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We've seen that classical logic is closely connected to the logic of subsets. For any set \( X \) we get a poset \( P(X) \), the
power set of \(X\), whose elements are subsets of \(X\), with the partial order being \( \subseteq \). If \( X \) is a set of "states" of the world, elements of \( P(X) \) are "propositions" about the world. Less grandiosely, if \( X \) is the set of states of any system, elements of \( P(X) \) are propositions about that system.
This trick turns logical operations on propositions - like "and" and "or" - into operations on subsets, like intersection \(\cap\) and union \(\cup\). And these operations are then special cases of things we can do in
other posets, too, like join \(\vee\) and meet \(\wedge\).
We could march much further in this direction. I won't, but try it yourself!
Puzzle 22. What operation on subsets corresponds to the logical operation "not"? Describe this operation in the language of posets, so it has a chance of generalizing to other posets. Based on your description, find some posets that do have a "not" operation and some that don't.
I want to march in another direction. Suppose we have a function \(f : X \to Y\) between sets. This could describe an
observation, or measurement. For example, \( X \) could be the set of states of your room, and \( Y \) could be the set of states of a thermometer in your room: that is, thermometer readings. Then for any state \( x \) of your room there will be a thermometer reading, the temperature of your room, which we can call \( f(x) \).
This should yield some function between \( P(X) \), the set of propositions about your room, and \( P(Y) \), the set of propositions about your thermometer. It does. But in fact there are
three such functions! And they're related in a beautiful way!
The most fundamental is this:
Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq Y \) define its inverse image under \(f\) to be
$$ f^{\ast}(S) = \{x \in X: \; f(x) \in S\} . $$ The pullback is a subset of \( X \).
The inverse image is also called the
preimage, and it's often written as \(f^{-1}(S)\). That's okay, but I won't do that: I don't want to fool you into thinking \(f\) needs to have an inverse \( f^{-1} \) - it doesn't. Also, I want to match the notation in Example 1.89 of Seven Sketches.
The inverse image gives a monotone function
$$ f^{\ast}: P(Y) \to P(X), $$ since if \(S,T \in P(Y)\) and \(S \subseteq T \) then
$$ f^{\ast}(S) = \{x \in X: \; f(x) \in S\}
\subseteq \{x \in X:\; f(x) \in T\} = f^{\ast}(T) . $$ Why is this so fundamental? Simple: in our example, propositions about the state of your thermometer give propositions about the state of your room! If the thermometer says it's 35°, then your room is 35°, at least near your thermometer. Propositions about the measuring apparatus are useful because they give propositions about the system it's measuring - that's what measurement is all about! This explains the "backwards" nature of the function \(f^{\ast}: P(Y) \to P(X)\), going back from \(P(Y)\) to \(P(X)\).
Propositions about the system being measured also give propositions about the measurement apparatus, but this is more tricky. What does "there's a living cat in my room" tell us about the temperature I read on my thermometer? This is a bit confusing... but there is an answer because a function \(f\) really does also give a "forwards" function from \(P(X) \) to \(P(Y)\). Here it is:
Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq X \) define its image under \(f\) to be
$$ f_{!}(S) = \{y \in Y: \; y = f(x) \textrm{ for some } x \in S\} . $$ The image is a subset of \( Y \).
The image is often written as \(f(S)\), but I'm using the notation of
Seven Sketches, which comes from category theory. People pronounce \(f_{!}\) as "\(f\) lower shriek".
The image gives a monotone function
$$ f_{!}: P(X) \to P(Y) $$ since if \(S,T \in P(X)\) and \(S \subseteq T \) then
$$f_{!}(S) = \{y \in Y: \; y = f(x) \textrm{ for some } x \in S \}
\subseteq \{y \in Y: \; y = f(x) \textrm{ for some } x \in T \} = f_{!}(T) . $$ But here's the cool part: Theorem. \( f_{!}: P(X) \to P(Y) \) is the left adjoint of \( f^{\ast}: P(Y) \to P(X) \). Proof. We need to show that for any \(S \subseteq X\) and \(T \subseteq Y\) we have
$$ f_{!}(S) \subseteq T \textrm{ if and only if } S \subseteq f^{\ast}(T) . $$ David Tanzer gave a quick proof in Puzzle 19. It goes like this: \(f_{!}(S) \subseteq T\) is true if and only if \(f\) maps elements of \(S\) to elements of \(T\), which is true if and only if \( S \subseteq \{x \in X: \; f(x) \in T\} = f^{\ast}(T) \). \(\quad \blacksquare\)
This is great! But there's also
another way to go forwards from \(P(X)\) to \(P(Y)\), which is a right adjoint of \( f^{\ast}: P(Y) \to P(X) \). This is less widely known, and I don't even know a simple name for it. Apparently it's less useful. Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq X \) define
$$ f_{\ast}(S) = \{y \in Y: x \in S \textrm{ for all } x \textrm{ such that } y = f(x)\} . $$ This is a subset of \(Y \).
Puzzle 23. Show that \( f_{\ast}: P(X) \to P(Y) \) is the right adjoint of \( f^{\ast}: P(Y) \to P(X) \).
What's amazing is this. Here's another way of describing our friend \(f_{!}\). For any \(S \subseteq X \) we have
$$ f_{!}(S) = \{y \in Y: x \in S \textrm{ for some } x \textrm{ such that } y = f(x)\} . $$This looks almost exactly like \(f_{\ast}\). The only difference is that while the left adjoint \(f_{!}\) is defined using "for some", the right adjoint \(f_{\ast}\) is defined using "for all". In logic "for some \(x\)" is called the
existential quantifier \(\exists x\), and "for all \(x\)" is called the universal quantifier \(\forall x\). So we are seeing that existential and universal quantifiers arise as left and right adjoints!
This was discovered by Bill Lawvere in this revolutionary paper:
By now this observation is part of a big story that "explains" logic using category theory.
Two more puzzles! Let \( X \) be the set of states of your room, and \( Y \) the set of states of a thermometer in your room: that is, thermometer readings. Let \(f : X \to Y \) map any state of your room to the thermometer reading.
Puzzle 24. What is \(f_{!}(\{\text{there is a living cat in your room}\})\)? How is this an example of the "liberal" or "generous" nature of left adjoints, meaning that they're a "best approximation from above"? Puzzle 25. What is \(f_{\ast}(\{\text{there is a living cat in your room}\})\)? How is this an example of the "conservative" or "cautious" nature of right adjoints, meaning that they're a "best approximation from below"?
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I am going ahead with asking what might appear as a debugging question. But,I am looking for something conceptual that I might have missed, and this question is somewhat related to my previous question: NVE MD simulation of inert gas: Problem maintaining equilibrium, although a different issue now. If the moderators decide to close this, I understand.
I am doing a constant energy (NVE) simulation of 49 Argon particles in a 100x100 2D box with Lennard Jones potential at temperature 240 K I am using the following:
1) Unit system: Natural unit system such that mass $m=1$, LJ parameters, $\sigma=1$, $\epsilon=1$, Boltzmann constant $k=1$ such that temperatureof 120 K is 1.
2) Initial Conditions: Square lattice position, Maxwell-Boltzmann velocities
3) Integration: Velocity-verlet with periodic boundary conditions. Time step of 0.001 (2fs in actual time)
4) I am checking for occurrence of a non-zero mean velocity (flying ice cubes): It does not occur.
I am unable to keep energy conservation within 1% for more than 2000 time steps (total actual time of 4ps). Most studies, lecture notes, online tutorials,etc., seem to be using these same values for time step, parameters, etc. While these never say how long they are able to conserve energy (and within what accuracy) I find research papers do 10ns on an average. I am failing 3 orders below. Velocity verlet seems like a fairly-accepted algorithm at least for something so simple!
