\displaystyle J =2}$, and ${\displaystyle m_{J}=-2,-1,0,1,2}$ for the five cones from bottom to top. Since ${\displaystyle \|L\|={\sqrt {L^{2}}}=\hbar {\sqrt {6}}}$, the vectors are all shown with length ${\displaystyle \hbar {\sqrt {6}}}$. The rings represent the fact that ${\displaystyle L_{z}}$ is known with certainty, but ${\displaystyle L_{x}}$ and ${\displaystyle L_{y}}$ are unknown; therefore every classical vector with the appropriate length and z-component is drawn, forming a cone. The expected value of the angular momentum for a given ensemble of systems in the quantum state characterized by ${\displaystyle \J}$ and ${\displaystyle m_{J }}$ could be somewhere on this cone while it cannot be defined for a single system (since the components of ${\displaystyle L}$ do not commute with each other).
I have a conic optimization of the form: $$\min_x \langle c, x \rangle,\ \text{s.t.}\ Ax = b,\ x \in K.$$ where $x \in \mathbb{R}^{n}$, $A$ is an $m \times n$ matrix, $b \in \mathbb{R}^m$, $K$ is a self dual cone in $\mathbb{R}^n$ and $\langle~,~\rangle$ is the standard Euclidean inner product on $\mathbb{R}^n$. I am looking for conditions under which the optimal value function is a continuous function of perturbations in the vector $b$. In particular, if we replace $b$ with $b + \Delta b$, so the linear constraint becomes $Ax = b + \Delta b$, and if we let $\phi ( \Delta b)$ denote the optimal value of this perturbed problem, I am interested in when $\| \phi(0) - \phi(\Delta b) \| \rightarrow 0$ as $\\| \Delta b\\|_{\infty} \rightarrow 0$. I have seen results that show $\phi(\Delta b)$ is a linear function in $\Delta b$ for perturbations that preserve the optimal partitions, but I am interested in the case when the optimal partition is not necessarily preserved.
Looking for this on the web I was provided with little information, only that as the pressure increases so does tension, and if there is a higher tension it results in a louder pop, how may that be expressed in the form of an equation? It is true that tension increases when you inflate a balloon, but the pressure has a peak value. Once the pressure reaches the peak value there is a sudden expansion of the balloon with a decrease of the internal pressure (see reference 1), this is due to the nonlinear behavior of the material. To obtain a mathematical expression let us assume the following: The material behaves as a neo-Hookean hyperelastic material and it behaves nearly incompressible. The shape of the balloon is spherical for the whole inflation process. This is not true but it simplifies things, so we can do it by hand. It is inflated slowly, so we can neglect inertial effects. The radius ($r$) and thickness ($t$) of the balloon at any instant are given by $$r = \lambda r_0\, ,\quad t = \frac{t_0}{\lambda^2}\, ,$$ where $r_0$ is the initial radius, and $t_0$ the initial thickness. $\lambda$ is termed the stretch, and the thickness should decrease quadratically because of the incompressibility. Since we have a neo-Hookean material we can say that the tension is given by $$\sigma_{11} = \sigma_{22} \equiv T = 2\alpha\left(\lambda^2 - \frac{1}{\lambda^4}\right)\, ,$$ and $\alpha$ is a material constant measuring how difficult is to inflate the balloon. Equilibrium between the tension (membrane stress) and the pressure gives $$P = \frac{2 T t}{r}\, ,$$ and, putting it all together, $$P = 2\alpha\left(\frac{1}{\lambda} - \frac{1}{\lambda^7}\right) \frac{2t_0}{r_0}\, .$$ And you end up with the behavior shown in the following figure. Now, we need to know the energy stored to estimate how loud it will sound when popped $$E = PV = P \frac{4}{3}\pi r^3 = 4\alpha\left(\lambda^2 - \frac{1}{\lambda^4}\right) t_0 r_0^2\, .$$ This expression can be rewritten in terms of the mass of the balloon and density of the material, as well. References Jame F. Doyle and C.T. Sun (2007). Theory of Elasticity, Purdue University.
Two Capacitors Are Better Than One I was looking for a good reference for some ADC-driving circuits, and ran across this diagram in Walt Jung’s Op-Amp Applications Handbook: And I smiled to myself, because I immediately remembered a circuit I hadn’t used for years. Years! But it’s something you should file away in your bag of tricks. Take a look at the RC-RC circuit formed by R1, R2, C1, and C2. It’s basically a stacked RC low-pass filter. The question is, why are there two capacitors? I discovered this type of approach myself a few months after I graduated from college and was working on some battery testing circuits. We needed a really slow RC filter, with a 10 second time constant. Really simple, right? I picked R = 10MΩ and C = 1μF, and followed the RC circuit with a good CMOS-input opamp follower with guaranteed maximum input bias current in the picoampere range. There’s a problem here, though. Let’s look at the specification for the Murata GRM219R71E105KA88, a 1μF 25V X7R 0805 capacitor. If you look at page two of the spec, under Insulation Resistance it says: C ≤ 0.047μF: More than 10000MΩ C > 0.047μF: More than 500 Ω⋅F For small capacitors, this means the insulation resistance is at least 10GΩ, but for large capacitors, they state it in terms of a resistance - capacitance product, equivalent to a time constant, of 500 seconds. For my 1μF capacitor, it would be 500MΩ minimum insulation resistance. That’s right, there’s parasitic parallel resistance: And what it means is we have a voltage divider. Worst case, we’d have R = 10MΩ and R p = 500MΩ, for a net DC gain of 500 / 510 ≈ 0.98. With a 12V battery, that would show up as V C = 11.76. Yuck! That’s almost a quarter-volt of error. That spec is worst-case; I don’t remember what I ran into when I was working on the battery tester project, it might have been maybe 10 or 20 millivolts sag due to leakage through the capacitor’s insulation resistance. A 5GΩ insulation resistance would have had a gain of 0.998, with a sag of 24mV on a 12V battery. (And with high-impedance nodes, you’ve also got to be careful about keeping your circuit board clean so there’s not a leakage path on the board through moisture or leftover solder flux.) And there’s really nothing you can do about it if you use this circuit topology, unless you’re willing to pay for ultra-low-leakage capacitors. You might think, hey, I can just raise the resistance and lower the capacitance by a factor of 5 or 10, but then the parasitic resistance will scale also, and in the end things won’t really change. But one solution to this is to use a second RC filter: What happens now, is that most of the DC voltage appears across the lower capacitor. There’s still leakage. The remaining DC voltage, which is very small, appears across the upper capacitor. For example, R 1 = R 2 = R = 10MΩ and R p1 = R p2 = R p = 500MΩ. Then the DC voltage gain from V in to V C1 is still about 0.98. The DC voltage gain from V in − V C1 to V C2 is also about 0.98, so we have $$ V_{out} = 0.98V_{in} + 0.98(V_{in} - 0.98V_{in}) \approx 0.9996 V_{in}$$ Now our output DC voltage for a 12V input is 11.995V, worst-case. (Typical insulation resistance will be higher so our gain is most likely going to be much closer to 1.) As far as the AC characteristics, we can basically neglect R p, and if I’ve done my algebra right, the transfer function from V in to V out is $$ \frac{V_{out}}{V_{in}} = \frac{3RCs + 1}{s^2 + 3RCs + 1} = \frac{3RCs + 1}{(\tau_{p1}s + 1)(\tau_{p2}s + 1)}$$ where $ \tau_{p1} = \frac{3-\sqrt{5}}{2}RC \approx 0.38197RC $ and $ \tau_{p2} = \frac{3+\sqrt{5}}{2}RC \approx 2.618RC $. What this means is that the highest frequency pole has a time constant that is only 0.38197 as much as our single-pole RC filter. The other pole and zero come into play at frequency about 7 times higher, and they’re close enough together that they don’t do much. So if we want the same effective time constant of 10 seconds, we need capacitor values that are 2.6x higher. If we choose C = 2.2μF and R = 12MΩ then that should suffice. In this case, the minimum insulation resistance should be 227MΩ and our factor of 0.98 turns into 227/(12+227) = 0.95, and the net DC gain is $$ V_{out} = 0.95V_{in} + 0.95(V_{in} - 0.95V_{in}) \approx 0.9975 V_{in}$$ For a 12V battery this would mean we’re sensing 11.97V at the output. Sigh. A 30mV sag. We could stack a third stage, which would help the DC gain but make the frequency response worse, driving us to higher capacitor values to get an equivalent time constant. I’d probably stop at two stages and just live with the 0.9975V worst-case DC gain, which is at least much better than the 0.98 worst-case DC gain of the one-stage RC filter with leakage. And anyway, that’s worst-case; the insulation resistance of capacitors is usually higher. But you get the idea. Two capacitors are better than one, when it comes to stopping leakage current. The same approach works for high-pass filters. Just switch the R’s and C’s around: This time the capacitors are supposed to get rid of DC voltage. Most of it is gone when we look across resistor R 1, and we add a second stage to get rid of most of the rest. The other point to make is that it’s really hard to use analog circuitry for very long time constants (more than a few seconds). The circuits become very prone to parasitic leakage. Much better to do this kind of long time-scale low-pass filtering with digital signal processing, if you have a microcontroller around anyway. Then all you have to do in analog is enough filtering for anti-aliasing, and it’s a tradeoff between how fast you want to sample and run your computations in the digital domain, and how large the components are in the analog domain. Summary Capacitors have leakage current through their insulation. It’s usually specified in terms of a minimum time constant (Ω⋅F). This forms a DC voltage divider, forming an error voltage on a low-pass or high-pass filter. You can decrease this error voltage by using a second RC stage. It essentially squares the relative error (0.02 → 0.0004). The only downside is that the RC time constant is decreased, and you need to adjust your RC values to compensate. But even after the compensation, the relative error is decreased by a second RC stage. In any case, keep this idea in your bag of tricks. © 2015 Jason M. Sachs, all rights reserved. Previous post by Jason Sachs: My Love-Hate Relationship with Stack Overflow: Arthur S., Arthur T., and the Soup Nazi Next post by Jason Sachs: Ten Little Algorithms, Part 1: Russian Peasant Multiplication I like your articles. Though many of them go over my head. I have a question related to this article if I may. How does temperature affect the performance of a capacitor? Does a capacitor work better/closer to its specification when it gets warm/hot? Does the life of a capacitor reduce if it's constantly running warm/hot? If I remember correctly, capacitors have a higher leakage at higher temperature. Not sure about the capacitance; from looking at the definitions of the standard temperature designations (NP0, X7R, Y5V, etc) https://en.wikipedia.org/wiki/Ceramic_capacitor#Temperature_dependence_of_capacitance it looks like capacitance can either go up or down. Lifetime definitely is reduced at elevated temperatures; Arrhenius's Law comes into play and the lifetime drops by about a factor of two for every 10 degree C rise. To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments. Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
If I pass individual photons through a M-Z interferometer with equal arms I will observe interference (eg only one detector will respond). As I increase the path length of one arm I will observe the two detectors responding alternately as I pass through each phase cycle. Eventually I suspect that at a certain point, the interference will disappear and the two detectors will respond with equal probability. What determines this point and what does this tell us about the 'length' of an individual photon. What does QM predict when the path difference is greater than this? There isn't a straight answer to this question, which sheds light onto the meaning of the subtle word coherence, because what we tend to call "decoherence" can have two main roots. A practical experimental meaning of the word coherence is "ability to show interference", and there are two ways whereby observable interference can disappear: ( 1) (energy) spectral spread within the pure quantum states in question; (genuine quantum decoherence. 2) Before we address these mechanisms, we need to be very clear on dispelling the notion of a "photon" as a little ball with schizophrenia and I'd urge you to read Daniel Sank's Most Wonderful Description Here and my commentary on his answer here: the only actors in the scene we'll discuss are ( 1) THE electromagnetic field and (the measuring instruments in your laboratory. We're talking about the electromagnetic field in a "one photon state" - this simply means the electromagnetic field has been raised one "notch" above its ground state and is undergoing unitary evolution such that statistics of the potential measurement events vary with time. These events are individual detections of an idealized photomultiplier tube, which tells you it has interacted with the EM field, and at the same time the EM field drops back to its ground state. 