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I'm trying to understand this paper on Maximum Entropy by Jaynes, and am stuck on something which should be rather simple.
We're attempting to maximize the entropy $-\sum_i p_i \ln(p_i)$ subject to the constraints $\langle f \rangle = \sum_i p_i f_i$ and $\sum_i p_i = 1$. Using Lagrange multipliers, we have:
$$L = -\sum_i p_i \ln(p_i) + \lambda \left( \sum_i p_i f_i - \langle f \rangle \right) + \mu \left( \sum_i p_i - 1 \right)$$
$$\frac{\delta L}{\delta p_i} = - \ln(p_i) - 1 + \lambda f_i + \mu = 0$$
So $p_i = e^{-1 + \mu + \lambda f_i}$. Plugging this back into the definition for $\langle f \rangle$:
$$\langle f \rangle = \sum_i p_i f_i = \sum_i f_i e^{-1 + \mu + \lambda f_i} = \sum_i f_i e^{-1 + \mu + \lambda f_i}$$
Equation (2-5) in Jaynes' paper states (ignoring sign conventions) that $\langle f \rangle = \frac{\partial}{\partial \lambda} \ln{Z}$, where $Z$ is the partition function, probably defined as $Z = \sum_i e^{-1 + \mu + \lambda f_i}$, or just simply $Z = \sum_i e^{\lambda f_i}$ if we ignore the normalisation. I tried to check this:
$$\langle f \rangle = \frac{1}{Z} \frac{\partial}{\partial \lambda} Z = \frac{1}{Z} \frac{\partial}{\partial \lambda} \sum_i e^{\lambda f_i} = \frac{1}{Z} \sum_i f_i e^{\lambda f_i} = \frac{1}{Z} \langle f \rangle$$
Which I doubt is correct. I'm probably making some obvious mistake, but for the life of me I can't find it. |
I will give an answer to my own question by showing the inverse case (as claimed by user my2cts): Given that there is a 4-vector $j$ that satisfies $\partial_{\mu} j^{\mu}$, one can show that the integral of the component of $j$, which is perpendicular to a spacelike 3-dimensional hypersurface, performed over this surface, will always be the same. The Proof only holds if all compononents of $j$ become 0 when we spatially go to infinity.
Use 2 sets of base vectors ($\{ e_0, e_1, e_2, e_3 \}$ and $\{ \tilde{e_0}, \tilde{e_1}, \tilde{e_2}, \tilde{e_3} \}$, that are connected by an active lorentz transformation $\Lambda^{-1}$ with $\Lambda^{-1}(e_{\mu}) = \tilde{e_{\mu}}$.
Both sets of base vectors give rise to 2 spatial 3 dimensional hypersurfaces, $S_1$ and $S_2$, which will be spanned by $\{e_1, e_2, e_3 \}$ and $\{, \tilde{e_1}, \tilde{e_2}, \tilde{e_3} \}$. Imagine hypersurface $S_1$ to be anchored at the point located at (using the first base) the coordinates (0, 0, 0, 0), while the 2nd one, $S_2$, is supposed to be located at (also using the first base) (T, 0, 0, 0). Now look at a 4-dimensional Volume, which on its "top" and its "bottom" side is bordered by $S_1$ and $S_2$, while its Sides" lie outside the region where $j\neq0$. The picture shows how it's meant to be, the black linings show the area where $j$ deviates from 0.
Now (indexed quantities like $j^{\mu}$ refer to components of the base-indepedent objects in the first base. $\omega$ is supposed to be the 4-dimensional volume form.
\begin{align}\int_V 0 \omega = \int_{V} 0 dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 = 0 = \int_{V} \partial_{\mu} j^{\mu} dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 \end{align}
Because of the (general) stokes-theorem for integration of differential forms, this is the same as
\begin{align}0=\int_{\partial V} j^0 dx^1 \wedge dx^2 \wedge dx^3 + j^1 dx^0 \wedge dx^2 \wedge dx^3 + j^2 dx^0 \wedge dx^1 \wedge dx^3 + j^3 dx^0 \wedge dx^1 \wedge dx^2\end{align}
We note at this point that $f = \omega(j, \cdot, \cdot, \cdot)= j^0 dx^1 \wedge dx^2 \wedge dx^3 + j^1 dx^0 \wedge dx^2 \wedge dx^3 + j^2 dx^0 \wedge dx^1 \wedge dx^3 + j^3 dx^0 \wedge dx^1 \wedge dx^2$ gives a 3-form, whiches exterior derivative $d f = \partial_{\mu} j^{\mu} dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$ is zero.
Because the regions at the sides don't contribute to the integral anyways, the integration must only be performed over the hyper surfaces $S_1$ and $S_2$. So we know:
\begin{align}\int_{S_1} f = \int_{S_2} f\end{align}
Let us now calculate those integrals:
\begin{align}\int_{S_1} f = \int dx dy dz f(e_1, e_2, e_3) = \int dx dy dz j^0 = c Q_1\end{align}
It is important to note here that the components of f, with respect to the base ($\tilde{dx}^1 \wedge \tilde[{dx}^2 \wedge \tilde{dx}^3$) transform like the components of a vector (which is intuitive, since the $j^{\mu}$ ARE the components of a vector!). So we have \begin{align}f = \omega(j, \dot, \dot, \dot)= \tilde{j}^0\tilde{ dx}^1 \wedge \tilde{dx}^2 \wedge \tilde{dx}^3 + \tilde{j}^1 \tilde{dx}^0 \wedge \tilde{dx}^2 \wedge \tilde{dx}^3 + \tilde{j}^2 \tilde{dx}^0 \wedge \tilde{dx}^1 \wedge tilde{dx}^3 +\tilde{ j}^3 \tilde{dx}^0 \wedge \tilde{dx}^1 \wedge \tilde{dx}^2\end{align}
Then the second integral computes to: \begin{align}\int_{S_2} f = \int dx dy dz f(\tilde{e_1},\tilde{ e_2}, \tilde{e_3}) = \int dx dy dz \tilde{j}^0 = c Q_2\end{align}
The total charge contained in the two hyper-surfaces is the same! Since charge conservation holds, this is not only true for this special choice of hyper-surfaces, but as well for $S_2$ to be anchored at any point. To give an answer to my own question, the invariance of charge holds for any spacelike hypersurface that covers the region with $j \neq 0$. This hypersurface can (in the right frame of reference) be regarded as a spatial volume. |
The Birkhoff–von Neumann theorem states that a doubly stochastic matrix (a matrix with non-negative entries in which rows and columns sum to 1) can be written as a convex combination of permutation matrices (0/1 matrices which contain precisely one 1 in each row and column). This immediately implies your result.
If you don't want to assume this theorem, you can use Hall's theorem directly. Given your table $A$, define a bipartite graph by having $n$ row-vertices, $n$ column-vertices, and connecting row $i$ and column $j$ if the $(i,j)$th entry is non-zero. Your goal is to find a perfect matching in the graph.
According to Hall's theorem, the graph contains a perfect matching if for all subsets $S$ of row vertices, the number of neighbors of $S$ is at least $|S|$. That is, given a set $S$ of rows, we need to show that the set$$T = \{ j : A_{ij} \neq 0 \text{ for some $i \in S$}\}$$contains at least $|S|$ columns. Indeed, the sum of the entries in the rows in $S$ is exactly $|S|$, and the sum of the entries in the columns in $T$ is exactly $|T|$. Therefore the sum of entries in $S \times T$ is at most $|T|$. However, by definition this sum must equal $|S|$ (since all other entries in the rows in $S$ are zero), and so $|S| \leq |T|$, which is what we wanted.
This argument shows that you can use standard algorithms for finding maximum matchings in bipartite graphs to actually find your $n$ entries.
We are now at an excellent position to prove the Birkhoff–von Neumann theorem. The proof is by induction on the number of non-zero entries. The argument above shows that there are at least $n$ non-zero entries, and in that case it's not hard to see that the matrix must be a permutation matrix, completing the base case.
Now consider an arbitrary doubly stochastic $A$ with more than $n$ non-zero entries, and find a matching $(i,\pi(i))$ in $A$. This matching corresponds to sum permutation matrix $P$. Define $\alpha = \min_i A(i,\pi(i)) > 0$. The matrix $B = \frac{A - \alpha P}{1 - \alpha}$ is also doubly stochastic (since subtracting $\alpha P$ removes exactly $\alpha$ from each row and column), with at least one more zero entry. By induction, $B$ is a convex combination $\sum_t \beta_t P_t$ of permutation matrices. Hence so is $A$: $$A = \alpha P + (1-\alpha) \sum_t \beta_t P_t.$$
The Birkhoff–von Neumann theorem can be stated in many different ways. It states that the set of doubly stochastic matrices is the convex hull of the permutation matrices. It also states (with some more work) that the vertices in the polytope of doubly stochastic matrices are the permutation matrices. |
What is Fluid Flow?
Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.
For example, if you are pouring water from a mug, the velocity of water is very high over the lip, moderately high approaching the lip, and very low at the bottom of the mug. The unbalanced force is gravity, and the flow continues as long as the water is available and the mug is tilted.
Types of Fluid
Following are the types of fluid:
Ideal fluid Real fluid Newtonian fluid Non-Newtonian fluid Ideal plastic fluid Incompressible fluid Compressible fluid Ideal fluid:A fluid is said to be ideal when it cannot be compressed and the viscosity doesn’t fall in the category of an ideal fluid. It is an imaginary fluid which doesn’t exist in reality. Real fluid:All the fluids are real as all the fluid possess viscosity. Newtonian fluid:When the fluid obeys Newton’s law of viscosity, it is known as a Newtonian fluid. Non-Newtonian fluid:When the fluid doesn’t obey Newton’s law of viscosity, it is known as Non-Newtonian fluid. Ideal plastic fluid:When the shear stress is proportional to the velocity gradient and shear stress is more than the yield value, it is known as ideal plastic fluid. Incompressible fluid:When the density of the fluid doesn’t change with the application of external force, it is known as an incompressible fluid. Compressible fluid:When the density of the fluid changes with the application of external force, it is known as compressible fluid. Below is the tabular representation of the fluid types:
Types of fluid Density Viscosity Ideal fluid Constant Zero Real fluid Variable Non zero Newtonian fluid Constant/ Variable \(T=u(\frac{\mathrm{d} u}{\mathrm{d} y})\) Non-Newtonian fluid Constant/ Variable \(T\neq u(\frac{\mathrm{d} u}{\mathrm{d} y})\) Incompressible fluid Constant Non zero/ zero Compressible fluid Variable Non zero/ zero Interested to learn more about other concepts related to fluid, below are the links: Classification of flows on the basis of Mach number: Incompressible flow has M<0.3. Compressible subsonic flow has M between 0.3 to 1. Graph of types of fluid flow: Types of Fluid Flow Steady or Unsteady Flow: Fluid flow can be steady or unsteady, depending on the fluid’s velocity: Steady:In steady fluid flow, the velocity of the fluid is constant at any point. Unsteady:When the flow is unsteady, the fluid’s velocity can differ between any two points. Viscous or Nonviscous Flow: Liquid flow can be viscous or nonviscous. Viscosity is a measure of the thickness of a fluid, and very gloppy fluids such as motor oil or shampoo are called viscous fluids. Fluid Flow Equation
ρ = density
V = Velocity
A = area
\(Flow \; rate = Area \times Velocity\)
Fluid Flow Through a Pipe
The general capacity of the pipes varies on its size. The table given below shows the capacity of the flow of fluid based on its size.
Pipe size (in inch) Maximum Flow (in gal/min) Velocity (in ft/s) Head Loss in (ft / 100 ft) 2 45 4.3 3.9 2.5 75 5.0 4.1 3 130 5.6 3.9 4 260 6.6 4.0 6 800 8.9 4.0 8 1600 10.3 3.8 10 3000 12.2 4.0 12 4700 13.4 4.0 14 6000 14.2 4.0 16 8000 14.5 3.5 18 10000 14.3 3.0 20 12000 13.8 2.4 24 18000 14.4 2.1 |
The following is an elementary question about circuit complexity. It is different from the kind of thing I have seen discussed, so I would be interested in any work that has been done on this kind of question. My apologies if this is easy or well known. By a Boolean function, I mean a function from $2^n\rightarrow2^m$ where $2$ is the set with two elements $0,1$. If $f:2^n\rightarrow2^m$ and $g:2^n\rightarrow2^k$ are Boolean functions, I denote by $\langle f,g\rangle$ the function $2^n\rightarrow2^{m+k}$ whose output is the output of $f$ followed by the output of $g$. Finally, let $|f|$ be the size (i.e. number of internal nodes) of the minimal circuit computing $f$. My question is the following: Suppose $| \langle f, \langle g,h \rangle \rangle | < |f|+|g|+|h|$ . Does it follow that $|\langle x,y \rangle | < |x|+|y|$ for some pair $x,y$ chosen from $f,g,h $ ? Naively, the idea is that we can build a circuit computing $\langle f,g \rangle$ from one computing $f$ and one computing $g$, and this has size $|f|+|g|$. But if $f$ and $g$ have something "in common", perhaps we can do better. Then the question is, suppose $f,g,h$ "have something in common". Does this mean two of these functions already "have something in common"? I would also be interested in any variations on this question, for example for considering only formulas or bounded circuits.
Suppose there is a 2-variable function $\phi: 2^n\times 2^n \to 2^m$. Take two constant vectors $c_1,c_2\in 2^n$ and define $f(x)=\phi(x,c_1)$, $g(x)=\phi(x,c_2)$ and $h(x)=\phi(x,c_1+c_2)$. For computing any two of these, the second arguments are arbitrary constant vectors, but to compute all three one might be able to use the fact that the three second arguments have a simple relationship.
For example (does this work?), take $m=1$ and define $\phi(x,y)$ to be the inner product of $x$ and $y$ (mod 2). Then $h(x)$ can be computed very quickly as $f(x)+g(x)$ but I don't see how to do any two of $f,g,h$ faster than one at at time.
A specific example that has a negative answer comes from Karatsuba multiplication in which one multiplies two polynomials $$(a_0 + a_1x)(b_0 + b_1x) =: c_0 + c_1x + c_2x^2$$ $$= (a_0b_0) + (a_0b_1 + a_1b_0)x + (a_1b_1)x^2$$ $$= (a_0b_0) + (a_0b_0 + (a_0+a_1)(b_0 + b_1) + a_1b_1)x + (a_1b_1)x^2$$ If we take $\mathbb{F}_2$-coefficients (or $2$-coefficients, in your notation), this can be viewed as three Boolean functions $2^4 \to 2^1$, or their concatenation a single function $2^4 \to 2^3$.
Karatsuba's trick allows one to turn the four multiplications and one addition into three multiplications and four additions. If you're counting all operations, and not just multiplications, it looks like you have increased from five to seven operations. But if you use this trick recursively, or take the $a_i,b_j$ to be polynomials themselves, then the reduction in multiplication wins out, and you can save overall operations.
I don't have a proof that two of the coefficients cannot be optimized without the third, but the reason I believe it is the case is because the Karatsuba trick exploits the fact that our coefficients of interest live in a rank-3 subspace of a 4-dimensional space. |
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Sat, 08 Jun 2019
I have pondered category theory periodically for the past 35 years, but not often enough to really stay comfortable with it. Today I was pondering again. I wanted to prove that !!1×A \cong A!! and I was having trouble. I eventually realized my difficulty: my brain had slipped out of category theory mode so that the theorem I was trying to prove did not even make sense.
In most of mathematics, !!1\times A!! would denote some specific entity and we would then show that that entity had a certain property. For example, in set theory we might define !!1\times A!! to be some set, say the set of all Kuratowski pairs !!\langle \varnothing, a\rangle!! where !!a\in A!!:
$$ 1×A =_{\text{def}} \{ z : \exists a\in A : z = \{\{\varnothing\}, \{\varnothing, a\}\} \} $$
and then we would explicitly construct a bijection !!f:A\leftrightarrow 1×A!!:
$$ f(a) = \{\{\varnothing\}, \{\varnothing, a\}\}$$
In category theory, this is not what we do. Everything is less concrete. !!\times!! looks like an operator, one that takes two objects and yields a third. It is not. !!1\times A!! does not denote any particular entity with any particular construction. (Nor does !!1!!, for that matter.) Instead, it denotes an unspecified entity, which happens to have a certain universal property, perhaps just one of many such entities with that property, and there is no canonical representative of the class. It's a mistake to think of it as a single thing, and it's also a mistake to ask the question the way I did ask it. You can't show that !!1×A!! has any specific property, because it's not a specific thing. All you can do is show that anything with the one required universal property must also have the other property. We should rephrase the question like this:
Maybe a better phrasing is:
The notation is still misleading, because it looks like !!1×A!! denotes the result of some operation, and it isn't. We can do a little better:
That it, that's the real theorem. It seems at first to be more difficult — where do we start? But it's actually easier! Because now it's enough to simply prove that !!A!! itself is a product of !!1!! and !!A!!, which is easily done: its projection morphisms are evidently !!! !! and !!{\text{id}}_A!!. And by a previous theorem that all products are isomorphic, any other product, such as !!B!!, must be isomorphic to this one, which is !!A!! itself.
(We can similarly imagine that any theorem that mentions !!1!! is preceded by the phrase “Let !!1!! be some terminal object.”) |
Summary
As orthocresol correctly notes, the presence of this kink is simply a feature of the $x$-shifted hyperbola that is the solution to the second-order kinetics problem, and the presence/characteristics of the kink will vary depending on the particular parameters of the problem.
However, this answer aims to demonstrate that:
The kink in the 2nd-order plot occurs due to a rapid change in the slope of the curve from "large" to "small" values. The kink only occurs in the 2nd-order plot if (a) the initial concentration and rate constant are both "high enough"; and (b) the time scale of interest is "short enough". A similar kink with first-order kinetics, again if the rate constant is "large enough" and the time scale is "short enough". also occurs
In order to generalize the solutions to the first- and second-order kinetics problems, coming from a chemical engineering background my standard approach is to non-dimensionalize, then solve.
1st-Order Problem
$$\ce{A ->[k_1] P} \\ ~ \\{\mathrm dC_\mathrm A\over \mathrm d t} = -k_1 C_\mathrm{A}~;\quad C_\mathrm A(0) = C_\mathrm{A}^\mathrm o$$
2nd-Order Problem
$$\ce{2A ->[k_2] Q} \\ ~ \\{\mathrm d C_\mathrm A\over \mathrm d t} = -k_2 C^2_\mathrm{A}~;\quad C_\mathrm{A}(0) = C_\mathrm{A}^\mathrm o$$
(Note that although I'm writing both reactions with $\ce A$ as the reactant, I'm assuming that these reactions do not occur simultaneously.)
To nondimensionalize, you define a new set of variables by dividing each dimensional variable by a scaling factor. Here, it makes the most sense to scale the concentration by the initial value:
$$\Theta = {C_\mathrm A \over C_\mathrm A^\mathrm o}~,~~\text{or}~~~ C_\mathrm{A} = C_\mathrm{A}^\mathrm{o}\,\Theta$$
The scaling factor for the time doesn't have an immediate, natural definition, so I'll use the generic variable $t^*$:
$$\tau = {t\over t^*}~,~~\text{or}~~~ t = t^*\,\tau$$
Then, these new variables are substituted into the original equations. The first-order equations become (using subscripts to indicate the order of the reaction under consideration):
$${C_\mathrm A^\mathrm o\over t^*} {\mathrm d \Theta_1\over \mathrm d \tau} = - k_1C_\mathrm A^\mathrm o\, \Theta_1$$
The scaling constants can then be grouped on the RHS of the differential equation:
$${\mathrm d \Theta_1\over \mathrm d \tau} = - k_1 t^* \Theta_1$$
For convenience, define the parameter $\eta = k_1 t^*$, yielding:
$${\mathrm d \Theta_1\over \mathrm d \tau} = -\eta\,\Theta_1~;\quad \Theta_1(0)=1 \tag{1}$$
The form of $\eta$ reveals that it is dimensionless, along with all of the other terms in the equation. This approach allows representation of any 1st-order kinetics problem to be represented by a single equation, once appropriate values for the system parameters $k_1$ and $t^*$, and thus $\eta$, are defined.
Non-dimensionalization of the 2nd-order equation proceeds as follows:
$${C_\mathrm A^\mathrm o\over t^*} {\mathrm d \Theta_2\over \mathrm d \tau} = - k_2\left(C_\mathrm A^\mathrm o\right)^2\, \Theta_2^2 \\{\mathrm d \Theta_2\over \mathrm d \tau} = - k_2 t^* C_\mathrm A^\mathrm o\,\Theta_2^2$$
$${\mathrm d \Theta_2\over \mathrm d \tau} = -\kappa\,\Theta_2^2~;\quad\Theta_2(0)=1 \tag{2}$$
The non-dimensional parameter $\kappa = k_2 t^* C_\mathrm{A}^\mathrm{o}$ is defined analogously to $\eta$, but has a different form due to the different reaction kinetics.
As an aside, note that in Eqs. $(1)$ and $(2)$ all of the information about the scale of each problem has been wrapped into $\eta$ and $\kappa$, respectively. Taking $\kappa$ as an example, if $C_\mathrm{A}^\mathrm{o}$ is 'large enough' relative to $t^*$ and $k_2$, then $\kappa$ will be large and the reaction will be "fast". Similarly, for a given $C_\mathrm{A}^\mathrm{o}$ and $k_2$, if you extend your time range of interest long enough ( viz., increase $t^*$), it will make the reaction term 'seem faster' compared to a shorter time scale.
The solutions to the above differential equations are:
$$\Theta_1(\tau) = e^{-\eta\,\tau} \tag{3}$$
$$\Theta_2(\tau) = {1\over 1 + \kappa\tau} \tag{4}$$
The derivations of the above are standard enough that I won't go into them in detail.
It will be useful later to have explicit expressions for the rates as functions of $\tau$. Substituting Eq. $(3)$ into Eq. $(1)$:
$${\mathrm d \Theta_1 \over \mathrm d \tau} = -\eta\, e^{-\eta\, \tau} \tag{5}$$
And, substituting Eq. $(4)$ into Eq. $(2)$:
$${\mathrm d \Theta_2 \over \mathrm d \tau} = -{\kappa\over \left(1 + \kappa\tau\right)^2} \tag{6}$$
Now ( finally!) to start tying these equations to the plots in the original question.
First, to demonstrate the utility of the scaled approach, consider the lower plot in the original question, where numerical values for the rate constant and initial concentration are given: $k_2 = 2~{1\over \mathrm{\mu M\, s}}$ (I'm assuming a typo on the units here) and $C_\mathrm{A}^\mathrm{o} = 10~\mathrm{\mu M}$. Further, the time scale has implicitly been defined by the choice of the domain of the abscissa: $t^* = 2~\mathrm{s}$. This leads to a value of $\kappa = 40$, which yields the following plot:
As can be seen, the visual appearance of the profile is essentially identical to that in the OP's plot.
Anticipating subsequent work, I'm actually going to work from here with a slightly higher value, $\kappa=54$:
The highlighted band in the figure marks the range where the (non-dimensional) slope decreases from $4$ to $0.25$, and falls right on the "kink" noted by the OP. Qualitatively, it marks the transition region from the "high slope" to the "low slope" portion of the curve.
It is the narrowness of this transition region $\left(\Delta\tau \ll 1 \right)$ that gives the visual impression of the "kink" in the curve.
There are a couple of ways one could compare this curve to a 1st-order curve. Naturally, one cannot define a 1st-order curve that lies exactly on the 2nd-order curve, as the two functional forms are different. However, after taking as given that $\Theta_1(0) = \Theta_2(0) = 1$, I see two intuitive ways of selecting a 'comparable' 1st-order curve: Constrain the initial slopes to be equal [cf. Eqs. $(5)$ and $(6)$]:
$$\left.{\mathrm d \Theta_1 \over \mathrm d \tau}\right|_{\tau=0} = \left.{\mathrm d \Theta_2 \over \mathrm d \tau}\right|_{\tau=0} \quad\longrightarrow\quad \eta = \kappa$$
Constrain the 'final values' (here, at $\tau = 1$) to be equal [cf. Eqs. $(3)$ and $(4)$]:
$$e^{-\eta} = {1\over 1+\kappa} \quad\text{or}\quad \eta = \ln{\left(1+\kappa\right)}$$
For Case #1, I'll plot $\eta = 54$ along with $\kappa = 54$:
As can be seen, in actuality, one
introduce a "kink" into a 1st-order curve, given high enough a value of $\eta$. In fact, when the initial slopes of the non-dimensional curves are made to be equal, the kink in the 1st-order curve is can than that in the 2nd-order curve! more dramatic
On the other hand, Case #2 looks much more like that depicted in the original question, without a visible kink in the 1st-order curve. Plotting for $\kappa=54$ and $\eta = \ln{\left(\kappa+1\right)} \approx 4$ gives:
(This is is why I picked $\kappa = 54$ above, so that the value for $\eta$ here would be approximately an integer value.)
In this case, the slope transition region for the 1st-order curve is broad $\left(\Delta\tau\sim 1\right)$, leading to no perceptible kink.
Further, it’s possible to plot a second-order curve where no kink is present. Taking $\kappa = 4$, comparable with the above $\eta = 4$ curve, gives the following:
As with the first-order chart with $\eta = 4$, the broader transition region above corresponds to the lack of an observable kink in the curve.
So: To kink, or not to kink? The scale is the question! |
In the paper ON A PAINLEVÉ-TYPE BOUNDARY-VALUE PROBLEM, the authors consider the BVP given by the ODE $$y''=y^2-x \tag{1} $$ with the boundary conditions $$\begin{align} y(0)&=0, \tag{2a} \\ y(x)& \sim \sqrt{x} \text{ as }x \to \infty \tag{2b}.\end{align}$$ They do so by studying the 1-parametric family of solutions $y_\alpha(x)$ of (1) with the initial conditions $$\begin{align} y(0)&=0, \tag{3a} \\ y'(0)&=\alpha. \tag{3b} \end{align}$$ It is conjectured* that the family $y_\alpha$ is classified into 5 different categories, depending on the value of $\alpha$ (please see attached plots):
If $\alpha > 0.924376$, $y_\alpha$ increases monotonically, and blows up in finite time (two "uppermost" red curves). If $\alpha \approx 0.924376$, $y_\alpha$ increases monotonically, and approaches $\sqrt{x}$ from below as $x \to \infty$ (upper green curve). If $-3.79199<\alpha<0.924376$, $y_\alpha$ oscillates about, and approaches $-\sqrt{x}$ as $x \to \infty$ (both blue curves). If $\alpha \approx -3.79199$, $y_\alpha$ goes down first, attains a minimum, and then goes up and approaches $\sqrt{x}$ from below as $x \to \infty$ (lower green curve). If $\alpha<-3.79199$, $y_\alpha$ goes down first, attains a minimum, and then intersecting $\sqrt{x}$ from below, resulting in a finite time blowup (two "lowermost" red curves).
I'm interested in finding those "critical slopes", which solve the BVP (1)-(2) (namely about 0.924376 and -3.79199). I don't understand how the authors obtained their decimal approximation to the upper critical slope, so I've decided to try and find them myself in the following way:
In order to find the upper slope for example, I have used Mathematica's numerical ODE solver. I noticed that using a slope of $1$ results in a blowup, and using a slope of $0$ results in oscillatory behaviour. From there I continued bisecting the interval $[0,1]$, closing in on the right slope. My problem with this method is that I'm at the mercy of the error tolerance of "NDSolve" and I'm not sure how good my approximation really is.
Here is my question:What is a numerical method that can find
both solutions of the BVP (1)-(2) to high precision (that is, determines both $\alpha$s up to $10^{-16}$, or preferably even better) efficiently?
To be clear, I am not interested as much in the actual solution $y_\alpha(x)$, the value of $\alpha$ itself suffices.
Thank you!
$*$ Part of this classification problem has been proven in a later paper. |
A particle is submitted to a time dependent force $$F(x,t)=\dfrac{k}{x^2}e^{-t/\tau}$$
Which is the Lagrangian of the particle?
I think that the force is derived from the potential $V$ and this potential has not explicit dependence of $\dot x$. So i can write
$$ \dfrac{d}{dt}\dfrac{\partial \mathcal L}{\partial \dot x} = m \ddot x$$
$$\mathcal L = T-\int \dfrac{\partial \mathcal L}{\partial x} dx$$ Then the lagrangian is $$\mathcal L = \dfrac{m}{2}\dot x^2 + \dfrac{k}{x}e^{-t/\tau}$$
Am i right? |
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Isomorphisms are very important in mathematics, and we can no longer put off talking about them. Intuitively, two objects are 'isomorphic' if they look the same. Category theory makes this precise and shifts the emphasis to the 'isomorphism' - the
way in which we match up these two objects, to see that they look the same.
For example, any two of these squares look the same after you rotate and/or reflect them:
An isomorphism between two of these squares is a
process of rotating and/or reflecting the first so it looks just like the second.
As the name suggests, an isomorphism is a kind of morphism. Briefly, it's a morphism that you can 'undo'. It's a morphism that has an inverse:
Definition. Given a morphism \(f : x \to y\) in a category \(\mathcal{C}\), an inverse of \(f\) is a morphism \(g: y \to x\) such that
and
I'm saying that \(g\) is 'an' inverse of \(f\) because in principle there could be more than one! But in fact, any morphism has at most one inverse, so we can talk about 'the' inverse of \(f\) if it exists, and we call it \(f^{-1}\).
Puzzle 140. Prove that any morphism has at most one inverse. Puzzle 141. Give an example of a morphism in some category that has more than one left inverse. Puzzle 142. Give an example of a morphism in some category that has more than one right inverse.
Now we're ready for isomorphisms!
Definition. A morphism \(f : x \to y\) is an isomorphism if it has an inverse. Definition. Two objects \(x,y\) in a category \(\mathcal{C}\) are isomorphic if there exists an isomorphism \(f : x \to y\).
Let's see some examples! The most important example for us now is a 'natural isomorphism', since we need those for our databases. But let's start off with something easier. Take your favorite categories and see what the isomorphisms in them are like!
What's an isomorphism in the category \(\mathbf{3}\)? Remember, this is a free category on a graph:
The morphisms in \(\mathbf{3}\) are paths in this graph. We've got one path of length 2:
$$ f_2 \circ f_1 : v_1 \to v_3 $$ two paths of length 1:
$$ f_1 : v_1 \to v_2, \quad f_2 : v_2 \to v_3 $$ and - don't forget - three paths of length 0. These are the identity morphisms:
$$ 1_{v_1} : v_1 \to v_1, \quad 1_{v_2} : v_2 \to v_2, \quad 1_{v_3} : v_3 \to v_3.$$ If you think about how composition works in this category you'll see that the only isomorphisms are the identity morphisms. Why? Because there's no way to compose two morphisms and get an identity morphism unless they're both that identity morphism!
In intuitive terms, we can only move from left to right in this category, not backwards, so we can only 'undo' a morphism if it doesn't do anything at all - i.e., it's an identity morphism.
We can generalize this observation. The key is that \(\mathbf{3}\) is a poset. Remember, in our new way of thinking a
preorder is a category where for any two objects \(x\) and \(y\) there is at most one morphism \(f : x \to y\), in which case we can write \(x \le y\). A poset is a preorder where if there's a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(x = y\). In other words, if \(x \le y\) and \(y \le x\) then \(x = y\). Puzzle 143. Show that if a category \(\mathcal{C}\) is a preorder, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(g\) is the inverse of \(f\), so \(x\) and \(y\) are isomorphic. Puzzle 144. Show that if a category \(\mathcal{C}\) is a poset, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then both \(f\) and \(g\) are identity morphisms, so \(x = y\).
Puzzle 144 says that in a poset, the only isomorphisms are identities.
Isomorphisms are a lot more interesting in the category \(\mathbf{Set}\). Remember, this is the category where objects are sets and morphisms are functions.
Puzzle 145. Show that every isomorphism in \(\mathbf{Set}\) is a bijection, that is, a function that is one-to-one and onto. Puzzle 146. Show that every bijection is an isomorphism in \(\mathbf{Set}\).
So, in \(\mathbf{Set}\) the isomorphisms are the bijections! So, there are lots of them.
One more example:
Definition. If \(\mathcal{C}\) and \(\mathcal{D}\) are categories, then an isomorphism in \(\mathcal{D}^\mathcal{C}\) is called a natural isomorphism.
This name makes sense! The objects in the so-called 'functor category' \(\mathcal{D}^\mathcal{C}\) are functors from \(\mathcal{C}\) to \(\mathcal{D}\), and the morphisms between these are natural transformations. So, the
isomorphisms deserve to be called 'natural isomorphisms'.
But what are they like?
Given functors \(F, G: \mathcal{C} \to \mathcal{D}\), a natural transformation \(\alpha : F \to G\) is a choice of morphism
$$ \alpha_x : F(x) \to G(x) $$ for each object \(x\) in \(\mathcal{C}\), such that for each morphism \(f : x \to y\) this naturality square commutes:
Suppose \(\alpha\) is an isomorphism. This says that it has an inverse \(\beta: G \to F\). This \(beta\) will be a choice of morphism
$$ \beta_x : G(x) \to F(x) $$ for each \(x\), making a bunch of naturality squares commute. But saying that \(\beta\) is the inverse of \(\alpha\) means that
$$ \beta \circ \alpha = 1_F \quad \textrm{ and } \alpha \circ \beta = 1_G .$$ If you remember how we compose natural transformations, you'll see this means
$$ \beta_x \circ \alpha_x = 1_{F(x)} \quad \textrm{ and } \alpha_x \circ \beta_x = 1_{G(x)} $$ for all \(x\). So, for each \(x\), \(\beta_x\) is the inverse of \(\alpha_x\).
In short: if \(\alpha\) is a natural isomorphism then \(\alpha\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\).
But the converse is true, too! It takes a
little more work to prove, but not much. So, I'll leave it as a puzzle. Puzzle 147. Show that if \(\alpha : F \Rightarrow G\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\), then \(\alpha\) is a natural isomorphism.
Doing this will help you understand natural isomorphisms. But you also need examples!
Puzzle 148. Create a category \(\mathcal{C}\) as the free category on a graph. Give an example of two functors \(F, G : \mathcal{C} \to \mathbf{Set}\) and a natural isomorphism \(\alpha: F \Rightarrow G\). Think of \(\mathcal{C}\) as a database schema, and \(F,G\) as two databases built using this schema. In what way does the natural isomorphism between \(F\) and \(G\) make these databases 'the same'. They're not necessarily equal!
We should talk about this. |
Existence of stable and unstable periodic solutions for semilinear parabolic problems
1.
School of Mathematics and Statistics, University of Sydney, N.S.W. 2006, Australia
2.
Department of Mathematics, Yokohama National University, 156 Tokiwadai Hodogaya-ku - Yokohama
$\frac{\partial u}{\partial t} - \Delta u = g(x,u) + h( t, x ),\quad \text{in} \quad (0,T) \times \Omega$
$u=0 ,\quad \text{on}\quad (0,T) \times \partial \Omega$
$u(0) = u(T),\quad \text{in} \quad \overline \Omega$
where $\Omega \subset R^N$ is a bounded domain with a smooth boundary, $g:\overline{\Omega} \times R \rightarrow R$ is a continuous function such that $g(x,\cdot )$ has a superlinear growth for each $x \in \overline{\Omega} $ and $h:(0,T) \times \Omega \to R$ is a continuous function.
Mathematics Subject Classification:Primary: 35K20, 35B1. Citation:E. N. Dancer, Norimichi Hirano. Existence of stable and unstable periodic solutions for semilinear parabolic problems. Discrete & Continuous Dynamical Systems - A, 1997, 3 (2) : 207-216. doi: 10.3934/dcds.1997.3.207
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Why and how is the Jordan Canonical form of a matrix in $M_3(\mathbb C)$ fully determined by its characteristic and minimal polynomials? And why does it fail for $n >3$?
Thanks.
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If $A\in M_3$ has one eigenvalue $a$, then its characteristic polynomial is $(x-a)^3$, and its minimal polynomial is $x-a$, $(x-a)^2$, or $(x-a)^3$. The only corresponding possibilities for Jordan form are, respectively, 3 blocks of size 1 ($A=aI$), a block of size 2 and hence another of size 1, and one block of size 3. Similarly, the possibilities are determined if there is more than one eigenvalue.
In $M_4$, consider the matrices $\begin{bmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\0 & 0 & 0 & 0\end{bmatrix}$ and $\begin{bmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}$.
From the characteristic polynomial $(x-a_1)^{n_1}\cdots (x-a_k)^{n_k}$ of a matrix $A\in M_n$ with distinct eigenvalues $a_1,\ldots,a_k$, you can see that the Jordan blocks for the eigenvalue $a_j$ have to have sizes adding to $n_j$. The largest size of one of these blocks is the exponent of $(x-a_j)$ appearing in the minimal polynomial of $A$. In case $n_j=3$, the only sizes possible for the blocks are $1+1+1$, $2+1$, and $3$, so the highest size of a block (the exponent in the minimal polynomial) determines the decomposition for $a_j$. If $n_j=2$ you have $1+1$ and $2$ as possible Jordan structure, so again it is determined by the largest size of a block. But when $n_j>3$, there are always decompositions like $2+2+[n-4\text{ ones}]$ and $2+1+1+[n-4\text{ ones}]$ which have the same largest size block (same exponent in the minimal polynomial), but are different Jordan forms.
In $M_3(\mathbb{C})$, if your characteristic polynomial gives three distinct complex roots your matrix will be diagonalisable. If you have one complex root and another double root, we can analyse it as follows:
Say your roots of the characteristic equation are $\lambda$ and $\lambda_1$ (twice). Then the block corresponding to the first eigenvalue is already completely determined. Then we only need to concentrate on the second block, which will be $2 \times 2$. This is because if the characteristic polynomial of your matrix is $(t - \lambda)(t - \lambda_1)^2$, there must be the factor $(t - \lambda)$ when $\lambda \neq 0$ in your minimal polynomial, otherwise there is no way to eliminate the block corresponding to $\lambda$ in the Jordan Form of the matrix when $\lambda \neq 0$ (Recall the definition of the minimal polynomial). To see this, observe that if your matrix $M$ is block diagonal, multiplication is done as follows:
$M = \left[\begin{array}{c|c} A & 0 \\ \hline 0 & B \end{array}\right]$, $M^2$ will be equal to $\left[\begin{array}{c|c} A^2 & 0 \\ \hline 0 & B^2 \end{array}\right].$
Having understood this, let us look at the factor in the minimal polynomial corresponding to that block. It will either be $(t - \lambda_1)$ or $(t - \lambda_1)^2$, if $\lambda_1 \neq 0$. Now the first case corresponds to the $2 \times 2$ block being
$\left[\begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_1 \end{array}\right]$ (why ?) while the second case to corresponds to a second block of the form $\left[ \begin{array}{cc} \lambda_1 & 1 \\ 0 & \lambda_1 \end{array} \right]$ (why)?
In the case when $\lambda$ or $\lambda_1$ is zero, you should know how the Jordan Form will look like given the minimal and characteristic polynomial just by looking at their definitions.
In the last case that you have three repeated complex roots, $\chi(t)$ will look like $(t - \lambda)^3$. Now we know that the minimal polynomial divides the characteristic polynomial, so that it will either be $(t - \lambda), (t - \lambda)^2$ or $(t - \lambda)^3$. This will then completely the determine the Jordan Form of your matrix, because in the first case, there are no $1's$ on the superdiagonal, the second case only one $1$ on the super diagonal up to rearrangement, and the third case two ones on the super diagonal.
From this we can see that in the case of $3 \times 3$ complex matrices, the characteristic and minimal polynomial of a matrix completely determines its Jordan Form. I believe Jonas has already told you to consider the $4 \times 4$ complex matrices he has given to see why this does not work when $n > 3$. |
Kumaran, V (1998)
Microscopic analysis of the coarsening of an interface in the spinodal decomposition of a binary fluid. In: Journal of Chemical Physics, 109 (8). pp. 3240-3244.
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Abstract
The coarsening of a random interface in a fluid of surface tension $\gamma$ and viscosity $\mu$ is analyzed using a curvature distribution function $A(K_m ,K_g ,t)$ which gives the distribution of the mean curvature $K_m$ and Gaussian curvature $K_g$ on the interface. There is a variation in the area distribution function due to the rate of change of $K_m,K_g$ and the compression of the interface due to tangential motion. The rates of change of mean and Gaussian curvature at a point are related to the rate of change of the normal velocity in the tangential directions along the interface. The fluid velocity is governed by the Stokes equation for a viscous flow, and the velocity field at a point is determined as an integral of the product of the Oseen tensor and the normal force at other points on the interface.Using a general form for this integral, it is shown that there is a characteristic variable $K_*=K_g / K_m^2-4K_g)^{1/2}$ which is independent of time even as $K_m$ and $K_g$ decrease proportional to $t^{-1}$ and $t^{-2}$, respectively. In the late stages, analytical forms for the distribution function are determined in the limit $K_m \ll K_*$ using a similarity variable $\eta = (\gamma K_mt/ \mu)$. Two reasonable approximations are used for the characteristic length for the correlation of the curvature and normal along the interface, and the results for these two approximations are quadratic polynomials in $|\eta|$ which are nonzero for a finite interval about $\eta = 0$. It is expected that the actual distribution function is in between these two limiting cases.
Item Type: Journal Article Additional Information: Copyright of this article belongs to The American Institute of Physics. Department/Centre: Division of Mechanical Sciences > Chemical Engineering Depositing User: Anka Setty Date Deposited: 04 Dec 2006 Last Modified: 19 Sep 2010 04:32 URI: http://eprints.iisc.ac.in/id/eprint/9034 Actions (login required)
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Wu Ki Tung's writing is not too precise mathematically, it is true. "Group Theory in Physics" is the title, and the level of mathematical rigor is not the highest, thus this sloppy text, actually sloppy two-letter word: "
Since the basis elements of the Lie algebra are generators of infinitesimal rotations, it is quite obvious that every representation of the group ". is automatically a representation of the corresponding Lie algebra
Representations of Lie groups
induce representations of Lie algebras, in the exact sense specified by Theorem IV of page 408 of the 2nd volume of Cornwell's "Group Theory in Physics" (abbreviated by me to the purpose of this text):
Let $\Gamma_{\mathcal{G}}$ be a d-dimensional analytic representation of linear Lie group $\mathcal{G}$, whose corresponding Lie algebra is $\mathcal{L}$. Then there exists a d-dimensional representation $\Gamma_{\mathcal{L}}$ of $\mathcal{L}$ defined for each $a\in\mathcal{L}$ by
$$\Gamma_{\mathcal{L}}(a) = \left[\frac{d}{dt}\Gamma_{\mathcal{G}}(\exp (ta))\right]_{t=0}$$
This is a relation between a representation of the group and a representation of its (real) algebra. Generalizations to complex group and complex representations of their Lie algebra follow easily.
The exact definition of a representation of a Lie algebra is here: https://en.wikipedia.org/wiki/Lie_algebra_representation |
Let $x_1, x_2... x_n$ be a random sample from a distribution with pdf:
$$f(x;\mu,\sigma)=\frac1{\sigma}\exp\left({-\frac{x-\mu}{\sigma}}\right)\,,-\infty<\mu<\infty;\, \sigma>0;\, x\ge\mu$$
How do I find the MLE for the parameters if both parameters are unknown?
I tried using the usual MLE with likelihood function:
$$L(\mu,\sigma|x_1...x_n)=\frac{1}{\sigma^n}\exp\left({-\frac{\sum{x_i}-n\mu}{\sigma}}\right)$$ But the derivative of this with respect of $\mu$ is a dead end.
I do know that if $\sigma$ is known, the MLE for $\mu$ is $\frac{\sum{x_i}}{n}$ and if $\mu$ is known, the MLE for $\sigma$ is $\frac{\sum{x_i}-n\mu}{n}$.
Do these help? What should be the approach? |
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Isomorphisms are very important in mathematics, and we can no longer put off talking about them. Intuitively, two objects are 'isomorphic' if they look the same. Category theory makes this precise and shifts the emphasis to the 'isomorphism' - the
way in which we match up these two objects, to see that they look the same.
For example, any two of these squares look the same after you rotate and/or reflect them:
An isomorphism between two of these squares is a
process of rotating and/or reflecting the first so it looks just like the second.
As the name suggests, an isomorphism is a kind of morphism. Briefly, it's a morphism that you can 'undo'. It's a morphism that has an inverse:
Definition. Given a morphism \(f : x \to y\) in a category \(\mathcal{C}\), an inverse of \(f\) is a morphism \(g: y \to x\) such that
and
I'm saying that \(g\) is 'an' inverse of \(f\) because in principle there could be more than one! But in fact, any morphism has at most one inverse, so we can talk about 'the' inverse of \(f\) if it exists, and we call it \(f^{-1}\).