It doesn't seem like a basic coding (i.e.,non-conceptual) mistake since the code does well until it gets to around 4ps. Some runs I can go upto 6 or 7ps, just lucky initial conditions I guess. Any helpful suggestions on where I could be going wrong? Thanks.
Update: 1) The error doesn't seem to scale with time-step $\Delta t$. Most papers seem to use $\Delta t$ =0.001, I have tried the range from 1e-2 to 1e-5.
2) My integrator code looks like this: { while(t
//Energy Calculation Ven(t)=ven; KE(U,V,ken,N); Ken(t)=ken; if(t==0){ Hen1=Ken(0)+Ven(0); cout<<"Total energy at t=0 is: "<<Hen1<<endl;} //Error Checking ErrMeanVel=(U.sum())/N; ErrEConserve=(abs(Ken(t)+Ven(t)-Hen1)*100)/Hen1; if(ErrMeanVel>1e-6){ cout<<"Flying ice cube at t="<<t*Deltat<<"ps. The mean velocity is"<<ErrMeanVel<<endl; break;} if(ErrEConserve>ETol) { cout<<"Energy conservation violated at t="<<t*Deltat<<"ps. The energy at this time is "<<(Ken(t)+Ven(t))<<" and the percentage error is "<<((abs(Ken(t)+Ven(t)-Hen1)*100)/Hen1)<<endl; break; } //Velocity Verlet for (i=0;i<N;i++){ X(0,i)=X(0,i)+(U(0,i)*Deltat)+0.5*Ax(i)*Deltat*(Deltat/mass); Y(0,i)=Y(0,i)+(V(0,i)*Deltat)+0.5*Ay(i)*Deltat*(Deltat/mass); U(0,i)=U(0,i)+(0.5*Ax(i)*Deltat/mass); V(0,i)=V(0,i)+(0.5*Ay(i)*Deltat/mass);} LJ(X,Y,N,ven,Fx,Fy,ctr); Ax=Fx.rowwise().sum(); Ay=Fy.rowwise().sum(); for (i=0;i<N;i++) { U(0,i)=U(0,i)+(0.5*Ax(i)*Deltat/mass); V(0,i)=V(0,i)+(0.5*Ay(i)*Deltat/mass); } //Boundary Conditions PeriodicBoundary(X,Y,L,N); t++;}
\Force calculator is: (no cutoff)
int i=0, j=0; double dx=0, dy=0,r2=0,F6=0,F12=0,F=0,Vcut=0; fx=MatrixXd::Zero(n,n); fy=MatrixXd::Zero(n,n); ven=0; MatrixXd ft(n,n); for (j=0;j<n;j++) { for(i=j+1;i<n;i++) { dx=(a(0,j)-a(0,i)); dy=(b(0,j)-b(0,i)); r2=dx*dx+dy*dy; if(r2<sigma) { ctr++; } F6=pow((sigma*sigma/r2),3); F12=pow((sigma*sigma/r2),6); F=(24*epsilon/r2)*(F6-2*F12); ven=ven+4*epsilon*(F12-F6); fx(i,j)=F*dx; fy(i,j)=F*dy; } } ft=fx.transpose(); fx=fx-ft; ft=fy.transpose(); fy=fy-ft;}
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The key to this question is to note that gunpowder doesn't technically explode -- it deflagrates. It doesn't have a super-sonic explosion, but rather a sub-sonic burn. To get the powerful kick needed to project a rifle or handgun bullet, we rely on the fact that gunpowder burns faster in a confined space. The tighter the space, the more temperature and pressure it can achieve.
Fired properly, the gunpowder is confined by the barrel, permitting it to reach the high pressures of a gun shot. Outside of a barrel, the brass case holding the gunpowder and the lightly set bullet provides surprisingly little containment. One can "cook off" a bullet over a fire, and the result is sudden and surprising, but far from lethal.
A single spark on the outside of the case would have a hard time setting off a bullet. Having done this once in a controlled setting to test the safety of such a bullet, it can take several seconds over the top of a torch to reach the critical temperature to deflagrate. A lone spark may have trouble. However, if your pyrokinetic can put the spark on the inside of the bullet, that'd be a very different story. That would be remarkably similar to what the primer actually does when firing the gun!
As mentioned earlier, the bullet would most certainly not escape the magazine. Most magazines are made of steel, and many tests will show just how little momentum the bullet actually picks up. Most of the time it's just shoved out of the case just far enough to give the gunpowder room to burn. However, we have to recognize that this is
still a confined space inside the handle of the gun. While it's not as small and well structured as the space inside a barrel, the gunpowder is still going to have to find an exit. It will build up pressure until it does find enough of an exit. This could be enough to cause damage.
The particular behavior is very dependent on the particular handgun and its construction. A revolver would most likely just shove the bullet out of the front of the cylinder, with little to no damage. An all steel handgun like a 1911, however, may contain the pressure better. This means it may fail in a more spectacular way. The small clip that holds the magazine into the gun would be my guess for "first to fail," causing the entire clip to pop out of the gun. If you had a "plastic" gun like a Glock 19, you could be in worse trouble. The bullets in the magazine are held in by a similar pin, but there's open access from the magazine to pressurize plastic all over. There's a decent chance that the force of the powder could rupture the plastic around the bottom edge of the slide (which is typically at a particularly nasty position for spraying plastic bits all over the gun's wielder).
Another question would be what happens to the other bullets? Depending on the exact mechanics of the rupture, you
might push one of the other bullets out of the way, exposing another case full of gunpowder. This would create a much larger effect, though it's not immediately clear what sort of mechanical topologies might cause this.
Ironically, rifle bullets might have a less extreme effect than handgun bullets. Many rifle calibers involve a large chamber for powder necked down to a smaller bullet. In a barrel that is shaped for this, this allows for devastating power. However, in a magazine, that space would simply be expansion room for the burning powder. That extra expansion room may keep the pressure down enough that the rifle round may never reach the high pressures that could cause serious damage to the handgun, despite having more gunpowder to work with.
All in all, I don't recommend experimenting to find the answer =) While there's still squabbling over whether guns kill people or people kill people,
everyone agrees that a misfired gun is a dangerous device and must be treated with respect until the misfire is resolved.
- From a long discussion in comments, it looks like the question of revolver rounds is of interest. Thanks to Deolater and Supercat for tugging at this thread, and Supercat for bringing data to the table! Edit
The key equation for determining the speed of a bullet is $F=p\cdot A$, the force propelling a bullet forward is the pressure behind the bullet times the cross sectional area of the bullet. Using the formula for work: $W=\int_0^LF\, dx$ where L is the length of the barrel, we can do some comparisons. Then, knowing that $E=\frac{1}{2}mv^2$, we can back out the velocity by noting that the velocity is proportional to the square root of E ($v\varpropto \sqrt E$)
We can consider two idealized cases for the powder burning. The first assumes constant pressure, and the second assumes the powder burns all at once, maximizing pressure at first. A realistic bullet will fall between one of these two extreme cases based on how fast the powder burns.
In the case of a constant pressure, we see $W=\int_0^LpA\, dx$ and thus $W \varpropto L$, where L is the length of the barrel. This means that $v \varpropto \sqrt L$ for the constant pressure case. In the case of an instantaneous burn, the pressure behind the bullet will obey some $p(x)=\frac{P_0}{x+C}$ where $P_0$ is the pressure at the start and $C$ is a constant capturing how much space is behind the bullet where pressure can be built before the bullet starts moving. This gives $W=\int_0^L\frac{P_0}{x+C}A\, dx$. If we cleverly choose units of length such that $C=1$ and thus $W\varpropto \ln(L+1)$.