2) 1. Energy Spectral Spread with Pure Quantum States If this mechanism is what is causing your inability to see fringes, then the answer to your question is very simple. Turn the laser up and remove all attenuators so that you can see interference fringes, where they fade out and other "coherence / incoherence" effects. Now turn your laser down so that there is "one photon in the kit at one time" and take zillions of measurements of single detection events at all places in space. The probability to detect at a given point is exactly proportional to the intensity of the high power field you saw before turning the light level down. Notwithstanding apparent "incoherence" effects cause the "fading" or lack of fringe visibility in the high power picture, you'll still see "lack of fringe visibility" show up extremely precisely in your probability densities even though we are dealing with pure quantum light states. Let's try to understand the underlying physics more deeply. In this scenario we have a fantastically ideal laser that always raises the electromagnetic field to the same pure quantum state. We have an idealized attenuator gating off your experimental kit, so that, at random, unpredictable times (that arise following a Poisson process) the field in the interferometer gets raised to the same, pure number state, but there are (almost) never any field excitations above those immediately above the EM field ground state ( i.e. the rate $\lambda$ events per unit time of the Poisson process, inversely proportional to the setting on your attenuator, is small compared to the rate at which your kit can measure). This situation is often colloquially described as "there's only one photon in the kit at a time". But a one photon number state is not a simple two dimensional state. The subspace of one photon states is itself infinite dimensional: there are two basis one photon number states, one for each polarization, for every wavevector $\vec{k}\in\mathbb{R}^3$. These are the energy / momentum eigenstates of the one-photon subspace. Their amplitudes evolve with time by taking on a phase of the form $\exp\left(i\,\hbar\,\frac{\|\vec{k}\|}{c}\,t\right)$. A one photon pure state can, like any quantum state, be in quantum superposition of the basis states. Now it is not emphasized enough that this includes one photon number states with a spread in energy. It is indeed possible to have "white" (broadband) perfectly coherent light, in the sense of its being in a pure quantum state (although it is very difficult to achieve very broadband pure states in practice). We'd write down our pure one photon state as something like this: $$\int_{\vec{k}} \left(\psi(\vec{k})_+\,\left\|\left.1\right>\right._{\vec{k}\,+} +\psi(\vec{k})_-\,\left\|\left.1\right>\right._{\vec{k}\,-}\right)\,\mathrm{d}^3 k$$ or a field theorist might write it down more like: $$\int_{\vec{k}} \left(\psi_+(\vec{k})\,a^\dagger_+(\vec{k})+\psi_-(\vec{k})\,a^\dagger_-(\vec{k})\right)\,\mathrm{d}^3 k\,\left\|\left.0\right>\right.$$ i.e. as a superposition of creation operators acting on the unique EM field ground state $\left\|\left.0\right>\right.$. Here the $\pm$ stand for left and right circular polarization states. $\psi_\pm(\vec{k})$ are the complex superposition co-efficients. Such a state has a definite spread in "frequency", but it's still a pure quantum state. It's simply now that energy / momentum gleaned from any measurement is uncertain. Let's write $\left\|\left.\Psi\right>\right.$ as a shorthand for the above monstrous superpostiions. Now, it can be shown that, given a pure one photon number state, you can write down the following quantities $$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right\| \mathbf{\hat{E}}^+(\vec{r},t)\left\|\left.\Psi\right>\right.\\ \vec{\phi}_B(\vec{r},\,t)&=&\left<\left.0\right.\right\| \mathbf{\hat{B}}^+(\vec{r},t)\left\|\left.\Psi\right>\right. \end{array}$$ for any one photon state and that they fulfill Maxwell's equations exactly. Conversely, for any properly normalized solution of Maxwell's equations, one can calculate a corresponding one-photon state $\left\|\left.\Psi\right>\right.$ ( i.e. the superposition co-efficients $\psi_\pm(\vec{k})$). Don't worry too much about the details: the key point is that a solution of Maxwell's equations, properly normalized, is the equivalent information to a one photon number state. It gets better. The solutions can be interpreted as probability amplitudes to make a detection with an idealized photomultiplier tube, because the time/space detection probability density is: $$p(\vec{r},\,t) = \frac{1}{2}\,\epsilon_0\,\|\vec{\phi}_E\|^2 + \frac{1}{2\,\mu_0}\,\|\vec{\phi}_B\|^2$$ So, what you have to do to calculate this probability density for your kit is the following: Solve Maxwell's equations for your kit Normalize them so that the integrated energy density over all space is unity The energy density $\frac{1}{2}\,\left(\epsilon_0\,\|\vec{E}\|^2 + \mu_0\,\|\vec{H}\|^2\right)$ of the normalized solution is the required probability density function of space and time. So now, with this Maxwellian picture in mind, you can see what's going on. If the pure EM field state has a spread of energies, the interference fringes in the Maxwell equation solutions will be at different places for different frequencies, because they correspond to fields with slightly different wavelengths. So you can get fringes "washing each other out" and giving apparent "incoherence" behavior even though the underlying quantum state is perfectly pure or "coherent". Even an atomic transition in cryogenically cooled atoms (so there is no Doppler-induced uncertainty in the transition energy) gives rise to a spread in photon energies. This is because the atom is coupled to a very wide range of frequencies in the coupled one-photon EM field number states. It "tries" to relax into them all equally, but destructive interference as it does so means that the overall amplitude for excitation of a given frequency is very low unless the frequency matches that of the perfectly sharp transition well. A perfectly sharp transition thus gives a nonzero linewidth Lorentzian spectral shape: indeed doing calculations to model the broadband coupling just spoken of foretells exactly the Lorentzian lineshape for the pure quantum light state. The linewidth is inversely proportional to the coupling strength. Strong coupling means that the destructive interference I just spoke of acts more swiftly and strongly throughout the relaxation process. 2. Genuine Quantum Decoherence In this case, before detection events, you can't think of the system as the evolution of the EM field alone. The quantum state space is now the tensor product $\mathscr{E}\otimes\mathscr{K}$ of the EM field one-photon state space $\mathscr{E}$ and that $\mathscr{K}$ of the experimental kit. Lasers' atoms get bumped around by their neighbors in thermalized systems. Cavities get vibrated. Optical tables get bumped by coffee cups being put down on them or shaken by the traffic on the freeway outside your lab. The light state becomes entangled with the hugely complicated state of the experimental kit. In this case, if you try to look at the state of the light alone in $\mathscr{E}$, you're looking at a reduced quantum state and this appears as a classical mixture of pure quantum states. It is modelled by the density matrix formalism. (Conversely and interestingly, such a classical mixture can always be thought of as a reduced pure quantum state in a superset quantum state space: see the notion of quantum purification for more details). But the density matrix formalism is equivalent to the incoherent adding of the effects of the constituent pure states, i.e. you can do the Maxwellian analysis above for each pure state in the mixture and them sum the classical probability densities weighted by the probabilities to be in each state in the mixture. This is genuine quantum decoherence (of the light state) and it is hard to tell what exactly the experimental outcome will be. If the whole experimental kit is genuinely time invariant, so that the probabilities of each pure state in the mixture are constant, then the answer to your question will be exactly the same as it was when we talked about energy spectral spread in pure quantum states: the probabilities of single detection events at low light levels will reflect the high light level intensities precisely. But in practice this time invariance is very difficult to achieve. You'll find that the notion of coherence length is very, very hard to measure rigorously, because the result will be highly dependent on the integration times, frequency responses, surface areas and so of in your detectors and processing electronics. You'll get a different (probably shorter) coherence length with increasing integration time, but this rule of thumb breaks down when you measure so fast that you begin to approach the "one photon in the kit at a time" régime. If you go to the wiki page for coherence length , the following is found as introduction: In physics, coherence length is the propagation distance over which a coherent wave (e.g. an electromagnetic wave) maintains a specified degree of coherence. Wave interference is strong when the paths taken by all of the interfering waves differ by less than the coherence length. A wave with a longer coherence length is closer to a perfect sinusoidal wave. Coherence length is important in holography and telecommunications engineering. This article focuses on the coherence of classical electromagnetic fields. In quantum mechanics, there is a mathematically analogous concept of the quantum coherence length of a wave function. Now you ask: Coherence length of a single photon A photon as a quantum mechanical entity will fall into the quantum coherence length of the wavefunction which is a different concept than for classical waves, and is what the comments have been trying to elucidate. The photon as measured in an experiment has one point of interaction, and the information it carries is only of its fourvector, energy and direction, and its spin which is either +1 or -1 . If it hits a grating, even one photon will fall on the appropriate "color" and its energy will be known, and one point on a screen or ccd will give its direction. If a magnetic field is used, the up or down of the spin may be also measured. That is all for an individual photon. The wavefunction used for describing photons has sinusoidal properties, but the wavefunction complex conjugate squared is what can be measured, and it is a probability distribution.. By construction a probability distribution becomes manifest by many samples, not by one measurement. Any wavelength built in the probability distribution will only appear when a number of samples is taken, not from an individual measurement. Which is what the commenters were trying to clarify. In conclusion, a single photon cannot have a coherence length. The probability of its manifesting has a coherence length and it is an attribute of collective behavior and can be measured by an ensemble of photons. The classical coherence length will also emerge in the behavior of the ensemble of photons, so, once measured, one can know how probable it is for a photon to manifest at that (x,y,z) in the experiment. Photons in ensembles build up the classical electromagnetic wave which will have its coherence length . I have found this image of how polarization is built up by individual photons that just have +1 or -1 spin useful in getting an intuition of how the classical wave is built up. Left and right handed circular polarization, and their associate angular momenta. A QED explanation of how individual photons build up a classical wave can be found here. CuriousOne makes a good point in comments and I will elaborate on that. While you are asking about the coherence length of a single photon, in the experiment that you describe you will have to detect many photons to judge if they are detected in both arms with equal probability. Where do this photons come from? If you assume that the photons are identical, then you would need some ideal light source with infinite coherence length. If you take a real light source, the difference between individual photons will define your decoherence length. You can see, then, that what you are actually measuring with your experiment is the coherence length of the light source. A single photon does not have such property. Now that discussion of this topic has ceased, may I offer my own simplified answer based on what I have learned? Firstly, no source of light can produce identical photons with exactly the same energies and direction. If the spread of energies is ΔE then the spread of wavelengths Δλ = ΔE λ 2/hc. The maximum number of wavelengths of path difference before interference effects disappear will be n= λ/Δλand the maximum path difference will be nλ= λ= 2/Δλ hc/ΔE. For a sodium atom at modest temperatures I calculate this to be of the order of 30 cm which seems to be about right. Additional effects may conspire to reduce this apparent coherence length even further. Secondly, it basically doesn't matter how weak the light source is; when you average the results over many photons you will see the same interference pattern. In other words, what we are measuring is the coherence length of the source of light, not the 'length' of the individual photons which it produces. By length maybe you mean wavelength. A single photon traveling at the speed of light and oscillating at a certain frequency will oscillate through one cycle every wavelength or say 500 nm. As you increase the length of one arm of the experiment the interference will go in and out of phase every one half cycle or every 250 nm.