Puzzle 140. Prove that any morphism has at most one inverse. Puzzle 141. Give an example of a morphism in some category that has more than one left inverse. Puzzle 142. Give an example of a morphism in some category that has more than one right inverse.
Now we're ready for isomorphisms!
Definition. A morphism \(f : x \to y\) is an isomorphism if it has an inverse. Definition. Two objects \(x,y\) in a category \(\mathcal{C}\) are isomorphic if there exists an isomorphism \(f : x \to y\).
Let's see some examples! The most important example for us now is a 'natural isomorphism', since we need those for our databases. But let's start off with something easier. Take your favorite categories and see what the isomorphisms in them are like!
What's an isomorphism in the category \(\mathbf{3}\)? Remember, this is a free category on a graph:
The morphisms in \(\mathbf{3}\) are paths in this graph. We've got one path of length 2:
$$ f_2 \circ f_1 : v_1 \to v_3 $$ two paths of length 1:
$$ f_1 : v_1 \to v_2, \quad f_2 : v_2 \to v_3 $$ and - don't forget - three paths of length 0. These are the identity morphisms:
$$ 1_{v_1} : v_1 \to v_1, \quad 1_{v_2} : v_2 \to v_2, \quad 1_{v_3} : v_3 \to v_3.$$ If you think about how composition works in this category you'll see that the only isomorphisms are the identity morphisms. Why? Because there's no way to compose two morphisms and get an identity morphism unless they're both that identity morphism!
In intuitive terms, we can only move from left to right in this category, not backwards, so we can only 'undo' a morphism if it doesn't do anything at all - i.e., it's an identity morphism.
We can generalize this observation. The key is that \(\mathbf{3}\) is a poset. Remember, in our new way of thinking a
preorder is a category where for any two objects \(x\) and \(y\) there is at most one morphism \(f : x \to y\), in which case we can write \(x \le y\). A poset is a preorder where if there's a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(x = y\). In other words, if \(x \le y\) and \(y \le x\) then \(x = y\). Puzzle 143. Show that if a category \(\mathcal{C}\) is a preorder, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(g\) is the inverse of \(f\), so \(x\) and \(y\) are isomorphic. Puzzle 144. Show that if a category \(\mathcal{C}\) is a poset, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then both \(f\) and \(g\) are identity morphisms, so \(x = y\).
Puzzle 144 says that in a poset, the only isomorphisms are identities.
Isomorphisms are a lot more interesting in the category \(\mathbf{Set}\). Remember, this is the category where objects are sets and morphisms are functions.
Puzzle 145. Show that every isomorphism in \(\mathbf{Set}\) is a bijection, that is, a function that is one-to-one and onto. Puzzle 146. Show that every bijection is an isomorphism in \(\mathbf{Set}\).
So, in \(\mathbf{Set}\) the isomorphisms are the bijections! So, there are lots of them.
One more example:
Definition. If \(\mathcal{C}\) and \(\mathcal{D}\) are categories, then an isomorphism in \(\mathcal{D}^\mathcal{C}\) is called a natural isomorphism.
This name makes sense! The objects in the so-called 'functor category' \(\mathcal{D}^\mathcal{C}\) are functors from \(\mathcal{C}\) to \(\mathcal{D}\), and the morphisms between these are natural transformations. So, the
isomorphisms deserve to be called 'natural isomorphisms'.
But what are they like?
Given functors \(F, G: \mathcal{C} \to \mathcal{D}\), a natural transformation \(\alpha : F \to G\) is a choice of morphism
$$ \alpha_x : F(x) \to G(x) $$ for each object \(x\) in \(\mathcal{C}\), such that for each morphism \(f : x \to y\) this naturality square commutes:
Suppose \(\alpha\) is an isomorphism. This says that it has an inverse \(\beta: G \to F\). This \(beta\) will be a choice of morphism
$$ \beta_x : G(x) \to F(x) $$ for each \(x\), making a bunch of naturality squares commute. But saying that \(\beta\) is the inverse of \(\alpha\) means that
$$ \beta \circ \alpha = 1_F \quad \textrm{ and } \alpha \circ \beta = 1_G .$$ If you remember how we compose natural transformations, you'll see this means
$$ \beta_x \circ \alpha_x = 1_{F(x)} \quad \textrm{ and } \alpha_x \circ \beta_x = 1_{G(x)} $$ for all \(x\). So, for each \(x\), \(\beta_x\) is the inverse of \(\alpha_x\).
In short: if \(\alpha\) is a natural isomorphism then \(\alpha\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\).
But the converse is true, too! It takes a
little more work to prove, but not much. So, I'll leave it as a puzzle. Puzzle 147. Show that if \(\alpha : F \Rightarrow G\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\), then \(\alpha\) is a natural isomorphism.
Doing this will help you understand natural isomorphisms. But you also need examples!
Puzzle 148. Create a category \(\mathcal{C}\) as the free category on a graph. Give an example of two functors \(F, G : \mathcal{C} \to \mathbf{Set}\) and a natural isomorphism \(\alpha: F \Rightarrow G\). Think of \(\mathcal{C}\) as a database schema, and \(F,G\) as two databases built using this schema. In what way does the natural isomorphism between \(F\) and \(G\) make these databases 'the same'. They're not necessarily equal!
We should talk about this. |
Spherical coordinates
Spherical coordinates can be a little challenging to understand at first. Spherical coordinates determine the position of a point in three-dimensional space based on the distance $\rho$ from the origin and two angles $\theta$ and $\phi$. If one is familiar with polar coordinates, then the angle $\theta$ isn't too difficult to understand as it is essentially the same as the angle $\theta$ from polar coordinates. But some people have trouble grasping what the angle $\phi$ is all about.
The following graphics and interactive applets may help you understand spherical coordinates better. On this page, we derive the relationship between spherical and Cartesian coordinates, show an applet that allows you to explore the influence of each spherical coordinate, and illustrate simple spherical coordinate surfaces.
Relationship between spherical and Cartesian coordinates
Spherical coordinates are defined as indicated in the following figure, which illustrates the spherical coordinates of the point $P$.
The coordinate $\rho$ is the distance from $P$ to the origin. If the point $Q$ is the projection of $P$ to the $xy$-plane, then $\theta$ is the angle between the positive $x$-axis and the line segment from the origin to $Q$. Lastly, $\phi$ is the angle between the positive $z$-axis and the line segment from the origin to $P$.
We can calculate the relationship between the Cartesian coordinates $(x,y,z)$ of the point $P$ and its spherical coordinates $(\rho,\theta,\phi)$ using trigonometry. The pink triangle above is the right triangle whose vertices are the origin, the point $P$, and its projection onto the $z$-axis. As the length of the hypotenuse is $\rho$ and $\phi$ is the angle the hypotenuse makes with the $z$-axis leg of the right triangle, the $z$-coordinate of $P$ (i.e., the height of the triangle) is $z=\rho\cos\phi$. The length of the other leg of the right triangle is the distance from $P$ to the $z$-axis, which is $r=\rho\sin\phi$. The distance of the point $Q$ from the origin is the same quantity.
The cyan triangle, shown in both the original 3D coordinate system on the left and in the $xy$-plane on the right, is the right triangle whose vertices are the origin, the point $Q$, and its projection onto the $x$-axis. In the right plot, the distance from $Q$ to the origin, which is the length of hypotenuse of the right triangle, is labeled just as $r$. As $\theta$ is the angle this hypotenuse makes with the $x$-axis, the $x$- and $y$-components of the point $Q$ (which are the same as the $x$- and $y$-components of the point $P$) are given by $x=r\cos\theta$ and $y=r\sin\theta$. Since $r=\rho\sin\phi$, these components can be rewritten as $x=\rho\sin\phi\cos\theta$ and $y=\rho\sin\phi\sin\theta$. In summary, the formulas for Cartesian coordinates in terms of spherical coordinates are \begin{align} x &= \rho\sin\phi\cos\theta\notag\\ y &= \rho\sin\phi\sin\theta\label{spherical_cartesian}\tag{1}\\ z &= \rho\cos\phi\notag. \end{align}
Exploring the influence of each spherical coordinate
The below applet allows you to see how the location of a point changes as you vary $\rho$, $\theta$, and $\phi$. The point $P$ corresponding to the value of the coordinates is shown as a large purple point. The green dot is the point $Q$, i.e., the projection of $P$ in the $xy$-plane.
Applet loading Spherical coordinates. Given the values for spherical coordinates $\rho$, $\theta$, and $\phi$, which you can change by dragging the points on the sliders, the large red point shows the corresponding position in
Cartesian coordinates. The green dot is the projection of the point in the $xy$-plane. You can visualize each of the spherical coordinates by the geometric structures that are colored corresponding to the slider colors. The length of the red line segment from the origin is $\rho$. The angle of the green portion of the disk in the $xy$-plane is $\theta$. The angle of the blue portion of the vertical disk is $\phi$. You can also move the large red point and the green projection of that point directly with the mouse.
Notice how you can obtain any point even though we restrict $\rho \ge 0$, $0 \le \theta < 2\pi$, and $0 \le \phi \le \pi$. Can you see why we only need $\phi$ to go up to $\pi$?
These restrictions removed much of the non-uniqueness of spherical coordinates. Notice there is still non-uniqueness at $\rho =0$, at $\phi = 0$ and at $\phi=\pi$. When any of these conditions are true, you can change the value of one or more of the other coordinates without moving the point.
Unfortunately, the convention for the notation of spherical coordinates is not standardized across disciplines. For example, in physics, the roles of $\theta$ and $\phi$ are typically reversed. In order to correctly understand someone's use of spherical coordinates, you must first determine what notational convention this are using. You cannot assume they follow the convention used here.
Simple spherical coordinate surfaces
These three next applets may help you understand what each of three spherical coordinates means. They show what the surfaces $\phi=$ constant, $\theta=$ constant, and $\rho=$ constant look like. The value of the constant is determined by the position of the sliders. In all cases, we restrict the surfaces to the region $\rho <5$.
Constant $\phi$
What does it mean for a point to have the spherical coordinate $\phi=\pi/3$? Take a look at the surfaces that are defined by the equation $\phi=$ constant.
Applet loading Surfaces of constant $\phi$ in spherical coordinates. The conical surface of $\phi=$ constant is shown, where the value of $\phi$ is determined by the blue point on the slider. Only the part of the surface where $\rho < 5$ is shown.
The surface $\phi=$ constant is simply a single cone, pointing either upward or downward. If you know that $\phi=\pi/3$, then you know the point is somewhere on a (wide) single cone that opens upward, i.e., the equation $\phi=\pi/3$ specifies a surface that is a single cone opening upward. The equation $\phi=\pi/2$ corresponds to the $xy$-plane.
The surface $\phi=$ constant is rotationally symmetric around the $z$-axis. Therefore it must depend on $x$ and $y$ only via the distance $\sqrt{x^2+y^2}$ from the $z$-axis. Using the relationship \eqref{spherical_cartesian} between spherical and Cartesian coordinates, one can calculate that \begin{align*} x^2+y^2 &= \rho^2\sin^2\phi(\cos^2\theta + \sin^2\theta)\\ &= \rho^2\sin^2\phi \end{align*} or $\sqrt{x^2+y^2} = \rho\sin\phi$. (Given that $0 \le \phi \le \pi$, we know that $\sin \phi \ge 0$ and the positive square root is $\rho\sin\phi$.) If we divide by $z=\rho\cos\phi$, we obtain a formula for $\phi$ in terms of Cartesian coordinates $$\frac{\sqrt{x^2+y^2}}{z} = \tan \phi.$$ We can rewrite the surface $\phi=$ constant as $$z = C \sqrt{x^2+y^2}$$ where $C=1/\tan \phi$, which is indeed the equation for a cone.
Constant $\theta$
The surface $\theta=$ constant is a half-plane off the $z$-axis. (It is plotted as a half-disk only because we restrict the plot to $\rho<5$.)
Applet loading Surfaces of constant $\theta$ in spherical coordinates. The half-plane surface of $\theta=$ constant is shown, where the value of $\theta$ is determined by the blue point on the slider. Only the part of the surface where $\rho < 5$ is shown, which makes the half-plane appear like a half-disk.
If a point has $\theta=\pi/2$, then you know the point is on the half of the $yz$-plane where $y$ values are positive. The equation $\theta=\pi/2$ is the equation for this half-plane.
From relationship \eqref{spherical_cartesian}, the ratio between $x$ and $y$ can be written, for example, as $y/x = \tan \theta$. If $\theta$ is held constant, then the ratio between $x$ and $y$ is constant. Thus, the equation $\theta=$ constant gives a line through the origin in the $xy$-plane. Since $z$ is unrestricted, we get a vertical plane. Looking back at relationship \eqref{spherical_cartesian}, we see it is only a half plane because $\rho\sin\phi$ cannot be negative.
Constant $\rho$
Most people don't have trouble understanding what $\rho=3$ means. It is the sphere of radius 3 centered at the origin. In general, the surface $\rho=$ constant is a sphere of radius $\rho$ centered at the origin.
Applet loading Surfaces of constant $\rho$ in spherical coordinates. The spherical surface of $\rho=$ constant is shown, where the value of $\rho$ is determined by the blue point on the slider.
From relationship \eqref{spherical_cartesian}, we can calculate that \begin{align*} x^2+y^2+z^2 &= \rho^2\sin^2\phi(\cos^2\theta + \sin^2\theta) + \rho^2\cos^2\phi\\ &= \rho^2(\sin^2\phi+\cos^2\phi)\\ &= \rho^2 \end{align*} verifying that $\rho=$ constant is the sphere of radius $\rho$ centered at the origin. |
Is my logic right?
Suppose there is a particle $p$ that can either decay into $ \{$a spin-1 and a spin-0 particle$\}$ or two spin-0 particles, then the lowest possible spin of $p$ is 2. This is because we need the spin to be even and large enough to accommodate the spin-1 product.
ADDED: $p$ is such that $p\to \pi^-+\rho^+$ and $p\to \xi+\xi$ where $\xi$ is a particle of spin 0. I want to know the lowest possible spin for $p$. I am assuming that conservation of total angular momentum and parity holds.
By the way, would the intrinsic parity of $p$ be negative, since I think each of $\rho^+$ and $\pi^-$ have negative intrinsic parity? |
With the set of parameters available to you, you cannot do this.
If you have the
actual track instead of the desired track, you will be able to calculate the wind.
The simplest way to do this is using vector math.
There are three vectors to consider:
ground speed vector $\vec{V_{gs}}$ air speed vector $\vec{V_{as}} $ wind speed vector $\vec{V_{ws}} $
$\vec{V_{gs}} =\vec{V_{as}} + \vec{V_{ws}} $
$\vec{V_{ws}} = \vec{V_{gs}} - \vec{V_{as}} $
I assume the actual track angle ($\phi$) and heading ($\psi$) are with respect to the true North.
The north component of your air speed is then: $V_{as} \cdot \cos(\psi)$ and the east component of your air speed is: $V_{as} \cdot \sin(\psi)$
For ground speed the decomposition is: north: $V_{gs} \cdot \cos(\phi)$ and the east component of your ground speed is: $V_{gs} \cdot \sin(\phi)$
$$\begin{bmatrix} V_{ws,north}\\ V_{ws,east} \end{bmatrix} = \begin{bmatrix} V_{gs} \cdot \cos(\phi) - V_{as} \cdot \cos(\psi) \\ V_{gs} \cdot \sin(\phi) - V_{as} \cdot \sin(\psi) \end{bmatrix}$$
You now have the north and east component of the wind vector. This you can change to a speed and direction, but I leave that last part up to you. Don't forget that wind direction is usually reported as the direction
which the wind is coming. from
To find the wind speed from the North and East components use the root of the sum of the squares:
$V_{ws}=\sqrt{V_{ws,north}^2 + V_{ws,east}^2}$
The wind direction can be found by
$\tan^{-1} (\frac {V_{ws,north}}{V_{ws,east} }) $
Note that this will give a division by 0 for winds exactly from north or south.
To implement it in a computer language the atan2 function can be used. This prevents division by zero and also returns the direction of the full range of the circle instead of semi-circular
wind_dir = atan2(-wind_north, -wind_east)
This should give the direction from which the wind is coming in radians. |
I'm trying to understand the Schoof algorithm for counting the number of points on elliptic curves in finite fields. I.e. the most basic algorithm to efficiently determine $\#E(F_p)$.
For literature, I'm referring mostly to the original Schoof (1985) and Gregg Musiker ("Schoofs Algorithm for Counting Points on $E(F_q)$", 2005), who has a nice summary of the Schoof algorithm. As an example implementation I'm looking at MIRACL's schoof.cpp and try to understand how it works in order to replicate the results.
I'm kind of stuck with the Frobenius Endomorphism, i.e. pg. 5 of Musiker, where he introduces the conversion between affine space (i.e. points on the curve) and polynoms. He says that for scalar multiplication of a point $P$ with an scalar $n$, the following holds:
$$nP = \big( \frac{\phi_n(x)}{\psi^2_n(x)}, \frac{\omega_n(x, y)}{\psi^3_n(x, y)} \big)$$
now on the left hand side is a point $(x, y)$, but on the right hand side there is what looks to me like projective representation of coordinates, i.e.
$$\big(\frac{X}{Z^2}, \frac{Y}{Z^3}\big)$$
only that $X$, $Y$ and $Z$ are polynomials this time. How can I convert between affine point representation and polynomial representation of points back and forth? |
Given $A_n\rightarrow \infty$ almost surely (a.s). Show that $\forall\ N > 0$, $P\left\{A_n<N\ \text{infinitely often}\right\} = 0$.
My thought: By the sake of contradiction, assume there exists an $\infty> N >0$ such that $P\left\{A_n<N\ \text{infinitely often}\right\} > 0$. This is equivalent to: $\lim_{n\rightarrow \infty} P\left\{\cup_{k\geq n} (A_k-N) < 0\right\} > 0$. But this means there exists a $k_1\geq n$ such that $A_{k_1} < N$ (1)
On the other hand, since $A_n\rightarrow \infty$ a.s, for every $\epsilon > 0$, $\lim_{n\rightarrow \infty} P\left\{\cup_{k\geq n} |A_k- \infty| \geq \epsilon\right\} = 0$. This means $|A_k-\infty|<\epsilon$ for every $k\geq n$. Since $k_1\geq n$, $\infty - A_{k_1} < |A_k- \infty| < \epsilon$ for any $\epsilon > 0$. Let $\epsilon = N$, we get $\infty < N+A_{k_1} < 2N$ (due to
(1)). This is a contradiction, since $\infty > N$. My question: I don't find my mathematical notation above legit, because I am doing subtraction with $\infty$. But I would like to know if my solution above is correct, so could someone please help verify? Also, I think there should be a shorter/slicker way to solve it, as my solution above is quite complicated and use more of real analysis. Anyone wants to give it a try? Any thoughts about my solution or this problem would be really appreciated anyway. |
I am finding it very difficult to find resources giving a overview of techniques that can be used to approximate a closed curve. For example, the wikipedia article on curve fitting has further links to linear/polynomial regression techniques that I am familiar with but their section for closed curve fitting is very brief and limited to mentions on fitting circles and ellipses. The references I can find here are also similar brief or specific:
My specific problem is as follows. I have a function $f(x,y)$ for which I have plotted some contour lines:
Each contour looks like part of a closed ellipse to me. To solve for the contour lines I considered solving the equation, in polar coordinates, $f(r \sin(\theta), r \cos(\theta)) = k$. However, this is not solvable as confirmed by Sage math.
Thus, I am looking for general approaches that will let me find an approximate closed curve to these contour lines. The only thing I can think of doing is to write down a regression equation for $r$ in terms of $\theta$, and then solve the least squares problem: i.e. find coefficients of the equation such that $(f(x, y) - k)^2)$ is minimised.
EDIT: As requested in the comment, $$f(x,y) = e^{-x - y} \sum_{m = 0}^\infty \frac{x^my^m}{(m!)^2}$$ and contour lines are obtained by considering rate parameters $x$ and $y$ such that the probability of a Skellam distributed random variable $Z$ taking value zero is held constant. While solutions specific to this equation are appreciated (such as in one of the previously linked posts where Stirling's approximation is used), please note that more generic methods of fitting closed curves would be more instructive for me in the future. |
This question arises when looking at a certain constant associated to (a certain Banach algebra built out of) a given compact group, and specializing to the case of finite groups, in order to try and do calculations for toy examples. It feels like the answer should be (more) obvious to those who play around with finite groups more than I do, or are at least know some more of the literature.
To be more precise: let $G$ be a finite group; let $d(G)$ be the maximum degree of an irreducible complex representation of $G$; and (with apologies to Banach-space theorists reading this) let $K_G$ denote the order of $G$ divided by the number of conjugacy classes.
Some easy but atypical examples:
if $G$ is abelian, then $K_G=1=d(G)$; if $G=Aff(p)$ is the affine group of the finite field $F_p$, $p$ a prime, then $$K_{Aff(p)}=\frac{p(p-1)}{p}=p-1=d(Aff(p))$$ Question.Does there exist a sequence $(G_n)$ of finite groups such that $d(G_n)\to\infty$ while$$\sup_n K_{G_n} <\infty ? $$
To give some additional motivation: when $d(G)$ is small compared to the order of $G$, we might regard this as saying that $G$ is not too far from being abelian. (In fact, we can be more precise, and say that $G$ has an abelian subgroup of small index, although I can't remember the precise dependency at time of writing.) Naively, then, is it the case that having $K_G$ small compared to the order of $G$ will also imply that $G$ is not too far from being abelian?
Other thoughts.Since the number of conjugacy classes in $G$ is equal to the number of mutually inequivalent complex irreps of $G$, and since $|G|=\sum_\pi d_\pi^2$, we see that $K_G$ is also equal to the mean square of the degress of complex irreps of $G$. Now it is very easy, given any large positive $N$, to find a sequence $a_1,\dots, a_m$ of strictly positive integers such that$$ \frac{1}{m}\sum_{i=1}^m a_i^2 \hbox{is small while} \max_i a_i > N $$so the question is whether we can do so in the context of degrees of complex irreps -- and if not, why not? The example of $Aff(p)$ shows that we can find examples with only one large irrep, but as seen above such groups won't give us a counterexample. |
Adamson, P. and Ader, C. and Andrews, M. and Anfimov, N. and Anghel, I. and Arms, K. and Arrieta-Diaz, E. and Aurisano, A. and Ayres, D. and Backhouse, C. and Baird, M. and Bambah, B. A. and Bays, K. and Bernstein, R. and Betancourt, M. and Bhatnagar, V. and Bhuyan, B. and Bian, J. and Biery, K. and Bocean, V. and Bogert, D. and Bolshakova, A. and Bowden, M. and Bower, C. and Broemmelsiek, D. and Bromberg, C. and Brunetti, G. and Bu, X. and Butkevich, A. and Capista, D. and Catano-Mur, E. and Chase, T. R. and Childress, S. and Choudhary, B. C. and Chowdhury, B. and Coan, T. E. and Coelho, J. A. B. and Colo, M. and Corwin, L. and Cronin-Hennessy, D. and Tutto, M. Del and Derwent, P. F. and Deepthi, K. N. and Demuth, D. and Desai, S. and Deuerling, G. and Devan, A. and Dey, J. and Dharmapalan, R. and Nowak, J. (2016)
First measurement of electron neutrino appearance in NOvA. Physical review letters, 116 (15). ISSN 1079-7114
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Abstract
We report results from the first search for $\nu_\mu\to\nu_e$ transitions by the NOvA experiment. In an exposure equivalent to $2.74\times10^{20}$ protons-on-target in the upgraded NuMI beam at Fermilab, we observe 6 events in the Far Detector, compared to a background expectation of $0.99\pm0.11$ (syst.) events based on the Near Detector measurement. A secondary analysis observes 11 events with a background of $1.07\pm0.14$ (syst.). The $3.3\sigma$ excess of events observed in the primary analysis disfavors $0.1\pi
Item Type: Journal Article Journal or Publication Title: Physical review letters Additional Information: © 2016 American Physical Society Uncontrolled Keywords: /dk/atira/pure/subjectarea/asjc/3100 Subjects: Departments: Faculty of Science and Technology > Physics ID Code: 78400 Deposited By: ep_importer_pure Deposited On: 26 Feb 2016 09:52 Refereed?: Yes Published?: Published Last Modified: 15 Oct 2019 01:36 URI: https://eprints.lancs.ac.uk/id/eprint/78400 Actions (login required)
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Definition:Isomorphism (Abstract Algebra)/Group Isomorphism Contents Definition
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: G \to H$ be a (group) homomorphism.
If $G$ is isomorphic to $H$, then the notation $G \cong H$ can be used (although notation varies). Also known as Isomorphism as defined here is known by some authors as simple isomorphism.
Let $S$ be the set defined as:
$S := \set {\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}: t \in \R}$ Then $\Z / 3 \Z$ is isomorphic to $A_4 / K_4$. Also see Results about group isomorphismscan be found here.
The word
isomorphism derives from the Greek morphe ( ) meaning μορφή formor structure, with the prefix iso-meaning equal.
Thus
isomorphism means equal structure. Sources 1964: Walter Ledermann: Introduction to the Theory of Finite Groups(5th ed.) ... (previous) ... (next): $\S 7$: Isomorphic Groups 1965: J.A. Green: Sets and Groups... (previous) ... (next): $\S 7.1$. Homomorphisms 1966: Richard A. Dean: Elements of Abstract Algebra... (previous) ... (next): $\S 1.5$ 1967: John D. Dixon: Problems in Group Theory... (previous) ... (next): Introduction: Notation 1967: George McCarty: Topology: An Introduction with Application to Topological Groups... (previous) ... (next): $\text{II}$: Quotient Groups 1971: Allan Clark: Elements of Abstract Algebra... (previous) ... (next): Introduction 1971: Allan Clark: Elements of Abstract Algebra... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 28 \gamma$ 1971: Allan Clark: Elements of Abstract Algebra... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 62$ 1978: John S. Rose: A Course on Group Theory... (previous) ... (next): $0$: Some Conventions and some Basic Facts 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra... (previous) ... (next): $\S 46$. Isomorphic groups 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra... (previous) ... (next): $\S 47.5 \ \text{(c)}$ Homomorphisms and their elementary properties 1982: P.M. Cohn: Algebra Volume 1(2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms 1996: John F. Humphreys: A Course in Group Theory... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Definition $8.8$ |
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Sat, 04 Jan 2014
There is a famous mistake of Augustin-Louis Cauchy, in which he is supposed to have "proved" a theorem that is false. I have seen this cited many times, often in very serious scholarly literature, and as often as not Cauchy's purported error is completely misunderstood, and replaced with a different and completely dumbass mistake that nobody could have made.
The claim is often made that Cauchy's
Cauchy never claimed to have proved any such thing, and it beggars belief that Cauchy could have made such a claim, because the counterexamples are so many and so easily located. For example, the sequence !! f_n(x) = x^n!! on the interval !![-1,1]!! is a sequence of continuous functions that converges everywhere on !![0,1]!! to a discontinuous limit. You would have to be a mathematical ignoramus to miss this, and Cauchy wasn't.
Another simple example, one that converges everywhere in !!\mathbb R!!, is any sequence of functions !!f_n!! that are everywhere zero, except that each has a (continuous) bump of height 1 between !!-\frac1n!! and !!\frac1n!!. As !!n\to\infty!!, the width of the bump narrows to zero, and the limit function !!f_\infty!! is everywhere zero except that !!f_\infty(0)=1!!. Anyone can think of this, and certainly Cauchy could have. A concrete example of this type is $$f_n(x) = e^{-x^{2}/n}$$ which converges to 0 everywhere except at !! x=0 !!, where it converges to 1.
Cauchy's controversial theorem is not what Wikipedia or nLab claim.It is that that the pointwise limit of a convergent
Here the counterexamples are not completely trivial. Probably the best-known counterexample is that a square wave (which has a jump discontinuity where the square part begins and ends) can be represented as a Fourier series.
(Cauchy was aware of this too, but it was new mathematics in 1821.Lakatos and others have argued that the theorem, understood in the waythat continuity was understood in 1821, is not actually erroneous, butthat the idea of continuity has changed since then. One piece ofevidence strongly pointing to this conclusion is that nobodycomplained about Cauchy's controversial theorem until 1847. But hadCauchy somehow, against all probability, mistakenly claimed that a
The confusion about Cauchy's controversial theorem arises from aperennially confusing piece of mathematical terminology: a convergent
The claim that Cauchy thought a sequence of continuous functions converges to a continuous limit is not only false but is manifestly so. Anyone making it has at best made a silly and careless error, and perhaps doesn't really understand what they are talking about, or hasn't thought about it.
[ I had originally planned to write about this controversial theorem in my series of articles about major screwups in mathematics, but the longer and more closely I looked at it the less clear it was that Cauchy had actually made a mistake. ] |
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In this chapter we learned about left and right adjoints, and about joins and meets. At first they seemed like two rather different pairs of concepts. But then we learned some deep relationships between them. Briefly:
Left adjoints preserve joins, and monotone functions that preserve enough joins are left adjoints.
Right adjoints preserve meets, and monotone functions that preserve enough meets are right adjoints.
Today we'll conclude our discussion of Chapter 1 with two more bombshells:
Joins
are left adjoints, and meets are right adjoints.
Left adjoints are right adjoints seen upside-down, and joins are meets seen upside-down.
This is a good example of how category theory works. You learn a bunch of concepts, but then you learn more and more facts relating them, which unify your understanding... until finally all these concepts collapse down like the core of a giant star, releasing a supernova of insight that transforms how you see the world!
Let me start by reviewing what we've already seen. To keep things simple let me state these facts just for posets, not the more general preorders. Everything can be generalized to preorders.
In Lecture 6 we saw that given a left adjoint \( f : A \to B\), we can compute its right adjoint using joins:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$ Similarly, given a right adjoint \( g : B \to A \) between posets, we can compute its left adjoint using meets:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ In Lecture 16 we saw that left adjoints preserve all joins, while right adjoints preserve all meets.
Then came the big surprise: if \( A \) has all joins and a monotone function \( f : A \to B \) preserves all joins, then \( f \) is a left adjoint! But if you examine the proof, you'l see we don't really need \( A \) to have
all joins: it's enough that all the joins in this formula exist:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$Similarly, if \(B\) has all meets and a monotone function \(g : B \to A \) preserves all meets, then \( f \) is a right adjoint! But again, we don't need \( B \) to have
all meets: it's enough that all the meets in this formula exist:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ Now for the first of today's bombshells: joins are left adjoints and meets are right adjoints. I'll state this for binary joins and meets, but it generalizes.
Suppose \(A\) is a poset with all binary joins. Then we get a function
$$ \vee : A \times A \to A $$ sending any pair \( (a,a') \in A\) to the join \(a \vee a'\). But we can make \(A \times A\) into a poset as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Then \( \vee : A \times A \to A\) becomes a monotone map, since you can check that
$$ a \le a' \textrm{ and } b \le b' \textrm{ implies } a \vee b \le a' \vee b'. $$And you can show that \( \vee : A \times A \to A \) is the left adjoint of another monotone function, the
diagonal
$$ \Delta : A \to A \times A $$sending any \(a \in A\) to the pair \( (a,a) \). This diagonal function is also called
duplication, since it duplicates any element of \(A\).
Why is \( \vee \) the left adjoint of \( \Delta \)? If you unravel what this means using all the definitions, it amounts to this fact:
$$ a \vee a' \le b \textrm{ if and only if } a \le b \textrm{ and } a' \le b . $$ Note that we're applying \( \vee \) to \( (a,a') \) in the expression at left here, and applying \( \Delta \) to \( b \) in the expression at the right. So, this fact says that \( \vee \) the left adjoint of \( \Delta \).
Puzzle 45. Prove that \( a \le a' \) and \( b \le b' \) imply \( a \vee b \le a' \vee b' \). Also prove that \( a \vee a' \le b \) if and only if \( a \le b \) and \( a' \le b \).
A similar argument shows that meets are really right adjoints! If \( A \) is a poset with all binary meets, we get a monotone function
$$ \wedge : A \times A \to A $$that's the
right adjoint of \( \Delta \). This is just a clever way of saying
$$ a \le b \textrm{ and } a \le b' \textrm{ if and only if } a \le b \wedge b' $$ which is also easy to check.
Puzzle 46. State and prove similar facts for joins and meets of any number of elements in a poset - possibly an infinite number.
All this is very beautiful, but you'll notice that all facts come in pairs: one for left adjoints and one for right adjoints. We can squeeze out this redundancy by noticing that every preorder has an "opposite", where "greater than" and "less than" trade places! It's like a mirror world where up is down, big is small, true is false, and so on.
Definition. Given a preorder \( (A , \le) \) there is a preorder called its opposite, \( (A, \ge) \). Here we define \( \ge \) by
$$ a \ge a' \textrm{ if and only if } a' \le a $$ for all \( a, a' \in A \). We call the opposite preorder\( A^{\textrm{op}} \) for short.
I can't believe I've gone this far without ever mentioning \( \ge \). Now we finally have really good reason.
Puzzle 47. Show that the opposite of a preorder really is a preorder, and the opposite of a poset is a poset. Puzzle 48. Show that the opposite of the opposite of \( A \) is \( A \) again. Puzzle 49. Show that the join of any subset of \( A \), if it exists, is the meet of that subset in \( A^{\textrm{op}} \). Puzzle 50. Show that any monotone function \(f : A \to B \) gives a monotone function \( f : A^{\textrm{op}} \to B^{\textrm{op}} \): the same function, but preserving \( \ge \) rather than \( \le \). Puzzle 51. Show that \(f : A \to B \) is the left adjoint of \(g : B \to A \) if and only if \(f : A^{\textrm{op}} \to B^{\textrm{op}} \) is the right adjoint of \( g: B^{\textrm{op}} \to A^{\textrm{ op }}\).
So, we've taken our whole course so far and "folded it in half", reducing every fact about meets to a fact about joins, and every fact about right adjoints to a fact about left adjoints... or vice versa! This idea, so important in category theory, is called
duality. In its simplest form, it says that things come on opposite pairs, and there's a symmetry that switches these opposite pairs. Taken to its extreme, it says that everything is built out of the interplay between opposite pairs.
Once you start looking you can find duality everywhere, from ancient Chinese philosophy:
to modern computers:
But duality has been studied very deeply in category theory: I'm just skimming the surface here. In particular, we haven't gotten into the connection between adjoints and duality!
This is the end of my lectures on Chapter 1. There's more in this chapter that we didn't cover, so now it's time for us to go through all the exercises. |
Difference between revisions of "Meissel-Lehmer method"
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Latest revision as of 23:39, 28 February 2012
The
Meissel-Lehmer method is a combinatorial method for computing [math]\pi(x)[/math] in time/space [math]x^{2/3+o(1)}[/math]. It is analysed at LMO???
Define an [math]x^{1/3}[/math]-
almost prime to be an integer less than x not divisible by any prime less than [math]x^{1/3}[/math]; such a number is either a prime, or a product of two primes which are between [math]x^{1/3}[/math] and [math]x^{2/3}[/math]. Each number of the latter form can be written in two ways as the product of a prime p between [math]x^{1/3}[/math] and [math]x^{2/3}[/math], and a prime between [math]x^{1/3}[/math] and [math]x/p[/math], except for the squares of primes between [math]x^{1/3}[/math] and [math]x^{1/2}[/math] which only have one such representation. The number [math]\pi(x,x^{1/3})[/math] of almost primes less than x is thus [math] \pi(x) - \pi(x^{1/3}) + \frac{1}{2} \sum_{x^{1/3} \leq p \leq x^{2/3}} [\pi( x/p ) - \pi(x^{1/3})] + \frac{1}{2} (\pi(x^{1/2})-\pi(x^{1/3}))[/math] (1)
(ignoring some floor and ceiling functions).
Using the sieve of Erathosthenes, one can compute [math]\pi(y)[/math] for all [math]y \leq x^{2/3}[/math] in time/space [math]x^{2/3+o(1)}[/math], so every expression in (1) is computable in this time except for [math]\pi(x)[/math]. So it will suffice to compute the number of [math]x^{1/3}[/math]-almost primes less than x in time [math]x^{2/3+o(1)}[/math].
If we let [math]\pi(x,a)[/math] denote the number of a-almost primes less than x (i.e. the number of integers less than x not divisible by any integer between 2 and a), we have the recurrence
[math]\pi(x,a) = \pi(x,a-1) - \pi(x/a, a)[/math] (2)
which reflects the basic fact that the a-1-almost primes are the union of the a-almost primes, and the a-1-almost primes multiplied by a. On the other hand, by factoring all the numbers up to [math]x^{1/3}[/math] by the sieve of Eratosthenes, we can store [math]\pi(y,a)[/math] for all [math]a, y \leq x^{1/3}[/math], to be retrievable in O(1) time. Meanwhile, by recursively expanding out (2) until the x parameter dips below [math]x^{1/3}[/math], one can express [math]\pi(x,x^{1/3})[/math] as an alternating sum of various [math]\pi(y,b)[/math] with [math]y,b \leq x^{1/3}[/math], with each [math]\pi(y,b)[/math] occurring at most once; summing using the stored values then gives [math]\pi(x,x^{1/3})[/math] as required. |
At one time, calories in foods were measured with a bomb calorimeter. A weighed amount of the food would be placed in the calorimeter and the system was then sealed and filled with oxygen. An electric spark ignited the food-oxygen mixture. The amount of heat released when the food burned would given an idea of the food calories present. Today calories are calculated from the protein, carbohydrate, and fat content of the food (all determined by chemical analysis). No more bombs needed.
Calorimetry Calorimetry is the measurement of the transfer of heat into or out of a system during a chemical reaction or physical process. A calorimeter is an insulated container that is used to measure heat changes. The majority of reactions that can be analyzed in a calorimetry experiment are either liquids or aqueous solutions. A frequently used and inexpensive calorimeter is a set of nested foam cups fitted with a lid to limit the heat exchange between the liquid in the cup and the air in the surroundings (see figure below). In a typical calorimetry experiment, specific volumes of the reactants are dispensed into separate containers and the temperature of each is measured. They are then mixed into the calorimeter, which starts the reaction. The reactant mixture is stirred until the reaction is complete, while the temperature of the reaction is continuously monitored.
Figure 17.7.1: A simple constant-pressure calorimeter.
The key to all calorimetry experiments is the assumption that there is no heat exchange between the insulated calorimeter and the room. Consider the case of a reaction taking place between aqueous reactants. The water in which the solids have been dissolved is the surroundings, while the dissolved substances are the system. The temperature change that is measured is the temperature change that is occurring in the surroundings. If the temperature of the water increases as the reaction occurs, the reaction is exothermic. Heat was released by the system into the surrounding water. An endothermic reaction absorbs heat from the surroundings, so the temperature of the water decreases as heat leaves the surroundings to enter the system.
The temperature change of the water is measured in the experiment and the specific heat of water can be used to calculate the heat absorbed by the surroundings \(\left( q_\text{surr} \right)\).
\[q_\text{surr} = m \times c_p \times \Delta T\]
In the equation, \(m\) is the mass of the water, \(c_p\) is the specific heat of the water, and \(\Delta T\) is \(T_f - T_i\). The heat absorbed by the surroundings is equal, but opposite in sign, to the heat released by the system. Because the heat change is determined at constant pressure, the heat released by the system \(\left( q_\text{sys} \right)\) is equal to the enthalpy change \(\left( \Delta H \right)\).
\[q_\text{sys} = \Delta H = -q_\text{surr} = - \left( m \times c_p \times \Delta T \right)\]
The sign of \(\Delta H\) is positive for an endothermic reaction and negative for an exothermic reaction.
Example 17.7.1
In an experiment, \(25.0 \: \text{mL}\) of \(1.00 \: \text{M} \: \ce{HCl}\) at \(25.0^\text{o} \text{C}\) is added to \(25.0 \: \text{mL}\) of \(1.00 \: \text{M} \: \ce{NaOH}\) at \(25.0^\text{o} \text{C}\) in a foam cup calorimeter. A reaction occurs and the temperature rises to \(32.0^\text{o} \text{C}\). Calculate the enthalpy change \(\left( \Delta H \right)\) in \(\text{kJ}\) for this reaction. Assume the densities of the solutions are \(1.00 \: \text{g/mL}\) and that their specific heat is the same as that of water.
Solution: Step 1: List the known quantities and plan the problem. Known \(c_p = 4.18 \: \text{J/g}^\text{o} \text{C}\) \(V_\text{final} = 25.0 \: \text{mL} + 25.0 \: \text{mL} = 50.0 \: \text{mL}\) \(\Delta T = 32.0^\text{o} \text{C} - 25.0^\text{o} \text{C} = 7.0^\text{o} \text{C}\) Density \(= 1.00 \: \text{g/mL}\) Unknown \(\Delta H = ? \: \text{kJ}\)
The volume and density can be used to find the mass of the solution after mixing. Then calculate the change in enthalpy by using \(\Delta H = q_\text{sys} = -q_\text{surr} = - \left( m \times c_p \times \Delta T \right)\).
Step 2: Solve.
\[\begin{align} m &= 50.0 \: \text{mL} \times \frac{1.00 \: \text{g}}{\text{mL}} = 50.0 \: \text{g} \\ \Delta H &= - \left( m \times c_p \times \Delta T \right) = - \left( 50.0 \: \text{g} \times 4.18 \: \text{J/g}^\text{o} \text{C} \times 7.0^\text{o} \text{C} \right) = -1463 \: \text{J} = -1.5 \: \text{kJ} \end{align}\]
Step 3: Think about the result.
The enthalpy change is negative because the reaction releases heat to the surroundings, resulting in an increase in temperature of the water.
Summary The process of calorimetry is described. Calculations involving enthalpy changes are illustrated. Contributors
CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon. |
I'm told to use Gauss's Theorem to compute the flux of a field $\vec F = <x,y^2,y+z>$ along the boundary of the cylindrical solid $x^2+y^2 \le 4$ below $z=8$ and above $z=x$.
I know by Gauss's Theorem that:
Net Flux = $\iint_{\partial D} \vec F \cdot \vec ndS = \iiint_D \nabla \cdot \vec FdV$
This computation is pretty straight forward. $\nabla \cdot \vec F = 2+2y$. But the region of integration is particularly difficult to map out.
I thought to use cylindrical coordinates and setting the bounds to $0 \le \theta \le 2 \pi$, $0 \le z \le 8$, and $0 \le r \le 4$, but this seems like it would just give me the area of the cylinder of height 8--and wouldn't include the part where z=x slices through the cylinder.
What would be the right way to go in terms of the bounds of integration? |
The critical value approach involves determining "likely" or "unlikely" by determining whether or not the observed test statistic is more extreme than would be expected if the null hypothesis were true. That is, it entails comparing the observed test statistic to some cutoff value, called the "
critical value." If the test statistic is more extreme than the critical value, then the null hypothesis is rejected in favor of the alternative hypothesis. If the test statistic is not as extreme as the critical value, then the null hypothesis is not rejected.