Now we can put some numbers to this. Thanks to supercat's find, we have a table for a .375 magnum. Now .357 is rather convenient in that $C$ is roughly 1 inch (the case is 1.29" and the bullet rests a bit inside, so 1" is actually probably very close to correct). If the powder were to burn with a constant pressure, the energy after moving 1/2" (escaping the edge of the chamber) would be 1/12th that of the energy when escaping a 6" barrel, and thus a velocity that was .00694 that of the 6" barrel shot. This would put its velocity around 10fps. If we instead assume an instantaneous burn, we can use the second set of equations to see that the energy would be about 20.8% of the energy escaping a 6" barrel. This second equation would instead put its velocity at 65fps. The actual speed would be somewhere between these extreme assumptions.
Sure enough, if we graph muzzle velocity from the page of .357 data, we see a sharp knee in the curve as the barrel length gets smaller. The data doesn't go below 2", but extrapolating the lines confirms that the velocity exiting the chamber would be very small with respect to that of a properly fired bullet.
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A positive-definite diagonal quadratic form$a_{1}x_{1}^{2}+\cdots +a_{n}x_{n}^{2}\;(a_{1},\ldots ,a_{n}\in \mathbb{N})$is said to be prime-universal if it is not universal and for every prime$p$there are integers$x_{1},\ldots ,x_{n}$such that$a_{1}x_{1}^{2}+\cdots +a_{n}x_{n}^{2}=p$. We determine all possible prime-universal ternary quadratic forms$ax^{2}+by^{2}+cz^{2}$and all possible prime-universal quaternary quadratic forms$ax^{2}+by^{2}+cz^{2}+dw^{2}$. The prime-universal ternary forms are completely determined. The prime-universal quaternary forms are determined subject to the validity of two conjectures. We make no use of a result of Bhargava concerning quadratic forms representing primes which is stated but not proved in the literature.
Suppose$a^{2}(a^{2}+1)$divides$b^{2}(b^{2}+1)$with$b>a$. We improve a previous result and prove a gap principle, without any additional assumptions, namely$b\gg a(\log a)^{1/8}/(\log \log a)^{12}$. We also obtain$b\gg _{\unicode[STIX]{x1D716}}a^{15/14-\unicode[STIX]{x1D716}}$under the abc conjecture.
For every integer$k\geq 2$and every$A\subseteq \mathbb{N}$, we define the$k$-directions sets of$A$as$D^{k}(A):=\{\boldsymbol{a}/\Vert \boldsymbol{a}\Vert :\boldsymbol{a}\in A^{k}\}$and$D^{\text{}\underline{k}}(A):=\{\boldsymbol{a}/\Vert \boldsymbol{a}\Vert :\boldsymbol{a}\in A^{\text{}\underline{k}}\}$, where$\Vert \cdot \Vert$is the Euclidean norm and$A^{\text{}\underline{k}}:=\{\boldsymbol{a}\in A^{k}:a_{i}\neq a_{j}\text{ for all }i\neq j\}$. Via an appropriate homeomorphism,$D^{k}(A)$is a generalisation of the ratio set$R(A):=\{a/b:a,b\in A\}$. We study$D^{k}(A)$and$D^{\text{}\underline{k}}(A)$as subspaces of$S^{k-1}:=\{\boldsymbol{x}\in [0,1]^{k}:\Vert \boldsymbol{x}\Vert =1\}$. In particular, generalising a result of Bukor and Tóth, we provide a characterisation of the sets$X\subseteq S^{k-1}$such that there exists$A\subseteq \mathbb{N}$satisfying$D^{\text{}\underline{k}}(A)^{\prime }=X$, where$Y^{\prime }$denotes the set of accumulation points of$Y$. Moreover, we provide a simple sufficient condition for$D^{k}(A)$to be dense in$S^{k-1}$. We conclude with questions for further research.
which was originally conjectured by Long and later proved by Swisher. This confirms a conjecture of the second author [‘A$q$-analogue of the (L.2) supercongruence of Van Hamme’, J. Math. Anal. Appl.466 (2018), 749–761].
Let$n$be a positive integer and$a$an integer prime to$n$. Multiplication by$a$induces a permutation over$\mathbb{Z}/n\mathbb{Z}=\{\overline{0},\overline{1},\ldots ,\overline{n-1}\}$. Lerch’s theorem gives the sign of this permutation. We explore some applications of Lerch’s result to permutation problems involving quadratic residues modulo$p$and confirm some conjectures posed by Sun [‘Quadratic residues and related permutations and identities’, Preprint, 2018, arXiv:1809.07766]. We also study permutations involving arbitrary$k$th power residues modulo$p$and primitive roots modulo a power of$p$.
We give a new formula for the number of cyclic subgroups of a finite abelian group. This is based on Burnside’s lemma applied to the action of the power automorphism group. The resulting formula generalises Menon’s identity.
For$\unicode[STIX]{x1D6FD}\in (1,2]$the$\unicode[STIX]{x1D6FD}$-transformation$T_{\unicode[STIX]{x1D6FD}}:[0,1)\rightarrow [0,1)$is defined by$T_{\unicode[STIX]{x1D6FD}}(x)=\unicode[STIX]{x1D6FD}x\hspace{0.6em}({\rm mod}\hspace{0.2em}1)$. For$t\in [0,1)$let$K_{\unicode[STIX]{x1D6FD}}(t)$be the survivor set of$T_{\unicode[STIX]{x1D6FD}}$with hole$(0,t)$given by
$$\begin{eqnarray}K_{\unicode[STIX]{x1D6FD}}(t):=\{x\in [0,1):T_{\unicode[STIX]{x1D6FD}}^{n}(x)\not \in (0,t)\text{ for all }n\geq 0\}.\end{eqnarray}$$
In this paper we characterize the bifurcation set$E_{\unicode[STIX]{x1D6FD}}$of all parameters$t\in [0,1)$for which the set-valued function$t\mapsto K_{\unicode[STIX]{x1D6FD}}(t)$is not locally constant. We show that$E_{\unicode[STIX]{x1D6FD}}$is a Lebesgue null set of full Hausdorff dimension for all$\unicode[STIX]{x1D6FD}\in (1,2)$. We prove that for Lebesgue almost every$\unicode[STIX]{x1D6FD}\in (1,2)$the bifurcation set$E_{\unicode[STIX]{x1D6FD}}$contains infinitely many isolated points and infinitely many accumulation points arbitrarily close to zero. On the other hand, we show that the set of$\unicode[STIX]{x1D6FD}\in (1,2)$for which$E_{\unicode[STIX]{x1D6FD}}$contains no isolated points has zero Hausdorff dimension. These results contrast with the situation for$E_{2}$, the bifurcation set of the doubling map. Finally, we give for each$\unicode[STIX]{x1D6FD}\in (1,2)$a lower and an upper bound for the value$\unicode[STIX]{x1D70F}_{\unicode[STIX]{x1D6FD}}$such that the Hausdorff dimension of$K_{\unicode[STIX]{x1D6FD}}(t)$is positive if and only if$t<\unicode[STIX]{x1D70F}_{\unicode[STIX]{x1D6FD}}$. We show that$\unicode[STIX]{x1D70F}_{\unicode[STIX]{x1D6FD}}\leq 1-(1/\unicode[STIX]{x1D6FD})$for all$\unicode[STIX]{x1D6FD}\in (1,2)$.
for$n\ges 0$. In this paper, we obtain the relation between the Jacobi continued fraction of the ordinary generating function of yn(q) and that of xn(q). We also prove that the transformation preserves q-TPr+1 (q-TP) property of the Hankel matrix$[x_{i+j}(q)]_{i,j \ges 0}$, in particular for r = 2,3, implying the r-q-log-convexity of the sequence$\{y_n(q)\}_{n\ges 0}$. As applications, we can give the continued fraction expressions of Eulerian polynomials of types A and B, derangement polynomials types A and B, general Eulerian polynomials, Dowling polynomials and Tanny-geometric polynomials. In addition, we also prove the strong q-log-convexity of derangement polynomials type B, Dowling polynomials and Tanny-geometric polynomials and 3-q-log-convexity of general Eulerian polynomials, Dowling polynomials and Tanny-geometric polynomials. We also present a new proof of the result of Pólya and Szegö about the binomial convolution preserving the Stieltjes moment property and a new proof of the result of Zhu and Sun on the binomial transformation preserving strong q-log-convexity.