(This post is the third in a series on geometry. (A geometric series, har har har.) They all assume you know about differential forms and such things. The first was on Liouville geometry, also known as exact symplectic geometry, on surfaces. The second went from them to contact geometry. So I’m assuming you know what those are.) We’ve seen that if you take a Liouville 1-form $\beta$ on a surface $S$ (i.e. such that $d\beta$ is nondegenerate, hence a symplectic form), then the 1-form $\alpha = \beta + dt$ on the 3-manifold $M = S \times [0,1]$ obtained by thickening $S$ is a contact form. (Here $t$ is the coordinate on $[0,1]$.) Moreover, we’ve seen that on each slice $S \times \{t\}$ of this thickening, the characteristic foliation (i.e. the pattern of how the slice intersects the contact planes) $\mathcal{F}$ coincides with $\ker \beta$. We’ve also noted that this contact form $\alpha$ is a vertically invariant contact form on $M$: it has no dependence on $t$. Indeed, the flow of the vertical vector field $\partial_t$ preserves $\alpha$, and hence is a contact vector field. Thus each slice $S \times \{t\}$ is transvserse to a contact vector field, and hence is a convex surface. Thus, starting from the simple but elegant structure of a Liouville 1-form on a surface, we have been led to 3-dimensional contact geometry, and convex surfaces. What we’re going to do now is go in the other direction, and start from a convex surface. We’re going to make a clear distinction now between a contact structure and a contact form. A contact form is a 1-form $\alpha$ such that $\alpha \wedge d\alpha$ is non-degenerate, i.e. so that $\ker \alpha$ is a non-integrable plane fielr. A contact structure $\xi$ is a non-integrable plane field. So any contact form $\alpha$ defines a contact structure $\xi$ by $\xi = \ker \alpha$, but a contact structure $\xi$ has many 1-forms defining it (at least locally). Given any contact form $\alpha$ such that $\ker \alpha = \xi$, we can multiply $\alpha$ by any smooth nonzero real-valued function $f$, and $f\alpha$ is then another contact 1-form, with $\ker(f\alpha) = \ker \alpha = \xi$. Well, let’s return to the definition of a convex surface: it’s an embedded surface $S$ in a contact 3-manifold for which there is a vector field $X$ transverse to $S$. Said tersely, a convex surface is a surface with a transverse contact vector field. Now, given a convex surface, we can introduce coordinates as we please. Let us define a coordinate $t$ by the transverse vector field $X$. So let $X = \partial_t$. We can then let $t=0$ on the surface $S$, and flowing along $X = \partial_t$, we obtain a coordinate $t$ which measures how far from $S$ we have flowed along $X$. Using this coordinate, we can describe a neighbourhood of $S$ as $S \times [-\varepsilon, \varepsilon]$, for some sufficiently small $\varepsilon$, where $S$ appears as $S \times \{0\}$ and the coordinate on the $[-\varepsilon, \varepsilon]$ factor is precisely $t$. For simplicitly, we can take $\varepsilon = 1$; by slowing down the vector field $X$ we can in fact fit this $S \times [-1,1]$ inside the previous $S \times [-\varepsilon, \varepsilon]$. So now we have a neighbourhood of $S$ given as $S \times [-1,1]$, and the transverse contact vector field is $X = \partial_t$. If we further denote by $x,y$ some local coordinates on $S$, then $x,y,t$ form some local coordinates on $S \times [-1,1]$. So the contact form $\alpha$ (or indeed any 1-form) can be written in the form \[ \alpha = f \; dx + g \; dy + u \; dt, \] where $f,g,u$ are real-valued functions on $S \times [-1,1]$. Now the functions $f,g,u : S \times [-1,1] \rightarrow \mathbb{R}$ might in general depend on $x,y,t$. But as $X = \partial_t$ s a contact vector field, the contact planes given by $\ker \alpha$ don’t depend on the $t$ coordinate at all. And hence we can take the contact form $\alpha$ not to depend on $t$ either. (Possibly $\alpha$ might depend on $t$, since multiplying $\alpha$ by any nonero real-valued function produces a 1-form with the same kernel; but for such an $\alpha$, we can “normalise” it, multiplying by a nonzero function, to make it independent of $t$. Or indeed replacing $f(x,y,t), g(x,y,t), u(x,y,t)$ with $f(x,y,0), g(x,y,0), u(x,y,0)$ would have the same effect.) In other words, since $S$ is a convex surface, there is a contact form $\alpha$ where $f,g,u$ only depend on $x,y$, and not $t$. We can write \[ \alpha = f(x,y) \; dx + g(x,y) \; dy + u(x,y) \; dt. \] Written in this way, the first two terms $f(x,y) \; dx + g(x,y) \; dy$ denote a 1-form purely on the surface $S$. Indeed, any 1-form on $S$ can be written this way. So let’s call it $\beta$. In a similar way, $u(x,y)$ can be regarded as a function purely on the surface $S$. We then have \[ \alpha = \beta + u \; dt \] where $\beta$ is a 1-form on $S$, and $u$ is a real-valued function on $S$. When $u \neq 0$, we can do even better. We can then divide the whole 1-form $\alpha$ by $u$ — and remember, multiplying the contact form by a nonzero function results in another contact form defining the same contact structure. So this allows us effectively to assume that $u=1$, and that $\alpha$ is of the form $\beta + dt$, where again $\beta$ is a 1-form and $u$ a real-valued function on $S$. Now if $\alpha = \beta + dt$ is a contact form, then it must satisfy the contact condition of being non-integrable, i.e. $\alpha \wedge d\alpha$ must be non-degenerate. Not every possible 1-form $\beta$ on $S$ and every function $u$ on $S$ will make $\beta + dt$ a contact form. Which possible 1-forms $\beta$ make a contact form? We can compute $\alpha \wedge d\alpha$ to find out: \[ \alpha \wedge d\alpha = (\beta + dt) \wedge d\beta = \beta \wedge d\beta + dt \wedge d\beta = dt \wedge d\beta. \] This is a contact form if and only if $d\beta$ is a non-degerate 2-form on $S$ — that is, if $\beta$ is a Liouville 1-form. Thus we have proved the following. PROPOSITION. Let $S$ be a convex surface in a 3-manfold with a contact structure $\xi$. Defining a transverse coordinate $t$ via the transverse contact vector field, $S$ has a neighbourhood on which $\xi$ has a contact form $\beta + u \; dt$, where $\beta$ is a 1-form and $u$ is a real-valued function on $S$. If further $u$ is nowhere zero, then $S$ has a neighbourhood on which $\xi$ has a contact form $\beta + dt$, where $\beta$ is a Liouville 1-form on $S$. In our previous episode, starting from Liouville structures on surfaces, we were led to convex surfaces in contact 3-manifolds. And now, we have gone back, from convex surfaces to Liouville structures. Now we know that not every surface has a Liouville structure: we saw previously that there can’t be one if $S$ is compact without boundary. And so a convex surface also can’t have a local contact form $\beta + dt$ if $S$ is compact without boundary. Such a convex surface $S$, compact without boundary, has a local contact form of the type $\beta + u \; dt$, as we’ve discussed. And remember we said that if $u$ is nowhere zero, then we could divide out by $u$ and obtain a local contact form of the type $\beta + dt$. But they can’t have a local contact form of the type $\beta + dt$. Hence for any convex surface $S$, compact without boundary, the contact form $\beta + u \; dt$ must have some zeroes of $u$. And as it turns out, the zeroes of $u$ are very interesting and important. What happens at the zeroes of $u$? They are precisely where the contact planes are vertical, i.e. where $\partial_t$, or the contact vector field $X$, lies in $\xi = \ker \alpha$. Indeed, \[ \alpha(X) = \alpha(\partial_t) = \beta(\partial_t) + u dt (\partial_t) = u. \] Here we used the fact that $\beta(\partial_t) = 0$, since $\beta$ is a 1-form on $S$, which is independent of the $t$ coordinate. So $\alpha(X) = 0$ precisely when $u=0$. The set of points where $u=0$ is called the dividing set (or decoupage, in the original French). It turns out that it’s a curve on $S$ and it splits $S$ into pieces where $u>0$ and where $u<0$. (This was proved by Giroux.) Note that when $u>0$, we have $\alpha(X)>0$; and when $u<0$, we have $\alpha(X)<0$. Suppose we paint one side of the contact planes white, and the other side black. We think of the black side as “positive”, and the white side as “negative”, in the following sense. Given any vector $V$, we will have $\alpha(V) > 0$ when $V$ points out of the white side, $\alpha(V) = 0$ when $V$ points along the plane, and $\alpha(V) < 0$ when $V$ points out of the black side. Thus, the contact planes are white side up when $u<0$, they become vertical along the dividing set $u=0$, where they flip over to be black side up when $u>0$. The standard notation is that the dividing set (i.e. where $u=0$) is denoted $\Gamma$; the region of $S$ where $u>0$ is denoted $R_+$; and the region of $S$ where $u<0$ is denoted $R_-$. The best thing is that, if you just consider the subset of $S \times [-1,1]$ where $u>0$ (say), i.e. $R_+ \times [-1,1]$, you can divide $\alpha$ out by $u$, and obtain a contact form of the type $\beta + dt$, where $\beta$ is Liouville. So the characteristic foliation on $R_+$ is a Liouville foliation, and there is a flow tangent to it which exponentially expands an area form. The same applies to $R_-$. So in fact a convex surface can be regarded as made up of two Liouville structures pieced together along a dividing set, where the contact planes flip over.
To make sense of this, you need to understand what the word “derivative” means here. A function $f \colon \mathbf{R}^n \to \mathbf{R}^m$ is said to be differentiable at the point $a \in \mathbf{R}^n$ if there is a linear transformation $L \colon \mathbf{R}^n \to \mathbf{R}^m$ such that$$f(a+h) - f(a) = L(h) + R(h)$$with a remainder term $R(h)$ which tends to zero faster than $\|h\|$ as $h \to 0$, i.e.,$$\lim_{h \to 0} \frac{R(h)}{\|h\|} = \lim_{h \to 0} \frac{f(a+h) - f(a) - L(h)}{\|h\|} = 0.\tag{$*$}$$If this is satisfied, the linear transformation $L$ is called the derivative of $f$ at $a$. (Sometimes it's called the total derivative, to distinguish it from the partial derivatives of $f$.) One can think of the linear transformation $L$ as just a matrix of size $n \times m$, and in fact it turns out that if the derivative exists at all, then this matrix has to be the Jacobian matrix (the $n \times m$ matrix of partial derivatives at the point $a$). But merely the fact that the Jacobian exists is not sufficient to guarantee that the derivative exists. There are counterexamples, like $f(x,y)=17$ if $x=0$ or $y=0$, and $f(x,y)=43$ otherwise; the partials of $f$ at the origin both exist (they are zero), but $f$ has no (total) derivative there, and in fact $f$ doesn't even fulfill the weaker condition of being continuous at the origin. However, there's a theorem that says that if the partials exist and are continuous (in some open set containing $a$), then the derivative exists. And this is what your teacher was referring to.
Suppose to have a chain (of size $L$) with bosons, and $\hat{a}_i^\dagger$,$\hat{a}_i$ are the associated creation and annihilation operators at site $i$. A Fock state can be written as: \begin{equation} \| n_1 \dots n_L \rangle = \prod_{i} \frac{1}{\sqrt{n_i!}} \left( \hat{a}_i^\dagger \right)^{n_i} \|\rangle \end{equation} where $\|\rangle$ is the empty state. Now we define a new set of bosons: \begin{align} \hat{b}_{k} &= \frac{1}{\sqrt{L}} \sum_{j} e^{-ikj} \hat{a}_j & \hat{b}_{k}^\dagger &= \frac{1}{\sqrt{L}} \sum_{j} e^{ikj} \hat{a}_j^\dagger \end{align} where $k$ are such that some boundary condition is fulfilled. Now a Fock state can be written as: \begin{equation} \| \dots \tilde{n}_k \dots \rangle = \prod_{k} \frac{1}{\sqrt{\tilde{n}_k!}} \left( \hat{b}_k^\dagger \right)^{\tilde{n}_k} \|\rangle \end{equation} The question is: is there a simple formula to express or compute the scalar product $\langle \dots \tilde{n}_k \dots \| n_1 \dots n_L \rangle$? Define the operators $\hat a(f)=\sum f_j\hat a_j$ and $ \|f_1,...,f_n\rangle:=\hat a(f_1)...\hat a(f_n)\|vac\rangle$. Then $\langle g_1,...,g_m\|f_1,...,f_n\rangle$ vanishes for $m\ne n$ and is a sum of the products $\langle g_1\|f_{j_1}\rangle...\langle g_n\|f_{j_n}\rangle$ for all possible permutations $(j_1,...,j_n)$ of $1,...,n$. Note that the $\langle g\|f\rangle$ are easy to compute. The formula you requested is a special case of this. I realise the question has already been answered, but just a suggestion to the topic starter: finite sums of the form $(\sum_i x_i)^N$ are expanded using objects called multinomial coefficients. They should make your life a little easier.
The aleph numbers, $\aleph_\alpha$ The aleph function, denoted $\aleph$, provides a 1 to 1 correspondence between the ordinal and the cardinal numbers. In fact, it is the only order-isomorphism between the ordinals and cardinals, with respect to membership. It is a strictly monotone ordinal function which can be defined via transfinite recursion in the following manner: $\aleph_0 = \omega$ $\aleph_{n+1} = \bigcap \{ x \in \operatorname{On} : \| \aleph_n \| \lt \|x\| \}$ $\aleph_a = \bigcup_{x \in a} \aleph_x$ where $a$ is a limit ordinal. To translate the formalism, $\aleph_{n+1}$ is the smallest ordinal whose cardinality is greater than the previous aleph. $\aleph_a$ is the limit of the sequence $\{ \aleph_0 , \aleph_1 , \aleph_2 , \ldots \}$ until $\aleph_a$ is reached when $a$ is a limit ordinal. Contents Aleph one $\aleph_1$ is the first uncountable cardinal. The continuum hypothesis The continuum hypothesis is the assertion that the set of real numbers $\mathbb{R}$ have cardinality $\aleph_{1}$. Gödel showed the consistency of this assertion with ZFC, while Cohen showed using forcing that if ZFC is consistent then ZFC+$\aleph_1<\|\mathbb R\|$ is consistent. Equivalent Forms The cardinality of the power set of $\aleph_{0}$ is $\aleph_{1}$ The is no set with cardinality $\alpha$ such that $\aleph_{0} < \alpha < \aleph_{1}$ Generalizations The generalized continuum hypothesis (GCH) states that if an infinite set's cardinality lies between that of an infinite set S and that of the power set of S, then it either has the same cardinality as the set S or the same cardinality as the power set of S. That is, for any infinite cardinal $\lambda$ there is no cardinal $\kappa$ such that $\lambda <\kappa <2^{\lambda}.$ GCH is equivalent to:\[\aleph_{\alpha+1}=2^{\aleph_\alpha}\] for every ordinal $\alpha.$ (occasionally called Cantor's aleph hypothesis) For more,see https://en.wikipedia.org/wiki/Continuum_hypothesis Aleph two Aleph hierarchy The $\aleph_\alpha$ hierarchy of cardinals is defined by transfinite recursion: $\aleph_0$ is the smallest infinite cardinal. $\aleph_{\alpha+1}=\aleph_\alpha^+$, the successor cardinal to $\aleph_\alpha$. $\aleph_\lambda=\sup_{\alpha\lt\lambda}\aleph_\alpha$ for limit ordinals $\lambda$. Thus, $\aleph_\alpha$ is the $\alpha^{\rm th}$ infinite cardinal. In ZFC the sequence $$\aleph_0, \aleph_1,\aleph_2,\ldots,\aleph_\omega,\aleph_{\omega+1},\ldots,\aleph_\alpha,\ldots$$ is an exhaustive list of all infinite cardinalities. Every infinite set is bijective with some $\aleph_\alpha$. Aleph omega The cardinal $\aleph_\omega$ is the smallest instance of an uncountable singular cardinal number, since it is larger than every $\aleph_n$, but is the supremum of the countable set $\{\aleph_0,\aleph_1,\ldots,\aleph_n,\ldots\mid n\lt\omega\}$. Aleph fixed point A cardinal $\kappa$ is an $\aleph$-fixed point when $\kappa=\aleph_\kappa$. In this case, $\kappa$ is the $\kappa^{\rm th}$ infinite cardinal. Every inaccessible cardinal is an $\aleph$-fixed point, and a limit of such fixed points and so on. Indeed, every worldly cardinal is an $\aleph$-fixed point and a limit of such. One may easily construct an $\aleph$-fixed point above any ordinal $\beta$: simply let $\beta_0=\beta$ and $\beta_{n+1}=\aleph_{\beta_n}$; it follows that $\kappa=\sup_n\beta_n=\aleph_{\aleph_{\aleph_{\aleph_{\ddots}}}}$ is an $\aleph$-fixed point, since $\aleph_\kappa=\sup_{\alpha\lt\kappa}\aleph_\alpha=\sup_n\aleph_{\beta_n}=\sup_n\beta_{n+1}=\kappa$. By continuing the recursion to any ordinal, one may construct $\aleph$-fixed points of any desired cofinality. Indeed, the class of $\aleph$-fixed points forms a closed unbounded class of cardinals.