Specifically, the four steps involved in using the critical value approach to conducting any hypothesis test are:
Specify the null and alternative hypotheses. Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic. To conduct the hypothesis test for the population mean μ, we use the t-statistic \(t^*=\frac{\bar{x}-\mu}{s/\sqrt{n}}\) which follows a t-distribution with n- 1 degrees of freedom. Determine the critical value by finding the value of the known distribution of the test statistic such that the probability of making a Type I error — which is denoted \(\alpha\) (greek letter "alpha") and is called the " significance level of the test" — is small (typically 0.01, 0.05, or 0.10). Compare the test statistic to the critical value. If the test statistic is more extreme in the direction of the alternative than the critical value, reject the null hypothesis in favor of the alternative hypothesis. If the test statistic is less extreme than the critical value, do not reject the null hypothesis. Example S.3.1.1 Mean GPA Section
In our example concerning the mean grade point average, suppose we take a random sample of
n = 15 students majoring in mathematics. Since n = 15, our test statistic t* has n - 1 = 14 degrees of freedom. Also, suppose we set our significance level α at 0.05, so that we have only a 5% chance of making a Type I error. Right-Tailed
The critical value for conducting the
right-tailed test H 0 : μ = 3 versus H A : μ > 3 is the t-value, denoted t \(\alpha\), n - 1, such that the probability to the right of it is \(\alpha\). It can be shown using either statistical software or a t-table that the critical value t 0.05,14 is 1.7613. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ > 3 if the test statistic t* is greater than 1.7613. Visually, the rejection region is shaded red in the graph. Left-Tailed
The critical value for conducting the
left-tailed test H 0 : μ = 3 versus H A : μ < 3 is the t-value, denoted -t (\(\alpha\), n - 1) , such that the probability to the leftof it is \(\alpha\). It can be shown using either statistical software or a t-table that the critical value -t 0.05,14is -1.7613. That is, we would reject the null hypothesis H 0: μ= 3 in favor of the alternative hypothesis H A: μ< 3 if the test statistic t* is less than -1.7613. Visually, the rejection region is shaded red in the graph. Two-Tailed
There are two critical values for the
two-tailed test H 0 : μ = 3 versus H A : μ ≠ 3 — one for the left-tail denoted -t (\(\alpha\) /2, n - 1) and one for the right-tail denoted t (\(\alpha\). The value /2, n- 1) - t (\(\alpha\)/2,is the n- 1) t-value such that the probability to the leftof it is \(\alpha\)/2, and the value t (\(\alpha\)/2,is the n- 1) t-value such that the probability to the rightof it is \(\alpha\)/2. It can be shown using either statistical software or a t-table that the critical value -t 0.025,14is -2.1448 and the critical value t 0.025,14is 2.1448. That is, we would reject the null hypothesis H 0: μ= 3 in favor of the alternative hypothesis H A: μ≠ 3 if the test statistic t* is less than -2.1448 or greater than 2.1448. Visually, the rejection region is shaded red in the graph. |
One way you can try to see it is through an analogy with the Laplace transform...
$$e^{-at}\stackrel{\mathcal{L}}\longleftrightarrow \frac{1}{s+a}$$
Where $a\in\mathbb{C}$. If $\Re\{a\}>0$, the exponential decays and the system is stable. If $\Re\{a\}=0$, there is no damping and the system is marginally stable. $\Re\{a\}<0$, the exponential grows infinitely and the system is unstable.
Therefore, the absolutely stable region is mapped as the left half of the s-Plane.
Now, for the Z-Transform:
$$a^n\stackrel{\mathcal{Z}}\longleftrightarrow \frac{z}{z-a}$$
We know that asequence is stable if it is absolutely summable. To simplify the analysis, this time, let us consider $a\in\mathbb{R}$. If $a<1$, $\sum_{n=0}^{\infty}a^{n}<\infty$ decays and converges. If $a=1$, in this case we have the discrete unit step, whose pole lay over the unit circle, characterizing marginal stability. If $a>1$, $\sum_{n=0}^{\infty}a^{n}$ increases indefinitely and does not converge.
Therefore, the absolute stable region is mapped inside the unit circle on the z-Plane.
You can also understand this through the relationship between the Region of Convergence of the Z-Transform ($ROC$) and the unit circle. For that, please refer to this answer on a question about region of convergence. |
From Wikipedia's Multiple Comparison
For hypothesis testing, the problem of multiple comparisons (also known as the multiple testing problem) results from the increase in type I error that occurs when statistical tests are used repeatedly. If
n independent comparisonsare performed, the experiment-wide significance level $\bar{\alpha}$, also termed FWER for familywise error rate, is given by $$ \bar{\alpha} = 1-\left( 1-\alpha_\mathrm{\{per\ comparison\}} \right)^n$$
I don't understand how the comparisons can be independent? Let the multiple tests be $\{H_i, K_i, T_i, c_i), i \in I\}$, where the $i$-th test is $H_i$ versus $K_i$, with testing statistic $T_i$ and critical value $c_i$. Now given a sample $X$, the test statistics $T_i(X), i\in I$ can't be independent, and therefore the testing rules $I_{T_i(X) \geq c_i}$ can't be independent either. Am I wrong? Thanks! |
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Isomorphisms are very important in mathematics, and we can no longer put off talking about them. Intuitively, two objects are 'isomorphic' if they look the same. Category theory makes this precise and shifts the emphasis to the 'isomorphism' - the
way in which we match up these two objects, to see that they look the same.
For example, any two of these squares look the same after you rotate and/or reflect them:
An isomorphism between two of these squares is a
process of rotating and/or reflecting the first so it looks just like the second.
As the name suggests, an isomorphism is a kind of morphism. Briefly, it's a morphism that you can 'undo'. It's a morphism that has an inverse:
Definition. Given a morphism \(f : x \to y\) in a category \(\mathcal{C}\), an inverse of \(f\) is a morphism \(g: y \to x\) such that
and
I'm saying that \(g\) is 'an' inverse of \(f\) because in principle there could be more than one! But in fact, any morphism has at most one inverse, so we can talk about 'the' inverse of \(f\) if it exists, and we call it \(f^{-1}\).
Puzzle 140. Prove that any morphism has at most one inverse. Puzzle 141. Give an example of a morphism in some category that has more than one left inverse. Puzzle 142. Give an example of a morphism in some category that has more than one right inverse.
Now we're ready for isomorphisms!
Definition. A morphism \(f : x \to y\) is an isomorphism if it has an inverse. Definition. Two objects \(x,y\) in a category \(\mathcal{C}\) are isomorphic if there exists an isomorphism \(f : x \to y\).
Let's see some examples! The most important example for us now is a 'natural isomorphism', since we need those for our databases. But let's start off with something easier. Take your favorite categories and see what the isomorphisms in them are like!
What's an isomorphism in the category \(\mathbf{3}\)? Remember, this is a free category on a graph:
The morphisms in \(\mathbf{3}\) are paths in this graph. We've got one path of length 2:
$$ f_2 \circ f_1 : v_1 \to v_3 $$ two paths of length 1:
$$ f_1 : v_1 \to v_2, \quad f_2 : v_2 \to v_3 $$ and - don't forget - three paths of length 0. These are the identity morphisms:
$$ 1_{v_1} : v_1 \to v_1, \quad 1_{v_2} : v_2 \to v_2, \quad 1_{v_3} : v_3 \to v_3.$$ If you think about how composition works in this category you'll see that the only isomorphisms are the identity morphisms. Why? Because there's no way to compose two morphisms and get an identity morphism unless they're both that identity morphism!
In intuitive terms, we can only move from left to right in this category, not backwards, so we can only 'undo' a morphism if it doesn't do anything at all - i.e., it's an identity morphism.
We can generalize this observation. The key is that \(\mathbf{3}\) is a poset. Remember, in our new way of thinking a
preorder is a category where for any two objects \(x\) and \(y\) there is at most one morphism \(f : x \to y\), in which case we can write \(x \le y\). A poset is a preorder where if there's a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(x = y\). In other words, if \(x \le y\) and \(y \le x\) then \(x = y\). Puzzle 143. Show that if a category \(\mathcal{C}\) is a preorder, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(g\) is the inverse of \(f\), so \(x\) and \(y\) are isomorphic. Puzzle 144. Show that if a category \(\mathcal{C}\) is a poset, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then both \(f\) and \(g\) are identity morphisms, so \(x = y\).
Puzzle 144 says that in a poset, the only isomorphisms are identities.
Isomorphisms are a lot more interesting in the category \(\mathbf{Set}\). Remember, this is the category where objects are sets and morphisms are functions.
Puzzle 145. Show that every isomorphism in \(\mathbf{Set}\) is a bijection, that is, a function that is one-to-one and onto. Puzzle 146. Show that every bijection is an isomorphism in \(\mathbf{Set}\).
So, in \(\mathbf{Set}\) the isomorphisms are the bijections! So, there are lots of them.
One more example:
Definition. If \(\mathcal{C}\) and \(\mathcal{D}\) are categories, then an isomorphism in \(\mathcal{D}^\mathcal{C}\) is called a natural isomorphism.
This name makes sense! The objects in the so-called 'functor category' \(\mathcal{D}^\mathcal{C}\) are functors from \(\mathcal{C}\) to \(\mathcal{D}\), and the morphisms between these are natural transformations. So, the
isomorphisms deserve to be called 'natural isomorphisms'.
But what are they like?
Given functors \(F, G: \mathcal{C} \to \mathcal{D}\), a natural transformation \(\alpha : F \to G\) is a choice of morphism
$$ \alpha_x : F(x) \to G(x) $$ for each object \(x\) in \(\mathcal{C}\), such that for each morphism \(f : x \to y\) this naturality square commutes:
Suppose \(\alpha\) is an isomorphism. This says that it has an inverse \(\beta: G \to F\). This \(beta\) will be a choice of morphism
$$ \beta_x : G(x) \to F(x) $$ for each \(x\), making a bunch of naturality squares commute. But saying that \(\beta\) is the inverse of \(\alpha\) means that
$$ \beta \circ \alpha = 1_F \quad \textrm{ and } \alpha \circ \beta = 1_G .$$ If you remember how we compose natural transformations, you'll see this means
$$ \beta_x \circ \alpha_x = 1_{F(x)} \quad \textrm{ and } \alpha_x \circ \beta_x = 1_{G(x)} $$ for all \(x\). So, for each \(x\), \(\beta_x\) is the inverse of \(\alpha_x\).
In short: if \(\alpha\) is a natural isomorphism then \(\alpha\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\).
But the converse is true, too! It takes a
little more work to prove, but not much. So, I'll leave it as a puzzle. Puzzle 147. Show that if \(\alpha : F \Rightarrow G\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\), then \(\alpha\) is a natural isomorphism.
Doing this will help you understand natural isomorphisms. But you also need examples!
Puzzle 148. Create a category \(\mathcal{C}\) as the free category on a graph. Give an example of two functors \(F, G : \mathcal{C} \to \mathbf{Set}\) and a natural isomorphism \(\alpha: F \Rightarrow G\). Think of \(\mathcal{C}\) as a database schema, and \(F,G\) as two databases built using this schema. In what way does the natural isomorphism between \(F\) and \(G\) make these databases 'the same'. They're not necessarily equal!
We should talk about this. |
I found the following exercise in Introduction to Metric and Topological Spaces by Sutherland (Chapter 10 Question 20).
Prove that the topology on a space X is discrete iff the diagonal $\Delta=\{ (x,x) \mid x\in X\}$ is open in the topological product $X \times X$.
I believe I could prove in the $implies$ direction. It is the converse that got me stuck.
So,I would want to prove that if the diagonal is open in the topological product then the topology on $X$ must be discrete.
I was thinking that I could achieve the result if I could show that every singleton $\{ x\}$ is open in $X$, but I couldn't think of a way to establish this. I tried considering $X-\{ x\}$ and try to show that it's closed but I couldn't continue to anything fruitful. Also, I think I could use projection maps but I am stumped as well.
Any help/hint is appreciated. Thanks in advance. |
Like JDługosz wrote, what will cause problems in the scenario you describe isn't so much your orbit as the fact that you are
within the gas giant's atmosphere.
I'm going to use Jupiter here to have some specific gas giant to use for examples. Feel free to look up the relevant data for any other gas giant, or come up with your own.
For the case we are interested in, a small mass orbiting a much larger mass where the radius of the orbit is equal to the larger body's radius (just dipping your toes into the Jovian atmosphere), orbital speed can be approximated as $$ v_o \approx \frac{v_e}{\sqrt{2}} $$
The escape velocity of Jupiter is approximately 59.5 km/s, so to dip our toes into the atmosphere we get an orbital velocity of approximately $$ v_o \approx \frac{59~500~\text{m/s}}{\sqrt{2}} \approx 42~100~\text{m/s} $$
To give an idea of how freakishly fast this is, it's equivalent to approximately 152,000 km/h or 94,200 miles per hour. It gets you between the Earth and the Moon in
2.5 hours. In mid-1976, an airplane managed to get to 3,530 km/h, which is about 1/43 of the orbital speed at the edge of Jupiter's atmosphere. The best we have managed on anything resembling a repeat basis is around 2,500 km/h, or 1/60 of what you would need.
For comparison, Jupiter's wind speeds peak in excess of 150 m/s. While quite a stiff gale, that's nowhere near orbital velocity; by the above estimate, about 1/280 (and that's assuming that top wind speeds occur in the uppermost layers of the atmosphere, which might not be the case). With such a large difference between orbital speeds and wind speeds, we can largely ignore wind speeds for the purposes of this question; even in a perfect situation, wind speed will contribute less than 0.36% of the required velocity. (Interestingly enough, according to the same source, Jupiter wind speeds have a peak very near the equator, which works well for us.)
Given that Jupiter has an equatorial diameter of 142,984 km and that the circumference of a circle is $\pi d$, 42.1 km/s gives an orbital period (if you can call it orbital) of $\frac{142984 \pi}{42.1} \approx 10~700~\text{seconds}$ or just under three hours. For comparison, Wikipedia gives Jupiter's sidereal rotation period ("day") of 9.925 hours (a shade over 9 hours 55 minutes).
For comparison, to get into a reasonably stable low Earth orbit you need a velocity of approximately 7.8 km/s (corresponding to an orbital period of about 90 minutes). To go to the Moon (which is pretty close to escape velocity), you need about 10.5 km/s relative to the Earth. Actual Earth escape velocity is 11,186 m/s. Compare Apollo by the numbers: Translunar Injection and look at particularly the Earth Fixed velocity figures for the various lunar missions.
Let's say you can somehow handwave the issue of absolute speed away. (After all, you got there somehow, and that already takes quite a bit of speed.) Let's also say that your craft is a very, very long, perfect cylinder with a forward cross section of 1 square meter, built to handle constant hurricane-level wind speeds. Every second, you are moving through 42,100 meters of atmosphere. That means that every second, your craft will need to push aside 42,100 cubic meters of atmospheric gases
while maintaining its speed (at least if you plan on staying at that altitude). Wikipedia gives the composition of Jupiter's atmosphere as approximately $89.8 \pm 2.0 \% ~\text{H}_2$ and $10.2 \pm 2.0 \% ~\text{He}$. Despite the fact that these two gases are among the lightest known, and that the density is going to still be low at the altitude we are talking about, pushing aside over 40,000 cubic meters of gas per second is going to cause some massive drag.
And that, my friend, is what will cause your craft to heat up, lose speed very quickly and eventually descend into the atmosphere, ruining your day. |
Let M be a deterministic Turing machine wich has the properties:
1) $\forall x,y \in \Sigma^* : t_M(xy) \ge t_M(x) + t_M(y)$
2) $\forall a \in \Sigma: t_M(a) \ge 1$ (Also 2) should be obvious for every DTM).
Then it follows that for all $x \in \Sigma^* : t_M(x) \ge |x| $. The graph $G_M$ induced by the transition function contains a cycle: To see this choose a word $w$ whose length $|w|$ is $> |Q|$ where $Q$ is the set of states of $M$. Then we have $t_M(w) \ge |w| > |Q|$. Since $M$ is at every time step on exactly one state, $M$ must visit in $t_M(w) > |Q|$ time steps one state at least twice, hence the graph $G_M$ must contain a cycle.
My question is this: Can we construct to every DTM $M'$ an equivalent DTM $M$ with the properties above? In my intuition this is possible: Just construct $M$ such that it reads all the input, writes what it has read, move the pointer to the beginning of the word and then gives control to $M'$. But is it possible to give a more formal proof for this? Or is my intuition wrong? |
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· A) Calculate the magnetic field within a wire carrying a current I uniformly distributed throughout the wire. B) A circular hole is now drilled through the wire, offset from the wire's center.
14 The Magnetic Field in Various Situations. 14–1 The vector potential. ... A rotating charged cylinder produces a magnetic field inside. A short radial wire rotating with the cylinder has charges induced on its ends. Now we can raise an interesting question.
· The magnetic field inside the wire should depend on the distance r from the center of the wire. So, you should recheck your calculation. For next question, magnetic field in space between cylinder and wire I believe would be the same as part a.
14 The Magnetic Field in Various Situations. 14–1 The vector potential. ... A rotating charged cylinder produces a magnetic field inside. A short radial wire rotating with the cylinder has charges induced on its ends. Now we can raise an interesting question.
Magnetic Field of a Spinning Cylinder. Find the magnetic field of a standard solenoid and compare it to the magnetic field produced by a spinning cylinder with a uniform surface charge. 8.02 Physics II: Electricity and Magnetism, Spring 2007
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The K&J Magnetic Field Calculator calculates the magnetic field strength in Gauss near a disc or cylinder shaped neodymium magnet. Enter the size of the magnet and a position to measure the field strength. All data is provided in gauss. (1 tesla = 10,000 gauss)
The outer cylinder is a thin cylindrical shell of radius 2R and current 2I in a direction opposite to the current in the inner cylinder. In a cross-sectional view the current is out of the page in the inner cylinder and into the page in the outer cylinder. What is the magnetic field everywhere?
Magnetic Field Intensity of a Uniformly Magnetized Cylinder The cylinder shown in Fig. 9.3.1 is uniformly magnetized in the z direction, M = M o i z . The first step toward finding the resulting H within the cylinder and in the surrounding free space is an evaluation of the distribution of magnetic …
A cylinder of mass m = 250 g and length l = 10 cm is placed on an inclined plane of a rake angle α = 30 °.A turn loop of wire is wound around the cylinder in the longitudinal orientation. The cylinder is situated in a magnetic field of an induction B = 0.5 T.The magnetic field …
Magnetic field of steady currents. Magnetic fields produced by electric currents can be calculated for any shape of circuit using the law of Biot and ... A solenoid of this kind can be made by wrapping some conducting wire tightly around a long hollow cylinder. The value of the field is. where n is the number of turns per unit length of the ...
The net magnetic field on the axis of the cylindrical coil is the sum of the magnetic fields of all the loops.Divide the length of the cylinder into small elements of length to get the total field.The number of coil turns in a length is:
What Is Electromagnetism? Test. STUDY. PLAY. What shape is the magnetic field produced when current is passed through a wire? ... Just as in a bar magnet, the magnetic field of a wire with a current forms a cylinder shape around the wire.
The result can be expressed in terms of elliptic functions with complicated arguments and the expressions for the magnetic field components and are even more complicated. It is much easier to do the computations numerically, taking advantage of Mathematica's very efficient NIntegrate routines. We thereby obtain: and.
Let us find a much simpler approach, at first for the "cylinder shell current distribution". The Magnetic Field of the "Cylinder Shell Current Distribution" However, if we look at the current distribution, we see that the curl of the magnetic field can only have an angular component .
the magnetized cylinder, we find the magnetic field for the ball is proportional to mu_0 times the magnetization but there is an additional factor of 1/3 for the sphere which was not present for the cylinder.
14 The Magnetic Field in Various Situations. 14–1 The vector potential. ... A rotating charged cylinder produces a magnetic field inside. A short radial wire rotating with the cylinder has charges induced on its ends. Now we can raise an interesting question.
Magnetic Field - The strength of the B-field in Gauss. It is a measure of the magnetic field's effects on its environment. Surface Field - The Magnetic Field as measured or calculated on the surface of a permanent magnet, typically on the center axis of magnetization.
The Field near an Infinite Cylinder. ... We used Gauss' Law to show that the field inside the shell was zero, and outside the shell the electric field was the same as the field from a point charge with a charge equal to the charge on the shell and placed at the center of the shell. ... and outside the cylinder the magnetic field is the same as ...
The magnetic field outside an infinitely long solenoid is equal to zero (see Example 9, Chapter 5 of Griffiths), and therefore also the field outside the magnetized cylinder will be equal to zero. The magnetic field inside an infinitely long solenoid can be calculated easily using Ampere's law (see Example 9, Chapter 5 of Griffiths).
Circular Magnetic Fields Distribution and Intensity . As discussed previously, when current is passed through a solid conductor, a magnetic field forms in and around the conductor. The following statements can be made about the distribution and intensity of the magnetic field.
Let us find a much simpler approach, at first for the "cylinder shell current distribution". The Magnetic Field of the "Cylinder Shell Current Distribution" However, if we look at the current distribution, we see that the curl of the magnetic field can only have an angular component .
Magnetic Field - The strength of the B-field in Gauss. It is a measure of the magnetic field's effects on its environment. Surface Field - The Magnetic Field as measured or calculated on the surface of a permanent magnet, typically on the center axis of magnetization.
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Physics 217 Practice Final Exam: Solutions Fall 2002 ... Calculate the magnetic field everywhere inside the hole, and sketch the lines of B on the figure. Superpose a narrow wire with ... Then, with B+ as the field from the wide cylinder and B ...
The magnetic field of a hollow wire is calculated. Electrodynamics . Electrostatics ... Will we encounter a magnetic field? Problem Statement. A wire with radius \(R\) shall carry a constant current density \(j_{0}\). ... A Cylinder Shell Current and its Magnetic Field;
Magnetic field formulae www.vaxasoftware.com Magnetic field due to an infinite, straight current filament d i B 2 π μ0 On-axis field due to N current loops and radius r N r i B 2 μ0 Axial field of a finite, straight, thin shell solenoid of length L and N loops. N L i B μ0 Force for a moving charge inside a magnetic field F qv B
Magnetic Field Intensity of a Uniformly Magnetized Cylinder The cylinder shown in Fig. 9.3.1 is uniformly magnetized in the z direction, M = M o i z . The first step toward finding the resulting H within the cylinder and in the surrounding free space is an evaluation of the distribution of magnetic …
For the part inside the wire, check to see if the function makes sense: for a uniformly distributed current, the magnetic field grows linearly with the distance from the axis, so it makes sense that for this current it would grow like the square of the distance from the axis.
Magnetic Field Cylinder Mounting Accessories. MFC Accesories Catalog. Request A Quote Ask a Question. Click to learn more about our product specifications. Configure a Part Number & Download CAD files. Display This Product Catalog. Order Cylinders & Accessories Online. Magnetic Field Cylinders CAD Configurator. Ask A Question. Name * Email * Phone.
Magnetic field in a cylinder with an off-axis hole. Ask Question 1. 2 $\begingroup$ ... calculate the magnetic field at the centre of the cable, and at the centre of the hole. I understand this to be a problem involving superposition. I have managed to calculate the magnetic field in the hole to be $ \frac{\mu_0 Jd}{2} $ ... I thought to ...
Examples of magnetic field calculations and applications Lecture 12 1 Example of a magnetic moment calculation We consider the vector potential and magnetic field due to the magnetic moment created by a rotating surface charge, σ, on a cylinder. The geometry is shown in Figure 1. The
Part of the field change recorded is due to distortion of the Earth’s field. The steel in the cylinder has a high permeability, so magnetic field flux lines are attracted to run along the cylinder rather than through the air next to it. Therefore the direction of the Earth’s field affects the field changes made by the cylinder.
A solenoid is a long conductor that is densely wound to form a cylindrical helix. The length of the solenoid is much longer than its diameter, thus we can neglect the imperfections of the field …
Here, this pi and that pi will cancel, and we can cancel one of these r squares with the r on the left hand side, and leaving b alone we will end up with magnetic field inside of the inner cylinder as neu0 i sub a divided by 2pia square times r.
I need to find out what is the magnetic field inside a rotating long cylinder with a charged density $\sigma$. ... Magnetic field inside of a long rotating cylinder. ... I have a simple question. I need to find out what is the magnetic field inside a rotating long cylinder with a charged density $\sigma$. The cylinder rotates around its axis ...
The net magnetic field on the axis of the cylindrical coil is the sum of the magnetic fields of all the loops.Divide the length of the cylinder into small elements of length to get the total field.The number of coil turns in a length is:
A very long non-conducting cylinder has N conduct- ing wires placed tightly together around its circumfer- ence and running parallel to its axis as shown below:
· A) Calculate the magnetic field within a wire carrying a current I uniformly distributed throughout the wire. B) A circular hole is now drilled through the wire, offset from the wire's center.
The Field Inside a Current-Carrying Wire. Let's use Ampere's Law to find the field inside a long straight wire of radius R carrying a current I. Assume the wire has a uniform current per unit area: J = I/πR 2. To find the magnetic field at a radius r inside the wire, draw a circular loop of radius r. |
I am told that the MV CLT can be proved using the Cramér–Wold device.
The theorem is as follows (from Flury's "A First Course in Multivariate Statistics") Suppose $\bf{X}_1, \bf{X}_2, \ldots$, $\bf{X}_n$ are independent, identically, distributed, p-variate random vector, with mean vectors $\bf{µ}=E[\bf{X}_i]$ and covariance matrices $\sigma = Cov[\bf{X}_i]$. Let $\bf{\overline{X}_N} = \frac{1}{N}\sum_{i=1}^n\bf{X}_i$. Then $\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu}) \overset{d}{\to} N(0, \sigma))$.
Cramér–Wold says I can show the above result by showing
$$a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu}) \overset{d}{\to} a^t N(0,\sigma) = N(0, a\sigma a^t) \forall a \in \mathbb{R}^p$$
Through properties of expectation and variance I have shown
$$E\left(a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu})\right) = 0$$ and $$Var\left(a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu})\right) = a\sigma a^t$$
So whatever $a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu})$ converges to, it must have the properties I want. How can I guarantee it is a univariate normal?
It occurs to me that it might be useful to write
$$a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu}) = \sum_{i=1}^p\left(a_i \sqrt{n} (\overline{x}_{ni} - \mu_i)\right)$$
Then by the univariate central limit theorem, each term converges in distribution to some normal. Can we then say that their sum converges in distribution to some normal (we already know it will have the mean and variance we are looking for)? If not, what else shouldI be looking at? |
Let $\alpha:R\to S$ be a map of unital rings, and let $M$ be an $R$-module. We have a canonical map of $R$-modules:
$$\begin{array}{rcl}i:M&\longrightarrow&S\otimes_RM\\[.05in]m&\longmapsto&1\otimes m\end{array}$$
where the $S$-module $S\otimes_RM$ obtained by extension of scalars is considered an $R$-module via restriction along $\alpha$. Contrary to appearances, this map is not necessarily injective, as seen in the following counterexample:
Let $R=\mathbb{Z}$, $S=\mathbb{Z}/2\mathbb{Z}$, and let $M$ be the $\mathbb{Z}$-module = abelian group of order $3$ generated by $t\in M$. Then $2\cdot t=t^2\ne 0$, but $$i(2\cdot t)=1\otimes (2\cdot t)=\alpha(2)\otimes t=0\otimes t=0_{S\otimes_RM}$$
What conditions should we require of $R,S,\alpha,$ and $M$ so that $i$ is an injective map of $R$-modules? It would be nice to find equivalent conditions, but I'd appreciate sufficient or necessary conditions alone as well.
The above example shows that a necessary condition is that $\ker\alpha\subset \operatorname{Ann}_R(M)$. A sufficient condition is that $\alpha$ is injective and $M$ is a flat $R$-module. Any other results in this direction? |
Fisher information for sample $x$ in experiment $(\Omega, \mathcal{F}, P_\theta)$ is defined as $$Var \left[\nabla_{\theta}\ell(\theta, x) \right] = \mathbb{E}\left[[\nabla_{\theta} \ell(\theta, x)] [\nabla_{\theta}\ell(\theta, x)]^T\right] $$ where $\ell(\theta, x) = \log(f(x|\theta)$.
I do not understand how this definition is applied to a very basic and well known example: Let $x \sim U(0,\theta)$. In this case the probability density of $x$ is $$f(x_i|\theta) = \begin{cases}\frac{1}{\theta} & x\in [0, \theta]\\0 & \text{otherwise}\end{cases}$$
It looks to me that the density it is not differentiable with respect to $\theta$ at $\theta = x$ , consequently $\nabla_{\theta}\ell(\theta, x)$ is not defined for $\theta = x$.
I know this is a very basic example, but for some reason I couldn't find a full derivation of fisher information for this case, all I see is the stated result that it is $\frac{1}{\theta^2}$.
I would appreciate anyone pointing out to what do I miss here... Thanks! |
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Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Multiplicity dependence of two-particle azimuthal correlations in pp collisions at the LHC
(Springer, 2013-09)
We present the measurements of particle pair yields per trigger particle obtained from di-hadron azimuthal correlations in pp collisions at $\sqrt{s}$=0.9, 2.76, and 7 TeV recorded with the ALICE detector. The yields are ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV
(Springer, 2015-09)
We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ...
Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2015-07-10)
The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ... |
A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content.
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
triple poster
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#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
someone who likes this
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#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
someone who probably broke the record for the most posts in a row
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#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
someone who is still doing this
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#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
someone who will stop now
Code: Select all
#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
fluffykitty
Posts: 638 Joined: June 14th, 2014, 5:03 pm
Someone who posted 7 times in a row
I like making rules
A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact:
Someone who responded to someone who was wrong about breaking the record for most posts in a row (see here
).
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
Moosey Posts: 2486 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
A person who is being responded to by a person who is writing what this link
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I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
testitemqlstudop Posts: 1186 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
Somebody who used a self-reference instead of a previous-person-reference when the reference relations are relaxed.
A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact:
Somebody who I replied to.
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
testitemqlstudop Posts: 1186 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
Somebody who made the same aforementioned error that I attempted to correct by the second-previous post.
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
this post before reading the post itself.
Last edited by PkmnQ
on May 19th, 2019, 5:42 am, edited 1 time in total.
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who can't use the new paste rle feature of LifeViewer.
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who likes to size stack every now and then.
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who has posted 4 times in a row, including this one
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is going for 12 posts (5)
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is wondering why he did this (6)
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is already past the halfway mark (7)
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Alternating rule (9)
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who wants to have a profile picture (10)
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is almost done (11)
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 666 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who has posted
12
times in a row and is now done
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x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
testitemqlstudop Posts: 1186 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
PkmnQ wrote:this post before reading the post itself.
Someone who either doesn't have or chooses to ignore the rules of English
Saka
Posts: 3138 Joined: June 19th, 2015, 8:50 pm Location: In the kingdom of Sultan Hamengkubuwono X
A person that is trick work to.
Airy Clave White It Nay
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x = 17, y = 10, rule = B3/S23
b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5b
o2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo!
(Check gen 2) |
I need a data structure for storing a number $n$ of elements, each of whom is associated with some different time $t_i$. $n$ varies and while it has a theoretical upper limit, this is many orders of magnitude larger than what is typically used.
Through my application I can ensure that:
Inserted elements are always newer than all existing elements, i.e., if an element associated with a time $\check{t}$ is inserted, then $\check{t}>t_i ∀ i ∈ {1,…,n}$. Elements are inserted one by one.
Only the oldest elements are removed, i.e., if element $j$ is removed, then $t_j < t_i ~∀ i ∈ \lbrace 1,…,n \rbrace \setminus \lbrace j \rbrace$. Removals happen mostly one by one, but there is no direct harm if an element’s removal is delayed, as long as the fraction of spuriously stored elements remains smaller than 1.
Apart from inserting and removing, the only thing I need to do is find the two neighbouring elements for some given time $\tilde{t}$ with $\min\limits_i t_i < \tilde{t} < \max\limits_{i} t_i$. With other words I need to find the two elements $j$ and $k$ such that $t_j<\tilde{t}<t_k$ and $∄ l ∈ \{1,…,n\}: t_j<t_l<t_k$.
My criteria for the data structure are:
Finding elements as described above should be as quick as possible. Inserting and removing should be quick. The data structure is comparably simple to implement.
As long as we are not talking about a small runtime offset, each criterion has priority over the next.
My research so far has yielded that the answer is likely some kind of self-balancing search tree, but I failed to find any information which of them is best for the case of one-sided inserting or deleting, and it will probably cost me a considerable time to find out myself. Also, I only found incomplete information about how well the trees self-organise and how quickly (e.g., AVL trees self-organise more rigidly than red-black trees), let alone how this is affected by one-sided inserting or deleting. |
Vanishing points are an important concept in 3D vision. Many papers related to them are using a notation called Gaussian sphere representation, which I find hard to understand at the beginning. This document will summarize what vanishing points (and their “Gaussian sphere representation”) are, how to represent them, what information they encode, how to find them, and why they are useful.
WARNING: This article is for people who want to learn more about vanishing points. Basic knowledge in 3D vision is assumed (e.g., you know the existance of camera calibraion (intrinsic) matrices).
Introduction
With perspective geometry,
all 3D parallel lines will intersect at the same point in 2D at their vanishing points when they are projected to an image. To see this, we can write down the equation of a 3D ray. Any point on a ray can be represented with \(\textbf{o}+\lambda \mathbf{d}\), in which \(\bf o\) is the origin and \(\mathbf{d}\) is the direction of the line. Let \(K\) be the camera calibration matrix, we have its 2D projection being
\[ z \begin{bmatrix} p_x\\p_y\\1 \end{bmatrix} = K \cdot (\mathbf{o}+\lambda \mathbf{d}). \]
The vanishing point \(\bf p\) is defined to be the
projection of a line at infinity. To findwhere \(\bf p\) is, we set \(\lambda \to \infty\). The contribution of \(\textbf{o}\) to theprojection point \(\bf p\) then becomes negligible. Therefore the vanishing point \(\mathbf{p}\sim K \cdot \mathbf{d}\) is uniquely determined by the line direction \(\mathbf{d}\).
The above derivation also tells us that if we can detect the vanishing point \(\bf p\) in an image, we cancompute the direction \(\mathbf{d}\) of those parallel lines in 3D! Isn't it amazing that wecan
infer the 3D information from single 2D images? In the next section, I will present an elegantand numerically stable way to compute \(\bf d\) from two 3D parallel lines. Gaussian Sphere Representation for Vanishing Points
Now let us say you have two parallel lines \(\ell_1 \mathbin{\!/\mkern-5mu/\!} \ell_2\) in 3D, whose endpoints are \((u_{1},u_{2})\) and \((v_{1},v_{2})\) when projected to the image plane \(z=1\), assuming a calibrated camera. Let the camera position \(o=(0,0,0)\). Then the normals of \(\triangle ou_{1}u_{2}\) and \(\triangle ov_{1}v_{2}\) are
\[ \begin{align} \mathbf{n}_{u} &= u_{1}\times u_{2}\\ \mathbf{n}_{v} &= v_{1}\times v_{2}. \end{align} \]
Any lines in the plane \(\triangle ou_{1}u_{2}\) must be orthogonal to its normal \(\mathbf{n}_{u}\). The same applies to the lines in the plane \(\triangle ov_{1}v_{2}\). Let \(\mathbf{d}\) be the direction vector of parallel line \(\ell_1\) and line \(\ell_2\). This orthogonal condition means that \(\mathbf{n}_{u} \perp \mathbf{d} \perp \mathbf{n}_{v}\), so we have
\[ \mathbf{d}=\mathbf{n}_{u}\times \mathbf{n}_{v}. \]
In addition, the intersection of \(\mathbf{d}\) with image plane \(z=1\) is also the intersectionpoint \(\bf p\) of line \((u_{1},u_{2})\) and line \((v_{1},v_{2})\) (read introduction sectionagain if you do not understand this). Traditionally, the intersection point $\bf p$ (in the imagespace) is referred as the
vanishing point, while the direction vector \(\mathbf{d}\) is calledthe Gaussian sphere representation of the vanishing point (IMHO a confusing name).
You can see that the vector \(\mathbf{d}\) is indeed a better representation compared to using the intersection point on the image plane. First, if \(\mathbf{d}\) does not intersect with the focal plane at all, e.g., when it is vertical, the vanishing point \(\bf p\) will be near the infinite, which is ill-condition. Second, it is hard to define a metric to measure the distance between two vanishing points. The common-used Euclidean distance is not a good metric if the camera calibration is known. With the direction vector \(\mathbf{d}\), we can use the angle between them
\[ D\left(\mathbf{d}_{1},\mathbf{d}_{2}\right)=\left\langle \frac{\mathbf{d}_{1}}{\left\Vert \mathbf{d}_{1}\right\Vert },\frac{\mathbf{d}_{2}}{\left\Vert \mathbf{d}_{2}\right\Vert }\right\rangle \]
as a more reasonable distance metric.
Vanishing Point Detection
Here, I give a RANSAC-based vanishing point detection algorithm that works with the Manhattan scenes, i.e., scenes with three orthogonal vanishing points, when the camera calibration is known. The algorithm starts with a set of detected 2D lines that can be found with LSD and then does RANSAC:
Random select three lines \(\ell_1\), \(\ell_2\), and \(\ell_3\). We assume that \(\ell_1 \mathbin{\!/\mkern-5mu/\!} \ell_2\) and \(\ell_1 \perp \ell_3 \perp \ell_2\) in 3D;
We compute the Gaussian sphere representation \(\mathbf{d}_1\) of the vanishing point for lines \(\ell_1\) and \(\ell_2\) with their endpoints using the method from previous section;
Compute \({\bf d}_2\) with the endpoints of \(\ell_3\). For the another line (that does not exist), we set normal of imaginary triangle \(\triangle ov_{1}v_{2}\) to be \(\mathbf{n}_v = \mathbf{d}_1\);
Compute \({\bf d}_3 = {\bf d}_1 \times {\bf d}_2\);
Check the consistency using RANSAC for vanishing points \(({\bf d}_1,{\bf d}_2,{\bf d}_3)\) with other lines;
Repeat 1-5 until a good set of vanishing points are found.
For scenes that are not Manhattan or the camera is not calibrated, a classic way is to use the combination of LSD + J-linkage. Readers can refer to the J-linkage clustering algorithm for more details.
Application: Photo Forensics (Camera Calibration)
Photo forensics is an art to detect whether a photo was tampered by Photoshop. In this example, wewill check if a photo was
cropped. This is equivalent to recovering the camera calibrationmatrix
\[ K=\begin{bmatrix} f & 0 & o_{x}\\ & f & o_{y}\\ & & 1 \end{bmatrix}. \]
Why \(K\) can tell us whether an image was cropped or not? This is because for common cameras and lens, the choice of the focal length \(f\) is often limited and the offset of the camera center \((o_x,o_y)\) should be around the center of the image. An image that has been arbitrarily cropped would violate those properties.
One way to recover \(K\) is from vanishing points. In order to successfully recover the cameracalibration, we need to know at least three
orthogonal vanishing points. Finding those vanishingpoints are not that hard, it requires the user to label three pair of parallel lines that areperpendicular to each other in 3D. Each pair of parallel lines uniquely determines a vanishingpoint. Three pair of parallel lines are easy to find in man-made environment, as objects tend to berectangular and cuboid.
Once we know three orthogonal vanishing points, we are able to recover the camera's calibration matrix \(K\). Let our 2D lines be \((\hat{u}_{1},\hat{u}_{2})\) and \((\hat{v}_{1},\hat{v}_{2})\) in the image space, in which the \(x\) and \(y\) coordinate of those variables are the pixel coordinates and \(z\) coordinates are 1. According to the camera's perspective projection model, we have
\[ \begin{align*} z_{u_i}\hat{u}_{i} & =Ku_{i}\\ z_{v_i}\hat{v}_{i} & =Kv_{i}, \end{align*} \]
where \(u_{i}\) and \(v_{i}\) are the coordinates in the calibrated image space for \(i \in \{1,2\}\). Then the direction vector
\[ \begin{align*} \mathbf{d} & =\mathbf{n}_{u}\times \mathbf{n}_{v}\\ & =\left(u_{1}\times u_{2}\right)\times\left(v_{1}\times v_{2}\right)\\ & \propto \left[\left(K^{-1}\hat{u}_{1}\right)\times\left(K^{-1}\hat{u}_{2}\right)\right]\times\left[\left(K^{-1}\hat{v}_{1}\right)\times\left(K^{-1}\hat{v}_{2}\right)\right]\\ & \stackrel{1}{=}\left[K^{T}\left(\hat{u}_{1}\times\hat{u}_{2}\right)\right]\times\left[K^{T}\left(\hat{v}_{1}\times\hat{v}_{2}\right)\right]\\ & \stackrel{2}{=}K^{-1}\left[\left(\hat{u}_{1}\times\hat{u}_{2}\right)\times\left(\hat{v}_{1}\times\hat{v}_{2}\right)\right]\\ & =: K^{-1}\left(\hat{\bf n}_{u}\times\hat{\bf n}_{v}\right)\\ & =: K^{-1}\hat{\bf d}. \end{align*} \]
Here, \(=:\) represents “define as” and in \(\stackrel{1}{=}\) and \(\stackrel{2}{=}\) we use a useful conclusion for cross product from Yi Ma's “An Invitation to 3D Vision” Exercise 2.4:
\[ K^{-T}\widehat{\omega}K^{-1}=\widehat{K\omega} \]
when \(K\) is invertible. See Yi's Book for more details. Notice that \(\hat{\mathbf{d}}\), \(\hat{n}_{u}\) and \(\hat{n}_{v}\) are known given the lines in the pixel coordinate. Now if we have three orthogonal vanishing points \(\mathbf{d}_{1}\perp \mathbf{d}_{2}\perp \mathbf{d}_{3}\), we can write down the equation
\[ \begin{align*} \hat{\bf d}^T_{1}K^{-T}K^{-1}\hat{\bf d}_{2} & =0\\ \hat{\bf d}^T_{2}K^{-T}K^{-1}\hat{\bf d}_{3} & =0\\ \hat{\bf d}^T_{3}K^{-T}K^{-1}\hat{\bf d}_{1} & =0. \end{align*} \]
Notice that
\[ S := K^{-T}K^{-1}=\frac{1}{f^{2}}\begin{bmatrix}1 & 0 & -o_{x}\\ 0 & 1 & -o_{y}\\ -o_{x} & -o_{y} & o_{x}^{2}+o_{y}^{2}+f^{2} \end{bmatrix} \]
so that the previous three equations are linear with respective to \(o_{x}\), \(o_{y}\) and \(o_x^{2}+o_y^{2}+f^{2}\). If we have more than three vanishing points, we can do a lasso optimization to optimize the L1 norm (you can use L2 for convenience, but L1 is generally more “robust”)
\[ \min_{f,v_{x},v_{y}}\sum_{i,j}\ \left|\hat{\bf d}^{T}_{i}S\hat{\bf d}_{j}\right| \]
to find the best camera parameters.
Acknowledgement
Thanks Cecilia Zhang for the discussion about vanishing points and photo forensics. Thanks Prof. Yi Ma for the discussion about the relationship between vanishing points and camera calibration. |
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hep-ph
Change to browse by: Bookmark(what is this?) General Relativity and Quantum Cosmology Title: Conical singularities and the Vainshtein screening in full GLPV theories
(Submitted on 21 Dec 2015 (v1), last revised 2 Mar 2016 (this version, v2))
Abstract: In Gleyzes-Langlois-Piazza-Vernizzi (GLPV) theories, it is known that the conical singularity arises at the center of a spherically symmetric body ($r=0$) in the case where the parameter $\alpha_{{\rm H}4}$ characterizing the deviation from the Horndeski Lagrangian $L_4$ approaches a non-zero constant as $r \to 0$. We derive spherically symmetric solutions around the center in full GLPV theories and show that the GLPV Lagrangian $L_5$ does not modify the divergent property of the Ricci scalar $R$ induced by the non-zero $\alpha_{{\rm H}4}$. Provided that $\alpha_{{\rm H}4}=0$, curvature scalar quantities can remain finite at $r=0$ even in the presence of $L_5$ beyond the Horndeski domain. For the theories in which the scalar field $\phi$ is directly coupled to $R$, we also obtain spherically symmetric solutions inside/outside the body to study whether the fifth force mediated by $\phi$ can be screened by non-linear field self-interactions. We find that there is one specific model of GLPV theories in which the effect of $L_5$ vanishes in the equations of motion. We also show that, depending on the sign of a $L_5$-dependent term in the field equation, the model can be compatible with solar-system constraints under the Vainshtein mechanism or it is plagued by the problem of a divergence of the field derivative in high-density regions. Submission historyFrom: Ryotaro Kase [view email] [v1]Mon, 21 Dec 2015 05:17:16 GMT (87kb) [v2]Wed, 2 Mar 2016 05:34:49 GMT (88kb) |
Pitch and Frequency
Sound is vibration transmitted through a medium, i.e. a solid, liquid or gas. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. The frequency is most often measured in
hertz(Hz). One hertz means that an event repeats once per second, in case of wave, one cycle per second. Twelve Tone Equal Temperament
Twelve-tone equal temperament divides the octave into 12 equal parts(semitones). The frequency ratio of two adjacent notes is the twelfth root of two:\(12\sqrt2=2^{1⁄12}\approx1.059463\).The frequency \(f\) of the n
th key on a piano is
$$f(n)=2^\frac{n-49}{12}\times(A=440Hz)$$
* While A=440Hz is widely used and accepted as standard, slightly different frequencies are used in some cases. See Concert pitch(Wikipedia) for reference.
Just Intonation
In just intonation, the frequencies of notes are defined by ratios of small whole numbers, e.g.