One of the open questions in the study of Carmichael numbers is whether, for a given$R\geq 3$, there exist infinitely many Carmichael numbers with exactly$R$prime factors. Chernick [‘On Fermat’s simple theorem’, Bull. Amer. Math. Soc.45 (1935), 269–274] proved that Dickson’s$k$-tuple conjecture would imply a positive result for all such$R$. Wright [‘Factors of Carmichael numbers and a weak$k$-tuples conjecture’, J. Aust. Math. Soc.100(3) (2016), 421–429] showed that a weakened version of Dickson’s conjecture would imply that there are an infinitude of$R$for which there are infinitely many such Carmichael numbers. In this paper, we improve on our 2016 result by weakening the required conjecture even further.
Let$n,r,k\in \mathbb{N}$. An$r$-colouring of the vertices of a regular$n$-gon is any mapping$\unicode[STIX]{x1D712}:\mathbb{Z}_{n}\rightarrow \{1,2,\ldots ,r\}$. Two colourings are equivalent if one of them can be obtained from another by a rotation of the polygon. An$r$-ary necklace of length$n$is an equivalence class of$r$-colourings of$\mathbb{Z}_{n}$. We say that a colouring is$k$-alternating if all$k$consecutive vertices have pairwise distinct colours. We compute the smallest number$r$for which there exists a$k$-alternating$r$-colouring of$\mathbb{Z}_{n}$and we count, for any$r$, 2-alternating$r$-colourings of$\mathbb{Z}_{n}$and 2-alternating$r$-ary necklaces of length$n$.
where$[x]$denotes the integral part of real$x$. The above summations were recently considered by Bordellès et al. [‘On a sum involving the Euler function’, Preprint, 2018, arXiv:1808.00188] and Wu [‘On a sum involving the Euler totient function’, Preprint, 2018, hal-01884018].
In this paper we study digit frequencies in the setting of expansions in non-integer bases, and self-affine sets with non-empty interior. Within expansions in non-integer bases we show that if$\unicode[STIX]{x1D6FD}\in (1,1.787\ldots )$then every$x\in (0,1/(\unicode[STIX]{x1D6FD}-1))$has a simply normal$\unicode[STIX]{x1D6FD}$-expansion. We also prove that if$\unicode[STIX]{x1D6FD}\in (1,(1+\sqrt{5})/2)$then every$x\in (0,1/(\unicode[STIX]{x1D6FD}-1))$has a$\unicode[STIX]{x1D6FD}$-expansion for which the digit frequency does not exist, and a$\unicode[STIX]{x1D6FD}$-expansion with limiting frequency of zeros$p$, where$p$is any real number sufficiently close to$1/2$. For a class of planar self-affine sets we show that if the horizontal contraction lies in a certain parameter space and the vertical contractions are sufficiently close to$1$, then every non-trivial vertical fibre contains an interval. Our approach lends itself to explicit calculation and gives rise to new examples of self-affine sets with non-empty interior. One particular strength of our approach is that it allows for different rates of contraction in the vertical direction.
In 1984, K. Mahler asked how well elements in the Cantor middle third set can be approximated by rational numbers from that set and by rational numbers outside of that set. We consider more general missing digit sets$C$and construct numbers in$C$that are arbitrarily well approximable by rationals in$C$, but badly approximable by rationals outside of$C$. More precisely, we construct them so that all but finitely many of their convergents lie in$C$.
Let$n$be a positive integer. We obtain new Menon’s identities by using the actions of some subgroups of$(\mathbb{Z}/n\mathbb{Z})^{\times }$on the set$\mathbb{Z}/n\mathbb{Z}$. In particular, let$p$be an odd prime and let$\unicode[STIX]{x1D6FC}$be a positive integer. If$H_{k}$is a subgroup of$(\mathbb{Z}/p^{\unicode[STIX]{x1D6FC}}\mathbb{Z})^{\times }$with index$k=p^{\unicode[STIX]{x1D6FD}}u$such that$0\leqslant \unicode[STIX]{x1D6FD}<\unicode[STIX]{x1D6FC}$and$u\mid p-1$, then
Let$k$be an arbitrary positive integer and let$\unicode[STIX]{x1D6FE}(n)$stand for the product of the distinct prime factors of$n$. For each integer$n\geqslant 2$, let$a_{n}$and$b_{n}$stand respectively for the maximum and the minimum of the$k$integers$\unicode[STIX]{x1D6FE}(n+1),\unicode[STIX]{x1D6FE}(n+2),\ldots ,\unicode[STIX]{x1D6FE}(n+k)$. We show that$\liminf _{n\rightarrow \infty }a_{n}/b_{n}=1$. We also prove that the same result holds in the case of the Euler function and the sum of the divisors function, as well as the functions$\unicode[STIX]{x1D714}(n)$and$\unicode[STIX]{x1D6FA}(n)$, which stand respectively for the number of distinct prime factors of$n$and the total number of prime factors of$n$counting their multiplicity.
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Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator
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Rich Trigonometry concepts, Compound angles
Trigonometry first established relationships between the angle and side measurements of a right angled triangle by defining trigonometric functions and thus extended the reach of mathematics wonderfully. Like any other branch of mathematics, from the basic trigonometric function definitions more advanced function relations are built stage by stage. Here we will deal with compound angle trigonometric function relations in more details.
You may refer to
for the foundational basic and rich trigonometry concepts. The rich concepts discussed here rest on the concepts introduced in the earlier discussion mentioned. Basic and rich Trigonometry concepts and its applications Proof of relations for $\sin (A+B)$, $\cos (A+B)$ and $\tan (A+B)$
The following figure will help the solution process.
In the explanation, instead of $\angle A$ and $\angle B$ we will be using $\angle \alpha$ and $\angle \theta$ respectively.
In the figure, $\triangle APB$, $\triangle PFC$, $\triangle AFD$, $\triangle DEF$ and $\triangle AEF$ are right triangles where $EF||PBC$ and $EF=BC$ and $BE=CF$.
AS $EF||PB$, $\angle CPF=\angle EFD=\alpha$. Again in right $\triangle EFD$, $\angle \alpha +\angle x=90^0$ and in right $\triangle AFD$, $\angle x +\angle DAF=90^0$ so that, $\angle DAF=\angle \alpha$.
With this background work we are now ready to derive the expanded relationship of $\sin (\alpha +\theta)$.
$\sin (\alpha + \theta) = \displaystyle\frac{AB}{AP}$
$=\displaystyle\frac{AE+BE}{AP}$
$=\displaystyle\frac{CF}{AP}+ \displaystyle\frac{AE}{AP}$
$=\displaystyle\frac{CF}{PF}.\displaystyle\frac{PF}{AP}+\displaystyle\frac{AE}{AF}.\displaystyle\frac{AF}{AP}$
$=\sin \alpha.{\cos \theta} + \cos \alpha.{\sin \theta}$.
Proof of $\cos (A+B)$
Using the same figure above we have,
$\cos (\alpha +\theta)=\displaystyle\frac{PB}{AP}$
$=\displaystyle\frac{PC-BC}{AP}$
$=\displaystyle\frac{PC}{AP}- \displaystyle\frac{EF}{AP}$
$=\displaystyle\frac{PC}{PF}.\displaystyle\frac{PF}{AP}-\displaystyle\frac{EF}{AF}.\displaystyle\frac{AF}{AP}$
$=\cos \alpha.{\cos \theta} - \sin \alpha.{\sin \theta}$.