This will be a talk for the conference: Ouroboros: Formal Criteria of Self-Reference in Mathematics and Philosophy, held in Bonn, February 16-18, 2018. Abstract. I shall give an elementary account of the universal algorithm, due to Woodin, showing how the capacity for self-reference in arithmetic gives rise to a Turing machine program $e$, which provably enumerates a finite set of numbers, but which can in principle enumerate any finite set of numbers, when it is run in a suitable model of arithmetic. Furthermore, the algorithm can successively enumerate any desired extension of the sequence, when run in a suitable top-extension of the universe. Thus, the algorithm sheds some light on the debate between free will and determinism, if one should imagine extending the universe into a nonstandard time scale. An analogous result holds in set theory, where Woodin and I have provided a universal locally definable finite set, which can in principle be any finite set, in the right universe, and which can furthermore be successively extended to become any desired finite superset of that set in a suitable top-extension of that universe. Ouroboros Bonn 2018 Conference Poster \| Slides Regula Krapf successfully defended her PhD dissertation January 12, 2017 at the University of Bonn, with a dissertation entitled, “Class forcing and second-order arithmetic.” I was a member of the dissertation examining committee. Peter Koepke was the dissertation supervisor. Regula Krapf, Class forcing and second-order arithmetic, dissertation 2017, University of Bonn. (Slides) Abstract. We provide a framework in a generalization of Gödel-Bernays set theory for performing class forcing. The forcing theorem states that the forcing relation is a (definable) class in the ground model (definability lemma) and that every statement that holds in a class-generic extension is forced by a condition in the generic filter (truth lemma). We prove both positive and negative results concerning the forcing theorem. On the one hand, we show that the definability lemma for one atomic formula implies the forcing theorem for all formulae in the language of set theory to hold. Furthermore, we introduce several properties which entail the forcing theorem. On the other hand, we give both counterexamples to the definability lemma and the truth lemma. In set forcing, the forcing theorem can be proved for all forcing notions by constructing a unique Boolean completion. We show that in class forcing the existence of a Boolean completion is essentially equivalent to the forcing theorem and, moreover, Boolean completions need not be unique. The notion of pretameness was introduced to characterize those forcing notions which preserve the axiom scheme of replacement. We present several new characterizations of pretameness in terms of the forcing theorem, the preservation of separation, the existence of nice names for sets of ordinals and several other properties. Moreover, for each of the aforementioned properties we provide a corresponding characterization of the Ord-chain condition. Finally, we prove two equiconsistency results which compare models of ZFC (with large cardinal properties) and models of second-order arithmetic with topological regularity properties (and determinacy hypotheses). We apply our previous results on class forcing to show that many important arboreal forcing notions preserve the $\Pi^1_1$-perfect set property over models of second-order arithmetic and also give an example of a forcing notion which implies the $\Pi^1_1$-perfect set property to fail in the generic extension. Regula has now taken up a faculty position at the University of Koblenz. This will be a talk for the University of Bonn Logic Seminar, Friday, January 13, 2017, at the Hausdorff Center for Mathematics. Abstract. Set-theoretic geology is the study of the set-theoretic universe $V$ in the context of all its ground models and those of its forcing extensions. For example, a bedrock of the universe is a minimal ground model of it and the mantle is the intersection of all grounds. In this talk, I shall explain some recent advances, including especially the breakthrough result of Toshimichi Usuba, who proved the strong downward directed grounds hypothesis: for any set-indexed family of grounds, there is a deeper common ground below them all. This settles a large number of formerly open questions in set-theoretic geology, while also leading to new questions. It follows, for example, that the mantle is a model of ZFC and provably the largest forcing-invariant definable class. Strong downward directedness has also led to an unexpected connection between large cardinals and forcing: if there is a hyper-huge cardinal $\kappa$, then the universe indeed has a bedrock and all grounds use only $\kappa$-small forcing. Slides This will be a talk January 10, 2017 for the Basic Notions Seminar, aimed at students, post-docs, faculty and guests of the Mathematics Institute, University of Bonn. Abstract. I shall give a general introduction to the theory of infinite games, using infinite chess — chess played on an infinite edgeless chessboard — as a central example. Since chess, when won, is won at a finite stage of play, infinite chess is an example of what is known technically as an open game, and such games admit the theory of transfinite ordinal game values. I shall exhibit several interesting positions in infinite chess with very high transfinite game values. The precise value of the omega one of chess is an open mathematical question. This talk will include some of the latest progress, which includes a position with game value $\omega^4$. It happens that I shall be in Bonn also for the dissertation defense of Regula Krapf, who will defend the same week, and who is one of the organizers of the seminar. Transfinite game values in infinite chess \| The mate-in-$n$ problem of infinite chess is decidable \| A position in infinite chess with game value $\omega^4$ \| more on infinite chess \| Slides I shall be very pleased to speak at the colloquium and workshop Infinity, computability, and metamathematics, celebrating the 60th birthdays of Peter Koepke and Philip Welch, held at the Hausdorff Center for Mathematics May 23-25, 2014 at the Universität Bonn. My talk will be the Friday colloquium talk, for a general mathematical audience. Abstract. I shall give a general introduction to the theory of infinite games, using infinite chess—chess played on an infinite edgeless chessboard—as a central example. Since chess, when won, is won at a finite stage of play, infinite chess is an example of what is known technically as an open game, and such games admit the theory of transfinite ordinal game values. I shall exhibit several interesting positions in infinite chess with very high transfinite game values. The precise value of the omega one of chess is an open mathematical question. Slides \| Schedule \| Transfinite game values in infinite chess \| The mate-in-n problem of infinite chess is decidable A four-lecture tutorial on the topic of Boolean ultrapowers at the Young Set Theory Workshop at the Hausdorff Center for Mathematics in Königswinter near Bonn, Germany, March 21-25, 2011. Boolean ultrapowers generalize the classical ultrapower construction on a power-set algebra to the context of an ultrafilter on an arbitrary complete Boolean algebra. Closely related to forcing and particularly to the use of Boolean-valued models in forcing, Boolean ultrapowers were introduced by Vopenka in order to carry out forcing as an internal ZFC construction, rather than as a meta-theoretic argument as in Cohen’s approach. An emerging interest in Boolean ultrapowers has arisen from a focus on the well-founded Boolean ultrapowers as large cardinal embeddings. Historically, researchers have come to the Boolean ultrapower concept from two directions, from set theory and from model theory. Exemplifying the set-theoretic perspective, Bell’s classic (1985) exposition emphasizes the Boolean-valued model $V^{\mathbb{B}}$ and its quotients $V^{\mathbb{B}}/U$, rather than the Boolean ultrapower $V_U$ itself, which is not considered there. Mansfield (1970), in contrast, gives a purely algebraic, forcing-free account of the Boolean ultrapower, emphasizing its potential as a model-theoretic technique, while lacking the accompanying generic objects. The unifying view I will explore in this tutorial is that the well-founded Boolean ultrapowers reveal the two central concerns of set-theoretic research–forcing and large cardinals–to be two aspects of a single underlying construction, the Boolean ultrapower, whose consequent close connections might be more fruitfully explored. I will provide a thorough introduction to the Boolean ultrapower construction, while assuming only an elementary graduate student-level familiarity with set theory and the classical ultrapower and forcing techniques. Article \| Abstract \| Lecture Notes
I am taking a course on quantum field theory where there is some confusion regarding the renormalisation scheme we are using (and a corresponding one in my mind). Apparently the lecturer meant MS-bar when he said MS. Anyway, on with the question. Consider the* 1-loop contribution to the 1PI 4-point correlation function of phi-4 theory in $d=4-\epsilon$ dimensions, i.e the Lagrangian is $$L=-\frac{1}{2}(\partial \phi)^2 -\frac{1}{2}m^2 \phi^2-\frac{1}{4!} \mu^\epsilon \lambda \phi^4$$ where $\mu$ is a mass scale required to make $\lambda$ dimensionless as we vary $\epsilon$. The Feynman rules yield an expression of the form $$\frac{(\mu^{ \epsilon}\lambda)^2}{2} \int \frac{d^d k}{(2 \pi)^d}\frac{1}{(k^2 +m^2)((k+p)^2+m^2)}$$where $p$ is a certain sum of external momenta, which eventually reduces to $$\frac{\lambda^2}{2 (4 \pi)^2} \int_0^1 d\alpha \ (m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}} \Gamma\left(\frac{\epsilon}{2}\right) \mu^{2 \epsilon} [4\pi]^{\frac{\epsilon}{2}}$$I can clearly pull one factor of $\mu^\epsilon$ into $ (m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}}$ to get $\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right)^{-\frac{\epsilon}{2}}$ so that the final, finite result at 1 loop looks like $$ -\frac{\lambda^2}{2 (4 \pi)^2} \int_0^1 d\alpha \log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right) +\text{2 other terms}$$ But where did the other factor of $\mu^\epsilon$ disappear to? Was it implicitly 'countertermed' away along with the factor $ [4\pi e^{-\gamma}]^{\frac{\epsilon}{2}}$? It looks as if this can be done at this order, but does this generalise to higher loop orders, i.e, is there a prescription telling us how many factors $\mu^\epsilon$ to counterterm away? Note that in phi-3 theory we get an identical loop integral by considering the 1-loop contribution to the 1PI 2-point correlation function, the difference being that $(\mu^{ \epsilon}\lambda)^2 \to (\mu^{ \frac{\epsilon}{2}} g)^2$ (where $g$ is the phi-3 coupling constant) in front of the integral. So that diagram does not have this problem somehow. There are two other contributions obtained by permuting the external momenta. These have identical integral expressions. EDIT: I'll write this down fully. The term inside the integral can be written as $$(m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}} \Gamma\left(\frac{\epsilon}{2}\right) \mu^{2 \epsilon} [4\pi]^{\frac{\epsilon}{2}} =\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^4}\right)^{-\frac{\epsilon}{2}}\Gamma\left(\frac{\epsilon}{2}\right) [4\pi]^{\frac{\epsilon}{2}}=\left(1-\frac{\epsilon}{2}\log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^4}\right)\right) \frac{2}{\epsilon} [4\pi e^{-\gamma}]^{\frac{\epsilon}{2}}$$ OR as $$=\left(1-\frac{\epsilon}{2}\log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right)\right) \frac{2}{\epsilon} [4\pi e^{-\gamma}\mu^2]^{\frac{\epsilon}{2}}$$ where the last two equalities are both asymptotic expansions in $\epsilon$ as $\epsilon \to 0$. I can absorb that last square bracket into the counterterm if I want to, at this order, and effectively ignore it.