Major 2nd: \(\frac98\) (= \(\frac32\times\frac32\times\frac12\) = an octave below the perfect 5th of the perfect 5th)
Major 3rd: \(\frac54\)
Perfect 4th: \(\frac43\) (= \(\frac21/\frac32\) = perfect 5th below the octave)
Perfect 5th: \(\frac32\)
Major 6th: \(\frac53\) (= \(\frac54\times\frac43\) = the perfect 4th of the major 3rd)
Major 7th: \(\frac{15}8\) (= \(\frac54\times\frac32\) = the perfect 5th of the major 3rd)
Octave: \(\frac21\) |
Research Open Access Published: Perturbational blowup solutions to the compressible Euler equations with damping SpringerPlus volume 5, Article number: 196 (2016) Article metrics
472 Accesses
Abstract Background
The
N-dimensional isentropic compressible Euler system with a damping term is one of the most fundamental equations in fluid dynamics. Since it does not have a general solution in a closed form for arbitrary well-posed initial value problems. Constructing exact solutions to the system is a useful way to obtain important information on the properties of its solutions. Method
In this article, we construct two families of exact solutions for the one-dimensional isentropic compressible Euler equations with damping by the perturbational method. The two families of exact solutions found include the cases \(\gamma >1\) and \(\gamma =1\), where \(\gamma\) is the adiabatic constant.
Results
With analysis of the key ordinary differential equation, we show that the classes of solutions include both blowup type and global existence type when the parameters are suitably chosen. Moreover, in the blowup cases, we show that the singularities are of essential type in the sense that they cannot be smoothed by redefining values at the odd points.
Conclusion
The two families of exact solutions obtained in this paper can be useful to study of related numerical methods and algorithms such as the finite difference method, the finite element method and the finite volume method that are applied by scientists to simulate the fluids for applications.
Background and Main results
The
N-dimensional isentropic compressible Euler equations with a damping term are written as
where \(\rho (t,x):[0,\infty )\times \mathbb {R}^N\rightarrow [0,\infty )\) and \(u(t,x):[0,\infty )\times \mathbb {R}^N\rightarrow \mathbb {R}^N\) represent the density and the velocity of the fluid respectively.
p represents the pressure function, which is given by
by the adiabatic \(\gamma\)-law. The constant \(\alpha \ge 0\) is the damping coefficient.
System (1) is one of the most fundamental equations in fluid dynamics. Many interesting fluid dynamic phenomena can be described by system (1) (Lions , 1998a; Lions , 1998b). The Euler equations \((\alpha =0)\) are also the special case of the noted Navier–Stokes equations, whose problem of whether there is a formation of singularity is still open and long-standing. Thus, the singularity formation in fluid mechanics has been attracting the attention of a number of researchers (Sideris 1985; Xin 1998; Suzuki 2013; Lei et al. 2013; Li and Wang 2006; Li et al. 2013).
Among others, we mention that in 2003, Sideris–Thomases–Wang (Sideris et al. 2003) obtained results for the three dimensional compressible Euler equations with a linear damping term with assumption \(\gamma >1\), that is, system (1) with \(N=3\) and \(\gamma >1\). They discovered that damping prevents the formation of singularities in small amplitude flows, but large solutions may still break down. They formulated the Euler system as a symmetric hyperbolic system, established the finite speed of propagation of the solution, and some energy estimates to obtain local existence as well as global existence of the solution. For larger solution, they showed that the solution will blow up in a finite time by establishing certain differential inequalities.
In this article, we consider the one dimensional case of system (1):
More precisely, we apply the perturbational method to obtain the following main results.
Theorem 1 For system (3) with \(\gamma >1\) and \(\alpha >0\), one has the following family of exact solutions with parameters \(\xi , \rho (0,0)>0, a_0>0\) and \(a_1\): where \(\rho ^{\gamma -1}(t,0)\) is given by a( t) and b( t) satisfy the following ordinary differential equations: Remark 2
The ordinary differential equation (O.D.E.) (6) will be analyzed in section 2 and it is well-known by the theory of ordinary differential equations that the solutions of system (7) exist and is \(C^2\) as long as
f and g, which are functions of \(\ddot{a}\), \(\dot{a}\) and a, are continuous. Theorem 3 For the family of exact solutions in Theorem 1, we have the following five cases. (i): If\(\xi >0\) and\(a_1\ge 0\), then the solution(4) is a global solution. (ii): If\(\xi >0, a_1<0\) and\(a_0>-a_1/\alpha\), then the solution(4) is a global solution. (iii): If\(\xi <0\), then the solution(4) blows up on a finite time. (iv): If\(\xi =0\) and\(a_1>0\), then the solution(4) blows up on the finite time\(T=\frac{1}{\alpha }\ln \frac{a_1}{a_1+a_0\alpha }>0\). (v): If\(\xi =0\), and\(a_1<0\) and\(a_0<-a_1/\alpha\), then the solution(4) blows up on the finite time\(T=\frac{1}{\alpha }\ln \frac{a_1}{a_1+a_0\alpha }>0\).
Moreover, we show that the singularity formations in the cases
iii), iv) and v) above are of essential type in the sense that the singularities cannot be smoothed by redefining values at the odd points. This is an improvement of the corresponding results in Yuen (2011).
For \(\gamma =1\), we obtain the following theorem.
Theorem 4 For system (3) with \(\gamma =1\) and \(\alpha >0\), one has the following family of exact solutions with parameters \(\xi , \rho (0,0)>0, a_0>0\) and \(a_1\). where a( t) and b( t) satisfy the following ordinary differential equations: Theorem 5 For the family of exact solutions in Theorem 4, we have the following five cases. (i): If\(\xi >0\) and\(a_1\ge 0\), then the solution(8) is a global solution. (ii): If\(\xi >0, a_1<0\) and\(a_0>-a_1/\alpha\), then the solution(8) is a global solution. (iii): If\(\xi <0\), then the solution(8) blows up on a finite time. (iv): If\(\xi =0\) and\(a_1>0\), then the solution(8) blows up on the finite time\(T=\frac{1}{\alpha }\ln \frac{a_1}{a_1+a_0\alpha }>0\). (v): If\(\xi =0\), and\(a_1<0\) and\(a_0<-a_1/\alpha\), then the solution(8) blows up on the finite time\(T=\frac{1}{\alpha }\ln \frac{a_1}{a_1+a_0\alpha }>0\). Analysis of an O.D.E.
Consider the following initial value problem.
where \(\gamma \ge 1\), \(\alpha >0\) and \(\xi \in \mathbb {R}\) are constants. We set
Lemma 6 For system (12), if \(T^*\) is finite, then the one-sided limit Proof
Note that we always have
Suppose \(\displaystyle \lim _{t\rightarrow T^*}a(t)>0\). Then we can extend the solution of (12) to \([0, T^*+\varepsilon )\) by solving the following system.
This contradicts the definition of \(T^*\). Thus, the lemma is established. \(\square\)
Lemma 7 For system (12), we have the following three cases. Case 1. If\(\xi >0 \quad \text {and} \quad a_1\ge 0, \quad \text{then} \quad T^*=+\infty\) . Case 2. If\(\xi >0, a_1<0 \quad \text {and} \quad a_0>-a_1/\alpha , \quad \text {then} \quad T^*=+\infty\) . Case 3. If\(\xi <0, \quad \text {then} \quad T^*<+\infty\). Proof
Suppose \(\xi >0\) and \(T^*<+\infty\). Then, for all \(t\in [0,T^*)\), we have
It follows that if \(a_1\ge 0\), then \(\displaystyle \lim _{t\rightarrow T^*}a(t)>0\), and if \(a_1<0\) and \(a_0>-a_1/\alpha\), then \(\displaystyle \lim _{t\rightarrow T^*}a(t)>0\). This is impossible by Lemma 6. Thus, Case 1. and Case 2. of the lemma are established.
Now, suppose \(\xi <0\). If \(T^*=+\infty\), then for all \(t>0\), we have
Thus,
Thus,
As \(A/\alpha <0\), we have \(a(t)<0\) for all sufficiently large
t. This is impossible as \(T^*=+\infty\). Thus, Case 3. is established. \(\square\) Remark 8
The case for \(\xi =0\) will be analyzed in the proof of Theorem 3.
Proofs of the Theorems Proof of Theorem 1
We divide the proof into steps.
Step 1. In the first step, we show a lemma. Lemma 9 For the 1- dimensional Euler equations with damping (3) with \(\gamma >1\) and \(\rho (0,0)>0\), we have the following relation. Proof of Lemma 9
It is well known that \(\rho\) is always positive if \(\rho (0,0)\) is set to be positive. From (3)\(_2\), we have, for \(\rho (0,0)>0\),
Thus, we have
Taking integration with respect to
x, we obtain
On the other hand, multiplying \(\rho ^{\gamma -2}\) on both sides of (3)\(_1\), we get
From (33), we have
and
Step 2. We set
where \(c:=c(t)\) and \(b:=b(t)\) are functions of
t. Then, (29) is transformed to
where we arrange the terms according to the coefficients of
x. Step 3. We use the Hubble transformation:
and set the coefficient of (38) to be zero. Thus,
Note that we have the novel identity
Multiplying the both sides of (40) by \(a^{\gamma +1}\), it becomes
for some constant \(\xi\).
where
where
Next, we prove Theorem 3 as follows.
Proof of Theorem 3
For \(\xi <0\), by Case 3. of Lemmas 6 and7, there exists a finite \(T^*>0\) such that the one-sided limit of
a( t) is zero as t approaches to \(T^*\). It remains to show \(T^*\) is not a removable singularity of \(\dot{a}/a\). To this end, suppose one has
Then,
Thus, the singularity is of essential type and case
iii) of Theorem 3 is proved.
For \(\xi =0\), (6)\(_1\) becomes
which can be solved by using integral factor. The solution is
Thus, \(a(T)=0\) if \(a_1>0\). Also, \(a(T)=0\) if \(a_1<0\) and \(a_0<-a_1/\alpha\), where \(T:=\frac{1}{\alpha }\ln \frac{a_1}{a_1+a_0\alpha }>0\). As
(
T, x) is an essential singularity of u( t, x) for any x. Thus, cases iv) and v) of Theorem 3 are established. The proof is complete. \(\square\) Proof of Theorems 4 and 5
The corresponding relation of Lemma 9 for \(\gamma =1\) is
With similar steps, one can obtain the family of exact solutions in Theorem 4.
Conclusion
The complicated Euler equations with a damping term (1) do not have a general solution in a closed form for arbitrary well-posed initial value problems. Thus, numerical methods and algorithms such as the finite difference method, the finite element method and the finite volume method are applied by scientists to simulate the fluids for applications in real world. Thus, our exact solutions in this article provide concrete examples for researchers to test their numerical methods and algorithms.
References
Lei Z, Du Y, Zhang QT (2013) Singularities of solutions to compressible Euler equations with vacuum. Math Res Lett 20:41–50
Li D, Miao CX, Zhang XY (2013) On the isentropic compressible Euler equation with adiabatic index \(\gamma =1\). Pac J Math 262:109–128
Li T, Wang D (2006) Blowup phenomena of solutions to the Euler equations for compressible fluid flow. J Diff Equ 221:91–101
Lions PL (1998) Mathematical topics in fluid mechanics, vol 1. Clarendon Press, Oxford
Lions PL (1998) Mathematical topics in fluid mechanics, vol 2. Clarendon Press, Oxford
Sideris TC (1985) Formation of singularities in three-dimensional compressible fluids. Commun Math Phys 101:475–485
Sideris TC, Thomases B, Wang D (2003) Long time behavior of solutions to the 3D compressible Euler equations with damping. Comm Partial Differ Equ 28:795–816
Suzuki T (2013) Irrotational blowup of the solution to compressible Euler equation. J Math Fluid Mech 15:617–633
Xin ZP (1998) Blowup of smooth solutions to the compressible Navier–Stokes equations with compact density. Commun Pure Appl Math LI:0229–0240
Yuen MW (2011) Perturbational blowup solutions to the compressible 1-dimensional Euler equations. Phys Lett A 375:3821–3825
Acknowledgements
This research paper is partially supported by the Grant: MIT/SRG02/15-16 from the Department of Mathematics and Information Technology of the Hong Kong Institute of Education.
Competing interests
The authors declare that they have no competing interests.
About this article Received Accepted Published DOI Keywords Blowup Global existence Euler equations Perturbational method Damping Singularity Mathematics Subject Classfication 35Q53 35B44 35C05 35C06 |
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Fri, 26 Apr 2019
What is the shed in “watershed”? Is it a garden shed? No.
I guessed that it meant a piece of land that sheds water into some stream or river. Wrong!
The Big Dictionary says that this shed is:
This meaning of “shed” fell out of use after the end of the 17th century.
This week I learned that there are no fewer than
I've said here before that I don't usually find written materialfunny, with very rare exceptions. But this story,
(Caution: sexual content.)
[ Addendum: However, I still demand to know: Where the hell is my Sonar Taxlaw fanfic? Fanfic writers of the world, don't think this gets you off the hook! ]
This is definitely the worst thing I learned this month. It's way worse than that picture of Elvis meeting Nixon.
Nobel Laureate and noted war criminal Henry Kissinger is also an honorary member of the Harlem Globetrotters.
As Maciej Cegłowski said, “And yet the cruel earth refuses to open and swallow up everyone involved.”
Katara was given the homework exercise of rationalizing the denominator of $$\frac1{\sqrt2+\sqrt3+\sqrt5}$$ which she found troublesome. You evidently need to start by multiplying the numerator and denominator by !!-\sqrt2 + \sqrt 3 + \sqrt 5!!, obtaining $$ \frac1{(\sqrt2+\sqrt3+\sqrt5)}\cdot \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{-\sqrt2 + \sqrt 3 + \sqrt 5} = \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{(-2 +3 + 5 + 2\sqrt{15})} = \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{6 + 2\sqrt{15}} $$ and then you go from there, multiplying the top and bottom by !!6 - 2\sqrt{15}!!. It is a mess.
But when I did it, it was much quicker. Instead of using !!-\sqrt2 + \sqrt 3 + \sqrt 5!!, I went with !!\sqrt2 + \sqrt 3 - \sqrt 5!!, not for any reason, but just at random. This got me: $$ \frac1{\sqrt2+\sqrt3+\sqrt5}\cdot \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{\sqrt2 + \sqrt 3 - \sqrt 5} = \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{(2 +3 - 5 + 2\sqrt{6})} = \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{2\sqrt{6}} $$
with the !!2+3-5!! vanishing in the denominator. Then the next step is quite easy; just get rid of the !!\sqrt6!!: $$ \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{2\sqrt{6}}\cdot \frac{\sqrt6}{\sqrt6} = \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12} $$
which is correct.
I wish I could take credit for this, but it was pure dumb luck.
It's often pointed out that women, even famous and accomplished women, are often described in newspaper stories as being someone's wife, but that the reverse rarely occurs. The only really well-known exception I could think of was Pierre Curie, who was a famous, prizewinning scientist (1903 Nobel Laureate, yo), but is often identified as having been the husband of Marie Skłodowska Curie (also 1903 Nobel Laureate).
But last week brought another example to my attention. There ware a great many news articles reporting that Salma Hayek's husband had pledged money to help rebuild Notre Dame cathedral. His name is François-Henri Pinault, and he is a billionaire. And the husband of Salma Hayek.
For example:
(etc.)
[ Addendum 20190808: Walt Mankowski brings up the excellent example of Sir Max Mallowan, CBE, a famous archaeologist and one of the original excavators of Ur. However, he is better known for having been the husband of Dame Agatha Christie from 1930 until he death in 1976. ]
Mon, 01 Apr 2019
This is the “rook” that is a sort of crow,
Okay, I've known all this for years, but today I had another thought.Why were there chariots in the
The earliest forerunner of chess was played in India. But I confirmed with Wikipedia that it didn't overlap with chariots:
Were the Guptas still using chariots in the 6th century? (And if so, why?) I think they weren't, but I'm not sure. Were the chariots intentionally anachronistic, even at the time the game was invented, recalling a time of ancient heroes? |
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Wed, 18 Sep 2019
Suppose you have a bottle that contains !!N!! whole pills. Each day you select a pill at random from the bottle. If it is a whole pill you eat half and put the other half back in the bottle. If it is a half pill, you eat it. How many half-pills can you expect to have in the bottle the day after you break the last whole pill?
Let's write !!E(N)!! for the expected number of half-pills. It's easily seen that !!E(N) = 0, 1, \frac32!! for !!N=0,1,2!!, and it's not hard to calculate that !!E(3) = \frac{11}{6}!!.
For larger !!N!! it's easy to use Monte Carlo simulation, and find that !!E(30)!! is almost exactly !!4!!. But it's also easy to use dynamic programming and compute that $$E(30) = \frac{9304682830147}{2329089562800}$$ exactly, which is a bit less than 4, only !!3.994987!!. Similarly, the dynamic programming approach tells us that $$E(100) = \frac{14466636279520351160221518043104131447711}{2788815009188499086581352357412492142272}$$ which is about !!5.187!!.
(I hate the term “dynamic programming”. It sounds so cool, but then you find out that all it means is “I memoized the results in a table”. Ho hum.)
As you'd expect for a distribution with a small mean, you're much more likely to end with a small number of half-pills than a large number. In this graph, the red line shows the probability of ending with various numbers of half-pills for an initial bottle of 100 whole pills; the blue line for an initial bottle of 30 whole pills, and the orange line for an initial bottle of 5 whole pills. The data were generated by this program.
The !!E!! function appears to increase approximately logarithmically. It first exceeds !!2!! at !!N=4!!, !!3!! at !!N=11!!, !!4!! at !!N=31!!, and !!5!! at !!N=83!!. The successive ratios of these !!N!!-values are !!2.75, 2.81,!! and !!2.68!!. So we might guess that !!E(N)!! first exceeds 6 around !!N=228!! or so, and indeed !!E(226) < 6 < E(227)!!. So based on purely empirical considerations, we might guess that $$E(N) \approx \frac{\log{\frac{15}{22}N}}{\log 2.75}.$$
(The !!\frac{15}{22}!! is a fudge factor to get the curves to line up properly.)
I don't have any theoretical justification for this, but I think it might be possible to get a bound.
I don't think modeling this as a Markov process would work well. There are too many states, and it is not ergodic.
[ Addendum 20190919: Ben Handley informs me that !!E(n)!! is simply the harmonic numbers, !!E(n) = \sum_1^n \frac1n!!. I feel a little foolish that I did not notice that the first four terms matched. The appearance of !!E(3)=\frac{11}6!! should have tipped me off. Thank you, M. Handley. ]
[ Addendum 20190920: I was so fried when I wrote this that I alsodidn't notice that the denominator I guessed, !!2.75!!, is almostexactly !!e!!. (The
[ Addendum 20191004: More about this ] |
Some mathematical elements change their style depending on the context, whether they are in line with the text or in an equation-type environment. This article explains how to manually adjust the display style.
Let's see an example
Depending on the value of $x$ the equation \( f(x) = \sum_{i=0}^{n} \frac{a_i}{1+x} \) may diverge or converge. \[ f(x) = \sum_{i=0}^{n} \frac{a_i}{1+x} \]
Superscripts, subscripts and fractions are formatted differently.
The maths styles can be set explicitly. For instance, if you want an in-line mathematical element to display as a equation-like element put
\displaystyle before that element. There are some more maths style-related commands that change the size of the text.
In-line maths elements can be set with a different style: \(f(x) = \displaystyle \frac{1}{1+x}\). The same is true the other way around: \begin{eqnarray*} \begin{eqnarray*} f(x) = \sum_{i=0}^{n} \frac{a_i}{1+x} \\ \textstyle f(x) = \textstyle \sum_{i=0}^{n} \frac{a_i}{1+x} \\ \scriptstyle f(x) = \scriptstyle \sum_{i=0}^{n} \frac{a_i}{1+x} \\ \scriptscriptstyle f(x) = \scriptscriptstyle \sum_{i=0}^{n} \frac{a_i}{1+x} \end{eqnarray*} \end{eqnarray*}
For more information see |
I am attempting to derive equations 2 and 6 from Xiao et al. paper "Valley contrasting physics in graphene" (Link to paper). The Hamiltonian for graphene with a staggered sublattice potential (in other words, a potential energy difference between the A and B sublattices) is given by: $$H = \frac{\sqrt{3}}{2}at(q_x\tau_z\sigma_x+q_y\sigma_y)+\frac{\Delta}{2}\sigma_z$$ Where $a=\text{lattice constant}$, $t=\text{hopping energy}\approx2.82eV$, $\sigma_x,\sigma_y,\sigma_z$ are our Pauli matrices and $\Delta$ is the energy difference between the A and B sublattice with the A sublattice having an additional on site energy of $\frac{\Delta}{2}$ and the B sublattice an additional on site energy of $\frac{-\Delta}{2}$. Also, $\tau_z$ means $\pm1$ and is used to select between the two valleys/Dirac points in the Brillouin zone.
From this Hamiltonian we should be able to get everything that we want to know about the Berry curvature and Bloch magnetic moment. These equations are given in the text (but not derived there or in any of their cited texts) as follows: ($m(k) \ \text{is the magnetic moment and,} \ \Omega(q) \ \text{is the Berry curvature}$):$$m(k)=\tau_z\frac{3ea^2{\Delta}t^2}{4\hbar({\Delta^2}+3q^2a^2t^2)}$$$$\Omega(q)=\tau_z\frac{3a^2{\Delta}t^2}{2({\Delta^2}+3q^2a^2t^2)^{3/2}}$$It should be noted that both the magnetic moment of the Bloch electron and the Berry curvature are vector quantities and that $q^2$ is the
magnitude of the crystal momentum. Also, the Berry curvature equation listed above is for the conduction band. I should also mention at this point that Xiao has a habit of switching between k and q, with q being the crystal momentum measured relative to the valley in graphene.
With this information in hand I will attempt to derive these equations with the help of Mathematica. However, I will fail and this is where I am hoping to get some help from all of you.
Step 1: First we need to determine the Eigenvalues/Dispersion relationship near the valleys (aka Dirac points of graphene). To do this all we need to do is find the eigenvalues of our Hamiltonian, which has the following form when represented as a matrix: $$H=\begin{pmatrix} \frac{\Delta}{2} & \frac{\sqrt{3}}{2}at({\tau_z}q_x-iq_y) \\ \frac{\sqrt{3}}{2}at({\tau_z}q_x+iq_y) & \frac{-\Delta}{2} \\ \end{pmatrix} $$ Using Mathematica (or I suppose you could also do this by hand relatively easily) we find that the eigenvalues of this matrix are: $${\pm}{\frac{1}{2}}\sqrt{\Delta^2+3a^2t^2q_x^2+3a^2t^2q_y^2}={\pm}{\frac{1}{2}}\sqrt{\Delta^2+3a^2t^2q^2}$$ This dispersion relationship (which is only valid near the Dirac points/Valleys) has the expected properties in that it opens up a gap of $\Delta$ between the valence and conduction bands and is also linear, as any Dirac cone should be.
Step 2: Next we need to find the eigenvectors. For this I followed the normal procedure to find the eigenvector of a matrix: (This is the eigenvector of the '+' valley and conduction band so we use $+q_x$ and the positive eigenvalue)$$\begin{bmatrix} 1 \\ \frac{-\Delta+{\sqrt{\Delta^2+3a^2t^2q_x^2+3a^2t^2q_y^2}}}{\sqrt{3}at(q_x-iq_y)} \\ \end{bmatrix}e^{ik\cdot{r}}$$I have included $e^{ik\cdot{r}}$ since this eigenvector is supposed to be Bloch solution to the Hamiltonian near the valley.
This is one of the places where I think I might have made a mistake. Now we need to normalize our eigenvectors since we are dealing with quantum mechanics:$$A^2\begin{bmatrix} 1 & \frac{-\Delta+{\sqrt{\Delta^2+3a^2t^2q_x^2+3a^2t^2q_y^2}}}{\sqrt{3}at(q_x+iq_y)} \\\end{bmatrix}\begin{bmatrix} 1 \\ \frac{-\Delta+{\sqrt{\Delta^2+3a^2t^2q_x^2+3a^2t^2q_y^2}}}{\sqrt{3}at(q_x-iq_y)} \\ \end{bmatrix}=1$$$$A^2=\frac{1}{1+\frac{\bigr(\Delta-\sqrt{\Delta^2+3a^2t^2(q_x^2+q_y^2)}\bigr)^2}{3a^2t^2(q_x^2+q_y^2)}}$$
Step 3: Now comes the calculation. I did this part in Mathematica because I didn't think it was worth my time to work through all of the algebra by hand. Others may disagree. In any case, the general formula for the Berry curvature is given by: (again, in this case $q$ is a vector quantity in 3 space)$$\Omega_n(q)=\nabla_q\times\left\langle u_n(q)\middle| i{\nabla}_q\middle| u_n(q) \right\rangle$$This is equation 1.27 from "Berry phase effects on electronic properties" by Xiao et al. (Link to paper) Since our function $u_n(q)$ is just our eigenvector without the added exponential (
This is another part where I could be wrong) and it only has components in $q_x$ and $q_y$ the equation for the Berry curvature can be re-written as:$$\Omega_n(q)=i\left\langle \frac{{\partial}u_n(q)}{{\partial}q_x}\middle| \frac{{\partial}u_n(q)}{{\partial}q_y}\right\rangle-\left\langle \frac{{\partial}u_n(q)}{{\partial}q_y}\middle| \frac{{\partial}u_n(q)}{{\partial}q_x} \right\rangle, \ \text{where} \ u_n(q)=\begin{bmatrix} 1 \\ \frac{-\Delta+{\sqrt{\Delta^2+3a^2t^2q_x^2+3a^2t^2q_y^2}}}{\sqrt{3}at(q_x-iq_y)} \\ \end{bmatrix}A^{1/2}$$Following this operation ought to give me the Berry curvature for +K valley in the conduction band. At this point I handed the calculation over to Mathematica and told it to simplify its result. The final formula that I obtained was:$$\Omega(q)=\frac{-3a^2t^2(q_x^2+q_y^2)\Delta-\Delta^3+\Delta^2\sqrt{\Delta^2+3a^2t^2(q_x^2+q_y^2)}}{(q_x^2+q_y^2)(\Delta^2+3a^2t^2(q_x^2+q_y^2))^{3/2}}$$This is very close to equation 6 in Xiao's paper but it's off by a a few terms in the numerator. What I am trying to figure out now is if I made an error in my derivation or if Xiao made a simplification step. I suspect that since we are only dealing with small $q$ that the last term in the numerator can be viwed as $\Delta^3$ and since only positive $\Delta$ makes physical sense the second and third terms cancel in the numerator. I will discuss the equation for the magnetic moment after I have heard back from all of you on whether or not my derivation so far is correct. Just so you don't have to scroll all the way back up here is the equation I was hoping for: (In my equation $\tau_z=-$)$$\Omega(q)=\tau_z\frac{3a^2{\Delta}t^2}{2({\Delta^2}+3q^2a^2t^2)^{3/2}}=\tau_z\frac{3a^2{\Delta}t^2}{2({\Delta^2}+3a^2t^2(q_x^2+q_y^2))^{3/2}}$$ |
The argument that the paper seems to be making appears strange to me.
According to the paper, the goal of CV is to estimate $\alpha_2$, the expected predictive performance of the model on new data, given that the model was trained on the observed dataset $S$. When we conduct $k$-fold CV, we obtain an estimate $\hat A$ of this number. Because of the random partitioning of $S$ into $k$ folds, this is a random variable $\hat A \sim f(A)$ with mean $\mu_k$ and variance $\sigma^2_k$. In contrast, $n$-times-repeated CV yields an estimate with the same mean $\mu_k$ but smaller variance $\sigma^2_k/n$.
Obviously, $\alpha_2\ne \mu_k$. This bias is something we have to accept.
However, the expected error $\mathbb E\big[|\alpha_2-\hat A|^2\big]$ will be larger for smaller $n$, and will be the largest for $n=1$, at least under reasonable assumptions about $f(A)$, e.g. when $\hat A\mathrel{\dot\sim} \mathcal N(\mu_k,\sigma^2_k/n)$. In other words, repeated CV allows to get a more precise estimate of $\mu_k$ and it is a good thing because it gives a more precise estimate of $\alpha_2$.
Therefore, repeated CV is strictly more precise than non-repeated CV.
The authors do not argue with that! Instead they claim, based on the simulations, that
reducing the variance [by repeating CV] is, in many cases, not very useful, and essentially a waste of computational resources.
This just means that $\sigma^2_k$ in their simulations was pretty low; and indeed, the lowest sample size they used was $200$, which is probably big enough to yield small $\sigma^2_k$. (The difference in estimates obtained with non-repeated CV and 30-times-repeated CV is always small.) With smaller sample sizes one can expect larger between-repetitions variance.
CAVEAT: Confidence intervals!
Another point that the authors are making is that
the reporting of confidence intervals [in repeated cross-validation] is
misleading.
It seems that they are referring to confidence intervals for the mean across CV repetitions. I fully agree that this is a meaningless thing to report! The more times CV is repeated, the smaller this CI will be, but nobody is interested in the CI around our estimate of $\mu_k$! We care about the CI around our estimate of $\alpha_2$.
The authors also report CIs for the non-repeated CV, and it's not entirely clear to me how these CIs were constructed. I guess these are the CIs for the means across the $k$ folds. I would argue that these CIs are also pretty much meaningless!
Take a look at one of their examples: the accuracy for
adult dataset with NB algorithm and 200 sample size. They get 78.0% with non-repeated CV, CI (72.26, 83.74), 79.0% (77.21, 80.79) with 10-times-repeated CV, and 79.1% (78.07, 80.13) with 30-times-repeated CV.
All of these CIs are useless, including the first one. The best estimate of $\mu_k$ is 79.1%. This corresponds to 158 successes out of 200. This yields 95% binomial confidence interval of (72.8, 84.5) -- broader even than the first one reported. If I wanted to report some CI, this is the one I would report.
MORE GENERAL CAVEAT: variance of CV.
You wrote that repeated CV
has become a popular technique for reducing the variance of cross-validation.
One should be very clear what one means by the "variance" of CV. Repeated CV reduces the variance of the estimate of $\mu_k$. Note that in case of leave-one-out CV (LOOCV), when $k=N$, this variance is equal to zero. Nevertheless, it is often said that LOOCV has actually the
highest variance of all possible $k$-fold CVs. See e.g. here: Variance and bias in cross-validation: why does leave-one-out CV have higher variance?
Why is that? This is because LOOCV has the highest variance as an estimate of $\alpha_1$ which is the expected predictive performance of the model on new data when built on a new dataset of the same size as $S$. This is a completely different issue. |
There is (at least) one way to prove unambiguity of a grammar $G = (N,T,\delta,S)$ for language $L$. It consists of two steps:
Prove $L \subseteq \mathcal{L}(G)$. Prove $[z^n]S_G(z) = |L_n|$.
The first step is pretty clear: show that the grammar generates (at least) the words you want, that is correctness.
The second step shows that $G$ has as many syntax trees for words of length $n$ as $L$ has words of length $n$ -- with 1. this implies unambiguity. It uses the structure function of $G$ which goes back to Chomsky and Schützenberger [1], namely
$\qquad \displaystyle S_G(z) = \sum_{n=0}^\infty t_nz^n$
with $t_n = [z^n]S_G(z)$ the number of syntax trees $G$ has for words of length $n$. Of course you need to have $|L_n|$ for this to work.
The nice thing is that $S_G$ is (usually) easy to obtain for context-free languages, although finding a closed form for $t_n$ can be difficult. Transform $G$ into an equation system of functions with one variable per nonterminal:
$\qquad \displaystyle \left[ A(z) = \sum\limits_{(A, a_0 \dots a_k) \in \delta} \ \prod\limits_{i=0}^{k} \ \tau(a_i)\ : A \in N \right] \text{ with } \tau(a) = \begin{cases} a(z) &, a \in N \\ z &, a \in T \\ \end{cases}.$
This may look daunting but is really only a syntactical transformation as will become clear in the example. The idea is that generated terminal symbols are counted in the exponent of $z$ and because the system has the same form as $G$, $z^n$ occurs as often in the sum as $n$ terminals can be generated by $G$. Check Kuich [2] for details.
Solving this equation system (computer algebra!) yields $S(z) = S_G(z)$; now you "only" have to pull the coefficient (in closed, general form). The TCS Cheat Sheet and computer algebra can often do so.
Example
Consider the simple grammar $G$ with rules
$\qquad \displaystyle S \to aSa \mid bSb \mid \varepsilon$.
It is clear that $\mathcal{L}(G) = \{ww^R \mid w \in \{a,b\}^*\}$ (step 1, proof by induction). There are $2^{\frac{n}{2}}$ palindromes of length $n$ if $n$ is even, $0$ otherwise.
Setting up the equation system yields
$\qquad \displaystyle S(z) = 2z^2S(z) + 1$
whose solution is
$\qquad \displaystyle S_G(z) = \frac{1}{1-2z^2}$.
The coefficients of $S_G$ coincide with the numbers of palindromes, so $G$ is unambiguous.
The Algebraic Theory of Context-Free Languages by Chomsky, Schützenberger (1963) On the entropy of context-free languages by Kuich (1970) |
Taking the derivative of $\ln(y) = b_0 + b_1 \cdot x$ with respect to $x$, you get
$$\frac{1}{y} \cdot \frac{dy}{dx}=b_1.$$ You can think of $$\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x},$$ the change in $y$ for a small change in $x$. This means that
$$b_1=\frac{\Delta y}{y} \cdot \frac{1}{\Delta x}=\frac{\frac{\Delta y}{y}}{\Delta x}.$$
Typically we would multiply $\frac{\Delta y}{y}$ by 100 to get a percentage change, which we can also do on the left hand side as well.
This means that $100 \cdot b_1$ is the approximate percentage change in $y$ for a small change in $x$. If you set $\Delta x=1$, that becomes the percentage change for a one unit change in $x$.
As long as your $x$ is not measured on a $[0,1]$ scale, the second interpretation is almost correct. Percentage points are used for arithmetic differences of two percentages. You have a mere percent change.
If your $x$ is measured on a $[0,1]$ scale, then a one unit change is a $100\%$ increase, not $1\%$, so then the effect is $0.2\%$ |
How would I solve the following.
An algorithm that is $O(n^2)$ takes 10 seconds to execute on a particular computer when n=100, how long would you expect to take it when n=500?
Can anyone help me answer dis.
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Since it is $O(n^2)$, then $t \leq c n^2$. Therefore, since $t = 10$, and $n = 100$, $c \geq t / n^2 = 10 / 100 00$.
Therefore, at $n = 500$. You have $t \leq 10 / 100 00 \times 500^2 = 250$.
But wait, the definition of the $O$-notation claims that the above formula works for large $n$'s only. I assumed that. Although honestly, I dont like this question. It is a bet weird.
Formally, there is absolutely no way to tell. O( ) notation is about the limiting behavior of a function (in this case, the running time of an algorithm) as its argument (in this case, the input size) grows to infinity. Without more information, it is absolutely impossible, even in principle, to predict behavior in the limit from a finite number of function values. |
I was playing with the idea of performing integration on a sound signal, since I've never heard of where it'd be used.
Is it used somewhere?
Here's something:
http://pcfarina.eng.unipr.it/Differentiation-Integration.htm
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I was playing with the idea of performing integration on a sound signal, since I've never heard of where it'd be used.
Is it used somewhere?
Here's something:
http://pcfarina.eng.unipr.it/Differentiation-Integration.htm
So, a sound signal $s$, in the DSP sense, is but a sequence of real-valued digital numbers:
$$ s = (1, 232, -12, 121, 0, 131, \dots) $$
Integrating that is nothing but adding these up, so the integrated signal $r$ would have the formula:
$$ r[n] = \sum\limits_{i=0}^n s[n] $$
or, writing that recursively, is the signal that it was at the last sampling time, plus the next input sample:
$$ r[n] = r[n-1] + s[n]\quad \text{ for }r>0\\ r[0] = 0$$
Now, that is what DSP people call a recursive system, or in this specific case, a single-pole IIR filter (IIR:
infinite impulse response, because you send one pulse in, you see it in the output forever). It's transfer function $H(z)$ is
$$H(z) = \frac{1}{1+a\cdot z^{-1}} $$
Your $a=1$, here, because the last value of $r$ doesn't get multiplied with a dampening factor. You will find a lot of literature out there explaining the properties of IIR filters, but for all practical things, you could build a
low-pass filter out of your integrator, if you use a factor $a<1$. |
Edit: So, my original question (stated below) was to find an error in my "proof" that immediate parabolic basins for rational maps are always simply connected. Since I have not received any answers as of yet I would ask alternatively if someone could point out an explicit example of a rational map with a parabolic fixed point which has a non-simply connected immediate basin, so that I could hopefully check by examining this example where my argument goes wrong. Kind regards an idiot
Hello!Sorry if this seems stupid. I know there must be an error in my thinking.
Let f be a rational map on the Riemann sphere with a parabolic fixed point $f(z_0)=z_0$, $f'(z_0)=e^{2\pi t}$ with $t\in \mathbb{Q}$. I will try to demonstrate that each parabolic immediate basin is simply connected. I know this is wrong. But I don't find the mistake in my "proof". So please help me out. By Leau-Fatou Flower theorem, in each immediate parabolic basin of $z_0$ there is an attractive petal $V$ such that each point in that immediate basin tends to $z_0$ via $V$. Let $V_0$ be such a petal, and for simplicity's sake let's say that $f(\overline{V_0})\subset V_0\cup{z_0}$ (i.e. no periodic jumping between different petals, can be achieved by simply taking an iterate $f^n$ instead of $f$ for suitable n). So we have: - $f(\overline{V_0})\subset V_0\cup{z_0}$ - $V_0\subset A^*(z_0)$ open and simply connected - For every $z\in A^*(z_0)$ there is some $n\in\mathbb{N}$ with $f^n(z)\in V_0$
We may slightly shrink $V_0$ if necessary such that $\partial V_0$ does not contain any postcritical points and $\overline{V_0}$ is homeomorphic to a closed disk.
Now for $k\in\mathbb{N}$ let $V_k$ be the component of $f^{-1}(V_0)$ that contains $V_0$. It's easy to see that $A^*(z_0)=\cup_{k=0}^{\infty}V_k$.
If $A^*(z_0)$ is not simply connected then there must be a minimal $m\in\mathbb{N}$ such that $V_m$ is not simply connected.
In that case let $B$ be a component of $ \hat{\mathbb{C}}-\overline{V_m}$, such that $\partial B$ does not contain $z_0$. Then $\partial B\subset \partial V_m$ and so $f(\partial B)\subset\partial V_{m-1}$. Since $\partial V_0$ contains no postcritical points, $\cup_{k=0}^m \partial V_m$ contains no critical points. Thus $f^m$ is locally injective on $\partial B\subset\partial V_m$ and $f^m(\partial B)$ is a full component of $\partial V_0$ (proper covering), hence $f^m(\partial B)=\partial V_0$, since $\partial V_0$ has only one component. But then there is $z\in\partial B\subset F(f)$ with $f^n(z)=z_0\in J(f)$. That's a contradiction.
Can someone help me see my mistake? I hope it's a simple one. |
I'm trying to calculate the momentum distribution of a 1D system of non-interacting identical fermions in a harmonic trap.
Given Feynman's answer (from his Statistical Mechanics book) for the position density matrix of a single trapped particle at $T>0$, $ \rho_1 (x, x'; \beta) = \sqrt{\cfrac{m \omega}{2 \pi \hbar \sinh (\beta \hbar \omega) }} \exp \left\{ \cfrac{-m \omega}{ 2 \hbar \sinh (\beta \hbar \omega) } \left[ (x^2 + x' ^2) \cosh (\beta \hbar \omega) - 2x x' \right] \right\} $ ,
the translationally invariant distribution of $ (x'-x) $ is $ \tilde{\rho} (s; \beta) = \int_{-\infty} ^\infty \mathrm{d}x \mathrm{d}x' \delta (s - (x' - x) ) \rho_1 (x, x'; \beta) = \cfrac{\mathrm{e}^{\frac{-m \omega s^2}{4\hbar} \coth \frac{\beta \hbar \omega}{2}}}{2 \sinh \frac{\beta \hbar \omega}{2}} $ .
Here, we define $\beta\equiv 1/(k_\text{B}T)$. The Fourier transform of $\tilde{\rho} (s ; \beta)$ is the momentum distribution of the system, which is also a Gaussian.
How would you construct a two-fermion version out of this? What about 3 fermions?
My first attempt was $ \tilde{\rho}_{2\text{F}} (s;\beta) = \frac{1}{2!} \left( 2 \tilde{\rho} (s; \beta) \tilde{\rho} (0; \beta) - 2 \tilde{\rho} (s; 2\beta) \right) $
using Feynman diagrams, keeping in mind the anti-periodicity $\rho_1 (x, x'; \beta + \beta) = -\rho_1 (x, x'; \beta)$ for fermions.
Visualizing the path along the imaginary time on a hypercylinder of circumference $\beta$ would look something like
Note here that diagrams with fermion paths intersecting each other have zero statistical weight.
I've written the full statement of my problem at http://mathb.in/1393 . Apparently my answer doesn't agree with two other numerical calculations... perhaps missing a normalizing factor in one of the terms.
Any comments appreciated. |
A model for this situation is to put 61000 ($n$) balls into an urn, of which 23000 ($n_1$) are labeled "A". 15000 ($k$) of these are drawn randomly without replacement. Of these, $m$ are found to be labeled "A". What is the chance that $m \ge 10000$?
The total number of possible samples equals the number of $k$-element subsets of an $n$-set, $\binom{n}{k}$. All are equally likely to be drawn, by hypothesis. Let $i \ge 10000$. The number of possible samples with $i$ A's is the number of subsets of an $n_1$-set having $i$ A's, times the number of subsets of an $n-n_1$-set having $k-i$ non-A's; that is, $\binom{n_1}{i}\binom{n-n_1}{k-i}$. Summing over all possible $i$ and dividing by the chance of each sample gives the probability of observing an overlap of $m = 10000$ or greater:
$$\Pr(\text{overlap} \ge m) = \frac{1}{\binom{n}{k}} \sum_{i=m}^{\min(n_1,k)} \binom{n_1}{i}\binom{n-n_1}{k-i}.$$
This answer is exact. For rapid calculation it can be expressed (in closed form) in terms of generalized hypergeometric functions; the details of this expression can be provided by a symbolic algebra program like
Mathematica. The answer in this particular instance is $3.8057078557887\ldots \times 10^{-1515}$.
We can also use a Normal approximation. Coding A's as 1 and non-A's as 0, as usual, the mean of the urn is $p = 23000/61000 \sim 0.377$. The standard deviation of the urn is $\sigma = \sqrt{p(1-p)}$. Therefore the standard error of the observed proportion, $u = 10000/15000 \sim 0.667$, is
$$se(u) = \sigma \sqrt{(1 - \frac{15000-1}{61000-1})/15000} \sim 0.003436.$$
(see http://www.ma.utexas.edu/users/parker/sampling/woreplshort.htm). Thus the observed proportion is $z = \frac{u - p}{se(u)} \sim 84.28$ standard errors larger than expected. Obviously the corresponding p-value is low (it computes to $1.719\ldots \times 10^{-1545}$). Although the Normal approximation is no longer very accurate at such extreme z values (it's off by 30 orders of magnitude!), it still gives excellent guidance. |
W.I.P.: Work in progress
Following p. 370 of Cramer's 1946
Mathematical Methods of Statistics, define $$\Xi_n = n(1 - \Phi(Z_n)) \,. $$ Here $\Phi$ is the cumulative distribution function of the standard normal distribution, $\mathscr{N}(0,1)$. As a consequence of its definition, we are guaranteed that $0\le \Xi_n \le n$ almost surely.
Consider a given realization $\omega \in \Omega$ of our sample space. Then in this sense $Z_n$ is both a function of $n$ and $\omega$, and $\Xi_n$ a function of $Z_n, n$, and $\omega$. For a fixed $\omega$, we can consider $Z_n$ a deterministic function of $n$, and $\Xi_n$ a deterministic function of $Z_n$ and $n$, thereby simplifying the problem. We aim to show results which hold for almost surely all $\omega \in \Omega$, allowing us to transfer our results from a non-deterministic analysis to the non-deterministic setting.