Proof of $\tan (A+B)$
Instead of deriving the relationship of $\tan (\alpha +\theta)$, using the results of $\sin (\alpha +\theta)$ and $\cos (\alpha +\theta)$ we will prove the relation,
$\tan (\alpha+\theta)=\displaystyle\frac{\tan \alpha +\tan \theta}{1-\tan \alpha.{\tan \theta}}$.
$\tan (\alpha+\theta)=\displaystyle\frac{\sin (\alpha +\theta)}{\cos (\alpha +\theta)}$
$=\displaystyle\frac{\sin \alpha.{\cos \theta} + \cos \alpha.{\sin \theta}}{\cos \alpha.{\cos \theta} - \sin \alpha.{\sin \theta}}$.
Taking out the factor $\cos \alpha.{cos \theta}$ from the numerator and denominaror of the RHS of the above relation we have,
$\tan (\alpha+\theta)=\displaystyle\frac{\tan \alpha +\tan \theta}{1-\tan \alpha.{\tan \theta}}$.
We need to understand the derivation of primarily $\sin (A +B)$ as derivation of $\cos (A + B)$ takes the same approach and from $\sin (A + B)$ and $\cos (A + B)$ the expression of $\tan (A +B)$ can easily be derived.
Note: Usually you won't need to derive the compound angle relations. Nevertheless understanding the proof enhances your belief on the important compound angle relationships that are used at various stages of Trigonometric and Geometric problem solving. Resources on Trigonometry and related topics
You may refer to our useful resources on Trigonometry and other related topics especially algebra.
Tutorials on Trigonometry Basic and Rich Concepts in Trigonometry part 2, proof of compound angle functions General guidelines for success in SSC CGL Efficient problem solving in Trigonometry A note on usability: The Efficient math problem solving sessions on School maths are equally usable for SSC CGL aspirants, as firstly, the "Prove the identity" problems can easily be converted to a MCQ type question, and secondly, the same set of problem solving reasoning and techniques have been used for any efficient Trigonometry problem solving.
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Dimensionality reduction is used to remove irrelevant and redundant features.
When the number of features in a dataset is bigger than the number of examples, then the probability density function of the dataset becomes difficult to calculate.
For example, if we model a dataset \(S = \{x^{(i)}\}_{i=1}^m,\ x \in R^{n}\) as a single Gaussian N(μ, ∑), then the probability density function is defined as: \(P(x) = \frac{1}{{(2π)}^{\frac{n}{2}} |Σ|^\frac{1}{2}} exp(-\frac{1}{2} (x-μ)^T {Σ}^{-1} (x-μ))\) such as \(μ = \frac{1}{m} \sum_{i=1}^m x^{(i)} \\ ∑ = \frac{1}{m} \sum_{i=1}^m (x^{(i)} – μ)(x^{(i)} – μ)^T\).
But If n >> m, then ∑ will be singular, and calculating P(x) will be impossible.
Note: \((x^{(i)} – μ)(x^{(i)} – μ)^T\) is always singular, but the \(\sum_{i=1}^m\) of many singular matrices is most likely invertible when m >> n.
Principal Component Analysis
Given a set \(S = \{x^{(1)}=(0,1), x^{(2)}=(1,1)\}\), to reduce the dimensionality of S from 2 to 1, we need to project data on a vector that maximizes the projections. In other words, find the normalized vector \(μ = (μ_1, μ_2)\) that maximizes \( ({x^{(1)}}^T.μ)^2 + ({x^{(2)}}^T.μ)^2 = (μ_2)^2 + (μ_1 + μ_2)^2\).
Using the method of Lagrange Multipliers, we can solve the maximization problem with constraint \(||u|| = μ_1^2 + μ_2^2 = 1\).\(L(μ, λ) = (μ_2)^2 + (μ_1 + μ_2)^2 – λ (μ_1^2 + μ_2^2 – 1) \)
We need to find μ such as \(∇_u = 0 \) and ||u|| = 1
After derivations we will find that the solution is the vector μ = (0.52, 0.85)
Generalization
Given a set \(S = \{x^{(i)}\}_{i=1}^m,\ x \in R^{n}\), to reduce the dimensionality of S, we need to find μ that maximizes \(arg \ \underset{u: ||u|| = 1}{max} \frac{1}{m} \sum_{i=1}^m ({x^{(i)}}^T u)^2\)\(=\frac{1}{m} \sum_{i=1}^m (u^T {x^{(i)}})({x^{(i)}}^T u)\) \(=u^T (\frac{1}{m} \sum_{i=1}^m {x^{(i)}} * {x^{(i)}}^T) u\)
Let’s define \( ∑ = \frac{1}{m} \sum_{i=1}^m {x^{(i)}} * {x^{(i)}}^T \)
Using the method of Lagrange Multipliers, we can solve the maximization problem with constraint \(||u|| = u^Tu\) = 1.\(L(μ, λ) = u^T ∑ u – λ (u^Tu – 1) \)
If we calculate the derivative with respect to u, we will find:\(∇_u = ∑ u – λ u = 0\)
Therefore u that solves this maximization problem must be an eigenvector of ∑. We need to choose the eigenvector with highest eigenvalue.
If we choose k eigenvectors \({u_1, u_2, …, u_k}\), then we need to transform the data by multiplying each example with each eigenvector.\(x^{(i)} := (u_1^T x^{(i)}, u_2^T x^{(i)},…, , u_k^T x^{(i)}) = U^T x^{(i)}\)
Data should be normalized before running the PCA algorithm:
1-\(μ = \frac{1}{m} \sum_{i=1}^m x^{(i)}\)
2-\(x^{(i)} := x^{(i)} – μ\)
3-\(σ_j^{(i)} = \frac{1}{m} \sum_{i=1}^m {x_j^{(i)}}^2\)
4-\(x^{(i)} := \frac{x_j^{(i)}}{σ_j^{(i)}}\)
To reconstruct the original data, we need to calculate \(\widehat{x}^{(i)} := U^T x^{(i)}\)
Factor Analysis
Factor analysis is a way to take a mass of data and shrinking it to a smaller data set with less features.
Given a set \(S = \{x^{(i)}\}_{i=1}^m,\ x \in R^{n}\), and S is modeled as a single Gaussian.
To reduce the dimensionality of S, we define a relationship between the variable x and a laten (hidden) variable z called factor such as \(x^{(i)} = μ + Λ z^{(i)} + ϵ^{(i)}\) and \(μ \in R^{n}\), \(z^{(i)} \in R^{d}\), \(Λ \in R^{n*d}\), \(ϵ \sim N(0, Ψ)\), Ψ is diagonal, \(z \sim N(0, I)\) and d <= n.
From Λ we can find the features that are related to each factor, and then identify the features that need to be eliminated or combined in order to reduce the dimensionality of the data.