As we know perturbative expansion of interacting QFT or QM is not a convergent series but an asymptotic series which generally is divergent. So we can't get arbitrary precision of an interacting theory by computing higher enough order and adding them directly. However we also know that we can use some resummation trick like Borel summation, Padé approximation and so on to sum a divergent series to restore original non-perturbative information. This trick is widely used in computing critical exponent of $\phi^4$ etc. My questions: Although it's almost impossible to compute perturbation to all order, is it true that we can get arbitrary precision of interacting system (like QCD) by computing higher enough order and using resummation trick like Borel summation? Is it true that in principle non-perturbative information like instanton, vortex can also be achieved by above methods? There is a solid example: $0$-dim $\phi^4$ theory, $$Z(g)\equiv\int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}e^{-x^2/2 -gx^4/4}$$ From the definition of $Z(g)$ above, $Z(g)$ must be a finite number for $g>0$. As usual we can compute this perturbatively, $$Z(g)= \int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}e^{-x^2/2}\sum_{n=0}^{\infty}\frac{1}{n!}(-gx^4/4)^n \sim \sum_{n=0}^{\infty} \int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}e^{-x^2/2} \frac{1}{n!}(-gx^4/4)^n \tag{1}$$ Note: In principle we can't exchange integral and infinite summation. It's why asymptotic series is divergent. $$Z(g)\sim \sum_{n=0}^{\infty} \frac{(-g)^n (4n)!}{n!16^n (2n)!} \tag{2}$$ It's a divergent asymptotic series. In another way, $Z(g)$ can be directly solved, $$Z(g)= \frac{e^{\frac{1}{8g}}K_{1/4}(\frac{1}{8g})}{2\sqrt{\pi g}} \tag{3}$$ where $K_n(x)$ is the modified Bessel function of the second kind. We see obviously that $Z(g)$ is finite for $g>0$ and $g=0$ is an essential singularity. But we can restore the exact solution $(3)$ by Borel resummation of divergent series $(2)$ First compute the Borel transform $$B(g)=\sum_{n=0}^{\infty} \frac{(-g)^n (4n)!}{(n!)^216^n (2n)!} = \frac{2K(\frac{-1+\sqrt{1+4g}}{2\sqrt{1+4g}})}{\pi (1+4g)^{1/4}} $$ where $K(x)$ is the complete elliptic integral of the first kind. Then compute the Borel Sum $$Z_B(g)=\int_0^{\infty}e^{-t}B(gt)dt=\frac{e^{\frac{1}{8g}}K_{1/4}(\frac{1}{8g})}{2\sqrt{\pi g}} \tag{4}$$ $$Z_B(g) = Z(g)$$ We see concretely by using the trick of Borel resummation, we can restore the exact solution from divergent asymptotic series.
Following (1) and (2), the Schwarzschild metric can be written as: $$ ds^2 = a(r)\cdot c^2 dt^2 - b(r)\cdot dr^2 - r^2\cdot(d\theta^2 + sin^2(\theta)\cdot d\varphi^2) $$ where: $$ a(r) = \begin{cases} \frac14\left(3\sqrt{1-\frac{r_s}{r_g}} - \sqrt{1-\frac{r^2 r_s}{r_g^3}}\right)^2 \qquad \text{if } 0\leq r \leq r_g\\ 1-\frac{r_s}{r}\qquad \text{if } r_g\leq r \end{cases} $$ $$ b(r) = \begin{cases} \left(1-\frac{r^2 r_s}{r_g^3}\right)^{-1} \qquad \text{if } 0\leq r \leq r_g\\ \left(1-\frac{r_s}{r}\right)^{-1} \qquad \text{if } r_g\leq r \end{cases} $$ Here : $r_s=2GM/c^2$ is the Schwarzschild radius of a spherical body of mass $M$ $r_g$ is the value of $r$ at the surface of the spherical body The part $r\in [0,r_g]$ is the interior Schwarzschild solution while the part $r\geq r_g$ is the exterior Schwarzschild solution. Assume $r_s < r_g$. Now, I graph $a(r)$ and $b(r)$. I see that $a(r)$ and $b(r)$ are continuous functions. I see that $a(r)$ looks fairly smooth. However, at the transition $r=r_g$, the function $b(r)$ does not look smooth. Is it normal? Why would $b(r)$ be not differentiable at the transition from $r<r_g$ to $r>r_g$? Here is a screenshot for $r_s=1$ and $r_g=2$.
Lethbridge Number Theory and Combinatorics Seminar: Adam Felix Date: 06/01/2015 Time: 11:00 KTH Royal Institute of Technology, Sweden University of Lethbridge How close is the order of a mod p to p−1? Let $a \in \mathbb{Z} \setminus \{0,\pm 1\}$, and let $f_{a}(p)$ denote the order of $a$ modulo $p$, where $p \nmid a$ is prime. There are many results that suggest $p - 1$ and $f_{a}(p)$ are close. For example, Artin's conjecture and Hooley's subsequent proof upon the Generalized Riemann Hypothesis. We will examine questions related to the relationship between $p - 1$ and $f_{a}(p)$. Location: C630 University Hall
The momentum equation can be simplified for the steady state condition as it was shown in example 6.3. The unsteady term (where the time derivative) is zero. Integral Steady State Momentum Equation \[ \label{mom:eq:govSTSF} \sum\pmb{F}_{ext} + \int_{c.v.} \pmb{g} \,\rho\, dV - \int_{c.v.}\pmb{P}\,dA + \int_{c.v.} \boldsymbol{\tau}\,dA = \int_{c.v.} \rho\, \pmb{U} U_{rn} dA \tag{15} \] Contributors Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.
Two representations of $R_z$ are equivalent if they are the same modulo only a global phase. $$\begin{pmatrix}1 & 0\\0 & e^{i\theta}\end{pmatrix}=e^{+i\theta/2}\begin{pmatrix}e^{-i\theta/2} & 0\\0 & e^{i\theta/2}\end{pmatrix}$$ If you apply this gate to any state $\|\psi\rangle$, the only difference in the outcome is a global (constant) phase of $e^{i\theta/2}$. This cannot be detected by any measurement. For example measurements on the state: $$\|\psi\rangle = \frac{1}{\sqrt{2}}\left(\|0\rangle + \|1\rangle \right)$$ will result in $\|0\rangle$ with a probability of $\left\|\frac{1}{\sqrt{2}}\right\|^2 = \frac{1}{2}$, and will result in $\|1\rangle$ with a probability of $\left\|\frac{1}{\sqrt{2}}\right\|^2 = \frac{1}{2}$. Now consider the state:$$\|\psi\rangle = \frac{e^{i\theta/2}}{\sqrt{2}}\left(\|0\rangle + \|1\rangle \right)$$ Measurements will: result in $\|0\rangle$ with a probability of $\left\|\frac{e^{i\theta/2}}{\sqrt{2}}\right\|^2 = \frac{1}{2}$, and will result in $\|1\rangle$ with a probability of $\left\|\frac{e^{i\theta/2}}{\sqrt{2}}\right\|^2 = \frac{1}{2}$. Two gates that are equivalent up to a global phase (one that's the same for all components) are equivalent for the purpose of anything you will be able to detect by measurement.
Electrochemical Impedance Spectroscopy: Experiment, Model, and App Electrochemical impedance spectroscopy is a versatile experimental technique that provides information about an electrochemical cell’s different physical and chemical phenomena. By modeling the physical processes involved, we can constructively interpret the experiment’s results and assess the magnitudes of the physical quantities controlling the cell. We can then turn this model into an app, making electrochemical modeling accessible to more researchers and engineers. Here, we will look at three different ways of analyzing EIS: experiment, model, and simulation app. Electrochemical Impedance Spectroscopy: The Experiment Electrochemical impedance spectroscopy (EIS) is a widely used experimental method in electrochemistry, with applications such as electrochemical sensing and the study of batteries and fuel cells. This technique works by first polarizing the cell at a fixed voltage and then applying a small additional voltage (or occasionally, a current) to perturb the system. The perturbing input oscillates harmonically in time to create an alternating current, as shown in the figure below. An oscillating perturbation in cell voltage gives an oscillating current response. For a certain amplitude and frequency of applied voltage, the electrochemical cell responds with a particular amplitude of alternating current at the same frequency. In real systems, the response may be complicated for components of other frequencies too — we’ll return to this point below. EIS experiments typically vary the frequency of the applied perturbation across a range of mHz and kHz. The relative amplitude of the response and time shift (or phase shift) between the input and output signals change with the applied frequency. These factors depend on the rates at which physical processes in the electrochemical cell respond to the oscillating stimulus. Different frequencies are able to separate different processes that have different timescales. At lower frequencies, there is time for diffusion or slow electrochemical reactions to proceed in response to the alternating polarization of the cell. At higher frequencies, the applied field changes direction faster than the chemistry responds, so the response is dominated by capacitance from the charge and discharge of the double layer. The time-domain response is not the simplest or most succinct way to interpret these frequency-dependent amplitudes and phase shifts. Instead, we define a quantity called an impedance. Like resistance in a static system, impedance is the ratio of voltage to current. However, it uses the real and imaginary parts of a complex number to represent the relation of both amplitude and phase to the input signal and output response. The mathematical tool that relates the impedance to the time-domain response is a Fourier transform, which represents the frequency components of the oscillating signal. To explain the idea of impedance more fully for a simple case, consider the input voltage as a cosine wave oscillating at an angular frequency ( ω): Then the response is also a cosine wave, but with a phase offset ( φ). Compared to the time shift in the image above, the phase offset is given as \phi = -\omega \,\delta t . The magnitude of the current and its phase offset depend on the physics and chemistry in the cell. Now, let’s consider the resistance from Ohm’s law: This quantity varies in time with the same frequency as the perturbing signal. It equals zero at times when the numerator also equals zero and becomes singular when the denominator equals zero. So unlike the resistance in a DC system, it’s not a very useful quantity! Instead, from Euler’s theorem, let’s express the time-varying quantities as the real parts of complex exponentials, so that: and We denote the coefficients V_0 and I_0\,\exp(i\phi) as quantities \bar{V} and \bar{I}, respectively. These are complex amplitudes that can be understood in terms of the Fourier transformation of the original time-domain sinusoidal signals. They express the distinct amplitudes and phase difference of the voltage and current. Because all of the quantities in the system are oscillating sinusoidally, we understand the physical effects by comparing these complex quantities, rather than the time-domain quantities. To describe the oscillating problem (often called phasor theory), we define a complex analogue of resistance as: This is the impedance of the system and, as the name suggests, it’s the quantity we measure in electrochemical impedance spectroscopy. It’s a complex quantity with a magnitude and phase, representing both resistive and capacitive effects. Resistance contributes the real part of the complex impedance, which is in-phase with the applied voltage, while capacitance contributes the imaginary part of the complex impedance, which is precisely out-of-phase with the applied voltage. EIS specialists look at the impedance in the form of a spectrum, normally with a Nyquist plot. This plots the imaginary component of impedance against the real component, with one data point for every frequency at which the impedance has been measured. Below is an example from a simulation — we’ll discuss how it’s modeled in the next section. Simulated Nyquist plot from an electrochemical impedance spectroscopy experiment. Points toward the top right are at lower frequencies (mHz), while those toward the bottom left are at higher frequencies (>100 Hz). In the figure above, the semicircular region toward the left side shows the coupling between double-layer capacitance and electrode kinetic effects at frequencies faster than the physical process of diffusion. The diagonal “diffusive tail” on the right comes from diffusion effects observed at lower frequencies. EIS experiments are useful because information about many different physical effects can be extracted from a single analysis. There is a quantitative relationship between properties like diffusion coefficients, kinetic rate constants, and dimensions of the features in Nyquist plots. Often, EIS experiments are interpreted using an “equivalent circuit” of resistors and capacitors that yields a similar frequency-dependent impedance to the one shown in the Nyquist plot above. This idea was discussed in my colleague Scott’s blog post on electrochemical resistances and capacitances. When there is a linear relation between the voltage and current, only one frequency will appear in the Fourier transform. This simplifies the analysis significantly. For the simple harmonic interpretation of the experiment in terms of impedance, we need the current response to oscillate at the same frequency as the voltage input. This means that the system must respond linearly. For an electrochemical cell, we can usually accomplish this by ensuring that the applied voltage is small compared to the quantity RT/F — the ratio of the gas constant multiplied by the temperature to the Faraday constant. This is the characteristic “thermal voltage” in electrochemistry and is about 25 mV at normal temperatures. Smaller voltage changes usually induce a linear response, while larger voltage changes cause an appreciably nonlinear response. Of course, with simulation to predict the time-domain current, we can always consider a nonlinear case and perform a Fourier transform numerically to study the effect on the impedance. In practice, the interpretation in terms of impedance illustrated above is best suited to the harmonic assumption. Impedance measurements are therefore often used in a complementary manner with transient techniques, such as amperometry or voltammetry, which are better suited for investigating nonlinear or hysteretic effects. Let’s look at a simple example of the physical theory that underpins these ideas to see how the impedance spectrum relates to the real controlling physics. Electrochemical Impedance Spectroscopy: The Model To model an EIS experiment, we must describe the key underlying physical and chemical effects, which are the electrode kinetics, double-layer capacitance, and diffusion of the electrochemical reactants. In electroanalytical systems, a large quantity of artificially added supporting electrolytes keeps the electric field low so that solution resistance can be neglected. In this case, we can describe the mass transport of chemical species in the system using the diffusion equation (Fick’s laws) with suitable boundary conditions for the electrode kinetics and