Following p. 374 of Cramer's 1946
Mathematical Methods of Statistics, assume for the moment (I aim to come back and supply a proof later) that we are able to show that (for any given $\omega \in \Omega$) the following asymptotic expansion holds (using integration by parts and the definition of $\Phi$):
$$ \frac{\sqrt{2\pi}}{n}\Xi_n = \frac{1}{Z_n}e^{-\frac{Z_n^2}{2}}\left( 1 + O \left( \frac{1}{Z_n^2} \right) \right) \quad ~~ as ~~ Z_n \to \infty \,. \tag{~}$$
Clearly we have that $Z_{n+1} \ge Z_n$ for any $n$, and $Z_n$ is almost surely an increasing function of $n$ as $n\to \infty$, therefore we claim in what follows throughout that for (almost surely all) fixed $\omega$: $$ Z_n \to \infty \quad \iff \quad n \to \infty \,. $$
Hence it follows that we have (where $\sim$ denotes asymptotic equivalence):
$$ \frac{\sqrt{2\pi}}{n} \Xi_n \sim \frac{1}{Z_n} e^{-\frac{1}{Z_n^2}} \quad ~~ as ~~ Z_n \to \infty \quad n \to \infty \,. $$
How we proceed in what follows amounts essentially to the method of dominant balance, and our manipulations will be formally justified by the following lemma:
Lemma: Assume that $f(n) \sim g(n)$ as $n \to \infty$, and $f(n) \to \infty$ (thus $g(n) \to \infty$). Then given any function $h$ which is formed via compositions, additions, and multiplications of logarithms and power laws (essentially any "polylog" function), we must have also that as $n \to \infty$: $$ h(f(n)) \sim h(g(n)) \,. $$ In other words, such "polylog" functions preserve asymptotic equivalence.
The truth of this lemma is a consequence of Theorem 2.1. as written here. Note also that what follows is mostly an expanded (more details) version of the answer to a similar question found here.
Taking logarithms of both sides, we get that:
$$\log ( \sqrt{2\pi} \Xi_n ) - \log n \sim -\log Z_n - \frac{Z_n^2}{2} \,. \tag{1}$$
This is where Cramer is somewhat cagey; he just says "assuming $\Xi_n$ is bounded", we can conclude blah blah blah. But showing that $\Xi_n$ is suitably bounded almost surely seems to be actually somewhat non-trivial. It seems that the proof of this may essentially be part of what's discussed on pp. 265-267 of Galambos, but I am not sure given that I am still working to understand the content of that book.
Anyway,
assuming one can show that $\log \Xi_n = o(\log n)$, then it follows (since the $-Z_n^2/2$ term dominates the $-\log Z_n$ term) that:
$$ - \log n \sim - \frac{Z_n^2}{2} \quad \implies \quad Z_n \sim \sqrt{2 \log n} \,. $$
This is somewhat nice, since it is already most of what we want to show, although again it is worthwhile to note that it is essentially only kicking the can down the road, since now we have to show some certain almost surely boundedness of $\Xi_n$. On the other hand, $\Xi_n$ has the same distribution for any maximum of i.i.d. continuous random variables, so this may be tractable.
Anyway, if $Z_n \sim \sqrt{2 \log n}$ a.s., then clearly one can also conclude that $Z_n \sim \sqrt{2 \log n}(1 + \alpha(n))$ for any $\alpha(n)$ which is $o(1)$ as $n \to \infty$. Using our lemma about polylog functions preserving asymptotic equivalence above, we can substitute this expression back into $(1)$ to get:
$$\log(\sqrt{2 \pi} \Xi _n)- \log n \sim -\log (1 + \alpha) - \frac{1}{2}\log 2 - \frac{1}{2}\log \log n - \log n - 2 \alpha \log n - \alpha^2 \log n \,. $$
$$ \implies -\log(\Xi_n \sqrt{2 \pi}) \sim \log(1 + \alpha) + \frac{1}{2} \log 2 + \frac{1}{2} \log \log n + 2\alpha \log n + \alpha^2 \log n \,. $$
Here we have to go even further, and
assume that $\log \Xi_n = o( \log \log n) ~~ as ~~ n \to \infty$ almost surely. Again, all Cramer says is "assuming $\Xi_n$ is bounded". But since all one can say a priori about $\Xi_n$ is that $0 \le Xi_n \le n$ a.s., it hardly seems clear that one should have $\Xi_n = O(1)$ almost surely, which seems to be the substance of Cramer's claim.
But anyway, assuming one believes that, then it follows that the dominant term which does not contain $\alpha$ is $\frac{1}{2} \log \log n$. Since $\alpha = o(1)$, it follows that $\alpha^2 = o(\alpha)$, and clearly $\log ( 1 + \alpha) = o (\alpha) = o(o(\alpha \log n))$, so the dominant term containing $\alpha$ is $2 \alpha \log n$. Therefore, we can rearrange and (dividing everything by $\frac{1}{2}\log\log n$ or $2 \alpha \log n$) find that
$$ - \frac{1}{2} \log \log n \sim 2 \alpha \log n \quad \implies \quad \alpha \sim - \frac{\log \log n}{4 \log n} \,. $$
Therefore, substituting this back into the above, we get that:
$$Z_n \sim \sqrt{2 \log n}- \frac{\log\log n}{2 \sqrt{2 \log n}} \,, $$
again, assuming we believe certain things about $\Xi_n$.
We rehash the same technique again; since $Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}}$, then it also follows that
$$ Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}} (1 + \beta(n)) = \sqrt{2 \log n} \left( 1 - \frac{\log \log n}{8 \log n}(1 + \beta(n)) \right) \,,$$
when $\beta(n)=o(1)$. Let's simplify a little before substituting directly back into (1); we get that:
$$ \log Z_n \sim \log(\sqrt{2 \log n}) + \underbrace{\log \left(1 - \frac{\log \log n}{8 \log n}(1 + \beta(n)) \right) }_{\log(O(1)) = o(\log n)} \sim \log (\sqrt{2 \log n}) \,.$$
$$ \frac{Z_n^2}{2} \sim \log n - \frac{1}{2} \log \log n (1 + \beta) + \underbrace{\frac{(\log \log n)^2}{8 \log n} ( 1 \beta)^2}_{o((1+ \beta) \log \log n)} \sim \log n - \frac{1}{2} (1 + \beta) \log \log n \,. $$
Substituting this back into (1), we find that:
$$ \log ( \sqrt{2 \pi} \Xi_n) - \log n \sim - \log(\sqrt{2 \log n}) - \log n + \frac{1}{2}(1 + \beta) \log \log n \quad \implies \quad \beta \sim \frac{\log (4 \pi \Xi_n^2)}{\log \log n} \,. $$
Therefore, we conclude that almost surely
$$Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}} \left(1 + \frac{\log(4 \pi) + 2 \log( \Xi_n)}{\log \log n} \right)\\ = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{ 2 \sqrt{2 \log n} } - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} \,. $$
This corresponds to the final result on p.374 of Cramer's 1946
Mathematical Methods of Statistics except that here the exact order of the error term isn't given. Apparently applying this one more term gives the exact order of the error term, but anyway it doesn't seem necessary to prove the results about the maxima of i.i.d. standard normals in which we are interested.
Given the result of the above, namely that almost surely:
$$Z_n \sim \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} \quad \implies \\ Z_n = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} + o(1)\,. \tag{$\dagger$}$$
2. Then by linearity of expectation it follows that:
$$ \mathbb{E}Z_n = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\mathbb{E}[\log (\Xi_n)]}{\sqrt{2 \log n}} + o(1) \quad \implies \\ \frac{\mathbb{E}Z_n}{\sqrt{2 \log n}} = 1 - \frac{\mathbb{E}[\log \Xi_n]}{2 \log n} + o(1) \,. $$
Therefore, we have shown that
$$ \lim_{n \to \infty } \frac{\mathbb{E} Z_n}{\sqrt{2 \log n}} = 1 \,,$$
as long as we can also show that
$$ \mathbb{E}[\log \Xi_n] = o(\log n) \,. $$
This might not be too difficult to show since again $\Xi_n$ has
the same distribution for every continuous random variable. Thus we have the second result from above.
1. Similarly, we also have from the above that almost surely:
$$\frac{Z_n}{\sqrt{2 \log n}} = 1 - \frac{\log(\Xi_n)}{2 \log n} +o(1),.$$
Therefore, if we can show that:
$$ \log(\Xi_n) = o(\log n) \text{ almost surely}, \tag{*}$$
then we will have shown the first result from above. Result (*) would also clearly imply a fortiori that $\mathbb{E}[\log (\Xi_n)] = o(\log n)$, thereby also giving us the first result from above.
Also note that in the proof above of ($\dagger$) we needed to assume anyway that $\Xi_n = o(\log n)$ almost surely (or at least something similar), so that if we are able to show ($\dagger$) then we will most likely also have in the process needed to show $\Xi_n = o(\log n)$ almost surely, and therefore if we can prove $(\dagger)$ we will most likely be able to immediately reach all of the following conclusions.
3. However, if we have this result, then I don't understand how one would also have that $\mathbb{E}Z_n = \sqrt{2 \log n} + \Theta(1)$, since $o(1) \not= \Theta(1)$. But at the very least it would seem to be true that $$\mathbb{E}Z_n = \sqrt{2 \log n} + O(1) \,.$$
So then it seems that we can focus on answering the question of how to show that $$ \Xi_n = o(\log n) \text{ almost surely.} $$
We will also need to do the grunt work of providing a proof for (~), but to the best of my knowledge that is just calculus and involves no probability theory, although I have yet to sit down and try it yet.
First let's go through a chain of trivialities in order to rephrase the problem in a way which makes it easier to solve (note that by definition $\Xi_n \ge 0$):
$$\Xi_n = o(\log n) \quad \iff \quad \lim_{n \to \infty} \frac{\Xi_n}{\log n} = 0 \quad \iff \quad \\ \forall \varepsilon > 0, \frac{\Xi_n}{\log n} > \varepsilon \text{ only finitely many times} \quad \iff \\ \forall \varepsilon >0, \quad \Xi_n > \varepsilon \log n \text{ only finitely many times} \,.$$
One also has that:
$$\Xi_n > \varepsilon \log n \quad \iff \quad n(1 - F(Z_n)) > \varepsilon \log n \quad \iff \quad 1 - F(Z_n) > \frac{\varepsilon \log n}{n} \\ \iff \quad F(Z_n) < 1 - \frac{\varepsilon \log n}{n} \quad \iff \quad Z_n \le \inf \left\{ y: F(y) \ge 1 - \frac{\varepsilon \log n}{n} \right\} \,. $$
Correspondingly, define for all $n$:
$$ u_n^{(\varepsilon)} = \inf \left\{ y: F(y) \ge 1 - \frac{\varepsilon \log n}{n} \right\} \,. $$
Therefore the above steps show us that:
$$\Xi_n = o(\log n) \text{ a.s.} \quad \iff \quad \mathbb{P}(\Xi_n = o(\log n)) = 1 \quad \iff \quad \\ \mathbb{P}(\forall \varepsilon > 0 , \Xi_n > \varepsilon \log n \text{ only finitely many times}) = 1 \\ \iff \mathbb{P}(\forall \varepsilon > 0, Z_n \le u_n^{(\varepsilon)} \text{ only finitely many times}) = 1 \\\iff \mathbb{P}(\forall \varepsilon >0, Z_n \le u_n^{(\varepsilon)} \text{ infinitely often}) =0 \,. $$
Notice that we can write:
$$ \{ \forall \varepsilon >0, Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} = \bigcap_{\varepsilon > 0} \{ Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} \,.$$
The sequences $u_n^{(\varepsilon)}$ become uniformly larger as $\varepsilon$ decreases, so we can conclude that the events $$\{ Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} $$ are decreasing (or at least somehow monotonic) as $\varepsilon$ goes to $0$. Therefore the probability axiom regarding monotonic sequences of events allows us to conclude that:
$$\mathbb{P}(\forall \varepsilon >0, Z_n \le u_n^{(\varepsilon)} \text{ infinitely often}) = \\\mathbb{P} \left( \bigcap_{\varepsilon > 0} \{ Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} \right) = \\\mathbb{P} \left( \lim_{\varepsilon \downarrow 0} \{ Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} \right) = \\\lim_{\varepsilon \downarrow 0} \mathbb{P}(Z_n \le u_n^{(\varepsilon)} \text{ infinitely often}) \,.$$
Therefore it
suffices to show that for all $\varepsilon >0$,
$$\mathbb{P}(Z_n \le u_n^{(\varepsilon)} \text{ infinitely often}) = 0 $$
because of course the limit of any constant sequence is the constant.
Here is somewhat of a sledgehammer result:
Theorem 4.3.1., p. 252 of Galambos, The Asymptotic Theory of Extreme Order Statistics, 2nd edition. Let $X_1, X_2, \dots$ be i.i.d. variables with common nondegenerate and continuous distribution function $F(x)$, and let $u_n$ be a nondecreasing sequence such that $n(1 - F(u_n))$ is also nondecreasing. Then, for $u_n < \sup \{ x: F(x) <1 \}$, $$\mathbb{P}(Z_n \le u_n \text{ infinitely often}) =0 \text{ or }1 $$
according as
$$\sum_{j=1}^{+\infty}[1 - F(u_j)]\exp(-j[1-F(u_j)]) < +\infty \text{ or }=+\infty \,. $$
The proof is technical and takes around five pages, but ultimately it turns out to be a corollary of one of the Borel-Cantelli lemmas. I may get around to trying to condense the proof to only use the part required for this analysis as well as only the assumptions which hold in the Gaussian case, which may be shorter (but maybe it isn't) and type it up here, but holding your breath is not recommended. Note that in this case $\omega(F)=+\infty$, so that condition is vacuous, and $n(1-F(n))$ is $\varepsilon \log n$ thus clearly non-decreasing.
Anyway the point being that, appealing to this theorem, if we can show that:
$$\sum_{j=1}^{+\infty}[1 - F(u_j^{(\varepsilon)})]\exp(-j[1-F(u_j^{(\varepsilon)})]) = \sum_{j=1}^{+\infty}\left[ \frac{\varepsilon \log j}{j} \right]\exp(-\varepsilon \log j) = \varepsilon \sum_{j=1}^{+\infty} \frac{ \log j}{j^{1 + \varepsilon}} < + \infty \,. $$
Note that since logarithmic growth is slower than any power law growth for any positive power law exponent (logarithms and exponentials are monotonicity preserving, so $\log \log n \le \alpha \log n \iff \log n \le n^{\alpha}$ and the former inequality can always be seen to hold for all $n$ large enough due to the fact that $\log n \le n$ and a change of variables), we have that:
$$ \sum_{j=1}^{+\infty} \frac{\log j}{j^{1 + \varepsilon}} \le \sum_{j=1}^{+\infty} \frac{j^{\varepsilon/2}}{j^{1 + \varepsilon}} = \sum_{j=1}^{+\infty} \frac{1}{j^{1 + \varepsilon/2}} < +\infty \,,$$
since the p-series is known to converge for all $p>1$, and $\varepsilon >0$ of course implies $1 + \varepsilon/2 > 1$.
Thus using the above theorem we have shown that for all $\varepsilon >0$, $\mathbb{P}(Z_n \le u_n^{(\varepsilon)} \text{ i.o.}) = 0$, which to recapitulate should mean that $\Xi_n = o(\log n)$ almost surely.
We need to show still that $\log \Xi_n = o(\log \log n)$. This doesn't follow from the above, since, e.g.,
$$ \frac{1}{n} \log n = o(\log n) \,, - \log n + \log \log n \not= o(\log n) \,. $$
However, given a sequence $x_n$, if one can show that $x_n = o( (\log n)^{\delta})$ for arbitrary $\delta >0$, then it does follow that $\log(x_n) = o(\log \log n)$. Ideally I would like to be able to show this for $\Xi_n$ using the above lemma (assuming it's even true), but am not able to (as of yet). |
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In this chapter we learned about left and right adjoints, and about joins and meets. At first they seemed like two rather different pairs of concepts. But then we learned some deep relationships between them. Briefly:
Left adjoints preserve joins, and monotone functions that preserve enough joins are left adjoints.
Right adjoints preserve meets, and monotone functions that preserve enough meets are right adjoints.
Today we'll conclude our discussion of Chapter 1 with two more bombshells:
Joins
are left adjoints, and meets are right adjoints.
Left adjoints are right adjoints seen upside-down, and joins are meets seen upside-down.
This is a good example of how category theory works. You learn a bunch of concepts, but then you learn more and more facts relating them, which unify your understanding... until finally all these concepts collapse down like the core of a giant star, releasing a supernova of insight that transforms how you see the world!
Let me start by reviewing what we've already seen. To keep things simple let me state these facts just for posets, not the more general preorders. Everything can be generalized to preorders.
In Lecture 6 we saw that given a left adjoint \( f : A \to B\), we can compute its right adjoint using joins:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$ Similarly, given a right adjoint \( g : B \to A \) between posets, we can compute its left adjoint using meets:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ In Lecture 16 we saw that left adjoints preserve all joins, while right adjoints preserve all meets.
Then came the big surprise: if \( A \) has all joins and a monotone function \( f : A \to B \) preserves all joins, then \( f \) is a left adjoint! But if you examine the proof, you'l see we don't really need \( A \) to have
all joins: it's enough that all the joins in this formula exist:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$Similarly, if \(B\) has all meets and a monotone function \(g : B \to A \) preserves all meets, then \( f \) is a right adjoint! But again, we don't need \( B \) to have
all meets: it's enough that all the meets in this formula exist:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ Now for the first of today's bombshells: joins are left adjoints and meets are right adjoints. I'll state this for binary joins and meets, but it generalizes.
Suppose \(A\) is a poset with all binary joins. Then we get a function
$$ \vee : A \times A \to A $$ sending any pair \( (a,a') \in A\) to the join \(a \vee a'\). But we can make \(A \times A\) into a poset as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Then \( \vee : A \times A \to A\) becomes a monotone map, since you can check that
$$ a \le a' \textrm{ and } b \le b' \textrm{ implies } a \vee b \le a' \vee b'. $$And you can show that \( \vee : A \times A \to A \) is the left adjoint of another monotone function, the
diagonal
$$ \Delta : A \to A \times A $$sending any \(a \in A\) to the pair \( (a,a) \). This diagonal function is also called
duplication, since it duplicates any element of \(A\).
Why is \( \vee \) the left adjoint of \( \Delta \)? If you unravel what this means using all the definitions, it amounts to this fact:
$$ a \vee a' \le b \textrm{ if and only if } a \le b \textrm{ and } a' \le b . $$ Note that we're applying \( \vee \) to \( (a,a') \) in the expression at left here, and applying \( \Delta \) to \( b \) in the expression at the right. So, this fact says that \( \vee \) the left adjoint of \( \Delta \).
Puzzle 45. Prove that \( a \le a' \) and \( b \le b' \) imply \( a \vee b \le a' \vee b' \). Also prove that \( a \vee a' \le b \) if and only if \( a \le b \) and \( a' \le b \).
A similar argument shows that meets are really right adjoints! If \( A \) is a poset with all binary meets, we get a monotone function
$$ \wedge : A \times A \to A $$that's the
right adjoint of \( \Delta \). This is just a clever way of saying
$$ a \le b \textrm{ and } a \le b' \textrm{ if and only if } a \le b \wedge b' $$ which is also easy to check.
Puzzle 46. State and prove similar facts for joins and meets of any number of elements in a poset - possibly an infinite number.
All this is very beautiful, but you'll notice that all facts come in pairs: one for left adjoints and one for right adjoints. We can squeeze out this redundancy by noticing that every preorder has an "opposite", where "greater than" and "less than" trade places! It's like a mirror world where up is down, big is small, true is false, and so on.
Definition. Given a preorder \( (A , \le) \) there is a preorder called its opposite, \( (A, \ge) \). Here we define \( \ge \) by
$$ a \ge a' \textrm{ if and only if } a' \le a $$ for all \( a, a' \in A \). We call the opposite preorder\( A^{\textrm{op}} \) for short.
I can't believe I've gone this far without ever mentioning \( \ge \). Now we finally have really good reason.
Puzzle 47. Show that the opposite of a preorder really is a preorder, and the opposite of a poset is a poset. Puzzle 48. Show that the opposite of the opposite of \( A \) is \( A \) again. Puzzle 49. Show that the join of any subset of \( A \), if it exists, is the meet of that subset in \( A^{\textrm{op}} \). Puzzle 50. Show that any monotone function \(f : A \to B \) gives a monotone function \( f : A^{\textrm{op}} \to B^{\textrm{op}} \): the same function, but preserving \( \ge \) rather than \( \le \). Puzzle 51. Show that \(f : A \to B \) is the left adjoint of \(g : B \to A \) if and only if \(f : A^{\textrm{op}} \to B^{\textrm{op}} \) is the right adjoint of \( g: B^{\textrm{op}} \to A^{\textrm{ op }}\).
So, we've taken our whole course so far and "folded it in half", reducing every fact about meets to a fact about joins, and every fact about right adjoints to a fact about left adjoints... or vice versa! This idea, so important in category theory, is called
duality. In its simplest form, it says that things come on opposite pairs, and there's a symmetry that switches these opposite pairs. Taken to its extreme, it says that everything is built out of the interplay between opposite pairs.
Once you start looking you can find duality everywhere, from ancient Chinese philosophy:
to modern computers:
But duality has been studied very deeply in category theory: I'm just skimming the surface here. In particular, we haven't gotten into the connection between adjoints and duality!
This is the end of my lectures on Chapter 1. There's more in this chapter that we didn't cover, so now it's time for us to go through all the exercises. |
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Isomorphisms are very important in mathematics, and we can no longer put off talking about them. Intuitively, two objects are 'isomorphic' if they look the same. Category theory makes this precise and shifts the emphasis to the 'isomorphism' - the
way in which we match up these two objects, to see that they look the same.
For example, any two of these squares look the same after you rotate and/or reflect them:
An isomorphism between two of these squares is a
process of rotating and/or reflecting the first so it looks just like the second.
As the name suggests, an isomorphism is a kind of morphism. Briefly, it's a morphism that you can 'undo'. It's a morphism that has an inverse:
Definition. Given a morphism \(f : x \to y\) in a category \(\mathcal{C}\), an inverse of \(f\) is a morphism \(g: y \to x\) such that
and
I'm saying that \(g\) is 'an' inverse of \(f\) because in principle there could be more than one! But in fact, any morphism has at most one inverse, so we can talk about 'the' inverse of \(f\) if it exists, and we call it \(f^{-1}\).
Puzzle 140. Prove that any morphism has at most one inverse. Puzzle 141. Give an example of a morphism in some category that has more than one left inverse. Puzzle 142. Give an example of a morphism in some category that has more than one right inverse.
Now we're ready for isomorphisms!
Definition. A morphism \(f : x \to y\) is an isomorphism if it has an inverse. Definition. Two objects \(x,y\) in a category \(\mathcal{C}\) are isomorphic if there exists an isomorphism \(f : x \to y\).
Let's see some examples! The most important example for us now is a 'natural isomorphism', since we need those for our databases. But let's start off with something easier. Take your favorite categories and see what the isomorphisms in them are like!
What's an isomorphism in the category \(\mathbf{3}\)? Remember, this is a free category on a graph:
The morphisms in \(\mathbf{3}\) are paths in this graph. We've got one path of length 2:
$$ f_2 \circ f_1 : v_1 \to v_3 $$ two paths of length 1:
$$ f_1 : v_1 \to v_2, \quad f_2 : v_2 \to v_3 $$ and - don't forget - three paths of length 0. These are the identity morphisms:
$$ 1_{v_1} : v_1 \to v_1, \quad 1_{v_2} : v_2 \to v_2, \quad 1_{v_3} : v_3 \to v_3.$$ If you think about how composition works in this category you'll see that the only isomorphisms are the identity morphisms. Why? Because there's no way to compose two morphisms and get an identity morphism unless they're both that identity morphism!
In intuitive terms, we can only move from left to right in this category, not backwards, so we can only 'undo' a morphism if it doesn't do anything at all - i.e., it's an identity morphism.
We can generalize this observation. The key is that \(\mathbf{3}\) is a poset. Remember, in our new way of thinking a
preorder is a category where for any two objects \(x\) and \(y\) there is at most one morphism \(f : x \to y\), in which case we can write \(x \le y\). A poset is a preorder where if there's a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(x = y\). In other words, if \(x \le y\) and \(y \le x\) then \(x = y\). Puzzle 143. Show that if a category \(\mathcal{C}\) is a preorder, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(g\) is the inverse of \(f\), so \(x\) and \(y\) are isomorphic. Puzzle 144. Show that if a category \(\mathcal{C}\) is a poset, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then both \(f\) and \(g\) are identity morphisms, so \(x = y\).
Puzzle 144 says that in a poset, the only isomorphisms are identities.
Isomorphisms are a lot more interesting in the category \(\mathbf{Set}\). Remember, this is the category where objects are sets and morphisms are functions.
Puzzle 145. Show that every isomorphism in \(\mathbf{Set}\) is a bijection, that is, a function that is one-to-one and onto. Puzzle 146. Show that every bijection is an isomorphism in \(\mathbf{Set}\).
So, in \(\mathbf{Set}\) the isomorphisms are the bijections! So, there are lots of them.
One more example:
Definition. If \(\mathcal{C}\) and \(\mathcal{D}\) are categories, then an isomorphism in \(\mathcal{D}^\mathcal{C}\) is called a natural isomorphism.
This name makes sense! The objects in the so-called 'functor category' \(\mathcal{D}^\mathcal{C}\) are functors from \(\mathcal{C}\) to \(\mathcal{D}\), and the morphisms between these are natural transformations. So, the
isomorphisms deserve to be called 'natural isomorphisms'.
But what are they like?
Given functors \(F, G: \mathcal{C} \to \mathcal{D}\), a natural transformation \(\alpha : F \to G\) is a choice of morphism
$$ \alpha_x : F(x) \to G(x) $$ for each object \(x\) in \(\mathcal{C}\), such that for each morphism \(f : x \to y\) this naturality square commutes:
Suppose \(\alpha\) is an isomorphism. This says that it has an inverse \(\beta: G \to F\). This \(beta\) will be a choice of morphism
$$ \beta_x : G(x) \to F(x) $$ for each \(x\), making a bunch of naturality squares commute. But saying that \(\beta\) is the inverse of \(\alpha\) means that
$$ \beta \circ \alpha = 1_F \quad \textrm{ and } \alpha \circ \beta = 1_G .$$ If you remember how we compose natural transformations, you'll see this means
$$ \beta_x \circ \alpha_x = 1_{F(x)} \quad \textrm{ and } \alpha_x \circ \beta_x = 1_{G(x)} $$ for all \(x\). So, for each \(x\), \(\beta_x\) is the inverse of \(\alpha_x\).
In short: if \(\alpha\) is a natural isomorphism then \(\alpha\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\).
But the converse is true, too! It takes a
little more work to prove, but not much. So, I'll leave it as a puzzle. Puzzle 147. Show that if \(\alpha : F \Rightarrow G\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\), then \(\alpha\) is a natural isomorphism.
Doing this will help you understand natural isomorphisms. But you also need examples!
Puzzle 148. Create a category \(\mathcal{C}\) as the free category on a graph. Give an example of two functors \(F, G : \mathcal{C} \to \mathbf{Set}\) and a natural isomorphism \(\alpha: F \Rightarrow G\). Think of \(\mathcal{C}\) as a database schema, and \(F,G\) as two databases built using this schema. In what way does the natural isomorphism between \(F\) and \(G\) make these databases 'the same'. They're not necessarily equal!
We should talk about this. |
CHAPTER 21-SURFACE AREA, VOLUME AND CAPACITY Question 1.
Find the volume and the total surface area of a cuboid, whose:
(i) Length = 15cm, breadth = 10cm and height = 8cm.
Solution:-
Volume of a cuboid \(=\text {Length} \times \text { Breadth } \times \text { Height }=15 \times 10 \times 8=1200 \mathrm{cm}^{3}\)
Total surface area of a cuboid \(2(l \times b+b \times h+h \times l)=2(15 \times 10+10 \times 8+8 \times 15) \) \(=2(150+80+120) 2 \times 350=700 \mathrm{cm}^{2}\)
(ii) l = 3.5m, b = 2.6m and h = 90cm,
Solution:-
Length = 3.5m breadth = 2.6m, height = 90cm \(=\frac{90}{100} m=0.9 m\).
Volume of a cuboid \(=l \times b \times h=3.5 \times 2.6 \times 0.9=8.19 m^{3}\)
Total surface area of a cuboid \(=2(l \times b+b \times h+h \times l)=2(3.5 \times 2.6+2.6 \times 0.9 \times 3.5) \)
=2(910+2.34+3.15)=2(14.59) \(=29.18 m^{2}\)
Question 2.
(i)The volume of a cuboid is 3456 \(\mathrm{cm}^{3}\). If its length =24 cm and breadth =18 cm; find its height.
Solution:
Volume of the given cuboid \(=3456 \mathrm{cm}^{3}\).
Length of the given cuboid =24 cm
Breadth of the given cuboid =18 cm
We know
Length × Breadth × Height = Volume of a cuboid
24×18× Height =3456
Height \(=\frac{3456}{24 \times 18}\)
Height \(=\frac{3456}{432}\)
Height =8cm
(ii) The volume of a cuboid is 7.68 \(m^{3}\). If its length = 3.2m and height =1.0m; find its breadth.
Solution:-
Volume of a cuboid =7.68 \(\mathrm{m}^{3}\)
Length of a cuboid =3.2 m
Height of a cuboid =1.0 m
We know
Length x Breadth x Height = Volume of a cuboid
3.2 × Breadth × 1.0=7.68\(\Rightarrow \text { Breadth }=\frac{7.68}{3.2 \times 1.0}\) \(\Rightarrow \text { Breadth }=\frac{7.68}{3.2}\)
⇒ Breadth =2.4 m
(iii) The breadth and height of a rectangular solid are 1.20 m and 80 cm respectively. If the volume of the cuboid is 1.92 \(m^{3}\); find its length.
Solution:-
Volume of a rectangular solid =1.92 \(\mathrm{m}^{3} \)
Breadth of a rectangular solid = 1.20 m
Height of a rectangular solid =80 cm=0.8 m
We know
Length × Breadth × Height = Volume of a rectangular solid (cubical)
Length × 1.20 × 0.8 = 1.92
Length × 0.96 = 1.92\(=\frac{1.92}{0.96}\) \(=\frac{192}{96}\)
Length =2 m
Question 3.
The length, breadth and height of a cuboid are in the ratio 5:3:2. If its volume is \(240 \mathrm{cm}^{3}\), find its dimensions. (Dimensions means: its length, breadth and height). Also find the total surface area of the cuboid.
Solution:-
Let length of the given cuboid =5x
Breadth of the given cuboid =3x
Height of the given cuboid =2x
Volume of the given cuboid \(=\text { Length } \times \text { Breadth } \times {height}\) \(=5 x \times 3 x \times 2 x=30 x^{3}\)
But we are given volume \(=240 \mathrm{cm}^{3}\) \(30 x^{3}=240 \mathrm{cm}^{3} \) \(x^{3}=\frac{240}{30}\) \(x^{3}=8\) \(x=8^{\frac{1}{3}}\) \(x=(2 \times 2 \times 2)^{\frac{1}{3}}\)
⇒x=2 cm
Length of the given cube \(=5 x=5 \times 2=10 \mathrm{cm} \)
Breadth of the given cube \(=3 x=3 \times 2=6 \mathrm{cm} \)
Height of the given cube \(=2 x=2 \times 2=4 \mathrm{cm} \)
Total surface area of the given cuboid \(=2(1 \times b+b \times h+h \times 1) \) \(=2(10 \times 6+6 \times 4+4 \times 10)=2(60+24+40)=2 \times 124=248 \mathrm{cm}^{2}\)
Question 4.
The length, breadth and height of a cuboid are in the ratio 6:5:3. If its total surface area is 504 c \(m^{2}\); find its dimensions. Also, find the volume of the cuboid.
Solution:-
Let length of the cuboid =6x
Breadth of the cuboid =5x
Height of the cuboid =3x
Total surface area of the given cuboid \(=2(1 \times b+b \times h+h \times l) \) \(=2(6 x \times 5 x+5 x \times 3 x+3 x \times 6 x)=2(30 \times 2+15 \times 2+18 \times 2) \) \(=2 \times 63 \times 2=126 x^{2}\)
But we are given total surface area of the given cuboid \(=504 \mathrm{cm}^{2} \) \(126 x^{2}=504 \mathrm{cm}^{2}\) \(\Rightarrow x^{2}=\frac{504}{126}\) \(\Rightarrow x^{2}=4\) \(\Rightarrow x=\sqrt{4}\)
⇒x=2 cm
Length of the cuboid \(=6 x=6 \times 2=12 \mathrm{cm} \)
Breadth of the cuboid \(=5 x=5 \times 2=10 \mathrm{cm}\)
Height of the cuboid \(=3 x=3 \times 2=6 \mathrm{cm}\)
Volume of the cuboid \(=l \times b \times h=12 \times 10 \times 6=720 \mathrm{cm}^{3}\)
Question 5.
Find the volume and total surface area of a cube whose edge is:
(i) 8 cm
Solution:-
Edge of the given cube =8cm
Volume of the given cube \(=(\text { Edge })^{3}=(8)^{3}=8 \times 8 \times 8=512 \mathrm{cm}^{3} \)
Total surface area of a cube \(=6(\text { Edge })^{2}=6 \times(8)^{2}=384 \mathrm{cm}^{2}\)
(ii) 2m 40 cm.
Solution:-
(ii)Edge of the given cube =2 m 40 cm=2.40 m
Volume of a cube \(=(\text { Edge })^{3}\)
Volume of the given cube \(=(2.40)^{3}=2.40 \times 2.40 \times 2.40=13.824 \mathrm{m}^{2}\)
Total surface area of the given cube \(=6 \times 2.4 \times 2.4=34.56 \mathrm{m}^{2}\)
Question 6.
Find the length of each edge of a cube, if its volume is:
(i) \(216 \mathrm{cm}^{3}\)
\( (\text { Edge })^{3} \)=Volume of a cube\( (\text { Edge })^{3}=216 \mathrm{cm}^{3}\)
Solution:-
Edge \(=(216)^{1 / 3}\)
Edge \(=(3 \times 3 \times 3 \times 2 \times 2 \times 2)^{1 / 3}\)
Edge \(=3 \times 2\)
Ans. Edge =6 cm.
(ii) \(1.728 \mathrm{m}^{3}\)
\( (Edge)^3\) = Volume of a cube \(∴(\mathrm{Edge})^{3}=1.728 \mathrm{m}^{3}\)\(\Rightarrow(\text { Edge })^{3}=\frac{1.728}{1000}=\frac{1728}{1000}\)\(Edge =\left(\frac{1728}{1000}\right)^{1 / 3}\)\(\mathrm{Edge}=\left(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}{10 \times 10 \times 10}\right)^{1 / 3}\)\(Edge =\frac{2 \times 2 \times 3}{10}\)\(Edge =\frac{12}{10} \mathrm{m}\)
Solution:-
Edge =1.2 m.
Question 7.
The total surface area of a cube is 216 cm
2. Find its volume.
\(6(\text {Edge})^{2}\)= Total surface area of a cube\(6(\text {Edge})^{2}=216 \mathrm{cm}^{2}\)\( (\text {Edge})^{2}=\frac{216}{6} \)\( (\text {Edge})^{2}=36\)
Solution:-
Edge \(=\sqrt{36}\)
Edge \(=\sqrt{36}\)
Volume of the given cube \(=(E d g e)^{3}=(6)^{3}=6 \times 6 \times 6=216 \mathrm{cm}^{3} \)
Question 8.
A solid cuboid of metal has dimensions 24 cm, 18 cm and 4 cm. Find its volume.
Solution:-
Length of the cuboid =24 cm
Breadth of the cuboid =18 cm
Height of the cuboid =4 cm\( Volume of the cuboid =l \times \mathrm{b} \times \mathrm{h}=24 \times 18 \times 4=1728 \mathrm{cm}^{3}\)
Question 9.
A wall 9 m long, 6 m high and 20 cm thick, is to be constructed using bricks of dimensions 30 cm, 15 cm and 10 cm. How many bricks will be required?
Solution:
Length of the wall \(=9 \mathrm{m}=9 \times 100 \mathrm{cm}=900 \mathrm{cm} \)
Height of the wall \(=6 \mathrm{m}=6 \times 100 \mathrm{cm}=600 \mathrm{cm}\)
Breadth of the wall =20 cm
Volume of the wall \(=900 \times 600 \times 20 \mathrm{cm}^{3}=10800000 \mathrm{cm}^{3} \)
Volume of one Brick \(=30 \times 15 \times 10 \mathrm{cm}^{3}=4500 \mathrm{cm}^{3}\)
Number of bricks required to construct the wall \(=\frac{\text { Volume of wall }}{\text { Volume of one brick}} \) \(=\frac{10800000}{4500}\)
=2400
Question 10.
A solid cube of edge 14 cm is melted down and recasted into smaller and equal cubes each of edge 2 cm; find the number of smaller cubes obtained.
Solution:-
Edge of the big solid cube = 14 cm
Volume of the big solid cube \(=14 \times 14 \times 14 \mathrm{cm}^{3}=2744 \mathrm{cm}^{3}\)
Edge of the small cube =2 cm
Volume of one small cube \(=2 \times 2 \times 2 \mathrm{cm}^{3}=8 \mathrm{cm}^{3} \)
Number of smaller cubes obtained \(=\frac{\text { Volume of big cube }}{\text { Volume of one small cube }}=\frac{2774}{8}=343\)
Question 11.
A closed box is cuboid in shape with length =40cm, breadth =30cm and height =50cm. It is made of thin metal sheet. Find the cost of metal sheet required to make 20 such boxes, if 1 \(m^{2} \) of metal sheet costs Rs. 45.
Solution:-
Length of closed box (1) =40cm
Breadth (b) =30cm
And height (h) =50cm
Total surface area \(=2(l \times b+b \times h+h \times l) \) \(=2(40 \times 30+30 \times 50+50 \times 40) \mathrm{cm}^{2}\) \(=2(1200+1500+2000) \mathrm{cm}^{2}\) \(=2 \times 4700=9400 \mathrm{cm}^{2}\)
Surface area of sheet used for 20 such boxes \(=9400 \times 20=188000 \mathrm{cm}^{2}=18.8 m^{2}\)
Cost of \(1 \mathrm{m}^{2} sheet = Rs.45\)
Total cost \(=18.8 \times 45=Rs.846\)
Question 12.
Four cubes, each of edge 9 cm, are joined as shown below:
Write the dimensions of the resulting cuboid obtained. Also, find the total surface area and the volume of the resulting cuboid.
Solution:-
Edge of each cube =9cm
(i) Length of the cuboid formed by 4 cubes (1) \(=9 \times 4=36 \mathrm{cm}\)
Breadth (b) =9cm and height (h) = 9cm
(ii) Total surface area of the cuboid = 2(lb + bh + hl)\(=2(36 \times 9+9 \times 9+9 \times 36) \mathrm{cm}^{2}\) \(=2(324+81+324) \mathrm{cm}^{2}\) \(=2 \times 729 \mathrm{cm}^{2}\) \(=1458 \mathrm{cm}^{2}\)
(iii) \(Volume =\quad l \times b \times h=36 \times 9 \times 9 cm^{2}=2916 cm^{3}\)
Question 13.
How many persons can be accommodated in a big-hall of dimensions 40 m, 25m and 15m; assuming that each person requires \(5 m^{3}\) of air?
Solution:-
No. of persons \(=\frac{\text { Vol. of the hall }}{\text { Vol. of air required for each person }}\)
Length of the hall =40m
Breadth =25m
Height =15m
Volume of the hall\( =1 \times \mathrm{b} \times \mathrm{h}=40 \times 25 \times 15=15000 \mathrm{m}^{3}\)
Volume of the air required for each person \(=5 \mathrm{m}^{3}\)
No. of persons who can be accommodated \(=\frac{\text { Volume of the hall }}{\text { Volume of air required for each person }}=\frac{15000 \mathrm{m}^{3}}{5 \mathrm{m}^{3}}=3000\)
Question 14.
The dimension of a class-room are; length = 15m, breadth =12m and height =7.5m. Find, how many children can be accommodated in this class-room; assuming 3.6 \(m^{3}\) of air is needed for each child.
Solution:-
Length of the room =15m
Breadth of the room =12m
Height of the room =7.5m
Volume of the room\( =L \times B \times H=15 \times 12 \times 7.5 \mathrm{m}^{3}=1350 \mathrm{m}^{3}\)
Volume of air required for each child \(=3.6 \mathrm{m}^{3}\)
No. of children who can be accommodated in the class room. \(=\frac{\text { Volume of class room }}{\text { Volume of air needed for each child }}=\frac{1350 \mathrm{m}^{3}}{3 \cdot 6 \mathrm{m}^{3}}\)
=375.
Question 15.
The length, breadth and height of a room are 6m, 5.4m and 4 m respectively. Find the area of:
(i) Its four-walls
(ii) Its roof.
Solution:-
Length of the room = 6m
Breadth of the room = 5.4m
Height of the room = 4m
(i) Area of four walls \(=2(L+B) \times H=2(6+5.4) \times 4=2 \times 11.4 \times 4=91.2 \mathrm{m}^{2}\)
(ii) Area of the roof \(=L \times B=6 \times 5.4=32.4 \mathrm{m}^{2}\) |
Definition:Real Interval/Unit Interval/Closed Definition
The closed interval from $0$ to $1$ is denoted $\mathbb I$ (or a variant) by some authors:
$\mathbb I := \left [{0 \,.\,.\, 1} \right] = \left\{{x \in \R: 0 \le x \le 1}\right\}$
This is often referred to as the
closed unit interval. Also denoted as
Sources which use the $\textbf{boldface}$ font for the number sets $\N, \Z, \Q, \R, \C$ tend also to use $\mathbf I$ for this entity.
Some sources merely use the ordinary italic font $I$.
An arbitrary interval is frequently denoted $\mathbb I$, although some sources use just $I$. Others use $\mathbf I$.
\(\displaystyle \openint a b\) \(:=\) \(\displaystyle \set {x \in \R: a < x < b}\) Open Real Interval \(\displaystyle \hointr a b\) \(:=\) \(\displaystyle \set {x \in \R: a \le x < b}\) Half-Open (to the right) Real Interval \(\displaystyle \hointl a b\) \(:=\) \(\displaystyle \set {x \in \R: a < x \le b}\) Half-Open (to the left) Real Interval \(\displaystyle \closedint a b\) \(:=\) \(\displaystyle \set {x \in \R: a \le x \le b}\) Closed Real Interval The term Wirth interval notation has consequently been coined by $\mathsf{Pr} \infty \mathsf{fWiki}$. |
Definition:Semigroup Endomorphism Definition
Let $\left({S, \circ}\right)$ be a semigroups.
Let $\phi: S \to S$ be a (semigroup) homomorphism from $S$ to itself.
Then $\phi$ is a semigroup endomorphism. Also see
The word
endomorphism derives from the Greek morphe ( ) meaning μορφή formor structure, with the prefix endo-(from ἔνδον') meaning inneror internal.
Thus
endomorphism means internal structure. |
Knowledge of the specific fact that $(\sin x)' = \cos x$ actually predates the general knowledge of calculus and derivatives. It was known in the following form: that for very small $\Delta x$, when you increase $x$ to $x + \Delta x$, the increase in value of the sine, from $\sin x$ to $\sin (x + \Delta x)$, is proportional to $\Delta x$ times $\cos x$. In other words, that$$\frac{\sin (x + \Delta x) - \sin x}{\Delta x} \approx \cos x$$The approximation being exact in the limit as $\Delta x \to 0$ is of course the modern definition of the derivative.
This happened historically in Indian mathematics, where Muñjala (around 932), Āryabhaṭa II (around 950), Prashastidhara (around 958) all give the above rule for calculating $\sin(x + \Delta x)$, and an explicit geometric reasoning / justification is given by Bhāskara II (around 1150) in his Siddhanta Shiromani. I have not found a perfectly good reference to these, but you can start with the following article:
Use of Calculus in Hindu Mathematics, by Bibhutibhusan Datta and Awadhesh Narayan Singh, revised by Kripa Shankar Shukla, Indian Journal of the History of Science, 19 (2): 95–104 (1984). (PDF)
It was first pointed out by Bapu Deva Shastri in
Bhaskara's knowledge of the Differential Calculus, Journal of the Asiatic Society of Bengal, Volume 27, 1858, pp. 213–6. |
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In this chapter we learned about left and right adjoints, and about joins and meets. At first they seemed like two rather different pairs of concepts. But then we learned some deep relationships between them. Briefly:
Left adjoints preserve joins, and monotone functions that preserve enough joins are left adjoints.
Right adjoints preserve meets, and monotone functions that preserve enough meets are right adjoints.
Today we'll conclude our discussion of Chapter 1 with two more bombshells:
Joins
are left adjoints, and meets are right adjoints.