Below the steps to estimate the parameters Ψ, μ, Λ.\(E[x] = E[μ + Λz + ϵ] = E[μ] + ΛE[z] + E[ϵ] = μ \) \(Var(x) = E[(x – μ)^2] = E[(x – μ)(x – μ)^T] = E[(Λz + ϵ)(Λz + ϵ)^T]\) \(=E[Λzz^TΛ^T + ϵz^TΛ^T + Λzϵ^T + ϵϵ^T]\) \(=ΛE[zz^T]Λ^T + E[ϵz^TΛ^T] + E[Λzϵ^T] + E[ϵϵ^T]\) \(=Λ.Var(z).Λ^T + E[ϵz^TΛ^T] + E[Λzϵ^T] + Var(ϵ)\)
ϵ and z are independent, then the join probability of p(ϵ,z) = p(ϵ)*p(z), and \(E[ϵz]=\int_{ϵ}\int_{z} ϵ*z*p(ϵ,z) dϵ dz\)\(=\int_{ϵ}\int_{z} ϵ*z*p(ϵ)*p(z) dϵ dz\) \(=\int_{ϵ} ϵ*p(ϵ) \int_{z} z*p(z) dz dϵ\) \(=E[ϵ]E[z]\)
So:\(Var(x)=ΛΛ^T + Ψ\)
Therefore \(x \sim N(μ, ΛΛ^T + Ψ)\) and \(P(x) = \frac{1}{{(2π)}^{\frac{n}{2}} |ΛΛ^T + Ψ|^\frac{1}{2}} exp(-\frac{1}{2} (x-μ)^T {(ΛΛ^T + Ψ)}^{-1} (x-μ))\)
\(Λ \in R^{n*d}\), if d <= m, then \(ΛΛ^T + Ψ\) is most likely invertible.
To find Ψ, μ, Λ, we need to maximize the log-likelihood function.\(l(Ψ, μ, Λ) = \sum_{i=1}^m log(P(x^{(i)}; Ψ, μ, Λ))\) \(= \sum_{i=1}^m log(\frac{1}{{(2π)}^{\frac{n}{2}} |ΛΛ^T + Ψ|^\frac{1}{2}} exp(-\frac{1}{2} (x^{(i)}-μ)^T {(ΛΛ^T + Ψ)}^{-1} (x^{(i)}-μ)))\)
This maximization problem cannot be solved by calculating the \(∇_Ψ l(Ψ, μ, Λ) = 0\), \(∇_μ l(Ψ, μ, Λ) = 0\), \(∇_Λ l(Ψ, μ, Λ) = 0\). However using the EM algorithm, we can solve that problem.
More details can be found in this video: https://www.youtube.com/watch?v=ey2PE5xi9-A
Restricted Boltzmann Machine
A restricted Boltzmann machine (RBM) is a two-layer
stochastic neural network where the first layer consists of observed data variables (or visible units), and the second layer consists of latent variables (or hidden units). The visible layer is fully connected to the hidden layer. Both the visible and hidden layers are restricted to have no within-layer connections.
In this model, we update the parameters using the following equations:
\(W := W + α * \frac{x⊗Transpose(h_0) – v_1 ⊗ Transpose(h_1)}{n} \\ b_v := b_v + α * mean(x – v_1) \\ b_h := b_h + α * mean(h_0 – h_1) \\ error = mean(square(x – v_1))\).
Deep Belief Network
A deep belief network is obtained by stacking several RBMs on top of each other. The hidden layer of the RBM at layer i becomes the input of the RBM at layer i+1. The first layer RBM gets as input the input of the network, and the hidden layer of the last RBM represents the output.
Autoencoders
An autoencoder, autoassociator or Diabolo network is a
deterministic artificial neural network used for unsupervised learning of efficient codings. The aim of an autoencoder is to learn a representation (encoding) for a set of data, typically for the purpose of dimensionality reduction. A deep Autoencoder contains multiple hidden units.
Loss function
For binary values, the loss function is defined as:
\(loss(x,\hat{x}) = -\sum_{k=1}^{size(x)} x_k.log(\hat{x_k}) + (1-x_k).log(1 – \hat{x_k})\).
For real values, the loss function is defined as:
\(loss(x,\hat{x}) = ½ \sum_{k=1}^{size(x)} (x_k – \hat{x_k})^2\).
Dimensionality reduction
Autoencoders separate data better than PCA.
Variational Autoencoder
Variational autoencoder (VAE) models inherit autoencoder architecture, but make strong assumptions concerning the distribution of latent variables. In general, we suppose the distribution of the latent variable is gaussian.
The training algorithm used in VAEs is similar to EM algorithm.
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I'll provide a description of $(\alpha)^{\mathfrak{A}}$ in the case of countable $\mathfrak{A}$. I base my answer on observations by Emil Jeřábek (see discussion below the initial question), but of course any mistakes here are due to me.
From Cantor's normal form theorem it is easy to conclude that each ordinal $\alpha\ge \omega^{\omega}$ have unique representation in the form $$\omega^{\omega}(1+\alpha')+\omega^{k_1}+\ldots+\omega^{k_n}\text{, where }k_1\ge k_2\ge\ldots\ge k_n.$$The first claim is that the order types of the form $(\alpha)^{\mathfrak{A}}$, where $\mathfrak{A}$ is countable non-standard model of $\mathsf{PA}$ are precisely the order types$$L+(\omega+(\omega^{\star}+\omega)\eta)^{k_1}+\ldots+(\omega+(\omega^{\star}+\omega)\eta)^{k_n},$$where $L$ is a recursively saturated linear order that is elementary equivalent to $(\omega^{\omega},<)$. The second claim is that, for given $(\alpha)^{\mathfrak{A}}$ the order type of the corresponding $L$ depends only on $\mathsf{SSy}(\mathfrak{A})$ and that we could recover $\mathsf{SSy}(\mathfrak{A})$ from the order type of $L$.
There is a classical result of A. Tarski and A. Mostowski [1,2] that all theories $\mathsf{Th}(\alpha,<)$ are decidable. A natural formalization of their decision procedure gives us $\Sigma_1$ formula $\mathsf{T}^\alpha(x)$ : "formula with Gödel number $x$ is true in $(\alpha,<)$". Their decidability result have been established via quantifier elimination in extended signature and all the used techniques were fairly elementary. A formalization of their proof in $\mathsf{PA}$ shows that $\mathsf{T}^{\alpha}$ acts as a truth-definition satisfying uniform Tarski's biconditionals: $$\mathsf{PA}\vdash \forall \vec{x}\prec \hat\alpha \;(\prec\upharpoonright_{\hat\alpha}\models\varphi(\vec{x})\mathrel{\leftrightarrow} \mathsf{T}^{\alpha}(\ulcorner\varphi(\dot{\vec{x}})\urcorner)),$$for all first-order formulas $\varphi(\vec{x})$ of the language of linear orders.
The presence of this truth definition implies that all the order types $(\alpha)^{\mathfrak{A}}$ would be recursively-saturated. If $t(\vec{x})$ is a recursive type in $(\alpha)^{\mathfrak{A}}$ then due to absoluteness of $\Delta_1$ properties for any standard $n$ model $\mathfrak{A}$ thinks that partial type $t(\vec{x})\upharpoonright_n$ is realized in $\prec\upharpoonright_{\hat\alpha}$. By overspill there are $a,\vec{b}\in\mathfrak{A}$ such that $\mathfrak{A}$ thinks that $\vec{b}$ realize the partial type $t(\vec{x})\upharpoonright_a$. But externally, again by absoluteness of $\Delta_1$ properties, tuple $\vec{b}$ is of the type $t(\vec{x})$ in $(\alpha)^{\mathfrak{A}}$.
Since for each $\alpha$, $\mathsf{PA}$ verifies that $\mathsf{T}^{\alpha}$ is $\Delta_1$, we have elementary equivalence $(\alpha)^{\mathfrak{A}}\equiv (\alpha,<)$. Additionally Tarski and Mostowski established that $(\omega^{\omega}\alpha,<)\equiv (\omega^\omega\beta,<)$, for any $1\le \alpha,\beta<\varepsilon_0$. Thus for any non-standar $\mathfrak{A}\models\mathsf{PA}$ and $1\le \alpha< \varepsilon_0$ the linear order $(\omega^{\omega}\alpha)^{\mathfrak{A}}$ is a recursively-saturated linear order elementary equivalent to $\omega^{\omega}$.