capacitance. In the COMSOL Multiphysics® software, we use the Electroanalysis interface together with an Electrode Surface boundary feature to describe these equations. For more details about how to set up this model, you can download the Electrochemical Impedance Spectroscopy tutorial example in the Application Library. Model tree for the Electroanalysis interface in an EIS model. Under Transport Properties, we can specify the diffusion coefficients of the redox species under consideration. We at least need the reduced and oxidized species for a single redox couple, such as the common redox couple ferro/ferricyanide, to use as an analytical reference. The Concentration boundary condition defines the fixed bulk concentrations of these species. The Electrode Reaction and Double Layer Capacitance subnodes for the Electrode Surface boundary feature contribute Faradaic and non-Faradaic current, respectively. For the double-layer capacitance, we typically use an empirically measured equivalent capacitance and specify the electrode reaction according to a standard kinetic equation like the Butler-Volmer equation. Note that we’re not referring to equivalent circuit properties at all here. In COMSOL Multiphysics, all of the inputs in the description of the electrochemical problem are physical or chemical quantities, while the output is a Nyquist plot. When analyzing the problem in reverse, we’re able to use an observed Nyquist plot from our experiments to make inferences about the real values of these physical and chemical inputs. In the settings for the Electrode Surface feature, we represent the impedance experiment by applying a Harmonic Perturbation to the cell voltage. Settings for the Electrode Surface boundary feature in an EIS model. Here, the quantity V_app is the applied voltage. The harmonic perturbation is applied with respect to a resting steady voltage (or current) on the cell. In this case, we have set this to a reference value of zero volts. With more advanced models, we might consider using the results of another COMSOL Multiphysics model, one that’s significantly nonlinear for example, to find the resting conditions to which the perturbation is applied. If you’re interested in understanding the mathematics of the harmonic perturbation in greater detail, my colleague Walter discussed them in a previous blog post. When studying lithium-ion batteries, for example, we can perform a time-dependent analysis of the cell’s discharge, studying its charge transport, diffusion and migration of the lithium electrolyte, and the electrode kinetics and diffusion of the intercalated lithium atoms. We can pause this simulation at various times to consider the impedance measured from a rapid perturbation. For further insight into the physics involved, you can read my colleague Tommy’s blog post on modeling electrochemical impedance in a lithium-ion battery. Electrochemical Impedance Spectroscopy: The Simulation App A frequent demand for electrochemical simulations is that they “fit” experimental data in order to determine unknown physical quantities or, more generally, to interpret the data at all. Even for experienced electroanalytical chemists, it can be difficult to intuitively “see” the physics and chemistry in the underlying graphs like the Nyquist plot. However, by simulating the plots under a range of conditions, the influence of different effects on the overall graph is revealed. Simulation is helpful for analyzing EIS, but it can also be time consuming for the experts involved. As was the case with my old research group, these experts can spend more time writing programs and running models to fit data together with experimental researchers than on the science. Wouldn’t it be nice if all electrochemical researchers could load experimental data into a simple interface, simulate impedance spectra for a given physical model and inputs, and even perform automatic parameter fitting? The good news is that we can! With the Application Builder in COMSOL Multiphysics, we can create an easy-to-use EIS app based on an underlying model. As a model can contain any level of physical detail, the app provides direct access to the physical data and isn’t confined to simple equivalent circuits. To highlight this, we have an EIS demo app based on the model available in the Application Library. The app user can set concentrations for electroactive species and tune the diffusion coefficients as well as the electrode kinetic rate constant and double-layer capacitance. After clicking the Compute button, the app generates results that can be visualized through Nyquist and Bode plots. The EIS simulation app in action. As well as enabling physical parameter estimation, this app is very helpful for teaching, since we can quickly change inputs and visualize the results that would occur in the experiment. A natural extension for the app is to import experimental data to the same Nyquist plot for direct comparison. We can also build up the underlying physical model to consider the influence of competing electrochemical reactions or follow-up homogeneous chemistry from the products of an electrochemical reaction. Concluding Thoughts Here, we’ve introduced electrochemical impedance spectroscopy and discussed some methods used to model it. We also saw how a simulation app built from a simple theoretical model can provide greater insight into the relationship between the theory of an electrochemical system and its behavior as observed in an experiment. Further Reading Explore other topics related to electrochemical simulation on the COMSOL Blog Comments (1) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
题目：Resolvent estimates for the Laplacian 报告人：Sanghyuk Lee (首尔大学) 时间：2019年2月28日（星期四）下午2：00-5：00 地点：工商楼105教室摘要： In this talk we are concerned with the resolvent estimates for the Laplacian $\Delta$ in Euclidean spaces: \[ \\| (-\Delta-z)^{-1}f \\|_{L^q}\le C(z)\\| f\\|_{L^p}, \ \ z \in \mathbb C\setminus (0,\infty).\] The resolvent estimates and its variants have applications to various related problems. Among them are uniform Sobolev estimates, unique continuation properties, limiting absorption principles, and eigenvalue bounds for the Schr\"odinger operators. Uniform resolvent estimates for $\Delta$ (i.e., the estimate with $C(z)$ independent of $z$) were first obtained by Kenig, Ruiz and Sogge, who completely characterized the Lebesgue spaces which allow such uniform resolvent estimates. But the precise bound $C(z)$ depending on $z$ has not been considered in general framework which admits all possible $p,q$. In this talk, we present a complete picture of sharp $L^p$--$L^q$ resolvent estimates, which may depend on $z$ and draw a connection to the Bochner-Riesz conjecture. The resolvent estimates in Euclidean space seem to be expected to behave in a simpler way compared with those on manifolds. However, it turns out that, for some $p,q$, the estimates exhibit unexpected behavior which is similar to those on compact Riemannian manifolds. We also obtain the non-uniform sharp resolvent estimates for the fractional Laplacians and discuss applications to related problems. 联系人：王成波老师（wangcbo@zju.edu.cn)
Forgot password? New user? Sign up Existing user? Log in What education websites do you people use other than Brilliant,AOPS,MITOCW,edX,Coursera,udacity?? Note by Eddie The Head 5 years, 5 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: Khan Academy!!! Log in to reply Also Function Space Wow, function space is a gr8 site! Thanks for the recommendation! :) @Happy Melodies – I agree @Happy Melodies me too Are you a member at Function space ? If so , can you explain how the environment there is ? @A Former Brilliant Member – i just got to know about function space i haven't created an ID there yet the 'me too' was for khan academy @Vaibhav Prasad – Ok Also try Quora. It's not an education site, but it really is awesome! Project Euler,Top Coder,uva,Talent Buddy and Code Chef for all you CS junkies. Do you use checkio?? No. I havent heard of it before. What is it? @Thaddeus Abiy – It's a way to learn Python by exploring a gaming world.....The tasks are structured almost like Project Euler .... just google it and you'll find an option to make an account.... I use Codecademy for programming. StackExchange It's a Q & A forum right???I've looked up answers there but never opened an account...time to change that..... Fuctionspace.org Try mathoverflow.com , physicsforidiots.com.P.S. - The latter website is really a good one.It is not a taunt because I also am one. Thanks for your suggestion...My Physics is also not that good......... Well it is much better than mine.{For instance just see my tag levels}.Can you just present some kind of a sequence for studying PCM so that I can plan my work better. I also started a discussion in this respect but it did not get any response .{see my feed you will get the started discussion named 'study sequence'. Thanks for starting this discussion.May God bless you. Glad that you like it! Like it!!I love it. Can't believe that I came across this note after such a long time !!! I'm surprised to see that nobody has mentioned AoPS. It is a great resource for learning and practicing maths problems! I asked for websites other than AoPS because everybody knows about that... :) Ah oops. Didn't read your note carefully, sorry. :P Nobody mentioned wolfram alpha Problem Loading... Note Loading... Set Loading...
It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced. Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit. @Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form. A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it. Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis. Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)? No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet. @MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it. Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow. @QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary. @Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer. @QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits... @QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right. OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ... So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study? > I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a... @MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really. When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.? @tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...) @MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences).
It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced. Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit. @Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form. A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it. Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis. Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)? No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet. @MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it. Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow. @QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary. @Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer. @QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits... @QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right. OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ... So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study? > I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a... @MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really. When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.? @tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...) @MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences).
Let's define a sequence of families of sets inside $[0, 1]$ interval: $$ F_n= \left\{ \left(\frac{j-1}{2^n},\frac{j}{2^n}\right] \mid \forall j \in \{ 1, \cdots, 2^n\} \wedge n \in \mathbb{N} \right\} $$ Now, we know that $[0, 1] = \{0\} \cup(0, 1]$. Now we are goint to define a new set $S_n$ as follow: $$ S_n = \bigcup_{X \in F_n} X $$ For me it's clear that $S_n = (0, 1]$ as long as $n$ is a natural fixed number, and so, it contains all irrational numbers in $(0, 1]$. The question is: does $S_\infty$ contain the irrationals? It is obvious that this set is not empty and contains rationals numbers. Now the questions: Does $S_\infty$ loses the irrationals? Are all rationals contained in the set $S_\infty$? Is $S_\infty = (0,1]$?? Many thanks in advance!!! (RE)$^2$EDIT: Lets define $S_\infty$. Let's start defining $F_\infty$: $$ F_\infty = \{ x \in (0, 1] \; \| \; x \in X_j \wedge X_j \in F_n \; \text{with} \; j \leq n \; \forall n, j \in \mathbb{N} \} $$ This implies $F_\infty$ would become a infinite (but countable?) partition of $(0,1]$ in disjoint subsets containing just a point. So, $S_\infty$ would just be: $$ S_n = \bigcup_{X \in F_\infty} X $$ Thanks @Zev Chonoles and @Asaf Karagila for the useful comments!
Let $f(x), g(x) \in \mathbb Z[x]$ be polynomials of positive degrees $r$ and $s$ respectively, and let $\operatorname{Res}(f,g)$ denote their resultant. Further, let $H(f)$ denote the naive height of a polynomial $f$; that is, $H(f)$ is equal to the maximum of absolute values of its coefficients. Now let us fix positive integers $r, s, H_0, R_0$. Then there are $(2H_0)^2(2H_0+1)^{r}(2H_0+1)^{s}$ pairs of integer polynomials $(f,g)$ such that $\deg f = r$, $\deg g = s$, and $\max\{H(f),H(g)\}\leq H_0$. My question is: what proportion of those pairs of polynomials satisfy the inequality $$\|\operatorname{Res}(f,g)\|\geq R_0?$$ I would like to derive a non-trivial lower bound on this quantity in terms of $r, s, H_0, R_0$. Was this question studied before? I would be thankful for any references. Also, if we set $g(x)$ to be equal to the derivative of $f(x)$, then this question essentially reduces to the question about the magnitude of the absolute value of a discriminant $D(f)$ of a polynomial $f(x)$. Then the question can be rephrased as follows: what proportion of polynomials $f(x)$ of degree $r$ and height at most $H_0$ satisfy the inequality $\|D(f)\| \geq R_0$? Perhaps, there are references to this special case? P.S. Note that it is a consequence of Hadamard's inequality that $$\|\operatorname{Res}(f,g)\| \leq (r+1)^{s/2}(s+1)^{r/2}H(f)^sH(g)^r,$$ so we need to demand $R_0 \leq (r+1)^{s/2}(s+1)^{r/2}H_0^{s+r}$, for otherwise the answer would trivially be zero.