Left adjoints are right adjoints seen upside-down, and joins are meets seen upside-down.
This is a good example of how category theory works. You learn a bunch of concepts, but then you learn more and more facts relating them, which unify your understanding... until finally all these concepts collapse down like the core of a giant star, releasing a supernova of insight that transforms how you see the world!
Let me start by reviewing what we've already seen. To keep things simple let me state these facts just for posets, not the more general preorders. Everything can be generalized to preorders.
In Lecture 6 we saw that given a left adjoint \( f : A \to B\), we can compute its right adjoint using joins:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$ Similarly, given a right adjoint \( g : B \to A \) between posets, we can compute its left adjoint using meets:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ In Lecture 16 we saw that left adjoints preserve all joins, while right adjoints preserve all meets.
Then came the big surprise: if \( A \) has all joins and a monotone function \( f : A \to B \) preserves all joins, then \( f \) is a left adjoint! But if you examine the proof, you'l see we don't really need \( A \) to have
all joins: it's enough that all the joins in this formula exist:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$Similarly, if \(B\) has all meets and a monotone function \(g : B \to A \) preserves all meets, then \( f \) is a right adjoint! But again, we don't need \( B \) to have
all meets: it's enough that all the meets in this formula exist:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ Now for the first of today's bombshells: joins are left adjoints and meets are right adjoints. I'll state this for binary joins and meets, but it generalizes.
Suppose \(A\) is a poset with all binary joins. Then we get a function
$$ \vee : A \times A \to A $$ sending any pair \( (a,a') \in A\) to the join \(a \vee a'\). But we can make \(A \times A\) into a poset as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Then \( \vee : A \times A \to A\) becomes a monotone map, since you can check that
$$ a \le a' \textrm{ and } b \le b' \textrm{ implies } a \vee b \le a' \vee b'. $$And you can show that \( \vee : A \times A \to A \) is the left adjoint of another monotone function, the
diagonal
$$ \Delta : A \to A \times A $$sending any \(a \in A\) to the pair \( (a,a) \). This diagonal function is also called
duplication, since it duplicates any element of \(A\).
Why is \( \vee \) the left adjoint of \( \Delta \)? If you unravel what this means using all the definitions, it amounts to this fact:
$$ a \vee a' \le b \textrm{ if and only if } a \le b \textrm{ and } a' \le b . $$ Note that we're applying \( \vee \) to \( (a,a') \) in the expression at left here, and applying \( \Delta \) to \( b \) in the expression at the right. So, this fact says that \( \vee \) the left adjoint of \( \Delta \).
Puzzle 45. Prove that \( a \le a' \) and \( b \le b' \) imply \( a \vee b \le a' \vee b' \). Also prove that \( a \vee a' \le b \) if and only if \( a \le b \) and \( a' \le b \).
A similar argument shows that meets are really right adjoints! If \( A \) is a poset with all binary meets, we get a monotone function
$$ \wedge : A \times A \to A $$that's the
right adjoint of \( \Delta \). This is just a clever way of saying
$$ a \le b \textrm{ and } a \le b' \textrm{ if and only if } a \le b \wedge b' $$ which is also easy to check.
Puzzle 46. State and prove similar facts for joins and meets of any number of elements in a poset - possibly an infinite number.
All this is very beautiful, but you'll notice that all facts come in pairs: one for left adjoints and one for right adjoints. We can squeeze out this redundancy by noticing that every preorder has an "opposite", where "greater than" and "less than" trade places! It's like a mirror world where up is down, big is small, true is false, and so on.
Definition. Given a preorder \( (A , \le) \) there is a preorder called its opposite, \( (A, \ge) \). Here we define \( \ge \) by
$$ a \ge a' \textrm{ if and only if } a' \le a $$ for all \( a, a' \in A \). We call the opposite preorder\( A^{\textrm{op}} \) for short.
I can't believe I've gone this far without ever mentioning \( \ge \). Now we finally have really good reason.
Puzzle 47. Show that the opposite of a preorder really is a preorder, and the opposite of a poset is a poset. Puzzle 48. Show that the opposite of the opposite of \( A \) is \( A \) again. Puzzle 49. Show that the join of any subset of \( A \), if it exists, is the meet of that subset in \( A^{\textrm{op}} \). Puzzle 50. Show that any monotone function \(f : A \to B \) gives a monotone function \( f : A^{\textrm{op}} \to B^{\textrm{op}} \): the same function, but preserving \( \ge \) rather than \( \le \). Puzzle 51. Show that \(f : A \to B \) is the left adjoint of \(g : B \to A \) if and only if \(f : A^{\textrm{op}} \to B^{\textrm{op}} \) is the right adjoint of \( g: B^{\textrm{op}} \to A^{\textrm{ op }}\).
So, we've taken our whole course so far and "folded it in half", reducing every fact about meets to a fact about joins, and every fact about right adjoints to a fact about left adjoints... or vice versa! This idea, so important in category theory, is called
duality. In its simplest form, it says that things come on opposite pairs, and there's a symmetry that switches these opposite pairs. Taken to its extreme, it says that everything is built out of the interplay between opposite pairs.
Once you start looking you can find duality everywhere, from ancient Chinese philosophy:
to modern computers:
But duality has been studied very deeply in category theory: I'm just skimming the surface here. In particular, we haven't gotten into the connection between adjoints and duality!
This is the end of my lectures on Chapter 1. There's more in this chapter that we didn't cover, so now it's time for us to go through all the exercises. |
I am working on "Elementary Number Theory" By Underwood Dudley and this is problem 13 in Section 4.
The question is "What can the last digit of a fourth power be?" I got the correct answer but I'm wondering if there is another more elegant way to do it. I'm also wondering if my argument is solid or if I just got the right answer by chance. My argument is as follows:
Any number A can be written as:
$A$ = $10^kn_k$ + $10^{k-1}$$n_{k-1}$ + $...$ + $10n_1$ + $n_0$,
where $n_k$ is the starting digit of the number and can take on any integer from 0 to 9 and so forth.
If we raise this to the fourth power and don't combine any of the terms, each term should be divisible by 10 except maybe not $n_0^4$. Now, $n_0^4$ must end must end in k where k satisfies:
$n_0^4$ $\equiv k\pmod {10}$.
And since $n_0^4$ ends in k, $A$ must end in k (since all of the other terms in the sum are divisible by 10). So, it suffices to just check what the last digit of the fourth powers of 1 through 9 are (since $0 \leq n_0 \leq 9$):
$0^4$ $\equiv 0\pmod{10}$
$1^4 \equiv 1\pmod{10}$
$2^4 \equiv 6\pmod{10}$
$3^4 \equiv 1\pmod{10}$
$4^4 \equiv 6\pmod{10}$
$5^4 \equiv 5\pmod{10}$
$6^4 \equiv 6\pmod{10}$
$7^4 \equiv 1\pmod{10}$
$8^4 \equiv 6\pmod{10}$
$9^4 \equiv 1\pmod{10}$.
So the fourth power of any integer must end in either a 0,1,5, or 6.
Did I get lucky or is this ok? Are there other more elegant ways?
Thank you! |
As with the review of Derivatives, it would be challenging to include a full review of Integrals. In this review, we try to include the most common integrals and rules used in STAT 414. There are many helpful websites as texts out there to help you review. We have provided links to Khan Academy for you to take a look at if you have difficulty recalling these methods.
For a function, \(f(x)\), its indefinite integral is:
\[\int f(x)\; dx=F(x)+C, \qquad \text{where } F^\prime(x)=f(x)\]
We provide a short list of common integrals and rules that are used in STAT 414. It is important to have a lot of practice and keep these skills fresh.
Common Integrals and Rules \(\int_a^a f(x)dx=0\) \(\int_a^b f(x)d(x)=-\int_b^a f(x)d(x)\) \(\int x^rdx=\frac{x^{r+1}}{r+1}+C\) The Fundamental Theorem of Calculus: Let \(f\) be integrable on \([a,b]\) and let \(F\) be any antiderivative of \(f\) there. Then, \(\int_a^b f(x)d(x)=F(b)-F(a)\). (FTC, Khan Academy) \(\int x^n dx=\dfrac{1}{n+1}x^{n+1}+C, \;\;n\neq(-1)\) \(\int \dfrac{1}{x}dx=\ln |x| +C\) \(\int e^x dx=e^x +C\) Integration Using Substitution: Let \(g\) have a continuous derivative on \([a,b]\) and let \(f\) be continuous on the range of \(g\). Then
\[\begin{equation}
\int_a^b f\left(g(x)\right)g^\prime(x)dx=\int_{g(a)}^{g(b)}f(u)du \end{equation}\]
where \(u=g(x)\). (u-Substitution, Khan Academy)
Integration by Parts
\[\begin{equation}
\int_a^b udv=\left[uv\right]_a^b-\int_a^b vdu \end{equation}\]. (Integration by Parts, Khan Academy) Example C.3.1
Integrate the following function from 0 to $t$
\[f(x)=\dfrac{2}{1000^2}xe^{-(x/1000)^2}\]
\[\int_0^t \frac{2}{1000^2}xe^{-(x/1000)^2} dx\label{eqn1}\]
Let \(u=\left(\frac{x}{1000}\right)^2\). Then \(du=\frac{2}{1000^2}xdx\). The equation becomes...
\[\begin{align*}
&= \int_0^{\left(\frac{t}{1000}\right)^2} e^{-u}du =-e^{-u}|_{0}^{\left(\frac{t}{1000}\right)^2}\\ &= -e^{-\left(\frac{t}{1000}\right)^2}-(-1)=1-e^{-\left(\frac{t}{1000}\right)^2}. \end{align*}\] Example C.3.2
Integrate the following:
\[\int_0^5 x^2e^{-x}dx\]
Let us begin by setting up integration by parts. Let
\[\begin{align*}
& u=x^2 \qquad dv=e^{-x}dx\\ & du=2xdx \qquad v=-e^{-x} \end{align*}\]
Then
\[\begin{align*}
uv|_0^5-\int_0^5 vdu &=-x^2e^{-x}|_0^5+2\int_0^5xe^{-x}dx\\ &= -x^2e^{-x}|_0^5+2\left[-xe^{-x}|_0^5+\int_0^5 e^{-x}dx\right]\\ &= -x^2e^{-x}|_0^5+2\left[-xe^{-x}|_0^5-e^{-x}|_0^5\right]\approx 1.75 \end{align*}\] Example C.3.3
Integrate the following from \(-\infty\) to \(\infty\).
\[f(y)=\frac{1}{2}e^{-|y|+ty}, \;\; \text{ for } -\infty<y<\infty\]
\begin{align*}
\int_{-\infty}^{\infty} \frac{1}{2} e^{ty-|y|}dy &= \int_{-\infty}^0 \frac{1}{2}e^{y+ty}dy+\int_0^{\infty} \frac{1}{2}e^{-y+ty}dy\\ & = \int_{-\infty}^0 \frac{1}{2}e^{y(1+t)}dy+\int_0^{\infty} \frac{1}{2}e^{-y(1-t)}dy\\ & = \frac{1}{2(1+t)}+ \frac{1}{2(1-t)}=\frac{1}{2}\left(\frac{1-t+t+1}{(1+t)(1-t)}\right)\\ & =\frac{1}{(1-t)(1+t)} \end{align*} |
Strategy
I would like to apply rational decision theory to the analysis, because that is one well-established way to attain rigor in solving a statistical decision problem. In trying to do so, one difficulty emerges as special: the alteration of SB’s consciousness.
Rational decision theory has no mechanism to handle altered mental states.
In asking SB for her credence in the coin flip, we are simultaneously treating her in a somewhat self-referential manner both as subject (of the SB experiment) and experimenter (concerning the coin flip).
Let’s alter the experiment in an inessential way: instead of administering the memory-erasure drug, prepare a stable of Sleeping Beauty clones just before the experiment begins. (This is the key idea, because it helps us resist distracting--but ultimately irrelevant and misleading--philosophical issues.)
The clones are like her in all respects, including memory and thought.
SB is fully aware this will happen.
We E. T. Jaynes replaces the question "how can we build a mathematical model of human common sense"--something we need in order to think through the Sleeping Beauty problem--by "How could we build a machine which would carry out useful plausible reasoning, following clearly defined principles expressing an idealized common sense?" Thus, if you like, replace SB by Jaynes' thinking robot, and clone that. can clone, in principle.
(There have been, and still are, controversies about "thinking" machines.
"They will never make a machine to replace the human mind—it does many things which no machine could ever do."
You insist that there is something a machine cannot do. If you will tell me precisely what it is that a machine cannot do, then I can always make a machine which will do just that!”
--J. von Neumann, 1948. Quoted by E. T. Jaynes in
Probability Theory: The Logic of Science, p. 4.)
--Rube Goldberg
The Sleeping Beauty experiment restated
Prepare $n \ge 2$ identical copies of SB (including SB herself) on Sunday evening. They all go to sleep at the same time, potentially for 100 years. Whenever you need to awaken SB during the experiment,
randomly select a clone who has not yet been awakened. Any awakenings will occur on Monday and, if needed, on Tuesday.
I claim that this version of the experiment creates exactly the same set of possible results, right down to SB's mental states and awareness, with exactly the same probabilities. This potentially is one key point where philosophers might choose to attack my solution. I claim it's the
last point at which they can attack it, because the remaining analysis is routine and rigorous.
Now we apply the usual statistical machinery. Let's begin with the sample space (of possible experimental outcomes). Let $M$ mean "awakens Monday" and $T$ mean "awakens Tuesday." Similarly, let $h$ mean "heads" and "t" mean tails. Subscript the clones with integers $1, 2, \ldots, n$. Then the possible experimental outcomes can be written (in what I hope is a transparent, self-evident notation) as the set
$$\eqalign{\{&hM_1, hM_2, \ldots, hM_n, \\&(tM_1, tT_2), (tM_1, tT_3), \ldots, (tM_1, tT_n), \\&(tM_2, tT_2), (tM_2, tT_3), \ldots, (tM_2, tT_n), \\&\cdots, \\&(tM_{n-1}, tT_2), (tM_{n-1}, tT_3), \ldots, (tM_{n-1}, tT_n) & \}.}$$
Monday probabilities
As one of the SB clones, you figure your chance of being awakened on Monday during a heads-up experiment is ($1/2$ chance of heads) times ($1/n$ chance I’m picked to be the clone who is awakened). In more technical terms:
The set of heads outcomes is $h = \{hM_j, j=1,2, \ldots,n\}$. There are $n$ of them.
The event where
you are awakened with heads is $h(i) = \{hM_i\}$.
The chance of any particular SB clone $i$ being awakened with the coin showing heads equals $$\Pr[h(i)] = \Pr[h] \times \Pr[h(i)|h] = \frac{1}{2} \times \frac{1}{n} = \frac{1}{2n}.$$
Tuesday probabilities
The set of tails outcomes is $t = \{(tM_j, tT_k): j \ne k\}$. There are $n(n-1)$ of them. All are equally likely, by design.
You, clone $i$, are awakened in $(n-1) + (n-1) = 2(n-1)$ of these cases; namely, the $n-1$ ways you can be awakened on Monday (there are $n-1$ remaining clones to be awakened Tuesday) plus the $n-1$ ways you can be awakened on Tuesday (there are $n-1$ possible Monday clones). Call this event $t(i)$.
Your chance of being awakened during a tails-up experiment equals $$\Pr[t(i)] = \Pr[t] \times P[t(i)|t] = \frac{1}{2} \times \frac{2(n-1}{n(n-1)} = \frac{1}{n}.$$
Bayes' Theorem
Now that we have come this far, Bayes' Theorem--a mathematical tautology beyond dispute--finishes the work. Any clone's chance of heads is therefore $$\Pr[h | t(i) \cup h(i)] = \frac{\Pr[h]\Pr[h(i)|h]}{\Pr[h]\Pr[h(i)|h] + \Pr[t]\Pr[t(i)|t]} = \frac{1/(2n)}{1/n + 1/(2n)} = \frac{1}{3}.$$
Because SB is indistinguishable from her clones--even to herself!--this is the answer she should give when asked for her degree of belief in heads.
Interpretations
The question "what is the probability of heads" has it can ask for the chance a fair coin lands heads, which is $\Pr[h] = 1/2$ (the Halfer answer), or it can ask for the chance the coin lands heads, conditioned on the fact that you were the clone awakened. This is $\Pr[h|t(i) \cup h(i)] = 1/3$ (the Thirder answer). two reasonable interpretations for this experiment:
In the situation in which SB (or rather any one of a set of identically prepared Jaynes thinking machines) finds herself, this analysis--which many others have performed (but I think less convincingly, because they did not so clearly remove the philosophical distractions in the experimental descriptions)--supports the Thirder answer.
The Halfer answer is correct, but uninteresting, because it is not relevant to the situation in which SB finds herself. This resolves the paradox.
This solution is developed within the context of a single well-defined experimental setup.
Clarifying the experiment clarifies the question. A clear question leads to a clear answer. Comments
I guess that, following Elga (2000), you could legitimately characterize our conditional answer as "count[ing] your own temporal location as relevant to the truth of h," but that characterization adds no insight to the problem: it only detracts from the mathematical facts in evidence. To me it appears to be just an obscure way of asserting that the "clones" interpretation of the probability question is the correct one.
This analysis suggests that the underlying philosophical issue is one of
identity: What happens to the clones who are not awakened? What cognitive and noetic relationships hold among the clones?--but that discussion is not a matter of statistical analysis; it belongs on a different forum. |
Stability of ground states for logarithmic Schrödinger equation with a $δ^{\prime}$-interaction
Department of Mathematics, IME-USP, Cidade Universitária, CEP 05508-090, São Paulo, SP, Brazil
$δ^{\prime}$
$i{\partial _t}u + \partial _x^2u + {\rm{ }}{\gamma ^\prime }(x)u + u{\mkern 1mu} {\rm{Log|}}u|2 = 0,(x,t) \in \mathbb{R} \times \mathbb{R} ,$ Mathematics Subject Classification:35Q51, 35Q55, 37K40, 34B37. Citation:Alex H. Ardila. Stability of ground states for logarithmic Schrödinger equation with a $δ^{\prime}$-interaction. Evolution Equations & Control Theory, 2017, 6 (2) : 155-175. doi: 10.3934/eect.2017009
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R. Adami and D. Noja, Existence of dynamics for a 1-d NLS equation perturbed with a generalized point defect
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R. Adami and D. Noja,
Stability and symmetry-breaking bifurcation for the ground states of a NLS with a
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R. Adami and D. Noja,
Exactly solvable models and bifurcations: The case of the cubic NLS with a
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R. Adami, D. Noja and N. Visciglia,
Constrained energy minimization and ground states for NLS with point defects,
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Stability of standing waves for a nonlinear Schrödinger equation with a repulsive {D}irac delta potential,
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R. Fukuizumi, M. Ohta and T. Ozawa,
Nonlinear Schrödinger equation with a point defect,
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R. Fukuizumi and A. Sacchetti,
Bifurcation and stability for nonlinear Schrödinger equations with double well potential in the semiclassical limit,
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A. Haraux,
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R.K. Jackson and M. Weinstein,
Geometric analysis of bifurcation and symmetry breaking in a {G}ross-{P}itaevskii equation,
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M. Kaminaga and M. Ohta,
Stability of standing waves for nonlinear {S}chrödinger equation with attractive delta potential and repulsive nonlinearity,
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E.W. Kirr, P. Kevrekidis and D. Pelinovsky,
Symmetry-breaking bifurcation in the nonlinear Schrödinger equation with symmetric potentials,
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A. Kostenko and M. Malamud,
Spectral theory of semibounded Schrödinger operators with $δ^{\prime}$-interactions,
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S. Le Coz, R. Fukuizumi, G. Fibich, B. Ksherim and Y. Sivan,
Instability of bound states of a nonlinear Schrödinger equation with a dirac potential,
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A. ~Sacchetti, Universal critical power for nonlinear Schrödinger equations with symmetric double well potential
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Logarithmic nonlinearity in theories of quantum gravity: {O}rigin of time and observational consequences,
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R. Adami and D. Noja, Existence of dynamics for a 1-d NLS equation perturbed with a generalized point defect
[2] [3]
R. Adami and D. Noja,
Stability and symmetry-breaking bifurcation for the ground states of a NLS with a
[4]
R. Adami and D. Noja,
Exactly solvable models and bifurcations: The case of the cubic NLS with a
[5]
R. Adami, D. Noja and N. Visciglia,
Constrained energy minimization and ground states for NLS with point defects,
[6] [7]
J. Angulo and A. H. Ardila, Stability of standing waves for logarithmic Schrödinger equation with attractive delta potential, Indiana Univ. Math. J., to appear.Google Scholar
[8] [9] [10] [11] [12]
T. Cazenave,
[13] [14]
R. Fukuizumi and L. Jeanjean,
Stability of standing waves for a nonlinear Schrödinger equation with a repulsive {D}irac delta potential,
[15]
R. Fukuizumi, M. Ohta and T. Ozawa,
Nonlinear Schrödinger equation with a point defect,
[16]
R. Fukuizumi and A. Sacchetti,
Bifurcation and stability for nonlinear Schrödinger equations with double well potential in the semiclassical limit,
[17]
A. Haraux,
[18] [19]
R.K. Jackson and M. Weinstein,
Geometric analysis of bifurcation and symmetry breaking in a {G}ross-{P}itaevskii equation,
[20]
M. Kaminaga and M. Ohta,
Stability of standing waves for nonlinear {S}chrödinger equation with attractive delta potential and repulsive nonlinearity,
[21] [22]
E.W. Kirr, P. Kevrekidis and D. Pelinovsky,
Symmetry-breaking bifurcation in the nonlinear Schrödinger equation with symmetric potentials,
[23]
A. Kostenko and M. Malamud,
Spectral theory of semibounded Schrödinger operators with $δ^{\prime}$-interactions,
[24]
S. Le Coz, R. Fukuizumi, G. Fibich, B. Ksherim and Y. Sivan,
Instability of bound states of a nonlinear Schrödinger equation with a dirac potential,
[25] [26]
A. ~Sacchetti, Universal critical power for nonlinear Schrödinger equations with symmetric double well potential
[27] [28] [29]
K. Zloshchastiev,
Logarithmic nonlinearity in theories of quantum gravity: {O}rigin of time and observational consequences,
[1]
Patricio Felmer, César Torres.
Radial symmetry of ground states for a regional fractional Nonlinear Schrödinger Equation.
[2]
Wen Feng, Milena Stanislavova, Atanas Stefanov.
On the spectral stability of ground states of semi-linear Schrödinger and Klein-Gordon equations with fractional dispersion.
[3]
Chuangye Liu, Zhi-Qiang Wang.
A complete classification of ground-states for a coupled nonlinear Schrödinger system.
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Zupei Shen, Zhiqing Han, Qinqin Zhang.
Ground states of nonlinear Schrödinger equations with fractional Laplacians.
[6]
Daniele Garrisi, Vladimir Georgiev.
Orbital stability and uniqueness of the ground state for the non-linear Schrödinger equation in dimension one.
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Jaime Angulo Pava, César A. Hernández Melo.
On stability properties of the Cubic-Quintic Schródinger equation with $\delta$-point interaction.
[8]
Giuseppe Maria Coclite, Helge Holden.
Ground states of the Schrödinger-Maxwell system with dirac
mass: Existence and asymptotics.
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Eugenio Montefusco, Benedetta Pellacci, Marco Squassina.
Energy convexity
estimates for non-degenerate ground states of nonlinear 1D
Schrödinger systems.
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Dongdong Qin, Xianhua Tang, Qingfang Wu.
Ground states of nonlinear Schrödinger systems with periodic or non-periodic potentials.
[11]
Zhitao Zhang, Haijun Luo.
Symmetry and asymptotic behavior of ground state solutions for schrödinger systems with linear interaction.
[12]
Alexander Komech, Elena Kopylova, David Stuart.
On asymptotic stability of solitons
in a nonlinear Schrödinger equation.
[13]
Kenji Nakanishi, Tristan Roy.
Global dynamics above the ground state for the energy-critical Schrödinger equation with radial data.
[14]
Chenmin Sun, Hua Wang, Xiaohua Yao, Jiqiang Zheng.
Scattering below ground state of focusing fractional nonlinear Schrödinger equation with radial data.
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Xiaoyan Lin, Yubo He, Xianhua Tang.
Existence and asymptotic behavior of ground state solutions for asymptotically linear Schrödinger equation with inverse square potential.
[16]
Silvia Cingolani, Mónica Clapp.
Symmetric semiclassical states to a magnetic nonlinear Schrödinger equation via equivariant Morse theory.
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Minbo Yang, Yanheng Ding.
Existence and multiplicity of semiclassical states for a quasilinear Schrödinger equation in $\mathbb{R}^N$.
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Reika Fukuizumi.
Stability and instability of standing waves for the nonlinear Schrödinger equation with harmonic potential.
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François Genoud.
Existence and stability of high frequency standing waves for a nonlinear
Schrödinger equation.
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There is a language: $L=\{w: w\in \{a,b\}^*, |w|_a\ \equiv |w|_b \equiv 0$ $(mod$ $5) \}$. My idea for DFA is - we count number of $a$$\pmod{5}$ and separately we count number of $b$$\pmod{5}$. So we end up with $5*5=25$ states, because each state have to keep track for both number of $a$ and $b$$\pmod{5}$. Is there any better aproach to this DFA ? Because for example when we have$\pmod{10}$ we end up with 100 states. I would appreciate any sugestions.
The minimal DFA for $L$ contains 25 states, as can easily be shown using Myhill–Nerode theory.
Indeed, consider the 25 words $\{a^ib^j : 0 \leq i,j < 5\}$. I will show that when a DFA reads any two different words from this list, it cannot end up in the same state. To show this, for any two different words $x,y$ from the list, I will give a word $z$ such that $xz \in L$ but $yz \notin L$.
Let $x = a^i b^j$ and $y = a^k b^\ell$ be two different words from the list. Since $x \neq y$, either $i \neq k$ or $j \neq \ell$ (or both). Assume without loss of generality that $i \neq k$. Take $z = a^{5-i} b^{5-j}$. Then $|xz|_a = |xz|_b = 5$, and so $xz \in L$. In contrast, $|yz|_a = 5-i+k \not\equiv 0 \pmod{5}$, since $i \not\equiv k \pmod{5}$ by assumption, showing that $yz \notin L$. |
I'm currently doing an exercise and faced with the following question:
We have the true form: $y_i=\beta_0 +\beta_1 d_i +u_i $
Where $d_i$ is a dummy variable. We have measured $d_i$ with measurement error such that 10% of those for whom $d_i=1$ have been recorded to have $d_i=0$ and similarly 10% of those for whom $d_i=0$ have been recorded to have $d_i=1$.
We have estimated $\hat\beta_1 =0.205$ with $SE=0.015$.
Compute the bias due to measurement error in the OLS regression.
I'm aware with measurement error in dummy variables we have a form of non-classical measurement error, where the measurement error is negatively correlated with the true value.
I've played around with it for a while but haven't really been able to figure out how to go about this. I attempted to use the reliability ratio:
Define $\tilde d_i=d_i + v_i$ & $p_d=Pr[d_i=1]$
Therefore
$\tilde p_d=0.9p_d+0.1(1-p_d)$ & $(1-\tilde p_d)=(1-p_d)0.9+0.1p_d$
$Var(d_i)=\ p_d(1-p_d)$
$Var(\tilde d_i)=\tilde p_d(1-\tilde p_d)=0.64(p_d(1-p_d))+0.09$
Subbing into:
$$\hat\beta_1=\beta_1*\frac{Var(d_i)}{Var(\tilde d_i)}$$
$$0.205=\beta_1\frac{p_d(1-p_d)}{0.64p_d(1-p_d)+0.09}$$
I'm aware this must be wrong, I imagine because perhaps this isn't classical measurement error. Any guidance on how to calculate the bias would be greatly appreciated. Thanks |
This work is based on Zengyi Dou's MS thesis. The oil well placement problem
The oil well placement problem is vital part of secondary oil production. Since the calculation of the net present value (NPV) of an investment depends on the solution of expensive partial differential equations that require tremendous computational resources, traditional methods are doomed to fail. The problem becomes exceedingly more difficult when we take into account the uncertainties in the oil price as well as in the ground permeability. In this study, we formulate the oil well placement problem as a global optimization problem that depends on the output of a finite volume solver for the two-phase immiscible flow (water-oil). Then, we employ the machinery of Bayesian global optimization (BGO) to solve it using a limited simulation budget. BGO uses Gaussian process regression (GPR) to represent our state of knowledge about the objective as captured by a finite number of simulations and adaptively selects novel simulations via the expected improvement (EI) criterion. Finally, we develop an extension of the EI criterion to the case of noisy objectives enabling us to solve the oil well placement problem while taking into account uncertainties in the oil price and the ground permeability. We demonstrate numerically the efficacy of the proposed methods and find valuable computational savings.
Problem specification
The objective of the oil well placement problem is to select the well locations (injection and production) that maximize the NPV of the investment. The NPV of the investment is:
\begin{equation} f_T(\mathbf{x}) = \int_0^T \left\{\sum_{\text{prod. wells}\;i}\left[c_o(t)q_{o,j}^-(\mathbf{x}, t) - c_{w,\text{disp}}q_{w,j}^-(\mathbf{x},t)\right] - \sum_{\text{inj. wells}\;j}c_{w,\text{inj}}q_{w,j}^+(\mathbf{x},t) \right\}(1 + r)^{-t/365}dt, \label{eq:npv} \end{equation} where \begin{equation} \mathbf{x}^* = \arg\max_{\mathbf{x}}\mathbb{E}\left[ f_T(\mathbf{x})\right], \end{equation} where the expectation is taken over the uncertainty in oil price as well as the uncertainty in the permeability field.
\begin{equation}
f_T(\mathbf{x}) = \int_0^T \left\{\sum_{\text{prod. wells}\;i}\left[c_o(t)q_{o,j}^-(\mathbf{x}, t) - c_{w,\text{disp}}q_{w,j}^-(\mathbf{x},t)\right]
- \sum_{\text{inj. wells}\;j}c_{w,\text{inj}}q_{w,j}^+(\mathbf{x},t)
\right\}(1 + r)^{-t/365}dt,
\label{eq:npv}
\end{equation}
where
\(\mathbf{x}\) is a design vector representing the location of all wells, e.g., if we consider one injection and one production well, then \(\mathbf{x}\) is four-dimensional. \(T\) is the time horizon of the investment. The first sum is over all production wells and the second sum over all injection wells. \(c_o(t)\) is the uncertainoil price time series, \(c_{w,\text{dis}}\) is the price for disposing contaminated water, and \(c_{w,\text{inj}}\) is the price of injecting water. \(q_{o,i}^-(\mathbf{x},t)\), \(q_{w,i}^-(\mathbf{x},t)\), and \(q_{w,j}^+(\mathbf{x},t)\) are the oil production rate, the water disposal rate, and the water injection rate, respectively, for a choice of design variables \(\mathbf{x}\) at time \(t\). Note that these rates depend on the solution of a coupled systems of partial differential equations that model two-phase flow through the ground. They also depend on the uncertainground permeability. \(r\) is a discount factor.
\begin{equation}
\mathbf{x}^* = \arg\max_{\mathbf{x}}\mathbb{E}\left[ f_T(\mathbf{x})\right],
\end{equation}
where the expectation is taken over the uncertainty in oil price as well as the uncertainty in the permeability field.
Quantifying uncertainties
For the
$$ \log c_o(t+1) = \log c_o(t) + \mu t + \sigma z_t, $$ where \(\mu\) is the growth rate, \(\sigma\) the volatility, and \(z_t \sim \mathcal{N}(0, 1)\). For the $$ \log K(\mathbf{x}_s) = \log K_0(\mathbf{x}_s) + G(\mathbf{x}_s), $$ where \(\mathbf{x}_s\) stands for the spatial location, \(K_0(\mathbf{x}_s)\) for a mean permeability profile, and \(G(\cdot)\) is a Gaussian process defined over the reservoir with zero mean and exponential covariance function. oil pricewe use a log-normal random walk with a constant drift. That is, we assume that:
$$
\log c_o(t+1) = \log c_o(t) + \mu t + \sigma z_t,
$$
where \(\mu\) is the growth rate, \(\sigma\) the volatility, and \(z_t \sim \mathcal{N}(0, 1)\).
For the
permeabilitywe use a log-Gaussian process model. That is, we assume that:
$$
\log K(\mathbf{x}_s) = \log K_0(\mathbf{x}_s) + G(\mathbf{x}_s),
$$
where \(\mathbf{x}_s\) stands for the spatial location, \(K_0(\mathbf{x}_s)\) for a mean permeability profile, and \(G(\cdot)\) is a Gaussian process defined over the reservoir with zero mean and exponential covariance function.
Solution strategy
When these quantities become uncertainty, this uncertainty is propagated to the NPV. The figure on the left demonstrates the broadening of the uncertainty in the NPV that is induced by this fact. Notice that including the permeability uncertainty more than doubles the risk associated with the investment.
We solved the oil well placement problem by employing the Bayesian global optimization (BGO) framework, albeit using a novel extension of the expected improvement that is robust under uncertainty.
The following video shows simulations that BGO picks at each iteration.
The final video shows the evolution of water saturation (water is blue) as a function of time for the best well locations found by our approach. |
A message to the aliens, part 23/23 (wat)
Earlier articles:IntroductionCommon featuresPage 1 (numerals)Page 2 (arithmetic)Page 3 (exponents)Page 4 (algebra)Page 5 (geometry)Page 6 (chemistry)Page 7 (mass)Page 8 (time and space)Page 9 (physical units)Page 10 (temperature)Page 11 (solar system)Page 12 (Earth-Moon system)Page 13 (days, months, and years)Page 14 (terrain)Page 15 (human anatomy)Page 16 (vital statistics)Page 17 (DNA chemistry)Page 18 (cell respiration and division)Pages 19-20 (map of the Earth)Page 21 (the message)Page 22 (cosmology)
This is page 23 (the last) of the
Cosmic Callmessage. An explanation follows.
This page is a series of questions for the recipients of the message.It is labeled with the glyph , which heretoforeappeared only on page 4 in the contextof solving of algebraic equations. So we might interpret it asmeaning a solution or a desire to solve or understand. I have chosen to translateit as “wat”.
I find this page irritating in its vagueness and confusion. Itslayout is disorganized. Glyphs are used inconsistent with their useselsewhere on the page and elsewhere in the message.For example, the mysterious glyph , whichhas something to do with the recipients of the message, and whichappeared only on page 21 is used here toask about both the recipients themselves and also about their planet.
The questions are arranged in groups. For easy identification, I havecolor-coded the groups.
Starting from the upper-left corner, and proceeding counterclockwise, wehave:
Kilograms, meters, and seconds, wat. I would have used the glyphs forabstract mass, distance, and time,and ,since that seems to be closer to the intended meaning.
Alien mathematics, physics, and biology, wat. Note that this asksspecifically about the recipients’ version of the sciences.None of these three glyphs has been subscripted before. Will themeaning be clear to the recipients? One also wonders why the messagedoesn't express a desire to understand human science, or sciencegenerally. One might argue that it does not make sense to ask therecipients about the human versions of mathematics and physics. But alater group expresses a desire to understand males and females, and therecipients don't know anything about that either.
Aliens wat. Alien [planet] mass, radius, acceleration wat. The meaning ofshifts here from meaning the recipients themselves to the recipients’planet. “Acceleration”is intended to refer to the planet's gravitational acceleration ason page 14. What if the recipientsdon't live on a planet? I suppose they will be familiar with planetsgenerally and with the fact that we live on a planet, which explainedback on pages 11–13, and will get the idea.
Fucking speed of light, how does it work?
Planck's constant, wat. Universal gravitation constant, wat?
Males and females, wat. Alien people, wat. Age of people,wat. This group seems to be about our desire to understand ourselves,except that the third item relates to the aliens. I'm not quite surewhat is going on. Perhaps “males and females” is intended to refer tothe recipients? But the glyphs are not subscripted, and there is nostrong reason to believe that the aliens have the same sexuality.
The glyph , already usedboth to mean the age of the Earth and the typical human lifespan, iseven less clear here. Does it mean we want to understand the reasonsfor human life expectancy? Or is it intended to continue the inquiryfrom the previous line and is asking about the recipients’ history orlifespan?
Land, water, and atmosphere of the recipients’ planet, wat.
Energy, force, pressure, power, wat. The usage here isinconsistent from the first group, which asked not about mass,distance, and time but about kilograms, meters, and seconds specifically.
Velocity and acceleration, wat. I wonder why these are in aseparate group, instead of being clustered with the previous group orthe first group. I also worry about the equivocation in acceleration,which is sometimes used to mean the Earth's gravitational accelerationand sometimes acceleration generally. We already said we want tounderstand mass ,!!G!! ,and the size of the Earth. The Earth's surface gravity can bestraightforwardly calculated from these, so there's nothing else tounderstand about that.
Alien planet, wat.The glyph hasheretofore been used only to refer to the planet Earth. It does not mean planetsgenerally, because it was not used in connection with Jupiter.Here, however, itseems to refer to the recipients’ planet.
The universe, wat. HUH???
That was the last page. Thanks for your kind attention.
[ Many thanks to Anna Gundlach, without whose timely email I might not have found the motivation to finish this series. ]
[Other articles in category /aliens/dd] permanent link
Math.SE report 2015-08
I only posted three answers in August, but two of them were interesting.
In why this !!\sigma\pi\sigma^{-1}!! keeps apearing in my grouptheory book? (cycledecomposition) thequerent asked about the “conjugation” operation that keeps croppingup in group theory. Why is it important? I sympathize with this;it wasn't adequately explained when I took group theory, and I hadto figure it out a long time later. Unfortunately I don't think Ipicked the right example to explain it, so I am going to try againnow.
Consider the eight symmetries of the square. They are of five types:
Rotation clockwise or counterclockwise by 90°. Rotation by 180°. Horizontal or vertical reflection Diagonal reflection The trivial (identity) symmetry
What is meant when I say that a horizontal and a vertical reflectionare of the same ‘type’? Informally, it is that the horizontalreflection looks just like the vertical reflection, if you turn yourhead ninety degrees. We can formalize this by observing that if werotate the square 90°, then give it a horizontal flip, then rotate itback, the effect is exactly to give it a vertical flip. In notation,we might represent the horizontal flip by !!H!!, the vertical flip by!!V!!, the clockwise rotation by !!\rho!!, and the counterclockwiserotation by !!\rho^{-1}!!; then we have
$$ \rho H \rho^{-1} = V$$
and similarly
$$ \rho V \rho^{-1} = H.$$
Vertical flips do not look like diagonal flips—the diagonal flip leaves two of the corners in the same place, and the vertical flip does not—and indeed there isno analogous formula with !!H!! replaced with one of the
diagonalflips. However, if !!D_1!! and !!D_2!! are the two diagonal flips,then we do have
$$ \rho D_1 \rho^{-1} = D_2.$$
In general, When !!a!! and !!b!! aretwo symmetries, and there is some symmetry !!x!! for which
$$xax^{-1} = b$$
we say that !!a!! is
conjugate to !!b!!.One can show thatconjugacy is an equivalence relation, which means that the symmetriesof any object can be divided into separate “conjugacy classes” such that twosymmetries are conjugate if and only if they are in the same class.For the square, the conjugacy classes are the five I listed earlier.
This conjugacy thing is important for telling when two symmetriesare group-theoretically “the same”, and have the samegroup-theoretic properties. For example, the fact that thehorizontal and vertical flips move all four vertices, while thediagonal flips do not. Another example is that a horizontal flip isself-inverse (if you do it again, it cancels itself out), but a 90°rotation is not (you have to do it
four times before it cancelsout.) But the horizontal flip shares all its properties with thevertical flip, because it is the same if you just turn your head.
Identifying this sameness makes certain kinds of arguments muchsimpler. For example, in countingsquares, I wanted tocount the number of ways of coloring the faces of a cube, and insteadof dealing with the 24 symmetries of the cube, I only needed to dealwith their 5 conjugacy classes.
The example I gave in my math.se answer was maybe less perspicuous. Iconsidered the symmetries of a sphere, and talked about how two rotations of the sphere by 17° are conjugate, regardless of what axisone rotates around. I thought of the square at the end, and threw itin, but I wish I had started with it.
How to convert a decimal to a fractioneasily? was themonth's big winner. OP wanted to know how to take a decimal like!!0.3760683761!! and discover that it can be written as!!\frac{44}{117}!!. The right answer to this is of course to usecontinued fraction theory, but I did not want to write a longtreatise on continued fractions, so I stripped down the theory toobtain an algorithm that is slower, but much easier to understand.
The algorithm is just binary search, but with a twist. If you are looking for afraction for !!x!!, and you know !!\frac ab < x < \frac cd!!, thenyou construct the mediant !!\frac{a+c}{b+d}!! and compare it with!!x!!. This gives you a smaller interval in which to search for!!x!!, and the reason you use the mediant instead of using!!\frac12\left(\frac ab + \frac cd\right)!! as usual is that if you use themediant you are guaranteed to exactly nail all the best rationalapproximations of !!x!!. This is the algorithm Idescribed a few years ago in your age as a fraction,again; there the binary search proceedsdown the branches of the Stern-Brocot tree to find a fraction closeto !!0.368!!.
I did ask a question this month: I was looking for a simpler versionof the dogbone spaceconstruction. Thedogbone space is a very peculiar counterexample of general topology,originally constructed by R.H. Bing. I mentioned it here in2007, and said, at the time:
[The paper] is on my desk, but I have not read this yet, and I may never.
I did try to read it, but I did not try very hard, and I did notunderstand it. So my question this month was if there was a simplerexample of the same type. I did not receive an answer, just afollowup comment that no, there is no such example.
[Other articles in category /math/se] permanent link
A message to the aliens, part 22/23 (cosmology)
Earlier articles:IntroductionCommon featuresPage 1 (numerals)Page 2 (arithmetic)Page 3 (exponents)Page 4 (algebra)Page 5 (geometry)Page 6 (chemistry)Page 7 (mass)Page 8 (time and space)Page 9 (physical units)Page 10 (temperature)Page 11 (solar system)Page 12 (Earth-Moon system)Page 13 (days, months, and years)Page 14 (terrain)Page 15 (human anatomy)Page 16 (vital statistics)Page 17 (DNA chemistry)Page 18 (cell respiration and division)Pages 19-20 (map of the Earth)Page 21 (the message)
This is page 22 of the
Cosmic Callmessage. An explanation follows.
The 10 digits are:
This page discusses properties of the entire universe. It is labeledwith a new glyph,,which denotes the universe or the cosmos. On this page I am onuncertain ground, because I know very little about cosmology. Myexplanation here could be completely wrong without my realizing it.
The page contains only five lines of text. In order, they state:
The Friedmannequationwhich is the current model for the expansion of theuniverse. This expansion is believed to be uniform everywhere, buteven if it isn't, the recipients are so close by that they willsee exactly the same expansion we do. If they have noticed the expansion,they may well have come to the same theoretical conclusions about it.The equation is:
$$H^2 = \frac{8\pi G}3\rho + \frac{\Lambda c^2 }3$$
where !!H!! is the Hubble parameter(which describes how quickly the universe is expanding),!!G!! is the universal gravitation constant(introduced on page 9),!!\rho!! is the density of the universe
(given on the next line),and !!\Lambda c^2!! ()is one of the forms of thecosmological constant (given on the following line).
The average densityof the universe ,given as !!2.76\times 10^{-27} \mathrm{kg}~\mathrm{m}^{-3}!!. The “density” glyph would have been more at homewith the other physics definitions of page9, but it wasn't needed until now, andthat page was full.
The cosmological constant !!\Lambda!! is about !!10^{-52}\mathrm{m}^{-2}!!. The related value given here, !!\Lambda c^2!!, is !!1.08\cdot10^{-35} \mathrm{s}^{-2}!!.
The calculated value of the Hubble parameter !!H!! is given here inthe rather strange form !!\frac1{14000000000}\mathrm{year}^{-1}!!.The reason it is phrased this way is that (assuming that !!H!! wereconstant) !!\frac1H!! would be the age of the universe, approximately14,000,000,000 years. So this line not only communicates ourestimate for the current value of the Hubble parameter, itexpresses it in units that may make clear our beliefs about the ageof the universe. It is regrettable that this wasn't stated moreexplicitly, using the glyph that was already used for the age of the Earth on page13. Therewas plenty of extra space, so perhaps the senders didn't think of it.