Let me now show that for any $1\le \alpha< \varepsilon_0$ and countable recursively-saturated $L\equiv (\omega^{\omega},<)$ there is countable $\mathfrak{A}\models \mathsf{PA}$ such that $L\simeq (\omega^\omega\alpha)^{\mathfrak{A}}$. Indeed, by a well-known result of Barwise and Ressayre [Theorem IV.5.7,3] all countable recursively-saturated models are resplendent and hence the existence of the desired $\mathfrak{A}$ follows from the fact that $\prec\upharpoonright_{\hat{(\omega^\omega\alpha)}}$ is an interpretation of $\mathsf{Th}(L)$ in $\mathsf{PA}$.
This concludes the proof of the first claim: the order $(\alpha)^{\mathfrak{A}}$ is $(\omega^{\omega}(1+\alpha'))^{\mathfrak{A}}+(\omega^{k_1}+\ldots+\omega^{k_n})^{\mathfrak{A}}$, we have already analysed the left summand and the order type of the right summand is trivially expressible from the order type of $\mathfrak{A}$. Now let me briefly sketch the situation with the connection between an order type of $L=(\omega^{\omega}(1+\alpha))^{\mathfrak{A}}$ and $\mathsf{SSy}(\mathfrak{A})$.
The key idea here is to assign to each element $a\in L$ the partial function $s_a\colon \omega\to \omega$. For all $n,m$ there are natural formulas $\varphi_{n,m}(x)$ expressing in all the structures structures $(\beta,<)$ the property of an ordinal $\gamma$ to be in the interval $[\omega^{n+1}\delta+\omega^{n}m,\omega^{n+1}\delta+\omega^{n}m)$ for some $\delta$. We put $s_a(n)=m$ if $L\models \varphi_{n,m}(a)$ and we put $s_a(n)$ to be undefined if $L\not \models \varphi_{n,m}(a)$, for all $m\in \omega$. We define $\mathsf{SF}(L)$ to be the set of all the partial functions $f\colon \omega\to \omega$ that are $s_a$, for some $a\in L$.
It is fairly easy to show that $f\in \mathsf{SF}((\omega^{\omega}\alpha)^{\mathfrak{A}})$ iff $f$ is a partial function which graph lies in $\mathsf{SSy}(\mathfrak{A})$. On the other hand it seems to be easy to show that countable recursively-saturated $L_1,L_2$ with $\mathsf{SF}(L_1)=\mathsf{SF}(L_2)$ are isomorphic (by providing a strategy for the Ehrenfeucht–Fraïssé game). Combining this two facts we prove the second claim.
[1] A. Tarski, A. Mostowski. "Arithmetical classes and types of well ordered systems. Preliminary report." Bull. Amer. Math. Sot., vol. 55 (1949), p. 65.
[2] J.E. Doner, A. Mostowski, and A. Tarski. "The Elementary Theory of Well-Odering—A Metamathematical Study—." Studies in Logic and the Foundations of Mathematics. Vol. 96. Elsevier, 1978. 1-54.
[3] J. Barwise. "Admissible Sets and Structures: An Approach to Definability Theory." Perspectivesin Mathematical Logic. Springer-Verlag, Berlin, 1975.
Update:I learned about a related paper by Harvey Friedman [4], where he studied order types of ordinals in ill-founded countable admissible sets. Unsurprisingly, the order types of ordinals $\mathsf{On}^{\mathfrak{M}}$ in countable admissible sets $\mathfrak{M}$ with ill-founded $\omega^{\mathfrak{M}}$ range precisely over the same order types as $(\omega^\omega)^{\mathfrak{A}}$, for countable non-standard $\mathfrak{A}\models \mathsf{PA}$. Friedman gave an explicit description of $\mathsf{On}^{\mathfrak{M}}$ in terms of $\mathsf{SSy}(\mathfrak{M})$ by giving (in my notations) description of orders $L$ with given $\mathsf{SF}(L)$.
[4] H. Friedman. "Countable models of set theories." Cambridge summer school in mathematical logic. Springer, Berlin, Heidelberg, 1973. 539-573.
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Mathtex Questions answered by this recipe Can PmWiki be used to display long mathematical sequences, logical demonstrations (in Fitch mode)?
This section is optional; use it to indicate the types of questions (if any) this recipe is intended to answer.
Description
MathTex replace MimeTex for those who have latex installed on ther server. MathTex recipe permits several lines logical or mathematical demonstrations.
Installation
GPL licenced written in C.
As MimeTex, MathTex run in cgi
for a simple usage, you can replace occurrences of mimtex by mathtex, after installation of the MathTex script in the cgi-bin directory.
Edit the c code to set dvipng or dvips+convert locations.
Compile the cgi by
cc mathtex.c -o mathtex.cgi Customization to add newpackages
You can add \usepackage{yourpackage} directly in the code after line 1023 and recompile the cgi
Simply replace MimeTex for simple usage
Put the mathtex.cgi in the cgi-bin directory of your server
For those who have LaTex installed on their server!
mv mimetex.cgi mimetex.old ln -s mathtex.cgi mimetex.cgi
(a symbolic link is set to mathtex.cgi, calling mimetex is calling mathtex)
Customisation to manage serveral line LaTeX scripts
I prepare a MathTex script from Pm MimeTex script.
The idea of this script is simply to replace the Markup {$ latex $} by
{$[= latex =]$} to manage several lines.
place the mathtex.php script into the cookbook directory
in the local/config.php, add the line
include_once('cookbook/mathtex.php');
set the values to the variables in the beginnine of the script (to run the cgi and to find the cache)
Sample
the following code: the code use the fitch and xcolor package
{$[= \begin{fitch} \fa Pp \to \neg\Diamond\neg{Pp} & A \\ \fa (p \to q) \to (\neg\Diamond q \to \neg\Diamond p) & B \\ \fa \color{magenta} p \to \neg{P\neg{Fp}} & f \\ \fj \color{green} \neg{p} \color{red} \to P\neg{F{p}} & f' \\ \fa\fh \neg{p} \land \neg{Fp} & \\ \fa\fa \color{green} \neg{p} & 5,\land\\ \fa\fa \color{red} P\neg{Fp} & MP 6,4\\ \fa\fa \color{red} P\neg{Fp} \color{black} \to \color{blue} \neg\Diamond\neg{P\neg{Fp}} & A, p/P\neg{Fp} \\ \fa\fa \color{blue} \neg\Diamond\neg{P\neg{Fp}} & MP 7,8 \\ \fa\fa ( \color{magenta} p \to \neg{P\neg{Fp}} \color{black} ) \to ( \color{blue} \neg\Diamond\neg{P\neg{Fp}} \color{black} \to \color{cyan} \neg\Diamond{p}) & B, q/\neg{P\neg{Fp}}) \\ \fa\fa \color{blue} \neg\Diamond\neg{P\neg{Fp}} \color{black} \to \color{cyan} \neg\Diamond{p} & MP 3,10 \\ \fa\fa \color{cyan} \neg\Diamond{p} & MP 9,11 \\ \fa \neg(\neg{p} \land \neg{Fp} \land\Diamond{p}) & 5,12 \\ \end{fitch} =]$}
give:
Release Notes
This first version is for testing. Maybe it is possible to do more simple Markup for several lines, I don't know.
If the recipe has multiple releases, then release notes can be placed here. Note that it's often easier for people to work with "release dates" instead of "version numbers".