$\newcommand\Tr{\text{Tr}}$One of the intriguing lessons we have learned in the past half-century of set-theoretic developments is that there is a surprisingly robust connection between infinitary game theory and fundamental set-theoretic principles, including large cardinals. Assertions about the existence of strategies in infinite games often turn out to have an unexpected set-theoretic power. In this post, I should like to give another specific example of this, which Thomas Johnstone and I hit upon yesterday in an enjoyable day of mathematics. Specifically, I’d like to prove that if we generalize the open-game concept from sets to classes, then assuming consistency, ZFC cannot prove that every definable open class game is determined, and indeed, over Gödel-Bernays set theory GBC the principle of open determinacy (and even just clopen determinacy) implies Con(ZFC) and much more. To review a little, we are talking about games of perfect information, where two players alternately play elements from an allowed space $X$ of possible moves, and together they build an infinite sequence $\vec x=\langle x_0,x_1,x_2,\ldots\rangle$ in $X^\omega$, which is the resulting play of this particular instance of the game. We have a fixed collection of plays $A\subset X^\omega$ that is used to determine the winner, namely, the first player wins this particular instance of the game if the resulting play $\vec x$ is in $A$, and otherwise the second player wins. A strategy for a player is a function $\sigma:X^{<\omega}\to X$, which tells a player how to move next, given a finite position in the game. Such a strategy is winning for that player, if he or she always wins by following the instructions, no matter how the opponent plays. The game is determined, if one of the players has a winning strategy. It is a remarkable but elementary fact that if the winning condition $A$ is an open set, then the game is determined. One can prove this by using the theory of ordinal game values, and my article on transfinite game values in infinite chess contains an accessible introduction to the theory of game values. Basically, one defines that a position has game value zero (for player I, say), if the game has already been won at that stage, in the sense that every extension of that position is in the winning payoff set $A$. A position with player I to play has value $\alpha+1$, if player I can move to a position with value $\alpha$, and $\alpha$ is minimal. The value of a position with player II to play is the supremum of the values of all the positions that he or she might reach in one move, provided that those positions have a value. The point now is that if a position has a value, then player I can play so as strictly to decrease the value, and player II cannot play so as to increase it. So if a position has a value, then player I has a winning strategy, which is the value-reducing strategy. Conversely, if a position does not have a value, then player II can maintain that fact, and player I cannot play so as to give it a value; thus, in this case player II has a winning strategy, the value-maintaining strategy. Thus, we have proved the Gale-Stewart theorem: every open game is determined. That proof relied on the space of moves $X$ being a set, since we took a supremum over the values of the possible moves, and if $X$ were a proper class, we couldn’t be sure to stay within the class of ordinals and the recursive procedure might break down. What I’d like to do is to consider more seriously the case where $X$ is a proper class. Similarly, we allow the payoff collection $A$ to be a proper class, and the strategies $\sigma:X^{<\omega}\to X$ are also proper classes. Can we still prove the Gale-Steward theorem for proper classes? The answer is no, unless we add set-theoretic strength. Indeed, even clopen determinacy has set-theoretic strength. Theorem. (GBC) Clopen determinacy for proper classes implies Con(ZFC) and much more. Specifically, there is a clopen game, such the existence of a winning strategy is equivalent to the existence of a satisfaction class for first-order truth. Proof. Let me first describe a certain open game, the truth-telling game, with those features, and I shall later modify it to a clopen game. The truth-telling game will have two players, which I call the challenger and the truth-teller. At any point in the game, the challenger plays by making an inquiry about a particular set-theoretic formula $\varphi(\vec a)$ with parameters. The truth-teller must reply to the inquiry by stating either true or false, and in the case that the formula $\varphi$ is an existential assertion $\exists x\,\psi(x,\vec a)$ declared to be true, then the truth teller must additionally identify a particular witness $b$ and assert that $\psi(b,\vec a)$ is true. So a play of the game consists of a sequence of such inquires and replies. The truth-teller wins a play of the game, provided that she never violates the recursive Tarskian truth conditions. Thus, faced with an atomic formula, she must state true or false in accordance with the actual truth or falsity of that atomic formula, and similarly, she must say true to $\varphi\wedge\psi$ just in case she said true to both $\varphi$ and $\psi$ separately (if those formulas had been issued by the challeger), and she must state opposite truth values for $\varphi$ and $\neg\varphi$, if both are issued as challenges. This is an open game, since the challenger will win, if at all, at a finite stage of play, when the violation of the Tarskian truth conditions is first exhibited. Lemma 1. The truth-teller has a winning strategy in the truth-telling game if and only if there is a satisfaction class for first-order truth. Proof. Clearly, if there is a satisfaction class for first-order truth, then the truth-teller has a winning strategy, which is simply to answer all questions about truth by consulting the satisfaction class. Since that class obeys the Tarskian conditions, she will win the game, no matter which challenges are issued. Conversely, suppose that the truth-teller has a winning strategy $\tau$ in the game. I claim that we may use $\tau$ to build a satisfaction class for first-order truth. Specifically, let $T$ be the collection of formulas $\varphi(\vec a)$ that are asserted to be true by $\tau$ in any play according to $\tau$. I claim that $T$ is a satisfaction class. We may begin by noting that since $T$ must correctly state the truth of all atomic formulas, it follows that the particular answers that $\tau$ gives on the atomic formulas does not depend on the order of the challenges issued by the challenger. Now, we argue by induction on formulas that the truth values issued by $\tau$ does not depend on the order of the challenges. For example, if all plays in which $\varphi(\vec a)$ is issued as a challenge come out true, then all plays in which $\neg\varphi(\vec a)$ is challenged will result in false, or else we would have a play in which $\tau$ would violate the Tarskian truth conditions. Similarly, if $\varphi$ and $\psi$ always come out the same way, then so does $\varphi\wedge\psi$. We don’t claim that $\tau$ must always issue the same witness $b$ for an existential $\exists x\,\psi(x,\vec a)$, but if it ever says true to this statement, then it will provide some witness $b$, and for that statement $\psi(b,\vec a)$, the truth value stated by $\tau$ is independent of the order of play by the challenger, by induction. Thus, by induction on formulas, the answers provided by the truth-teller strategy $\tau$ gives us a satisfaction predicate for first-order truth. QED Lemma 2. The challenger has no winning strategy in the truth-telling game. Proof. Suppose that $F$ is a strategy for the challenger. So $F$ is a proper class function that directs the challenger to issue certain challenges, given the finite sequence of previous challenges and truth-telling answers. By the reflection theorem, there is a closed unbounded proper class of cardinals $\theta$, such that $F”V_\theta\subset V_\theta$. That is, $V_\theta$ is closed under $F$, in the sense that if all previous challenges and responses come from $V_\theta$, then the next challenge will also come from $V_\theta$. Since $\langle V_\theta,\in\rangle$ is a set, we have a satisfaction predicate on it. Consider the play, where the truth-teller replies to all inquires by consulting truth in $V_\theta$, rather than truth in $V$. The point is that if the challenger follows $\tau$, then all the inquiries will involve only parameters $\vec a$ in $V_\theta$, provided that the truth-teller also always gives witnesses in $V_\theta$, which in this particular play will be the case. Since the satisfaction predicate on $V_\theta$ does satisfy the Tarskian truth conditions, it follows that the truth-teller will win this instance of the game, and so $F$ is not a winning strategy for the challenger. QED Thus, if open determinacy holds for classes, then there is a satisfaction predicate for first-order truth. This implies Con(ZFC) for reasons I explained on my post KM implies Con(ZFC) and much more, by appealing to the fact that we have the collection axiom relative to the class for the satisfaction predicate itself, and this is enough to verify that the nonstandard instances of collection also must be declared true in the satisfaction predicate. But so far, we only have an open game, rather than a clopen game, since the truth-teller wins only by playing the game out for infinitely many steps. So let me describe how to modify the game to be clopen. Specifically, consider the version of the truth-telling game, where the challenger must also state on each move a specific ordinal $\alpha_n$, which descend during play $\alpha_0>\alpha_1>\cdots>\alpha_n$. If the challenger gets to $0$, then the truth-teller is declared the winner. For this modified game, the winner is known in finitely many moves, because either the truth-teller violates the Tarskian conditions or the challenger hits zero. So this is a clopen game. Since we made the game harder for the challenger, it follows that the challenger still can have no winning strategy. One can modify the proof of lemma 1 to say that if $\tau$ is a winning strategy for the truth teller, then the truth assertions made by $\tau$ in response to all plays with sufficiently large ordinals for the challenger all agree with one another independently of the order of the formulas issued by the challenger. Thus, there is a truth-telling strategy just in case there is a satisfaction class for first-order truth. So clopen determinacy for class games implies the existence of a satisfaction class for first-order truth, and this implies Con(ZFC) and much more. QED One may easily modify the game by allowing a fixed class parameter $B$, so that clopen determinacy implies that there is a satisfaction class relative to truth in $\langle V,\in,B\rangle$. Furthermore, we may also get iterated truth predicates. Specifically, consider the iterated truth-telling game, which in addition to the usual language of set theory, we have a hierarchy of predicates $\Tr_\alpha$ for any ordinal $\alpha$. We now allow the challenger to ask about formulas in this expanded language, and the truth teller is required to obey not only the usual Tarskian recursive truth conditions, but also the requirements that $\Tr_\alpha(\varphi(\vec a))$ is declared true just in case $\varphi(\vec a)$ uses only truth predicates $\Tr_\beta$ for $\beta<\alpha$ and also $\varphi(\vec a)$ is declared true (if this challenge was issued). The main arguments as above generalize easily to show that the challenger cannot have a winning strategy in this iterated truth-telling game, and the truth-teller has a strategy just in case there is a satisfaction predicate for truth-about-truth iterated through the ordinals. Thus, the principle of open determinacy for proper class games implies Con(Con(ZFC)) and $\text{Con}^\alpha(\text{ZFC})$ and so on. Let me finish by mentioning that Kelley-Morse set theory is able to prove open determinacy for proper class games in much the same manner as we proved the Gale-Stewart theorem above, using well-ordered class meta-ordinals, rather than merely set ordinals, as well as in other ways. If there is interest, I can make a further post about that, so just ask in the comments!