The average temperature of the universe, about 2.736 kelvins. This is based on measurements of the cosmic microwave background radiation, which is the same in every direction, so if the recipients have noticed it at all, they have seen the same CMB that we have.
The nextarticle will discuss the final page, shown atright. (Click to enlarge.) Try to figure it out before then.
[Other articles in category /aliens/dd] permanent link
A message to the aliens, part 21/23 (the message)
Earlier articles: IntroductionCommon featuresPage 1 (numerals)Page 2 (arithmetic)Page 3 (exponents)Page 4 (algebra)Page 5 (geometry)Page 6 (chemistry)Page 7 (mass)Page 8 (time and space)Page 9 (physical units)Page 10 (temperature)Page 11 (solar system)Page 12 (Earth-Moon system)Page 13 (days, months, and years)Page 14 (terrain)Page 15 (human anatomy)Page 16 (vital statistics)Page 17 (DNA chemistry)Page 18 (cell respiration and division)Pages 19-20 (map of the Earth)
This is page 21 of the
Cosmic Callmessage. An explanation follows.
The 10 digits are:
This page discusses the message itself. It is headed with the glyphfor “physics” .
Theleftmost part of the page has a cartoon of the Yevpatoria RT-70 radiotelescopethat was used to send the message, labeled “Earth” . Coming out thethe telescope is a stylized depiction of a radio wave. Two rulersmeasure the radio wave. The smaller one measures a single wavelength,and is labeled “frequency =5,010,240,000 Hz ” and “wavelength =0.059836 meters ”; these are thefrequency and the wavelength of the radio waves used to send themessage. The longer ruler has the notation “127×127×23”, describingthe format of the message itself, 23 pages of 127×127 bitmaps, andalso “43000 people ”, which I do notunderstand at all. Were 43,000 people somehow involved with sendingthe message? That seems far too many. Were there 43,000 people inYevpatoria in 1999? That seems far too few; the current population isover 100,000. I am mystified.
At the other end of the radio wave isthe glyph , which ishard to decipher, because it appears only on this page and on theunhelpful page 23. I
guess it is intended to refer to therecipients of the message.
[ Addendum 20151219: Having reviewed page 23, I am still in thedark.References to the mass and radius ofsuggest that it refers to the recipients’
planet, but references to the mathematics, physics, and biology ofsuggests that it refers to the recipients themselves. ]
In the lower-right corner of the page is another cartoon of the RT-70,this time with a ruler underneath showing its diameter, 70 meters.Above the cartoon is the power output of the telescope, 150 kilowatts.
The nextarticle will discuss page 22, shown atright. (Click to enlarge.) Try to figure it out before then.
[Other articles in category /aliens/dd] permanent link |
This is a question about generalizing trace defined for positive operators to trace class operators, in the terminology below, in the context of functional analysis.
Let $T \in B(H)$ (i.e. $T$ is a bounded linear operator on a Hilbert space $H$). Say $T$ is positive (denoted $T \geq 0$) if $T$ is self-adjoint and $\langle Tx, x\rangle \geq 0$ for all $x$ (where $\langle\cdot ,\cdot \rangle$ is the inner product on $H$). Define $$\mathrm{tr}(T)=\sum_j \left<Te_j, e_j \right>$$
the trace of $T$ for $T$ positive, where $e_j$ is an orthonormal basis for $H$. This definition is independent of the choice of basis. Let $B_0(H)$ be the compact operators in $B(H)$. Define
$B^1(H) = \mathrm{span} \{ T \in B_0(H) : T \geq 0, \mathrm{tr}\,T<\infty \}$
the trace class operators.
Why is it that for $T \in B^1(H)$, $T=\sum_0^3 i^k T_k$ for some $T_k \geq 0$ (where $i \in \mathbb{C}$), and why is the extension of trace to $B^1(H)$ defined as
$\mathrm{tr}\,T = \sum_0^3 i^k \mathrm{tr} T_k$
well-defined? (This is from Pedersen,
Analysis Now.) |
I have seen that if $G$ is a finite group and $H$ is a proper subgroup of $G$ with finite index then $ G \neq \bigcup\limits_{g \in G} gHg^{-1}$. Does this remain true for the infinite case also?
Not in general. Every matrix in $\text{GL}_2(\mathbf C)$ is conjugate to an invertible upper triangular matrix (use eigenvectors), and the invertible upper triangular matrices are a proper subgroup.
In your original question you require $H$ to have finite index in $G$. Most of the other answers are treating this as unintended. If you actually did want to require this, then the result is true for infinite groups as well.
Proof: Let $[G:H]=n$. The action of $G$ on $G/H$ gives a map $G \to S_n$; let $K$ be the kernel of this map. Then $H/K$ and $G/K$ are finite groups, so we know $G/K \neq \bigcup g (H/K) g^{-1}$. In particular, there is some coset $fK$ which is not in any $g H g^{-1}$. So $f$ is not in any $g H g^{-1}$.
If you really want an example with a discrete (i.e. "almost-finite") feel, consider $G=$ union of all symmetric groups $S_n$ acting on $\{1,2,3,\dots\}$ and $H$ the stabilizer of $1$. Since every element of $G$ belongs to some $S_n$ it can be conjugated inside $S_{n+1}$ to an element of $H$. The usual swindle...
(Let me drop the finite index requirement as in the other answers)
This remains true for discrete virtually solvable groups. Indeed the property of having no proper subgroup containing a conjugate of every element in the group is stable by extension.
I let you check that if the property holds for G_1 and G_2, and G is such that there is a sequence 0->G_1->G->G_2->0, then the property also holds for G.
Note that this property is not stable by direct limit. If K is the algebraic closure of a finite field, SL_n(K) is the direct limit of finite groups, yet the property is not verified. This gives an example of an amenable group for which the property does not hold.
It is not yet known if there are amenable groups of finite type which do not satisfy the property.
On the other hand, the property does not hold for non-abelian free groups, or, in the realm of uncountable groups, for connected semi-simple complex lie groups that are not solvable.
Given an arbitrary countable group $H$ containing an element of large enough order but no involutions, there exists a two-generator simple group $G$ such that $H$ is a proper subgroup of $G$ and $G=\bigcup \limits _{g\in G} gHg^{-1}$. This is Theorem 17 of
S.V. Ivanov and A.Y. Ol'shanskii, Some applications of graded diagrams in combinatorial group theory, Groups—St. Andrews 1989, Vol. 2, London Math. Soc. Lecture Note Ser., vol. 160, Cambridge Univ. Press, Cambridge, 1991, pp. 258–308. |
My book states that if we perturb a given Hamiltonian for the Schrödinger Equation
$$ H = \frac{p^2}{2m} +V(x) $$
to
$$ H' = \frac{p^2}{2m} + V(x) + \frac{\lambda p}{m} $$
then we can rewrite the perturbed Hamiltonian in the form
$$ H' = \frac{(p+\lambda)^2}{2m} + V(x) - \frac{\lambda^2}{2m} = \frac{p'^2}{2m} + V(x) - \frac{\lambda^2}{2m}. $$
Furthermore, it goes on to say that if we know the eigenvalues and eigenfunctions of the unperturbed Hamiltonian, $E_n^{(0)}$ and $\psi_n^{(0)}$ respectively, we can readily use the fact that the wave number is now $k' = \frac{p'}{\hbar} = \frac{(p+\lambda)}{\hbar}$ to say that the eigenfunctions of the perturbed Hamiltonian must be
$$\psi_n = \psi^{(0)}_n e^{-i\lambda x/\hbar}$$
and thus the new energies are
$$E_n = E_n^{(0)} - \frac{\lambda^2}{2m}.$$
Can someone please explain how the energy eigenfunctions can be so easily obtained considering the momentum operator has been shifted by a constant? I believe that the answer has to do with the fact that momentum space representation of $\psi$ is the Fourier Transform of $\psi(x)$, but I'm not sure how to prove this. |
Permanent of an $m \times n$-matrix $A = \left\Vert a_{ij} \right\Vert$
The function $$ \mathrm{per}(A) = \sum_\sigma a_{1\sigma(1)}\cdots a_{m\sigma(m)} $$
where $a_{ij}$ are elements from a commutative ring and summation is over all one-to-one mappings $\sigma$ from $\{1,\ldots,m\}$ into $\{1,\ldots,n\}$. If $m=n$, then $\sigma$ represents all possible permutations, and the permanent is a particular case of the Schur matrix function (cf. Immanant) $$ d_\chi^H (A) = \sum_{\sigma\in H} \chi(\sigma) \prod_{i=1}^n a_{i\sigma(i)} $$ for $H \subseteq S_n$, where $\chi$ is a character of degree 1 on the subgroup $H$ (cf. Character of a group) of the symmetric group $S_n$ (one obtains the determinant for $H=S_n$, $\chi =\pm 1$, in accordance with the parity of $\sigma$).
The permanent is used in linear algebra, probability theory and combinatorics. In combinatorics, a permanent can be interpreted as follows: The number of systems of distinct repesentatives for a given family of subsets of a finite set is the permanent of the incidence matrix for the incidence system related to this family.
The main interest is in the permanent of a matrix consisting of zeros and ones (a $(0,1)$-matrix), of a matrix containing non-negative real numbers, in particular doubly-stochastic matrices (in which the sum of the elements in any row and any column is 1), and of a complex Hermitian matrix. The basic properties of the permanent include a theorem on expansion (the analogue of Laplace's theorem for determinants) and the Binet–Cauchy theorem, which gives a representation of the permanent of the product of two matrices as the sum of the products of the permanents formed from the cofactors. For the permanents of complex matrices it is convenient to use representations as scalar products in the symmetry classes of completely-symmetric tensors (see, e.g., [3]). One of the most effective methods for calculating permanents is provided by Ryser's formula: $$ \mathrm{per}(A) = \sum_{t=0}^{n-1} (-1)^t \sum_{X \in \Gamma_{n-t}} \prod_{i=1}^m r_i(X) $$ where $\Gamma_k$ is the set of submatrices of dimension $m \times k$ for the square matrix $A$, $r_i = r_i(X)$ is the sum of the elements of the $i$-th row of $X$ and $i,k=1,\ldots,m$.
As it is complicated to calculate permanents, estimating them is important. Some lower bounds are given below.
a) If $A$ is a $(0,1)$-matrix with $r_i(A) \ge t$, $i=1,\ldots,m$, then $$ \mathrm{per}(A) \ge \frac{t!}{(t-m)!} $$ for $t \ge m$, and $$ \mathrm{per}(A) \ge t! $$ if $t < m$ and $\mathrm{per}(A) > 0$.
b) If $A$ is a $(0,1)$-matrix of order $n$, then $$ \mathrm{per}(A) \ge \prod_{i=1}^n \{ r_i^* + i - n \} $$ where $r_1^* \ge \cdots \ge r_n^*$ are the sums of the elements in the rows of $A$ arranged in non-increasing order and $\{ r_i^* + i - n \} = \max(0, r_i^* + i - n )$.
c) If $A$ is a positive semi-definite Hermitian matrix of order $n$, then $$ \mathrm{per}(A) \ge \frac{n!}{s(A)^n} \prod_{i=1}^n |r_i|^2 $$ where $s(A) = \sum_{i,j} a_{ij}$ if $s(A) > 0$.
Upper bounds for permanents:
1) For a -matrix of order ,
2) For a completely-indecomposable matrix of order with non-negative integer elements,
3) For a complex normal matrix with eigen values ,
The most familiar problem in the theory of permanents was van der Waerden's conjecture: The permanent of a doubly-stochastic matrix of order is bounded from below by , and this value is attained only for the matrix composed of fractions . A positive solution to this problem was obtained in [4].
Among the applications of permanents one may mention relationships to certain combinatorial problems (cf. Combinatorial analysis), such as the "problème des rencontresproblème de rencontres" and the "problème d'attachementproblème d'attachement" (or "hook problemhook problem" ), and also to the Fibonacci numbers, the enumeration of Latin squares (cf. Latin square) and Steiner triple systems (cf. Steiner system), and to the derivation of the number of -factors and linear subgraphs of a graph, while doubly-stochastic matrices are related to certain probability models. There are interesting physical applications of permanents, of which the most important is the dimer problem, which arises in research on the adsorption of di-atomic molecules in surface layers: The permanent of a -matrix of a simple structure expresses the number of ways of combining the atoms in the substance into di-atomic molecules. There are also applications of permanents in statistical physics, the theory of crystals and physical chemistry.
References
[1] H.J. Ryser, "Combinatorial mathematics" , Wiley & Math. Assoc. Amer. (1963) Zbl 0112.24806 [2] V.N. Sachkov, "Combinatorial methods in discrete mathematics" , Moscow (1977) (In Russian); translated by V. Kolchin: Encyclopedia of Mathematics and Its Applications 55. Cambridge University Press (1995) Zbl 0845.05003 [3] H. Minc, "Permanents" , Addison-Wesley (1978) [4] G.P. Egorichev, "The solution of van der Waerden's problem on permanents" , Krasnoyarsk (1980) (In Russian); Adv. Math. 42 (1981) 299-305. Zbl 0478.15003. [5] D.I. Falikman, "Proof of the van der Waerden conjecture regarding the permanent of a doubly stochastic matrix" Math. Notes , 29 : 6 (1981) pp. 475–479 Mat. Zametki , 29 : 6 (1981) pp. 931–938. Zbl 0475.15007 Comments
The solution of the van der Waerden conjecture was obtained simultaneously and independently of each other in 1979 by both O.I. Falikman, [5], and G.P. Egorichev, [4], [a4]. For some details cf. also [a2]–[a5].
References
[a1] D.E. Knuth, "A permanent inequality" Amer. Math. Monthly , 88 (1981) pp. 731–740 [a2] J.C. Lagarias, "The van der Waerden conjecture: two Soviet solutions" Notices Amer. Math. Soc. , 29 : 2 (1982) pp. 130–133 [a3] J.H. van Lint, "Notes on Egoritsjev's proof of the van der Waerden conjecture" Linear Algebra Appl. , 39 (1981) pp. 1–8 [a4] G.P. [G.P. Egorichev] Egorychev, "The solution of van der Waerden's problem for permanents" Adv. in Math. , 42 : 3 (1981) pp. 299–305 [a5] J.H. van Lint, "The van der Waerden conjecture: Two proofs in one year" Math. Intelligencer , 4 (1982) pp. 72–77 [a6] R.M. Wilson, "Non-isomorphic triple systems" Math. Zeitschr. , 135 (1974) pp. 303–313 [a7] A. Schrijver, "A short proof of Minc's conjecture" J. Comb. Theory (A) , 25 (1978) pp. 80–83 [a8] H. Minc, "Nonnegative matrices" , Wiley (1988) How to Cite This Entry:
Permanent.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Permanent&oldid=35188 |
Here is how I like to understand the Weil pairing.
The dual of an abelian variety $A$ is the scheme $\hat{A} = \mathrm{Hom}(A, B\mathbf{G}_m)$. Here $B\mathbf{G}_m$ is the stack of line bundles and $\mathrm{Hom}$ refers to homomorphisms of group stacks. One therefore has a perfect pairing
$A \times \hat{A} \rightarrow B\mathbf{G}_m$.
This actually means something relatively concrete: For each pair of points $(a, a') \in A \times \hat{A}$ one has a one dimensional vector space $L(a,a')$, together with isomorphisms
$L(a_1 + a_2, a') \simeq L(a_1, a') \otimes L(a_2, a')$,
$L(a, a'_1 + a'_2) \simeq L(a, a'_1) \otimes L(a, a'_2)$
satisfying some compatibility conditions (you can find the details of this definition under the heading of biextensions; there is a nice explanation in SGA7). One of these compatibilities is that the two isomorphisms
$L(a_1 + a_2, a'_1 + a'_2) \simeq L(a_1, a'_1) \otimes L(a_1, a'_2) \otimes L(a_2, a'_1) \otimes L(a_2, a'_2)$
should coincide.
One consequence of the definition is that $L(0, a') = L(0 + 0, a') \simeq L(0, a') \otimes L(0, a')$ which means that $L(0, a')$ is canonically trivialized, as is $L(a, 0)$ by symmetry. Moreover, the two trivializations of $L(0,0)$ must be the same.
If we choose an $n$-torsion point $a$ of $A$ then $L(a, a')$ will be a line bundle with a trivialization of its $n$-th tensor power. Similarly, if $a'$ is also an $n$-torsion point then $L(a,a')^{\otimes n}$ will come with
two trivializations coming from its identification with the canonically trivialized line bundles $L(na, a') = L(0,a')$ and $L(a, na') = L(a,0)$. Comparing these two trivializations, we get an element of $\mathbf{G}_m$ and therefore a pairing
$A[n] \times \hat{A}[n] \rightarrow \mathbf{G}_m$.
However, we notice that the induced trivializations of $L^{\otimes n^2} \simeq L(na, na')$ must coincide. Therefore the image of this map actually lands in $\mathbf{G}_m[n] = \mu_n$. This gives the Weil pairing
$A[n] \times \hat{A}[n] \rightarrow \mu_n$.
I don't know of a way to interpret the Hilbert symbol in quite this way, but I'd be very interested if someone could suggest one! |
Question in brief
Let $a$ and $b$ be unit vectors in $\mathbb{R}^d$. Let $f$ be the $1-step$ transition function of a random walk on the $d$ dimensional unit sphere. I am interested in evaluating $\mathrm{E}[\exp(a^T f(b))]$ and one of things I want to know is any transition function/distribution exist where this quantity can be computed in closed form? The transition function should not be "trivial" e.g. it should be dependent on the input and it should not be identity.
More Details
I will also love to know which papers/textbooks deal with this problem? If there is no way to get an exact form, then what methods have people used to study the behavior of $\mathrm{E}[\exp(a^Tf(b))]$ as a function of $a$ and $b$? Is it possible to compute this or approximate this when $f$ is a random orthonormal matrix from the Haar measure?
What I did
I tried to evaluate this quantity in two "simple" cases but both of them turned out to be intractable. In the first case, I assumed $f$ to be a random householder reflection, so I assumed that the walk proceeded by first sampling a vector $x$ from $U_{S^{d-1}}$, i.e. the uniform distribution over $S^{d-1}$. Then $f = I - 2xx^T$. But finding the marginal of a single projection of $U_{S^{d-1}}$ is itself difficult, and I could not figure out the value of $\mathrm{E}[\exp( (a^Tx)(b^Tx))]$ which requires a joint distribution of two projections of $x$.
The second case I tried was with random givens rotations. I chose the indices $i,j$ uniformly randomly from the $\binom d2$ possible pairs and then chose the rotation angle $\theta$ from a distribution that was symmetric about $\pi$. But this formulation boils down to the following computation $\mathrm{E}_{\theta}[\exp(\alpha \cos(\theta) + \beta \sin(\theta))]$. I couldn't think of any distribution over $\theta$ such that this expression could be evaluated. |
From Ravenel's article "Localization and Periodicity in Homotopy Theory":
Two spectra $E$ and $F$ are said to be
Bousfield equivalentwhen they give the same localization functor, or equivalently when $E_\ast (X)=0$ iff $F_\ast (X)=0$. The equivalence class of $E$ is denoted by $\langle E \rangle$. There is a partial ordering on the set of Bousfield classes. We say that $\langle E \rangle \geq \langle F \rangle$ if $E_\ast (X)=0$ implies that $F_\ast (X)=0$. Thus $\langle S^0 \rangle$ is the biggest class and $\langle pt \rangle $ is the smallest. Smash products and wedges are well defined on Bousfield classes. A class $\langle F \rangle$ is the complementof $\langle E \rangle$ if $\langle E \rangle \vee \langle F \rangle = \langle S^0 \rangle$ and $\langle E \rangle \wedge \langle F \rangle = \langle pt \rangle$. A class may or may not have a complement. It is easy to find examples of classes (e.g., that of an integer Eilenberg-Mac Lane spectrum) that do not.
I was trying to figure out why this last statement is true, and at first I wanted to apply cohomotopy to a hypothetical equivalence $H\mathbb{Z} \vee F \simeq S^0$, but then I realized that of course there's no reason that we should have such an equivalence. Is there some other easy approach? |
I am reading the paper by Griffin and Brown (2010) where at one step in their MCMC procedure they need to sample from the following conditional posterior:
$$ p(\lambda|\gamma, \Psi)\propto \pi(\lambda)\frac{1}{(2\gamma^2)^{p\lambda}(\Gamma(\lambda))^p}\left(\prod_{i=1}^p\Psi_i\right)^\lambda $$
They say that $\lambda$
can be updated using a Metropolis-Hastings random walk update on $\log\lambda$. We propose $λ' = \exp\{\sigma^2_\lambda z\}\lambda$, where $z$ is standard normal then $\lambda'$ is accepted with probability $$ \min\left\{1, \frac{\pi(\lambda')}{\pi(\lambda)}\left(\frac{\Gamma(\lambda)}{\Gamma(\lambda')}\right)^p\left((2\gamma^2)^{-p}\prod_{i=1}^p\Psi_i\right)^{\lambda'-\lambda}\right\} $$
My question:
why is there no term in the acceptance probability that accounts for the fact that we are proposing new draws on the logarithmic scale?
See for example here: Sampling on a logarithmic scale The paper, however, has more than 300 citations and I know several successful papers that have used the same type of Metropolis-Hastings procedure; therefore, I'm inclined to think I am missing something. But what?
Griffin and Brown (2010) Inference with normal-gamma prior distributions in regression problems, Bayesian Analysis, https://projecteuclid.org/euclid.ba/1340369797 |
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Tue, 08 Aug 2017
I should have written about this sooner, by now it has been so long that I have forgotten most of the details.
I first encountered Paul Erdős in the middle 1980s at a talk by János Pach about almost-universal graphs. Consider graphs with a countably infinite set of vertices. Is there a "universal" graph !!G!! such that, for any finite or countable graph !!H!!, there is a copy of !!H!! inside of !!G!!? (Formally, this means that there is an injection from the vertices of !!H!! to the vertices of !!G!! that preserves adjacency.) The answer is yes; it is quite easy to construct such a !!G!! and in fact nearly all random graphs have this property.
But then the questions become more interesting. Let !!K_\omega!! be thecomplete graph on a countably infinite set of vertices. Say that!!G!! is “almost universal” if it includes a copy of !!H!! for everyfinite or countable graph !!H!!
I enjoyed the talk, and afterward in the lobby I got to meet RonGraham and Joel Spencer and talk to them about their Ramsey theorybook, which I had been reading, and about a problem I was working on.Graham encouraged me to write up my results on the problem and submitthem to
I find the almost-universal graph thing very interesting. It is still an open research area. But none of this was what I was planning to talk about. I will return to the point. A couple of years later Erdős was to speak at the University of Pennsylvania. He had a stock speech for general audiences that I saw him give more than once. Most of the talk would be a description of a lot of interesting problems, the bounties he offered for their solutions, and the progress that had been made on them so far. He would intersperse the discussions with the sort of Erdősism that he was noted for: referring to the U.S. and the U.S.S.R. as “Sam” and “Joe” respectively; his ever-growing series of styles (Paul Erdős, P.G.O.M., A.D., etc.) and so on.
One remark I remember in particular concerned the $3000 bounty he offered for proving what is sometimes known as the Erdős-Túran conjecture: if !!S!! is a subset of the natural numbers, and if !!\sum_{n\in S}\frac 1n!! diverges, then !!S!! contains arbitrarily long arithmetic progressions. (A special case of this is that the primes contain arbitrarily long arithmetic progressions, which was proved in 2004 by Green and Tao, but which at the time was a long-standing conjecture.) Although the $3000 was at the time the largest bounty ever offered by Erdős, he said it was really a bad joke, because to solve the problem would require so much effort that the per-hour payment would be minuscule.
I made a special trip down to Philadelphia to attend the talk, with the intention of visiting my girlfriend at Bryn Mawr afterward. I arrived at the Penn math building early and wandered around the halls to kill time before the talk. And as I passed by an office with an open door, I saw Erdős sitting in the antechamber on a small sofa. So I sat down beside him and started telling him about my favorite graph theory problem.
Many people, preparing to give a talk to a large roomful of strangers,would have found this annoying and intrusive. Some people might not want totalk about graph theory with a passing stranger. But most people arenot Paul Erdős, and I think what I did was probably just the rightthing; what you
After a little while it was time to go down to the auditorium for the the talk, and afterward one of the organizers saw me, perhaps recognized me from the sofa, and invited me to the guest dinner, which I eagerly accepted. At the dinner, I was thrilled because I secured a seat next to Erdős! But this was a beginner mistake: he fell asleep almost immediately and slept through dinner, which, I learned later, was completely typical. |
I have heard vague information about point groups and symmetry before but I am unsure how they actually help me predicting physical properties of molecules. How do I assign point groups and what good does it do me?
Please answer at a low level.
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I have heard vague information about point groups and symmetry before but I am unsure how they actually help me predicting physical properties of molecules. How do I assign point groups and what good does it do me?
Please answer at a low level.
Point groups are a very valuable tool for analysing molecules without knowing much about them. They rely on analysing a molecule’s symmetry and deducing both physical parameters and the shape of molecular orbitals from the information gathered. The downside is that one must know a molecule’s geometry before one can apply group theory to it, but for most molecules the geometry is known anyway.
Picture a molecule, for example ammonia $\ce{NH3}$. Remember that it has a trigonal pyramidal shape with $\angle(\ce{H-N-H}) \approx 107^\circ$ Consider all atoms of the same type (only applicable to hydrogen here) identical. Look at this molecule naively and think what kind of symmetry it has.
One of the easiest symmetry elements that you can find is probably a mirror plane that transects the molecule along one $\ce{N-H}$ bond and places the image of the second hydrogen onto the third hydrogen. They are shown in figure two by the blue-ish ovals. A plane of symmetry is typically called $\sigma$.
You may also see, that if you stick an axis into the nitrogen atom through its lone pair (the thick, blue axis in figure 2), you can rotate the molecule $120^\circ$ around this axis and it will again look the same. Three rotations along this axis make $360^\circ$, so due to some formalism (which is really helpful if you think about it), the rotation is called $C_3$.
All we have gotten so far is a simple set of symmetry operations but they don’t help much. We need to put them together. We can define the connection of two symmetry operations by saying ‘we’ll perform one first, the other after it’; since neither change the molecule the final molecule will remain unchanged. Now to prove that the symmetry elements present in a molecule constitute a mathematical group, we need to prove the group axioms:
Closure The connection $A \circ B$ of two elements of a group must always be an element of the group.
I just proved this before. A symmetry operation will transform all hydrogens onto a hydrogen position, no two hydrogens on the same and nitrogen will always remain unchanged. We will always gain the same molecule. We just need to make sure we find
all possibilities correctly when searching. Associativity $(A \circ B) \circ C = A \circ (B \circ C)$.
If you think about this practically, it makes sense. Why should it matter whether I rotate first and apply a mirror plane later or the other way around?
Existence of an identity element One element must exist so that $E \circ A = A$ and $A \circ E = A$.
Well, this is easily done; we just include the identity, i.e.
no transformation as one of our symmetry operations. It is labelled $E$. Inverse elements For each element $A$ there must exist an element $A^{-1}$ so that $A \circ A^{-1} = E$.
This requires a bit of working but in the end we can prove this. For $\sigma$, the same $\sigma$ is always inverse element. For $C_3$, $C_3^2$, the rotation around $240^\circ = -120^\circ$ is the inverse element.
The good thing is that if we didn’t identify all the possible operations initially, we can use the connection $\circ$ to indentify the ones we missed.
Now because there is only a certain number of inherently different groups around, they have been given names for easier identification. Our ammonia example would be the point group $\bf\it C_\mathrm{3v}$. $C$, because we have a rotation axis, $_3$ because rotation is $\frac{1}{3}$ of the full circle and $_\mathrm{v}$ because the planes of symmetry are
vertical to the rotation axis (rather than horizontal).
(Naming point groups is rather simple. Find the most important axis — the rotation with the highest index. If that is the only rotation, it is a $C$ group. If there are rotations perpendicular to that one, the group is called $D$. If there is an improper rotation, it is labelled $S$. The highest index of rotation is also the group’s index. $_\mathrm{v}, _\mathrm{h}$ and $_\mathrm{d}$ denote planes of symmetry vertical, horizontal or diagonal to this axis. Finally, $C_s$ is the group that contains only one plane of symmetry, $C_i$ contins only the inversion and finally, tetrahedra ($T_\mathrm{d}$), octahedra ($O_\mathrm{h}$) and icosahedra ($I_\mathrm{h}$) have their own names due to their high symmetry.)
Part of the beauty of point groups is already apparant here. If we know what symmetry operations transform the atoms onto each other, this must also be true for some physical properties — those that can be expressed as vectors. Most notably, a dipole moment
must transform onto itself during any symmetry operation, so ammonia’s dipole moment must be along the rotation axis. Molecules that display inversion symmetry cannot be a dipole. Molecules can only be chiral (but are always chiral) if their point groups do not contain inversion, planes of symmetry or improper rotation (a rotation followed by a mirror image perpendicular to the rotation axis). There are more examples, but these are probably the most important.
Finally, for each point group a
character table exists. This groups sets of similar symmetry operations together according to a set of specific rules (e.g. the two rotations $C_3$ and $C_3^2$ will be grouped together) whose details I won’t go into, and also includes irreducible representations. Any spec of the molecule (and all its symmetry-equivalent specs) must transform as one of these irreducible representations. The $\mathrm{p}_z$ orbital of nitrogen in ammonia (the one along the $C_3$ axis) transforms according to the irreducible representation $\mathrm{a_1}$ — essentially, $\mathrm{a_1}$ means that something will always be the same. The $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals together will transform as $\mathrm{e}$. Knowing the irreducible representations of atomic orbitals helps to predict molecular orbitals, because only those of the same irreducible representation will mix and combine (for others, the constructive overlap is always $0$).
You can google character tables to find lists of them, and thereby also which point groups actually exist.
Point groups are even more powerful than shown here; but I tried to keep the application parts low-level. There are a number of questions here on Stack Exchange tagged group-theory that go more into depth. |
Let consider a steady state CSTR in which occurs the following reaction:
$$\ce{aA + bB \leftrightarrow cC}$$
Notations:
$F_i$ is the molar flow of $i$, $r$ is the rate of reaction, $V$ is the volume, $X$ is the conversion at the end of the reactor, $\nu_i$ is the stoichiometric coefficient of $i$ and $\xi$ is the extent of reaction.
Mass balance:
$$\ce{IN + PROD - CONS = OUT + VAR}$$
Therefore,
$$ \begin{cases} F_{A,in} + 0 - arV = F_{A,out} \\ F_{B,in} + 0 - brV = F_{B,out} \\ F_{C,in} + crV - 0 = F_{C,out} \\ \end{cases} $$
As $$\xi = -rV = -X\frac{F_{i,in}}{\nu_i}$$ we have for $\ce{A}$:
$$\begin{alignat}{} F_{A,out} &= F_{A,in} - arV \\ & = F_{A,in} - a(-\xi) \\ & = F_{A,in} - a\left(X\frac{F_{A,in}}{a}\right) \\ & = \left(1 - X \right)F_{A,in} \end{alignat}$$
For $B$ with can find as well:
$$F_{B,out} = \left(1 - X \right)F_{B,in} = \frac{b}{a}\left(1 - X \right)F_{A,in}$$
My problem with C:
Now if I follow my calculus for $\ce{C}$, here is what I get:
$$\begin{alignat}{} F_{C,out} &= F_{C,in} + crV \\ & = F_{C,in} + c(-\xi) \\ & = F_{C,in} + c\left(X\frac{F_{C,in}}{c}\right) \\ & = \left(1 + X \right)F_{C,in} \end{alignat}$$
Which states that if there is no $C$ at the beginning, there is no $C$ produced at all! Which is wrong. Can you tell me where I am wrong?
NB: I have no clue, which tags fit best for this question. |
I'm told that this is true, but I can't imagine why. It seems like the fact that there is less air would make the engines less efficient... But that probably just shows how little I know about jet engines.
For a quick explanation, you need to know that
Thrust is the difference between the entry impulse of the air entering the engine and the exit impulse of the heated fuel-air mixture leaving the engine. Impulse is mass times velocity, and expressed with a mass flow $\dot m$, thrust T is $$T = \dot m \cdot (v_{exit} - v_{entry})$$ The exit impulse is increased by accelerating the airflow through the engine, and the acceleration is achieved by heating the air. Each gram of fuel heats up a given mass of air by a certain number of centigrades. The definition of the energy content of fuels is given as the capacity to heat one pound of water by one degree Fahrenheit. The definition of one Calorie is similar but in metric units. Since the heat capacity of both water and air are almost constant at moderate temperatures, the starting temperature makes little difference to the absolute temperature increase when a given amount of energy is added. Thermal efficiency
Thermal efficiency is the ratio between the mechanical work extracted as thrust and the heat energy spent on heating the air, and it is affected indirectly by the flight altitude. Please see the Wikipedia article on the Carnot cycle. This and similar cycles describe the workings of all combustion engines in thermodynamic terms. Basically, it says that the efficiency of a combustion engine cannot be greater than the temperature ratio between the temperature increase from ambient ($t_{amb}$) to the maximum temperature $t_{max}$ of the process, divided by the maximum temperature. All temperatures must be expressed as total temperatures, where 0° means 0 K or -273.15°C. Operating in colder air makes the ratio bigger and improves efficiency. $$\eta_t = \frac{t_{max} - t_{amb}}{t_{max}} $$
If $t_{amb}$ is 290 K (16.85°C or 62°F) and the fuel heats up the air to 1400 K (2060°F), the thermal efficiency according to the formula above is 79.3%.
At cruise altitude $t_{amb}$ is only 220 K (-53.15°C or -63.7°F), and the same fuel flow relative to air flow will lift the maximum temperature only to 1320 K (in reality even less; for more precise reasoning see below). Now the thermal efficiency is 83.33%! If the maximum temperature is maintained, both thrust and thermal efficiency will go up; the latter to 84.3%.
In reality, the total efficiency will be lower because we have not included propulsive efficiency, friction effects or power offtake by bleed air, pumps and generators. Propulsive efficiency describes how well the acceleration of air is performed.
Heating the fuel-air mixture
Burning a fuel-air mixture will add thermal energy to it, about 43 MJ for every kilogram of kerosene (if we assume complete combustion). The isobaric heat capacity or specific heat of air (close enough, the mixture has very little fuel but lots of air in it) is 29 J per mol and per K, so those 43 MJ will heat 1000 mol of air by 1483 K. The heat capacity changes slightly with humidity and temperature, but little enough that we can consider it constant for this purpose. If the air starts at 220 K, precompression in the intake will heat it to approx. 232 K, further compression in the engine will heat it up to approx. 600 K if we assume a compression ratio of 25, and this is the temperature at the entry of the combustion chamber.
Those 1000 mol of air weigh about 29 kg, and adding a full kilo of fuel and burning the mixture will heat it to 2083 K. If you want more details about the parameters in a typical jet engine, please see the diagram in this answer. Since the mixture picks up speed as it burns, the fuel mass is also heated and combustion is never complete, the maximum temperature given here will not be reached in reality.
If we start on the ground with an air temperature of 290 K, the temperature in the intake would drop slightly because we will not be flying fast enough for any precompression to happen in the intake. Now the compressor will heat the air to 730 K, and again adding and burning that kilo of kerosene will heat 1000 mol of air to 2213 K. Ideally.
In reality, the engine control will see that the limit temperatures are not exceeded, but here we can play with the numbers as we like. The exact values will certainly be slightly different (more frictional heating in the compressor, heat loss to the outside, slight drift in specific heat with temperature), but the gist of the explanation is correct.
Explanation in layman's terms
Burning the fuel-air mixture heats it and makes the gas expand. This happens at nearly constant pressure and in a restricted volume, so the only way to make room for this expansion is for the gas to flow faster. Nearly constant pressure means that the density of the gas must decrease. The density ratio between the heated and the unburnt gas is proportional to its
temperature ratio, measured in absolute temperature.
However, the amount of fuel burnt determines the
absolute temperature increase, the difference in degrees between the burnt gas inside the combustion chamber and the unburnt gas at the intake. For a given amount of fuel, the temperature ratio which can be achieved with an absolute temperature increase becomes smaller the higher the temperature of the unburnt gas is. Thus, efficiency decreases with a higher temperature of the intake air.
What matters for a jet engine is the pressure and temperature differentials between the exhaust gas and the ambient atmosphere. It is the expansion and high kinetic energy of the exhaust gas as it exits the engine that provides the thrust (and noise) of a jet (note this does not take into account the bypass portion of a turbofan).
The ambient pressure is atmospheric pressure, which for example at the surface is roughly 1000 hPa and at cruise might be 200 hPa or roughly a fifth of the pressure at the surface. The temperature at that altitude is also typically around -50 C.
The exhaust gas pressure and temperature is controlled by a few things:
The compression by the N2 compressor stages -- Increases temperature and pressure The hot section -- Greatly increases temperature and pressure The N1/N2 turbine stages -- slight decrease in temp/pressure (work done on moving the turbines).
As the outside pressure is dropping as we climb, to maintain the same pressure differential in the engine, we need less temperature and pressure in the engine, and one way to do this is to reduce airflow into the engine and fuel added to that air. The atmosphere takes care of reducing airflow (there is just less of it up at cruise, though this also depends on airspeed) and the FADEC takes care of adjusting the fuel flow. The net result is less fuel needed to produce the same pressure differential when the air outside has a lower pressure, e.g. cruise flight.
EDIT:
Some of the other answers/comments make reference to mass flow through the jet, and particularly the mass flow through the exhaust nozzle. I agree with that, but I didn't mention it directly because that mass flow is setup by the pressure gradient within the engine. I should also clarify that pressure at the nozzle will be at or very close to ambient atmospheric pressure and it is the pressure gradient between that ambient pressure and that in the hot section that establishes the mass flow rate out of the engine.
Lastly, to address the bypass ratio comment, see Lnafziger's comment. The turbofan engines on the EMB-145 are similar in that the bypass provides more thrust at sea level than cruise. This perhaps relates to increased fuel efficiency at cruise in that the N1 fan is doing less work and thus the N1 turbine is extracting less energy from the engine.
They work better at high altitude firstly because the air is cooler. Cool air expands more when heated than warm air. It is the expansion of the air that drives combustion engines.
The second reason is the low density of the air. Low density causes low drag and therefore the aircraft flies much faster at high altitude than on low altitude when it is given the same thrust. At this high speed, the mass flow through the engine is comparable to the mass flow at low speed in high density air (low altitude). The amount of energy needed is to heat the air to exhaust temperature is comparable between high and low altitudes. But since the aircraft at high altitude flies much faster, the amount of power generated is higher $(Power=Thrust\times{Speed})$ at altitude.
The difference with propeller aircraft is that at high speeds the propeller loses efficiency, and therefore the available power decreases with altitude.
For a non-math approach:
Let's think how a jet engine works and compare low altitude with high altitude flights. The engine takes air from the intake situated at the front. As you are climbing, the air becomes less dense (there is less air mass in a volume) so you need to go a little faster just so that the mass of the air coming in through the intake is the same in a given second. You will actually get the same mass flow of air at high altitudes as you will get at low altitudes, but you are actually travelling faster.
Then you compress that air, remembering that as you are now travelling faster higher up, the ram effect will help you out and compress some of that air for you, just by 'ramming' your engines into it at high speed. As you compress it you pass it to the combustion chamber where it burns. This combustion stage is the same for both high and low altitudes, although the fact that at higher altitudes air is colder actually helps a little bit, as we can burn more fuel without reaching dangerous temperatures, so that's nice.
After burning it up, the air is passed through a turbine then expelled out the back. Now here it gets a bit complicated: You see, it is more efficient to accelerate a lot of air (mass) a little bit(small dv), than to accelerate a little bit of air(small mass) to a very fast speed(dv). This means in turn that the faster the airplane moves, the better the jet propulsive efficiency gets. So as you climb, you go faster and the flow gets better efficiency, plus the lower air pressure behind you means there is less of a force pushing against your outflow.
So what do we have in low vs high flight:
Same amount of air intake, same amount of combustion, same amount of fuel used, better jet propulsion at higher altitudes and better speed at higher altitudes. You just get more bangs for your bucks at higher altitude.
For a mathematical approach:
This is because the air is cooler and less dense which means that there is less fuel to air mixture at higher altitudes giving it a better fuel efficiency
The higher the altitude the thinner the atmoshere means less air resistance or drag on the airplane so it needs less engine thrust to push it. Thats lucky because the engine loses thrust with altitude at almost the same rate because as less air is available to the engine, the fuel system must reduce fuel to maintain the correct air/fuel ratio to support combustion and keep the engine alight. Its a win win situation.
The engine of an airliner is designed to be as efficient as possible over a journey containing a take-off, a climb, and most of its time at 35000 to 40000 feet where air pressure is about 1/4 to 1/5 of ground level. The engine has a few extra compression stages to work efficiently at normal cruise at the expense of overheating if flown for long at full power near ground level due to too much compression at the intake. Look up water injection for an interesting way to get take-off boost in a medium altitude engine in the 707.
I think most are simply thinking too much. The easiest and likely the most complete answer is resistance (or friction). High altitude air is less dense, making it easier to pass through. Oxygen content in high altitude is exactly the same as sea level. While the air up there is the same air we breath, there is less of that air in the same volume container. Space vehicles do not use jet engines. To turn or do any movement actually they have "jets" in various locations around the shuttle. "Jets" in this case are not jet engines, they are simply small nozzles that pressurized gases gets released through. Keep in mind that with zero air, there is no resistance to movement, remember newton's laws of motion: every action has an equal an opposite reaction.
As you know as altitude increasing pressure and temp both are reducing up to stratosphere after that temp remain constant pressure drop continues so density of air reducing so its create less drag aircraft travel at high speed this pressure loss is overcome by ram pressure rise at the inlet of engine and aircraft required less power to move faster at 36,000 ft to 40,000 ft above more power full engine required to run faster so the blade tip does not get stall.
I'm not even close to an expert. But here it goes.
Air like water is thick. Submarines are slower than boats. Jets are faster than boats. Automobiles are faster than boats.
Space has no friction because there is no matter. But I think jets still work in space. Of course they need oxygen. Just like Superman doesn't need ground friction to run fast, while some other super heroes do. Which is why I find it unrealistic how a super heroes that need ground friction are able to run so fast and make sharp turns without doing serious damage to floors.
So I'm guessing that being that the air is less thick up higher, that the easier it is to travel through. The jets don't depend on friction like propellers do. Superman doesn't need friction like Flash Gordon or Wonder Woman. So in space, Wonder Woman would be helpless because her propellers don't work while Superman's jets will work just fine.
Of course jet's need oxygen. So I'm not sure how all that works.
And something I didn't think of was what was mentioned in another post. Sound needs air. So yeah. Sound can increase friction.
protected by DeltaLima Oct 20 '14 at 19:47
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When I went for interview at Imperial College, my interviewer (Dr. P. Kelly) asked me quite a nice question, which (for some unearthly reason) I decided to follow up on afterwards.
The question was a simple one: given a string (e.g.) "examplesgnome" consisting of two substrings (in this case, "examples" and "gnome"), how can we swap the two substrings in-place? In the interview, we covered the third method in more detail, but touched upon the other two. Afterwards, I calculated their computational efficiency, and determined while they all have the same upper bound, one has a lower average running time.
Method one: reversal
This is one Dr. Kelly suggested, which is really rather neat. It consists of two steps: reversing the entire string, then reversing both substrings. For example:
examplesgnome emongselpmaxe gnomeexamples
If we take
n
to equal the total number of characters,
a
to equal the number of characters in the smaller substring, and
b
to equal the number of characters in the larger substring, this has:
\left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{a}{2} \right\rfloor + \left\lfloor \frac{b}{2} \right\rfloor
swaps, so we may as well say it's of linear complexity
O(n)
.
Method two: cycles
This one was also suggested by Dr. Kelly, and involves taking the first character of the larger substring out to a temporary location, then cyclically moving characters
b
places along into the generated space. Once the strings have been swapped (
n
swaps have taken place), we put the stored character back in the final space. For example:
examplesgnome _xamplesgnome gxamples_nome gxa_plesmnome gxamplesmno_e gxampl_smnoee g_amplxsmnoee gnamplxsm_oee gnam_lxsmpoee gnamelxsmpoe_ gnamelx_mpoes gn_melxampoes gnomelxamp_es gnome_xamples gnomeexamples
Since each character moves to its final position in one move, the algorithm is intuitively of complexity
O(n)
— but we can go further and say it's of absolute complexity
\sim (n + 2)
, taking the steps to remove the first character and fill the last space into account.