The primary way ECNs determine if a liquidity taker's flow is 'toxic' or not is by looking at aftermath charts. The aftermath chart shows the average mark-to-market profit of trades done by the liquidity taker as a function of either time or number of top-of-book updates (optionally broken down by currency pair). The trade profit is usually viewed from the ... Art markets typically have huge transaction costs of the order of 10%, caused by buyers premium and auction fees. Therefore long holding periods are unavoidable, with long-term returns somewhere between those of bonds and equities. By its very nature, art is not easily replicated so arbitrage or derivatives are out. The rationality of agents (aka collectors) ... Mispricing can only be measured relative to some asset pricing modelFama (1970) famously defined an efficient market as, "a market in which prices always 'fully reflect' all available information."A perhaps less widely understood point of Fama is that any test of market efficiency is a joint test of: (1) market efficiency and (2) an asset pricing model! ... The only way you can use limit orders to provide liquidity is to post prices that are the same as the prices of the limit orders, and then you will not be earning any spread. In other words, what you are asking is not possible, since to earn spread you would have to quote a bid that is lower than the price of your client's limit buy order, or an offer that ... Whereas when you marketmake on a last-look basis:- You, the marketmaker, are sending indicative prices to the ECN- The ECN sends orders to you and is at risk (since you have the option to reject, hopefully rarely)When you marketmake on a no-last-look (NLL) basis:- the ECN is sending indicative prices- You, the marketmaker, send orders to the ECN and ... I think one should look at the problem from two different angles to get an answer to this.Firstly, you can look (as you said you did) look at $\hat{\epsilon}$ in terms of a disturbance like you said, meaning the returns $R_{it}$ are depending linearly on the $R_{mt}$ - the market or factor returns. Then you can figure there is some regression involved an ... I would say the financial- and the art market is very different, only the roots of the market / auctions is the same.As the art market is unique and very illiquid, alot of the strategies from the modern financial market simply does not apply.I have been building (and still maintains) a toolbox of models, which mostly try to find trends based on multiple ... The general effect of quantitative analysis of the markets is to enforce randomness.Suppose a strategic quant finds a predictable pattern where a stock always rises on Tuesdays. His institution will commence buying the stock every Monday, and selling on Tuesday. The trading itself pushes the stock price up on Monday and down on Tuesday (in general), so if ... For Q1, the function $a(t)$ is the instantaneous correlation. The form given by (2) is basically the Cholesky decomposition. Of course, you may directly show, uisng Levy's characterization, that$$\widetilde{W}(t) = \int_0^t\bigg[\frac{1}{\sqrt{1-\|\|a(t)\|\|^2}} dZ(t) -\frac{a(t)^T}{\sqrt{1-\|\|a(t)\|\|^2}} dW^B(t) \bigg]$$is a standard scalar Brownian motion ... Q1: $$(1)\rightarrow(2)$$(1): $a(t)$ is the instantaneous correlation of $\rho(Z_t,W_t)$ because:$$\rho(dZ_t,dW_t)=\dfrac{Cov(dZ_t,dW_t)}{\sigma_{dZ_t}\sigma_{dW_t}}=\dfrac{E(dZ_t\cdot dW_t)}{\sqrt{dt} \sqrt{dt}}=\dfrac{\langle dZ_t, dW_t\rangle}{t}=a(t)$$$\Rightarrow$ (2) holds as following, in the 1-dim case:$dZ_t\sim N(0,dt),$$dW_t,\tilde{dW_t}\... I have asked myself the very same question when I first read the book. As far as I can tell, the "scalability" condition is only imposed for technical reasons. It simplifies the subsequent proof of the Fundemental Theorem of Asset Pricing in constrained markets.There are several papers that have shown that the theorem is valid for conic constraints. ... Calibrating to swaption prices would give you the right volatilities for your model, but you have to use the floating notes (or similar instruments, as swaps) in order to get the right drifts. In any case, your model have to be able to exactly replicate the floating notes prices in order to be considered a valid model, and you can feel comfortable to use it ... For LMM I thing the Rebonato's book 2002 is a good reference. He has explained the condition of vol quotation that allow existence of calibration solution.LMM parameters and inputs are quite complexe, calibrator not work maybe caused by your implementation's bugs but not only data input. I think it is better if you calibrate virtually before true market ... it's difficult to say that they are not popular. Some people definitely use them for live pricing. I'd say the real question is "why are they not popular in the academic literature"?One answer would simply be that most the questions that arise in their use are ones of fiddliness which do not make good papers. You should make your borrow cost sufficient to dissuade unlimited short selling. In practice, each short would require you to borrow shares from your broker. This is usually handled when computing transaction cost. You should account for this in your trading algorithm or in the factor model itself. A simple method would make shorts some N% more expensive ... Idiosyncratic volatility is NOT included in the regressors, so it should not be and actually cannot be part of your matrix X. Idiosyncratic volatility is the volatility (of Y) your matrix X (explanatory variables) cannot explain (i.e. remaining unexplained part), so it is the error term of your regression equation.Just compute the standard deviation of ... To answer your question:I mean in a theoretical sense: If we have a particular market model (which I guess we may assume is complete or frictionless if need be) where shorting and fractional purchases are allowed, does presence of arbitrage necessarily make all kinds of derivatives have zero value?The answer is no. See example below. Went over your ... With a resting order the market maker's client takes (buys) at the offer, or gives (sells) at the bid. The market maker prices the deal that way as with any other order type. A resting order is simply another type of order and the client pays the market maker's spread like with any order type. I am not sure I agree with @CPL593H if you're a market maker and ... To answer your answer: Suppose you are the holder of the open contract. You hedge it by executing a vanilla forward at 1.1679 for date 92. You now have an arbitrage, for if the fx forward for one of the dates 88 to 91 becomes higher than that for date 92, you can switch the hedge to that other date, This means that the true price of your open contract ... Yes I used one in the early 2000s. At the time, US interest rates were quite high (5 or 6pct) and the market skew was such that -100bp receivers were more expensive than +100 payers. The lognormal model is very inappropriate for this skew regime , but the normal model is much closer, having symmetric skew. As a byproduct of this we noticed that the model ... This paper Limit Order Strategic Placement with Adverse Selection Risk and the Role of Latency (by L and Mounjid) is probably exactly what you expect.It is explained how (given a model of orderbook dynamics close to the Queue reactive model) to decide to cancel or not a limit order with respect of the state of the bid and ask queue sizes and your position ... This reference price is also sometimes called intrinsic price. One of the simplest ways to improve it in regards to the mid-price (assuming you have the depth data) is the following:define a parameter: the size of a hypothetical market order. Let's say it's about the typical sum of first 3-10 order book levels of the instrument;execute a Buy order with ... In practice, most derivatives traded on Fed Funds rates are linear(i.e. Forwards) rather than non-linear (options and exotics). As such, there has not been a strong case for precise modeling of the full distribution of a Fed Funds rate for a particular day. In contrast , there is a large market for derivatives on 3month USD Libor , which is less sensitive ... As a starting point I would make the assumption that your new orders are negligibly small, in order that their market impact does not affect the trajectory of the price. This will provide a reasonable way to test strategies that do not possess any forward looking or snooping data.When you are in the position that your actions are believed to impact the ... The first response is wrong, you can do this providing the venue is operating a central limit order book and you can match with their client flow. There are several venues or liquidity pools that allow this. If your limit order is at the top of book and someone wants to trade at that price then you will be filled at your price. Making money purely from the ... What if you write$$P[R_{n+1} = d\|F_n] = 1 - P[R_{n+1} = u\|F_n] ?$$Let us write $P(u) = P[R_{n+1} = u\|F_n]$Then the part to show is$$u \bar{S}_n P(u) + d \bar{S}_n (1-P(u))$$and this$$\bar{S}_n \left(d +(u-d)P(u) \right),$$where we just expanded terms and then extracted the coefficients. I am also interested in the answer to this question, and would like to expand a little bit on it as well.First of all, let me add some value in terms of a partial answer:There are restrictions on when short selling is allowed. According to the SEC, and the "Alternative Uptick Rule" short selling is not allowed on "a stock that has dropped more than 10 ...
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 ...
marty cohen Martin Cohen is a retired computer programmer who loves dancing (favorites are West Coast Swing, Waltz, Foxtrot, and Salsa), writing (but not revising) poems, and solving math problems (that's why I'm here). He is currently trying to learn improv at the Westside Comedy Theater (westsidecomedy.com) in Santa Monica, CA. He can also be reached at mjcohen@acm.org. Member for 8 years, 3 months 67 profile views Last seen Oct 7 at 23:28 Communities (7) Top network posts 92 What is the most unusual proof you know that $\sqrt{2}$ is irrational? 81 Why is $\frac{987654321}{123456789} = 8.0000000729?!$ 57 What proportion of positive integers have two factors that differ by 1? 53 Are the sums $\sum_{n=1}^{\infty} \frac{1}{(n!)^k}$ transcendental? 45 Show that this sum is an integer. 39 Is there a simple, constructive, 1-1 mapping between the reals and the irrationals? 37 What is the most surprising result that you have personally discovered? View more network posts →
There are answers in -as I think- in mathoverflow for "what are physical interpretations of zeta at negative arguments" (or the like, I don't have the actual words in mind). Just browse a bit through MO or MSE using that terms as filter. 0) Need of a new concept - and a basic requirement: Make the general term explicite For the interpretation and assignment of a meaningful value in your series you must define, how the general term, say $a_k$, shall be constructed, i.e. how the value of $a_k$ is dependend on its index $k$. This is a requirement which shall be demanded in all discussions of divergent series (and should not been forgot when such series, written only in their "obvious" numerical forms, are considered at all). In this example the value of the general term, say $a_k$, reduces simply to the index $k$ if taken from $k=1 \ldots \infty $ and we have actually the formal statement $S = \sum_{k=1}^\infty k$. But because the summing procedure is obviously divergent if we want assign something meaningful to it. we must introduce some new concept 1) Geometric series A useful idea here was (for instance already by L. Euler) that we can cofactor a variable $x$ at the terms $a_k$ such that for some $x$ the series becomes convergent and evaluatable to some finite value and see what happens with the sequence of resulting sum-values, when the expressions of $x$ approximate $1$ . For instance, we could redefine your sum as$$ S(x) = 1 + 2x + 3x^2 + 4x^3 + ... \\ S = \lim_{x \to 1^-} S(x) $$because there are some $x$ (actually it is important, that there exists a of $x$) where this is convergent, for instance for $x=0 \ldots \frac12$. We can then even observe that with that definition the sum $S(x)$ has a closed form $$S(x) = {d\over dx}{1\over 1-x } = {1\over (1-x)^2 }$$This is already a very nice representation, because it allows now even negative $x$ which result in finite values for the alternating sum, even if divergent, because it is accepted to assign that fraction's value also to its formal power series-expression even if the latter is divergent - except, well, except if $x=1$. continuous interval But what we want here is actually just the latter, so we have not yet a satisfying answer. 2) Dirichlet series Another idea is to apply a function of $x$ to the exponent of the terms $a_k$ and then let $x$ go to $1$ . Thus we consider the notation$$ S(x) = 1^x + 2^x + 3^x +4^x + ... $$ which is convergent for a continuous interval of $x$, for instance of $x = -\infty \ldots -2$ (and even a bit more). But still, for $x=1$ we get no obvious answer. But interestingly, for the whole interval of convergence we have also the functional relation$$ \begin{matrix} \text{let } &A(x) &=& 1^x - 2^x + 3^x - 4^x + \ldots \\ \text{then } &S(x) &=& A(x) + 2 \cdot 2^x \cdot S(x) \\ \text{and} & S(x) &=& A(x)/(1-2 \cdot 2^x) \end{matrix}$$And now we can assign a meaningful value to $S = \lim_{x \to 1^- }$. Either by evaluating $A(1)$ as conditionally converging series in the given form, or by the above geometric-series interpretation and its derivative at $x=-1$ where both ways of evaluation give the same rational expression $A(1)=\frac 14$. After that nothing more divergent is there and we get$$ S = \lim_{x \to 1^-} S(x) = \frac 14 / (1-2\cdot 2^x) = \frac 14 / (1-4)= - \frac 1{12}$$having now a proposal for a meaningful assignment of a finite value for the infinite divergent series. 3) Caveat Of course that value must conform with all and any cases where such series occur in mathematics, and one example of a contradiction with known results taken from the mathematic would invalidate that procedure! without divergent summation Moreover, we would hope that even in the physical world, where we model some observations with that series, such an assignement of values to a divergent series would hold. Interestingly such observations exist in the real world and it seems that the whole process and also the final value meets the modeling of that observations. (Examples are given to the according questions either in MSE here or in mathoverflow (I've just found (1)(2)), you can do a search for the important words "zeta" and "at negative arguments" or similar)
Taylor series A Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point $$f(a) \approx \sum\limits_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(x-a)^n}$$ where $f^{n}(a)$ donates the $n^{th}$ derivative of $f$ evaluated at the point $a$. And here’s a very intuitive example: The exponential function $e^x$ (in blue), and the sum of the first $n + 1$ terms of its Taylor series at $0$ (in red). Newton’s method Overview In calculus, Newton’s method is an iterative method for finding the roots of a differentiable function $f$. In machine learning, we apply Newton’s method to the derivative $f’$ of the cost function $f$. One-dimension version In the one-dimensional problem, Newton’s method attempts to construct a sequence ${x_n}$ from an initial guess $x_0$ that converges towards some value $\hat{x}$ satisfying $f'(\hat{x})=0$. In another word, we are trying to find a stationary point of $f$. Consider the second order Taylor expansion $f_T(x)$ of $f$ around $x_n$ is $$f_T(x)=f_T(x_n+\Delta x) \approx f(x_n) + f'(x_n) \Delta x + \frac{1}{2}f”(x_n) \Delta x^2$$ So now, we are trying to find a $\Delta x$ that sets the derivative of this last expression with respect to $\Delta x$ equal to zero, which means $$\frac{\rm{d}}{\rm{d} \Delta x}(f(x_n) + f'(x_n) \Delta x + \frac{1}{2}f”(x_n) \Delta x^2) = f'(x_n)+f”(x_n)\Delta x = 0$$ Apparently, $\Delta x = -\frac{f'(x_n)}{f”(x_n)}$ is the solution. So it can be hoped that $x_{n+1} = x_n+\Delta x = x_n – \frac{f'(x_n)}{f”(x_n)}$ will be closer to a stationary point $\hat{x}$. High dimensional version Still consider the second order Taylor expansion $f_T(x)$ of $f$ around $x_n$ is $$f_T(x)=f_T(x_n+\Delta x) \approx f(x_n) + \Delta x^{\mathsf{T}} \nabla f(x_n) + \frac{1}{2} {\rm H} f(x_n) \Delta x (\Delta x)^{\mathsf{T}}$$ where ${\rm H} f(x)$ donates the Hessian matrix of $f(x)$ and $\nabla f(x)$ donates the gradient. (See more about Taylor expansion at https://en.wikipedia.org/wiki/Taylor_series#Taylor_series_in_several_variables) So $\Delta x = – [{\rm H}f(x_n)]^{-1}\nabla f(x_n)$ should be a good choice. Limitation As is known to us all, the time complexity to get $A^{-1}$ given $A$ is $O(n^3)$ when $A$ is a $n \times n$ matrix. So when the data set is of too many dimensions, the algorithm will work quite slow. Advantage The reason steepest descent goes wrong is that the gradient for one weight gets messed up by simultaneous changes to all the other weights. And the Hessian matrix determines the sizes of these interactions so that Newton’s method minimize these interactions as much as possible. References
I am looking at the following image (from http://www.schoolphysics.co.uk/age16-19/Atomic%20physics/X%20rays/text/X_ray_spectra/index.html): and I want to know, of the two curves, which one represents the element with the higher atomic number. That is, I understand that X-rays are scattered and that the peaks are characteristic of a material. What I am less clear on i why one curve should be above another relative to their atomic numbers (I am going to assume the cutoff voltage on the left is identical even though in this picture it isn't). The reason I am confused is that I am told that that $\frac{1}{\sqrt{\lambda}} \propto Z$ which is all very well, but on this graph it would seem to indicate that the upper line is a heavier (higher Z) material. Is hat the case, or is intensity actually proportional to $\frac{1}{\sqrt{\lambda}}$ as well? I would have thought the lower curve would be a higher Z since it usually take less energy to knock electrons off of elements further up the periodic table. Anyhow, what clarification might be offered is appreciated.

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