Method three: reduction
This is the method I came up with during the interview and, with some help, made work. It basically involves swapping the first smaller substring into its final position, then applying the same algorithm to the resulting subsubstrings of the second substring. For example:
examplesgnome gnomelesexamp gnomeampex les gnomeexpam les gnomeexma ples gnomeexamples
This is quite a neat one to implement, and will perform
O(n)
swaps in the worst case (when the smaller substring is only one character long, as one character moves to its final position with each swap).
However, what if we consider the case "foobar", where the substrings are "foo" and "bar"? In this case, method three only takes 3 swaps. Obviously, this method is limited in the upper bound to
O(n)
swaps, but it seems it can do the operation in far fewer — as few as
\Omega (a)
.
I haven't got a concrete proof for this, but I believe this is actually of complexity:
\sim (a \left\lfloor \frac{b}{a} \right\rfloor + (b \bmod{a}) \left\lfloor \frac{a}{b \bmod{a}} \right\rfloor)
as the lengths of the substring for each iteration
i
depend on the remainder of
\frac{b}{a}
for the
i - 1
th iteration. This complexity can be manipulated to prove that it's also of complexity
O(n)
:
All-in-all, I'm quite pleased with the interview: even if I'm not successful, it's given me an interesting problem to consider. Corrections and further thoughts on the problem are welcomed. |
Math/Display
< Math >
Contents Display Math
The famous result (once more) is given by \startformula c^2 = a^2 + b^2. \stopformula
This, when typeset, produces the following:
Numbering Formulae
The famous result (once more) is given by \placeformula \startformula c^2 = a^2 + b^2. \stopformula
This, when typeset, produces the following:
The
\placeformula command is optional, and produces the equation number; leaving it off produces an unnumbered equation.
Changing format of numbers
You can use
\setupformulas to change the format of numbers. For example to get bold numbers inside square brackets use
\setupformulas[left={[},right={]},numberstyle=bold] which gives
To get equations also numbered by section, add the command:
\setupnumber[formula][way=bysection]
To the start of your document.
To get alphabets instead of numbers, use
\setupformulas[numberconversion=Character] which gives
Changing Formula alignment
Normally a formula is centered, but in case you want to align it left or right, you can set up formulas to behave that way. Normally a formula will adapt its left indentation to the environment:
\setuppapersize[A5] \setuplayout[textwidth=8cm] \setupformulas[align=left] \startformula c^2 = a^2 + b^2 \stopformula \setupformulas[align=middle] \startformula c^2 = a^2 + b^2 \stopformula \setupformulas[align=right] \startformula c^2 = a^2 + b^2 \stopformula
Or in print:
With formula numbers the code is:
\setuppapersize[A5] \setuplayout[textwidth=8cm] \setupformulas[align=left] \placeformula \startformula c^2 = a^2 + b^2 \stopformula \setupformulas[align=middle] \placeformula \startformula c^2 = a^2 + b^2 \stopformula \setupformulas[align=right] \placeformula \startformula c^2 = a^2 + b^2 \stopformula
And the formulas look like:
When tracing is turned on (
\tracemathtrue) you can visualize the bounding box of the formula,
As you can see, the dimensions are the natural ones, but if needed you can force a normalized line:
\setuppapersize[A5] \setuplayout[textwidth=8cm] \setupformulas[align=middle,strut=yes] \tracemathtrue \placeformula \startformula c^2 = a^2 + b^2 \stopformula
This time we get a more spacy result. [Ed. Note: For this example equation, there appears to be no visible change.]
We will now show a couple of more settings and combinations of settings. In centered formulas, the number takes no space
\setuppapersize[A5] \setuplayout[textwidth=8cm] \tracemathtrue \setupformulas[align=middle] \startformula c^2 = a^2 + b^2 \stopformula \placeformula \startformula c^2 = a^2 + b^2 \stopformula
You can influence the placement of the whole box with the parameters
leftmargin and
rightmargin.
\setuppapersize[A5] \setuplayout[textwidth=8cm] Some example text, again, to show where the right and left margins of the text block are. \tracemathtrue \setupformulas[align=right,leftmargin=3em] \startformula c^2 = a^2 + b^2 \stopformula \placeformula \startformula c^2 = a^2 + b^2 \stopformula \setupformulas[align=left,rightmargin=1em] \startformula c^2 = a^2 + b^2 \stopformula \placeformula \startformula c^2 = a^2 + b^2 \stopformula
You can also inherit the margin from the environment.
\setuppapersize[A5] \setuplayout[textwidth=8cm] Some example text, again, to show where the right and left margins of the text block are. \tracemathtrue \setupformulas[align=right,margin=standard] \startformula c^2 = a^2 + b^2 \stopformula \placeformula \startformula c^2 = a^2 + b^2 \stopformula
The distance between the formula and the number is only applied when the formula is left or right aligned.
\setuppapersize[A5] \setuplayout[textwidth=8cm] \tracemathtrue \setupformulas[align=left,distance=2em] \startformula c^2 = a^2 + b^2 \stopformula \placeformula \startformula c^2 = a^2 + b^2 \stopformula
Referencing formulae
The famous result (and again) is given by \placeformula[formulalabel] \startformula c^2 = a^2 + b^2. \stopformula And now we can refer to formula \ref[][formulalabel].
This, when typeset, produces the following:
Note, that
\ref expects two arguments, therefore you need the brackets twice.
By default, only the formula number appears as a reference. This can be changed by using
\definereferenceformat. For example, to create a command
\eqref which shows the formula number in brackets, use
\definereferenceformat[eqref][left=(,right=)] Sub-Formula Numbering Automatic Sub-Formula Numbering
Examples: \startsubformulas[eq:1] \placeformula[eq:first] \startformula c^2 = a^2 + b^2 \stopformula \placeformula[eq:second] \startformula c^2 = a^2 + b^2 \stopformula \stopsubformulas Formula (\in[eq:1]) states the Pythagora's Theorem twice, once in (\in[eq:first]) and again in (\in[eq:second]).
The Manual Method
Sometimes, you need more fine grained control over numbering of subformulas. In that case one can make use of the optional agument of
\placeformula command and the related
\placesubformula commands which can be used to produce sub-formula numbering. For example:
Examples: \placeformula{a} \startformula c^2 = a^2 + b^2 \stopformula \placesubformula{b} \startformula c^2 = a^2 + b^2 \stopformula
What's going on here is simpler than it might appear at first glance. Both
\placeformula and
\placesubformula produce equation numbers with the optional tag added at the end; the sole difference is that the former increments the equation number first, while the latter does not (and thus can be used for the second and subsequent formulas that use the same formula number but presumably have different tags).
This is sufficient for cases where the standard ConTeXt equation numbers suffice, and where only one equation number is needed per formula. However, there are many cases where this is insufficient, and
\placeformula defines
\formulanumber and
\subformulanumber commands, which provide hooks to allow the use of ConTeXt-managed formula numbers with plain TeX equation numbering. These, when used within a formula, simply return the formula number in properly formatted form, as can be seen in this simple example with plain TeX's
\eqno. Note that the optional tag is inherited from
\placeformula.
More examples: \placeformula{c} \startformula \let\doplaceformulanumber\empty c^2 = a^2 + b^2 \eqno{\formulanumber} \stopformula
In order for this to work properly, we need to turn off ConTeXt's automatic formula number placement; thus the
\let command to empty
\doplaceformulanumber, which must be placed
after the start of the formula. In many practical examples, however, this is not necessary; ConTeXt redefines
\displaylines and
\eqalignno to do this automatically.
For more control over sub-formula numbering,
\formulanumber and
\subformulanumber have an optional argument parallel to that of
\placeformula, as demonstrated in this use of plain TeX's
\eqalignno, which places multiple equation numbers within one formula.
\placeformula \startformula \eqalignno{ c^2 &= a^2 + b^2 &\formulanumber{a} \cr c &= \left(a^2 + b^2\right)^{\vfrac{1}{2}} &\subformulanumber{b}\cr a^2 + b^2 &= c^2 &\subformulanumber{c} \cr d^2 &= e^2 &\formulanumber\cr} \stopformula
Note that both
\formulanumber and
\subformulanumber can be used within the same formula, and the formula number is incremented as expected. Also, if an optional argument is specified in both
\placefigure and
\formulanumber, the latter takes precedence.
More examples for left-located equation number: \setupformulas[location=left] \placeformula{d} \startformula \let\doplaceformulanumber\empty c^2 = a^2 + b^2 \leqno{\formulanumber} \stopformula and \placeformula \startformula \leqalignno{c^2 &= a^2 + b^2 &\formulanumber{a} \cr a^2 + b^2 &= c^2 &\subformulanumber{b} \cr d^2 &= e^2 &\formulanumber\cr} \stopformula
-- 23:46, 15 Aug 2005 (CEST) Prinse Wang
List of Formulas
You can have a list of the formulas contained in a document by using
\placenamedformula instead of
\placeformula. Only the formulas written with
\placenamedformula are not put in the list, so that you can control precisely the content of the list.
Example:
\subsubject{List of Formulas} \placelist[formula][criterium=text,alternative=c] \subsubject{Formulas} \placenamedformula[one]{First listed Formula} \startformula a = 1 \stopformula \endgraf \placeformula \startformula a = 2 \stopformula \endgraf \placenamedformula{Second listed Formula}{b} \startformula a = 3 \stopformula \endgraf
Gives:
Shaded background for part of a displayed equation
(see also Framed)
To highlight part of a formula, you can give it a gray background using
\mframed: the following is the code you can use in mkii (see below what one has to do in mkiv):
\setuppapersize[A5] \setupcolors[state=start] \def\graymath{\mframed[frame=off, background=color, backgroundcolor=gray, backgroundoffset=3pt]} \startformula \ln (1+x) =\, \graymath{x - {x^2\over2}} \,+ {x^3\over3}-\cdots. \stopformula
\setuppapersize[A5] \definemathframed[graymath] [ frame=off, location=mathematics, background=color, backgroundcolor=lightgray, backgroundoffset=2pt ] \starttext Since for $|x| < 1$ we have \startformula \log(1+x) = \graymath{x- \displaystyle{x^2\over2}} + {x^3 \over 3} + \cdots \stopformula we may write $\log(1+x) = x + O(x^2)$. \stoptext
The result is shown below (possibly the framed part of the formula is not aligned correctly with the remainder of the formula because the mkiv engine on Context Garden is not up to date…). |
Although András' comment already answers the question, I think it is worthwile to give a few more details explicitely here, in order to point out that the analyticity is in fact a consequence of the
resolvent identity only and has nothing to do with the operator whose resolvent we consider:
Let $X$ denote a complex Banach space, let $[X]$ denote the space of all bounded linear operators on $X$ and let $U \subseteq \mathbb{C}$ be non-empty and open.
Definition. A mapping $\mathcal{R}: U \to [X]$ is called a pseudo-resolvent if it fulfils the so-called resolvent identity\begin{align*} \mathcal{R}(\lambda) - \mathcal{R}(\mu) = (\mu - \lambda) \mathcal{R}(\lambda)\mathcal{R}(\mu)\end{align*}for all $\lambda, \mu \in U$.
Note that the resolvent of every closed linear operator on $X$ is a pseudo-resolvent, but there are also pseudo-resolvents which are not the resolvent of a closed linear operator (for instance the constant mapping $\lambda \mapsto 0$).
Proposition. Let $\mathcal{R}: U \to [X]$ be a pseudo-resolvent, let $\lambda,\lambda_0 \in U$ and assume that $|\lambda-\lambda_0| < \|\mathcal{R}(\lambda_0)\|^{-1}$ (where we define $0^{-1} := \infty$). Then\begin{align*} \mathcal{R}(\lambda) = \sum_{n=0}^\infty (\lambda_0 - \lambda)^n \mathcal{R}(\lambda_0)^{n+1},\end{align*}where the series converges absolutely in $[X]$ (which is endowed with the operator norm). In particular, the mapping $\mathcal{R}: U \to [X]$ is analytic.
Proof. It follows from the resolvent identity that\begin{align*} \mathcal{R}(\lambda) \big[\operatorname{id} - (\lambda_0 - \lambda) \mathcal{R}(\lambda_0)\big] = \mathcal{R}(\lambda_0).\end{align*}The operator norm of $(\lambda - \lambda_0) \mathcal{R}(\lambda_0)$ is strictly smaller than $1$ by assumption, so we conclude from the Neumann series that the operator in square brackets is invertible and that its inverse is given by $\sum_{n=0}^\infty (\lambda_0 - \lambda)^n \mathcal{R}(\lambda_0)^n$. This proves the assertion.
A few references:
For the case where the pseudo-resolvent is actually a resolvent of a closed operator:
"K.-J. Engel and R. Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)", Proposition IV.1.3 (as already pointed out by András in the comments)
"W. Arendt, C. Batty, M. Hieber, F. Neubrander: Vector-Valued Laplace Transforms and Cauchy Problems (2011)", Corollary B.3 in the appendix
For pseudo-resolvents:
"M. Haase: The Functional Calculus for Sectorial Operators (2006)": in the appendix of this book, pseudo-resolvents are treated as resolvents of so-called multi-valued operators.
Personal remark concerning the Math.SE post mentioned in the question: To derive the analyticity of the resolvent as an "immediate" consequence of the resolvent identity without noting that this "proof" requires continuity to be shown first is, at least in my experience, an easily made mistake. I recall falling into this trap myself some time ago. |
I'll try with some calculations : please, check it and the formulae used ...
A solid ball with a mass $m$ of $1$ kg falls (with the usual approxiamtions : no drag, etc.) with an acceleration $a$ that is about $10 \ m/sec^2$.
This means that falling from a tower $80$ meters heigh, it will touch ground after $4$ sec, with a final velocity of about $40 \ m/sec$.
You are right : in the same time, the Earth will "fall towards" the ball, attracted by the same gravitational force.
The mass $M$ of the Earth is about $6 \times 10^{24} \ kg$.
The acceleration $A$ of the Earth that is proportional to $a$ as the reciprocal of the masses; i.e. :
$A = a \times m/M = 10/(6 \times 10 ^{24}) \approx 2 \times 10 ^{-24} \ m/sec^2$.
After $4$ seconds, the Earth reachs a velocity of fall of $8 \times 10 ^{-24} \ m/sec$ and it has traversed a space $s \approx 16 \times 10 ^{-24} \ m$.
This means that, due to the reciprocal attraction, the two bodies will touch each other in slightly less than $4 \ sec$ and that the "real" space traversed by the falling ball with respect to the Earth is about :
$80$ meters minus the space traversed by the "falling" Earth during the short time of the fall.
It can be useful to recall that :
the size of atoms is measured in
picometers : trillionths ($10 ^{-12}$) of a meter.
If the ball has a mass $m'$ of $1000 \ kg$, the force with which it "pulls" the Earth will be $1000$ times greater, producing an acceleration $A' \approx 2 \times 10 ^{-21} \ m/sec^2$.
This implies that in this second case the space traversed by the Earth duing the fall of the heavier ball will be $s' \approx 16 \times 10 ^{-21} \ m$.
According to Aristotle (
Physiscs, Book VII) the "law" of dynamics is :
"if a power $\alpha$ moves a body $\beta$ during time $\delta$ for a distance $\gamma$, then an equal power $\alpha$ will move a body half of $\beta$ along a distance twice as $\gamma$ in the same time".
In an anachronistic way, we can say :
$F \propto V$.
Thus, if we apply this law to "our model" of free fall, with the weight of the body as the force, we have that - assuming that after an initial short time of acceleration the falling body will reach a constant "terminal velocity" - in the second case the acquired speed $v'$must be $1000$ times the first one : $v$.
This means that after $4$ seconds the heavier ball has traversed a space : $s'= v' \times t$, i.e. $s' = 1000 \times v \times t = 1000 \times s$, where $s$ is the space traversed by the lighter body after a fall of $4$ seconds.
Conclusion
As you can see the "same factor" : $1000$ acts in a completely different way in the two models.
Due to the huge mass of the Earth, for bodies of "normal" size (meaning body of our daily life experience) it has no
experimentally verifiable effect on the behaviour of different bodies falling due to the gravitational force.
In the Aristotelian model, that factor has an evident
experimentally verifiable effect on the behaviour of different bodies falling due to the "tendency towards the centre".
This is exactly where is the "conceptual" difference : to perform some sort of experimental test.
Thus, the answer to :
Was Aristotle really wrong about gravity?
is a definitive
YES if we try to answer the question from the point of view of modern science, a point of view that is not that of aristotelian natural philosophy.
If instead we want to compare two different qualitative "worldviews", things are different (see at least the philosophical debate involving : Thomas Kuhn, The Incommensurability of Scientific Theories, Scientific Revolutions, Historicist Theories of Scientific Rationality, Imre Lakatos and Paul Feyerabend).
Notes
i) In our computations we made approximations; approximation is a modern concept.
Without precise mathematical laws there are no possible approximations.
Aristotle's "natural laws" are
not approximations in the modern sense; they are qualitative description of facts, like the one made by the same Aristotle on botany and zoology (whih were impressively accurate, by the way).
ii) Writing the "aristotelian equation" we make an "historical mistake" : he never thinked in terms of mathematical formuale.
Please, note that the mathematics used by Galileo in his analysis of the free falling body problem was only the theory of proportions of Eudoxus/Euclid, that was already available in Aristotle's time.
Thus, the "tools" available to ancient natural philosophers were more or less the same compared to the ones available to Galileo (not so with Newton ...).
Postscriptum
The above "experiment" was alredy discussed by Galileo, in advance of the correct formulation of the law of universal gravitation. See :
SALV. We infer therefore that large and small bodies move with the same speed provided they are of the same specific gravity.
SIMP. Your discussion is really admirable; yet I do not find it easy to believe that a bird-shot falls as swiftly as a cannon ball.
SALV. Why not say a grain of sand as rapidly as a grindstone? But, Simplicio, I trust you will not follow the example of many others who divert the discussion from its main intent and fasten upon some statement of mine which lacks a hair’s-breadth of the
truth and, under this hair, hide the fault of another which is as big as a ship’s cable. Aristotle says that “an iron ball of one hundred pounds falling from a height of one hundred cubits reaches the ground before a one-pound ball has fallen a single cubit.” I say that they arrive at the same time. You find, on making the experiment, that the larger outstrips the smaller by two finger-breadths, that is, when the larger has reached the ground, the other is short of it by two finger-breadths; now you would not hide behind these two fingers the ninety-nine cubits of Aristotle, nor would you mention my small error and at the same time pass over in silence his very large one. Aristotle declares that bodies of different weights, in the same medium, travel (in so far as their motion depends upon gravity) with speeds which are proportional to their weights; this he illustrates by use of bodies in which it is possible to perceive the pure and unadulterated effect of gravity, eliminating other considerations, for example, figure as being of small importance, influences which are greatly dependent upon the medium
which modifies the single effect of gravity alone.Thus we observe that gold, the densest of all substances, when beaten out into a very thin leaf, goes floating through the air; the same thing happens with stone when ground into a very fine powder.
But if you wish to maintain the general proposition you will have to show that the same ratio of speeds is preserved in the case of all heavy bodies, and that a stone of twenty pounds moves ten times as rapidly as one of two; but I claim that this is false and that, if they fall from a height of fifty or a hundred cubits, they will reach the earth at the same moment. |
Dynamics of { $\lambda tanh(e^z): \lambda \in R$\ ${ 0 }$ }
1.
Department of Mathematics, Indian Institute of Technology Guwahati, Guwahati - 781039, India, India
$\mathcal{M} = { f_{\lambda}(z) = \lambda f(z) : f(z) = \tanh(e^{z}) \mbox{for} z \in \mathbb{C} \mbox{and} \lambda \in \mathbb{R} \setminus \{ 0 \} }$
is studied. We prove that there exists a parameter value $\lambda^$* $\approx -3.2946$ such that the Fatou set of $f_{\lambda}(z)$ is a basin of attraction of a real fixed point for $\lambda > \lambda^$* and, is a parabolic basin corresponding to a real fixed point for $\lambda = \lambda^$*. It is a basin of attraction or a parabolic basin corresponding to a real periodic point of prime period $2$ for $\lambda < \lambda^$*. If $\lambda >\lambda^$*, it is proved that the Fatou set of $f_{\lambda}$ is connected and, is infinitely connected. Consequently, the singleton components are dense in the Julia set of $f_{\lambda}$ for $\lambda >\lambda^$*. If $\lambda \leq \lambda^$*, it is proved that the Fatou set of $f_{\lambda}$ contains infinitely many pre-periodic components and each component of the Fatou set of $f_{\lambda}$ is simply connected. Finally, it is proved that the Lebesgue measure of the Julia set of $f_{\lambda}$ for $\lambda \in \mathbb{R} \setminus \{ 0 \}$ is zero.
Mathematics Subject Classification:Primary: 37F45, 37F5. Citation:M. Guru Prem Prasad, Tarakanta Nayak. Dynamics of { $\lambda tanh(e^z): \lambda \in R$\ ${ 0 }$ }. Discrete & Continuous Dynamical Systems - A, 2007, 19 (1) : 121-138. doi: 10.3934/dcds.2007.19.121
[1] [2]
Hiroki Sumi.
Dynamics of postcritically bounded polynomial semigroups I:
Connected components of the Julia sets.
[3] [4]
Luiz Henrique de Figueiredo, Diego Nehab, Jorge Stolfi, João Batista S. de Oliveira.
Rigorous bounds for polynomial Julia sets.
[5] [6] [7]
Jun Hu, Oleg Muzician, Yingqing Xiao.
Dynamics of regularly ramified rational maps: Ⅰ. Julia sets of maps in one-parameter families.
[8] [9] [10] [11] [12]
Ranjit Bhattacharjee, Robert L. Devaney, R.E. Lee Deville, Krešimir Josić, Monica Moreno-Rocha.
Accessible points in the Julia sets of stable exponentials.
[13]
Koh Katagata.
Transcendental entire functions whose Julia sets contain any infinite collection of quasiconformal copies of quadratic Julia sets.
[14]
Suzanne Lynch Hruska.
Rigorous numerical models for the dynamics of complex Hénon mappings on their chain recurrent sets.
[15]
Weiyuan Qiu, Fei Yang, Yongcheng Yin.
Quasisymmetric geometry of the Cantor circles as the Julia sets of rational maps.
[16]
Rich Stankewitz, Hiroki Sumi.
Random backward iteration algorithm for Julia sets of rational semigroups.
[17] [18]
Alexander Blokh, Lex Oversteegen, Vladlen Timorin.
Non-degenerate locally connected models for plane continua and Julia sets.
[19]
Hiroki Sumi, Mariusz Urbański.
Measures and dimensions of Julia sets of semi-hyperbolic
rational semigroups.
[20]
2018 Impact Factor: 1.143
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Elementary discrete dynamical systems biology problems Problem 1
The mass of a fish is increasing by 100 grams per year.
Set up a dynamical system model that describes the evolution of the fish's mass. Be sure to define your notation, including the meaning of any variable for time. If the mass of a fish in year 0 is 45 grams, calculate the mass in the next 4 years. Be sure to show how you used the formula in part (a) and to write your results using the notation you defined. Find all equilibria of your dynamical system. Problem 2
In a population of elk, each year the number of babies is 20% of the population size at the beginning of the year. However, a disease is killing one quarter of the population each year.
Set up a dynamical system model that describes the evolution of the population size. Be sure to define your notation, including the meaning of any variable for time. If in one year, there are 18,000 elk, calculate the number of elk for the next 4 years. (Don't worry about having fractional elk.) Be sure to show how you used the formula in part (a) and to write your results using the notation you defined. Find all equilibria of your dynamical system. Problem 3
The human population of the earth is currently increasing by about 1.1% per year. If it continues at this rate, how many years will it take for the population to double? If the current population is 7 billion people, in how many years will the population reach 56 billion people?
Problem 4
The population of Steller sea lions is decreasing at a rate of about 5% per year. If it continues at this rate, how many years will it take for the population to decline by one-half? If the current population is 40,000, in how many years will the population reach 10,000 Steller sea lions?
Problem 5
In the first days of life, the cells in a human embryo divide into two cells approximately every day.
Assuming the number of cells doubles every day, write down a dynamical system for the evolution of the number of cells. Be sure to define your notation, including the meaning of any variable for time. After fertilization, the embryo consists of a single cell. Solve the dynamical system to obtain an expression for the number of cells as a function of the number of days since fertilization. If a pregnancy lasts 40 weeks and the cell division continued at the same rate, how many cells would the baby have upon being born? How does this number compare to the number of atoms in the observable universe, which is estimated to be about $10^{80}$? What can you conclude about the rate of cell division during the course of the pregnancy? Problem 6 In every 5 week period, the population size of a group of rabbits increases by a fixed fraction. Set up a dynamical system model that describes the evolution of the rabbit population. Be sure to define your notation, including the meaning of any variable for time. Your model should include an unknown parameter, which you should clearly define.
You count the number of rabbits in a population every 5 weeks and obtain the following data. You also calculate how much the population changed from each observation until the next.
Time (weeks) Population size Population change 0 101 20 5 121 44 10 165 73 15 238 55 20 293 83 25 376 113 30 489 138 35 627
You plot population change versus population size, fit the result with a line through the origin, and obtain a slope of 0.3.
Based on this result, determine the unknown parameter from your model. Write down your resulting dynamical system model that describes the data. Be sure to include the initial condition equation.
Solve the dynamical system model to obtain an expression for the population size in any time interval. Your final result should not contain any parameter, but only the variable used for time. Rewrite your solution to give the population size as a function of the number of weeks since the beginning. Based on your model, estimate the rabbit population after 35 weeks and after two years (104 weeks). Problem 7
Bacteria are growing in a beaker so that the population size increases by 14.87% every minute.
Set up a dynamical system model that describes the evolution of the population size. Be sure to define your notation, including the meaning of any variable for time. How long does it take the population size to double? If the population continues to grow at this rate, by what factor does the population size increase in one hour? In two hours? In four hours? An experiment is begun at midnight with just a few bacteria so that the fraction of the beaker that the bacteria occupy is approximately $0.00000005959 = 5.959 \times 10^{-8}$. With this initial condition, the bacteria completely fill the beaker after two hours, at 2 AM. At what time was the beaker half full? Imagine the researchers realized before 2AM that the bacteria were about to overflow the beaker. They found three more empty beakers of the same size as the original beaker so that they had a total of four beakers to hold the bacteria. At what time did the bacteria fill all four beakers? Problem 8
Without hunting, a population of deer would be growing by a factor of $a$ each year, where the parameter $a$ is a number greater than one.
Write down a dynamical system for the unhunted population of the deer and find all equilbiria. Are the equilibria stable or unstable? Modify the dynamical system of part (a) so that $b$ deer are hunted each year, where the parameter $b$ is a positive number. Find all equilibria. Modify the dynamical system of part (a) so that a fraction $c$ of the deer are hunted each year, where the parameter $c$ is a positive number. Find all equilibria. Calculate a condition on $c$ for the equilibria to be stable. Calculate a condition on $c$ for there to be more than one equilibrium. Problem 9
Consider the dynamical system \begin{align*} z_{n+1}=f(z_n) \quad \text{for $n=0,1,2,3,\ldots$ ,} \end{align*} where $f$ is defined by \begin{align*} f(\bigstar)=\frac{3\bigstar^2}{2+\bigstar^2}. \end{align*} The graph of $z_{n+1}=f(z_n)$ is shown below along with the diagaonal $z_{n+1}=z_n$.
Write down an equation for the equilibria and show that the system has the equilibria 0, 1, and 2.
Use cobwebbing to determine what happens to the system if the initial condition $z_0$ is slightly less than 1 and if the initial condition $z_0$ is slightly greater than 1.
Imagine that $z_n$ represents the firing rate in time step $n$ of a neuron in a monkey's brain. At time step zero, either a banana or a stick is shown briefly to the monkey. If the banana was shown, the neuron starts firing at a rate $z_0$ of about 1.2. If a stick was shown, the neuron starts firing at a rate $z_0$ of about 0.8.
Imagine that you don't know if the monkey was shown a stick or a banana at time step zero. However, after one of these was briefly shown to the monkey, the firing rate of the neuron has evolved for many time steps, and now the neuron is firing at a rate $z_n$ of about 2. Was the monkey shown a banana or a stick? Justify your answer using the results from the first two parts.
One you have worked on a few problems, you can compare your solutions to the ones we came up with. |
I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention)
First of all, let's make it empirically clear that there is no hard and fast constraint on how energy will behave in a collision. Imagine different materials involved in a head-on collision (billiard balls, basketballs, you know the drill) and we're done with that.
However, conservation of momentum is much, much more straightforward. Let's recapitulate the basics: momentum is $\vec p_0 = m\vec v_0$. When a force $\vec F$ acts on our body for $\Delta t$ seconds, the new momentum will be $\vec p = \vec p_0 + \vec F\Delta t$.
$\vec F\Delta t$ is called
impulse and it even has the same units as momentum (that's why we can sum them): $[m][v]=kg\cdot\frac{m}{s}=\frac{kg\cdot m}{s^2}\cdot s=[F][\Delta t]$.
So, what happens during a collision is that, during that brief amount of contact time, two objects exert force on one another. According to
Newton's Third Law, the two forces are of equal magnitude and opposite directions. This is true in every instant of that brief amount of time (would be true if it was a large amount of time too).
So, for every infinitesimal amount of time $\mathrm dt$ during contact, if object $A$'s momentum gains the impulse $\vec F\mathrm dt$, then it is apparent that object $B$'s momentum will be decreased of $-\vec F\mathrm dt$! And, in the end of the collision, the variation of momentum in $A$ will be the same as in $B$ (but reversed). After all, both objects suffered the exact same force $F$ (but reversed), for the exact same amount of time!
We can state that as (if $\vec F$ is the force acted upon object $A$):$$\Delta\vec p_A = \int_0^T\vec F(t)\mathrm dt$$$$\Delta\vec p_B = \int_0^T-\vec F(t)\mathrm dt$$$$\Rightarrow\Delta\vec p_A+\Delta\vec p_B=\vec0$$
SUMMING UP: momentum conservation is an obvious consequence of Newton's Third Law, while energy conservation, although it exists, develops in more subtle and elusive ways, like conversion from one kind to another (in collisions, usually it goes from kinetic to thermal). |
The method by which we calculate wavefunctionals is actually remarkably similar to the methods by which we calculate the position space representations of states in standard quantum mechanics. The details of this will depend heavily on your exact theory, but for simplicity I will assume we are working with a free scalar field theory.
As always, we start with the Hamiltonian for our theory. In momentum space, this is given by
$$H=\frac{1}{2}\int\frac{\mathrm{d}^3\textbf{p}}{(2\pi)^3}\left(\hat{\pi}^2+\omega_{\textbf{p}}^2\hat{\phi}^2\right),$$
where $\pi$ is the momentum conjugate to the field $\phi$, and $\omega_{\textbf{p}}=\sqrt{\textbf{p}^2+m^2}$. In the "coordinate" representation (that is, when acting on wavefunctionals $\Psi[\phi]$), the field and momentum operators act as
$$\hat{\phi}(\textbf{p})\Psi[\phi]=\phi(\textbf{p})\Psi[\phi]$$$$\hat{\pi}(\textbf{p})\Psi[\phi]=-i\frac{\delta}{\delta\phi(\textbf{p})}\Psi[\phi],$$
in perfect analogy with one-dimensional quantum mechanics.
Now, just like in standard quantum mechanics, we define
$$a(\textbf{p})=\sqrt{\frac{\omega_{\textbf{p}}}{2}}\left(\hat{\phi}(\textbf{p})+\frac{i}{\omega_{\textbf{p}}}\hat{\pi}(\textbf{p})\right),$$
for which the Hamiltonian takes the form
$$H=\int\frac{\mathrm{d}^3\textbf{p}}{(2\pi)^3}\,\omega_{\textbf{p}}a^{\dagger}(\textbf{p})a(\textbf{p})+E_0,$$
where $E_0$ is the ground-state energy.
So far, this is all rather standard. Now, let us recall that the ground state of the theory is the state $|0\rangle$ such that $a(\textbf{p})|0\rangle=0$ for all $\textbf{p}$. This implies that
$$a(\textbf{p})\Psi_0[\phi]=\sqrt{\frac{\omega_{\textbf{p}}}{2}}\left(\phi(\textbf{p})+\frac{1}{\omega_{\textbf{p}}}\frac{\delta}{\delta\phi(\textbf{p})}\right)\Psi_0[\phi]=0,$$
where $\Psi_0[\phi]=\langle\phi|0\rangle$. It is easily checked that the state
$$\Psi_0[\phi]=\exp\left(-\frac{1}{2}\int\mathrm{d}^3\textbf{p}\,\omega_{\textbf{p}}\,\phi(\textbf{p})^2\right)$$
satisfies this requirement.
Now that we have found the ground state, finding single-particle states is rather simple. We simply note that single particle states are made by acting on the ground state with the creation operator. This gives
$$\Psi_{\textbf{k}}[\phi]=a^{\dagger}(\textbf{k})\Psi_{0}[\phi]\propto\phi(\textbf{k})\exp\left(-\frac{1}{2}\int\mathrm{d}^3\textbf{p}\,\omega_{\textbf{p}}\,\phi(\textbf{p})^2\right).$$
Multiparticle states are constructed similarly, with repeated application of $a^{\dagger}$. For instance, we have
$$\Psi_{\textbf{k},\textbf{q}}[\phi]=a^{\dagger}(\textbf{q})\Psi_{\textbf{k}}[\phi]=\left[2\phi(\textbf{k})\phi(\textbf{q})-\frac{1}{\omega_{\textbf{q}}}\delta(\textbf{k}-\textbf{q})\right]\exp\left(-\frac{1}{2}\int\mathrm{d}^3\textbf{p}\,\omega_{\textbf{p}}\,\phi(\textbf{p})^2\right).$$
This demonstrates explicitly that one cannot simply write a multiparticle state as a tensor product of two single particle states. Indeed, multiparticle states are more complicated than that.
I hope this helps! |
Despite my utmost effort the book has errors: typos, grammatical and substantive. This page is for reporting substantive errors not typos. I am however interested in knowing about all errors so please email me at rvc@petercorke.com and provide enough information to locate the error, and tell me what you think the error is. By default I will acknowledge you, let me know if you don’t want this.
The convention here is that new text is shown in
bold, and text to be deleted is shown in strike through font. Versions
There are currently three versions of the book – you can identify the version from the copyright page (page iv):
The second editionpublished in 2017. The first edition, second printing, published in early 2013. The first edition, first printing published in 2011.
All errata for the first edition can be found here.
Errata for second edition page 50
The equation at the top of the page has two terms swapped, it should be
\(\mathbf{c}^\prime_2 =\mathbf{c}^\prime_3 \times \mathbf{c}^\prime_1 \)
thanks Pekka Forsman, added 20180821. page 60
In the 2D column of the table:
Orientation is a
3×3 2×2 rotation matrix
Pose is a
4×4 3×3 transformation matrix thanks Erwin Coumans, added 20180110. page 62 Q11 is the same as Q9 Q14, the app is no longer available. thanks Haidar Obeid, added 20190806. page 65
The results on this page are computed based on
>> TB=SE3(1,2,0)*SE3. Rz Rx(pi/2)
a rotation about the x-axis, not z-axis as in the book.
Further down the page we should use the adjoint of TCB not TBC
>> vc = TBC TCB.Ad * vb; thanks Willem Remie, added 20171203. page 68
The code line
w = w + wd*dt; attitude = attitude .* UnitQuaternion.omega(wd*dt);
should read
w = w + wd*dt; attitude = attitude .* UnitQuaternion.omega(w*dt);
since w*dt is the infinitesimal angular displacement which is converted to a unit quaternion and compounded with the attitude from the previous timestep.
thanks Chen Peng, added 20190811. page 71
The equation for velocity is wrong, it should be
\(\dot{s}(t) = 5At^4 + 4Bt^3 + 3Ct^2 + 2Dt + E\)
The fourth term is \(2Dt\) not \(2D\) as printed.
thanks Ahmed Mohamed, added 20181202. page 77
The last equation in the first row of equations on the page has some terms swapped, it should be
\( \beta^\prime = \beta + \theta^* \)
thanks Haidar Obeid, added 20180723. page 87
Equation (3.25) has a notational error, a superscript ended up on the wrong side of the dot operator. It should read
thanks Ronghao Zheng, added 20190519. page 105
In Fig 4.6, the top left shaded box should be labelled “distance control”, not “speed control”.
page 107
In the topmost equation, there is an error in the 3-vector on the left-hand side. The middle element, shown as \( \dot{\omega} \) should be \( \dot{y} \). The correct equation is:
\[\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{\theta} \end{pmatrix} =\begin{pmatrix} \cos \theta & 0 \\ \sin \theta & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} v \\ \omega \end{pmatrix}\]
thanks Andrew Vardy, added 20190114. page 108
The second matrix equation has a missing term, the expression \( -k_\rho \cos \alpha \) should be \( -k_\rho \rho \cos \alpha \). The correct equation is:
\[ \begin{pmatrix} \dot{\rho} \\ \dot{\alpha} \\ \dot{\beta} \end{pmatrix} =\begin{pmatrix} -k_\rho \rho \cos \alpha \\ k_\rho \sin \alpha – k_\alpha \alpha – k_\beta \beta \\ -k_\rho \sin \alpha \end{pmatrix} \]
thanks Iain McCulloch, added 20190407. page 142
The variables start and goal are not set by the code example.
>> lp.plan('cost', [1 10 10]) >> lp.query(start, goal); >> lp.query([1 2 pi/2], [2 -2 0]);A* path cost 35 >> lp.plot() page 146 >> plot_vehicle(p, 'box', 'size', [20 30], 'fill', 'r', 'alpha', 0.1) >> plot_vehicle(p, 'box', 'size', [20 30], 'fillcolor', 'r', 'alpha', 0.1) page 196 >> a1 = 1; >> E = Rz('q1') * Tx(a1) * Tz('q2')>> E = Rz('q1') * Tx(a1) * Tx('q2') page 198
There’s an error in (7.3), the element in the second row and fourth column \( \alpha_j \sin \theta_j \) should be \( a_j \sin \theta_j \)
thanks Willem Remie, added 20180506. page 273
Fig 9.21 shows a Simulink model for computed-torque control
sl_ctorque. The torque is computed from the desired position and velocity rather than the actual position and velocity as described in (9.13). If the controller is tracking well then these will be close and the discrepancy minor but the error dynamics of (9.14) will likely not be obeyed. To be true to (9.13) the model should be changed: the q and qd inputs to the RNE block should come from the q and qd outputs of the Robot block. thanks Edoardo Farniolli. page 509 >> plot_poly(p1, 'wo', 'fill', 'b', 'alpha', 0.2); >> plot_poly(p1, 'wo', 'fillcolor', 'b', 'alpha', 0.2); page 510 >> plot_poly(p2, 'k', 'fill', 'r', 'alpha', 0.2) >> plot_poly(p2, 'k', 'fillcolor', 'r', 'alpha', 0.2) Page 543
The first equation in (15.2) should be
\[ \dot{X} = Y \omega_z – Z \omega_y – v_x \]
thanks Farrokh Janabi-Sharifi, added 20190929. Page 551
The command
>> plot2(p)
won’t work because p has dimensions 2x4x501, and the indexes are respectively: u/v ordinate, corner and timestep. To plot the trajectory of corner 1 over time in the uv-plane use
>> plot2(squeeze(P(:,1,:))')
To plot the trajectories of all 4 corners you need to use a loop
>> clf >> for i=1:4 hold on plot2(squeeze(p(:,i,:))') end Page 571 >> T_C0 = SE3(0.3, 0.3, -2)*SE3.Rz(0.4); >> vs = IBVS_sph(cam, 'T0', T_C0, 'verbose') >> vs = IBVS_sph(cam, 'pose0', T_C0, 'verbose') Page 605
Section C.2.1, the definitions of the vectors parallel and normal to the line are swapped. The non homogeneous vector \((-\ell_2,\ell_1)\) is a normal to the line, and \((\ell_1,\ell_2)\) is parallel to the line.
thanks Parker Lusk, added 20180121. |
Newform invariants
Coefficients of the \(q\)-expansion are expressed in terms of a basis \(1,\beta_1,\beta_2,\beta_3\) for the coefficient ring described below. We also show the integral \(q\)-expansion of the trace form.
Basis of coefficient ring in terms of a root \(\nu\) of \(x^{4}\mathstrut -\mathstrut \) \(x^{3}\mathstrut -\mathstrut \) \(9\) \(x^{2}\mathstrut +\mathstrut \) \(2\) \(x\mathstrut +\mathstrut \) \(15\):
\(\beta_{0}\) \(=\) \( 1 \) \(\beta_{1}\) \(=\) \( \nu \) \(\beta_{2}\) \(=\) \( \nu^{2} - \nu - 5 \) \(\beta_{3}\) \(=\) \( \nu^{3} - 2 \nu^{2} - 5 \nu + 5 \)
\(1\) \(=\) \(\beta_0\) \(\nu\) \(=\) \(\beta_{1}\) \(\nu^{2}\) \(=\) \(\beta_{2}\mathstrut +\mathstrut \) \(\beta_{1}\mathstrut +\mathstrut \) \(5\) \(\nu^{3}\) \(=\) \(\beta_{3}\mathstrut +\mathstrut \) \(2\) \(\beta_{2}\mathstrut +\mathstrut \) \(7\) \(\beta_{1}\mathstrut +\mathstrut \) \(5\)
For each embedding \(\iota_m\) of the coefficient field, the values \(\iota_m(a_n)\) are shown below.
For more information on an embedded modular form you can click on its label.
This newform does not admit any (nontrivial) inner twists.
\( p \) Sign \(2\) \(-1\) \(3\) \(-1\) \(13\) \(-1\) \(103\) \(1\)
This newform can be constructed as the intersection of the kernels of the following linear operators acting on \(S_{2}^{\mathrm{new}}(\Gamma_0(8034))\):
\(T_{5}^{4} \) \(\mathstrut -\mathstrut 2 T_{5}^{3} \) \(\mathstrut -\mathstrut 11 T_{5}^{2} \) \(\mathstrut +\mathstrut 23 T_{5} \) \(\mathstrut +\mathstrut 4 \) \(T_{7}^{4} \) \(\mathstrut -\mathstrut 5 T_{7}^{3} \) \(\mathstrut -\mathstrut 11 T_{7}^{2} \) \(\mathstrut +\mathstrut 82 T_{7} \) \(\mathstrut -\mathstrut 87 \) |
Let $\mu$ be a regular Borel Measure on $\mathbb{R}$ and suppose that there exists a $C > 0$ such that for every $x \in \mathbb{R}$ and $r > 0$\begin{align*}\mu((x - r,x+r)) \leq Cr.\end{align*}Let
m be the Lebesgue measure on $\mathbb{R}$. I have shown that $\mu << m$ ($\mu$ is absolutely continuous wrt m). I now want to show that the Radon-Nikodym derivative of $\mu$ wrt m is in $L^{\infty}(\mathbb{R},m)$.
I know that $\mu$ and
m are two $\sigma$-finite measures and $\mu << m$. So, by the Radon-Nikdoym Theorem, there exists a positive non-negative measure function f so that\begin{align*}\mu(E) = \int_E f dm\end{align*}for every measurable set E. So, I need to show that $f \in L^{\infty}(\mathbb{R},m)$. So I need to show that for every $N \in \Sigma$ with $m(N) = 0$ there exists $C > 0$ so that\begin{align*}\left\vert{f(s)}\right\vert \leq C\end{align*}for every $s \in N^c$.
I am not sure how to proceed/how to relate this to the above integral or the first equation. Any ideas or recommendations would be greatly appreciated. Thank you in advance. |
Image: Probability distribution for the sum of two six-sided dice
A bar chart illustrating the probability distribution for a random variable $X$ that is given by the sum of the result of rolling two six-sided dice.
The probability distribution is \begin{gather*} P(x) = \begin{cases} \frac{1}{36} & \text{if $x \in \{2,12\}$}\\ \frac{2}{36}=\frac{1}{18} & \text{if $x \in \{3,11\}$}\\ \frac{3}{36}=\frac{1}{12} & \text{if $x \in \{4,10\}$}\\ \frac{4}{36}=\frac{1}{9} & \text{if $x \in \{5,9\}$}\\ \frac{5}{36} & \text{if $x \in \{6,8\}$}\\ \frac{6}{36} =\frac{1}{6} & \text{if $x = 7$}\\ 0 & \text{otherwise.} \end{cases} \end{gather*}
Image file: two_dice_distribution.png
Source image file: two_dice_distribution.py
Source image type: Python Image links
This image is found in the pages |
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