See Also Cookbook / AMmathjax Add markup to embed math using ASCII MATH by MathJax. (Stable) ASCIIMath Display MathML rendered ascii formula into PmWiki 2.x pages (Stable) Formula Lightweight rendering of mathematical formulas in wiki pages (Beta) FractionsPlus Adds simple markup to write fractions. (Maintained) JsMath Add markup to display and embed mathematical formulas in wiki pages using TeX. (Obsolete - use MathJax) LaTeXMathJax MathJax support for PmWiki (stable) LaTeXMathML Translating LaTeX math notation dynamically to Presentation MathML (Stable) LinuxTex Enable TeX markup LuaExecute Lets Lua scripts dynamically generate parts of your wiki page (Beta) MathJax Add markup to embed math using TeX syntax. (Stable) MimeTeX Add markup to display and embed mathematical formulas in wiki pages (Stable) SageCell This allows you to embed an interactive SageCell into your webpage. Sage is an open source alternative to MathLab, Mathematica, and Maple. (Stable) SimpleJsMath Notes on modifying the jsMath for a skin (not a complete recipe) TrueLatex Enables interpretation and rendering of real LaTeX markups in PmWiki. Support for XeLaTex included in v1.8. (Working perfectly. Very simple to use, customizable and comes with lots of options.) Contributors
felie
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In the laboratory, in your body, and in the outside environment, the majority of chemical reactions take place in solutions. Macroscopically a
solution is defined as a homogeneous mixture of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or molecule with respect to another.
There are a number of reasons why solutions are so often encountered both in nature and in the laboratory. The most common type of solution involves a liquid
solvent which dissolves a solid solute. (The term solvent usually refers to the substance present in greatest amount. There may be more than one solute dissolved in it.) Because a liquid adopts the shape of its container but does not expand to fill all space available to it, liquid solutions are convenient to handle. You can easily pour them from one container to another, and their volumes are readily measured using graduated cylinders, pipets, burets, volumetric flasks, or other laboratory glass-ware. Moreover, atoms or molecules of solids dissolved in a liquid are close together but still able to move past one another. They contact each other more frequently than if two solids were placed next to each other. This “intimacy” in liquid solutions often facilitates chemical reactions.
Pictured above is a familiar solution: salt water. In this solution, water serves as the solvent, dissolving the solute, salt.
Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react with a given quantity of another. Examples in other sections have shown that the amount of substance is the quantity which determines how much of one material will react with another. The ease with which solution volumes may be measured suggests that it would be very convenient to know the amount of substance dissolved per unit volume of solution. Then by measuring a certain volume of solution, we would also be measuring a certain amount of substance.
One measure of the
concentration c of a solute in a solution is often called molarity, but it is probably better to call it "the concentration in molar units" or "molar concentration" (keeping the parameter concentration, and its unit, M for molar distinct). The molar concentration is the amount of the substance per unit volume (L or dm 3) of solution(not solvent):
\[\text{Concentration of solute, M}=\frac{\text{amount of solute, mol}}{\text{volume of solution, L}}\]
\[c_{\text{solute, M}}~=~\frac{n_{\text{solute, mol}}}{V_{\text{solution,L}}} \label{2}\]
The units moles per liter (mol liter
–1) or moles per cubic decimeter (mol dm –3) are used to express molar concentration. They are equivalent (since 1 dm –3 = 1 liter).
If a pure substance is soluble in water, it is easy to prepare a solution of known concentration. A container with a sample of the substance is weighed accurately, and an appropriate mass of sample is poured through a funnel into a volumetric flask, as shown in the figure. The container is then reweighed. Any solid adhering to the funnel is rinsed into the flask, and water is added until the flask is about three-quarters full. After swirling the flask to dissolve the solid, water is added carefully until the bottom of the meniscus coincides with the calibration mark on the neck of the flash. This process is shown in detail in Figure \(\PageIndex{1}\) :
a) a weight boat is zeroed on the balance.
b) 58.441g of NaCl (~1 mole) is measured.
c) The NaCl is quantitatively added to water.
d) Mixing the solution dissolves NaCl.
e) The solution is added to volumetric flask by funnel.
f) Water is added to up the neck of the volumetric flask.
g) A dropper is used to dilute to the 1 liter line.
h) The meniscus reaches the mark.
Example \(\PageIndex{1}\): Concentration
A solution of KI was prepared as described above. The initial mass of the container plus KI was 43.2874 g, and the final mass after pouring was 30.1544 g. The volume of the flask was 250.00 ml. What is the concentration of the solution?
Solution: The concentration can be calculated by dividing the amount of solute by the volume of solution [Eq. \(\ref{2}\)]:
\[ c_{\text{KI}} = \frac{n_{\text{KI, mol}}}{V_{\text{solution, L}}}\]
We obtain n
KI from the mass of KI added to the flask:
\[ m_{\text{KI}} = 43.2874 - 30.1544 \text{ g} = 13.1330 \text{ g} \]
\[ n_{\text{KI}} = 13.1330 \text{ g} \times \frac{\text{1 mol}}{\text{166}\text{.00 g}} = 7.9115 \times 10^{-2} \text{ mol} \]
The volume of solution is 250.00 ml, or
\[ V_{\text{solution}} = 250.00 \text{cm}^{3} \times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} = 2.5000 \times 10^{-1} \text{dm}^{3} \]
Thus
\[c_{\text{KI}}=\frac{n_{\text{KI}}}{V_{\text{solution}}}=\frac{\text{7}\text{.9115}\times \text{10}^{\text{-2}}\text{ mol }}{\text{2}\text{.50 }\times \text{10}^{\text{-1}}\text{ dm}^{\text{3}}}=\text{3}\text{.1645 }\times 10^{^{\text{-1}}}\text{mol dm}^{\text{-3}}\]
Note that the definition of concentration is entirely analogous to the definitions of density, molar mass, and stoichiometric ratio that we have previously encountered. Concentration will serve as a conversion factor relating the volume of solution to the amount of dissolved solute.
\[\text{Volume of solution}\overset{concentration}{\longleftrightarrow}\text{amount of solute} ~~~~~~~~~~~~~ V\overset{c}{\longleftrightarrow}n\]
Because the volume of a liquid can be measured quickly and easily, concentration is a much-used quantity. The next two examples show how this conversion factor may be applied to commonly encountered solutions in which water is the solvent (
aqueous solutions).
Example \(\PageIndex{2}\) : Amount of HCl
An aqueous solution of HCl [represented or written HCl(
aq)] has a concentration of 0.1396 mol dm –3. If 24.71 cm 3 (24.71 ml) of this solution is delivered from a buret, what amount of HCl has been delivered? Solution
Using concentration as a conversion factor, we have
\[V \text{ } \rightarrow {c} \text{ } n \]
\[n_{\text{HCl}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.1396 mol}}{\text{1 dm}^{\text{3}}}\]
The volume units will cancel if we supply a unity factor to convert cubic centimeters to cubic decimeters:
\[ \begin{align} n_{\text{HCl}} & =\text{24} \text{.71 cm}^{\text{3}} \times \frac{\text{0} \text{.1396 mol}}{\text{1 dm}^{\text{3}}} \times ( \frac{\text{1 dm}}{\text{10 cm}} )^{\text{3}} \\ & =\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.1396 mol}}{\text{1 dm}^{\text{3}}}\times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \\ & = 0.003 450 \text{ mol} \end{align}\]
The concentration units of moles per cubic decimeter are often abbreviated
M, pronounced molar. That is, a 0.1- M (one-tenth molar) solution contains 0.1 mol solute per cubic decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles and for writing textbook problems; however, when doing calculations, it is difficult to see that
\[\text{1 dm}^{\text{3}}\times \text{1 }\text{M}=\text{1mol}\]
Therefore we recommend that you
always write the units in full when doing any calculations involving solution concentrations. Also, it is sometimes easier to use the unit liter, which is equivalent to cubic deciliters:
\[\text{1 dm}^{\text{3}}\times \text{1 }\dfrac{\text{mol}}{\text{dm}^{\text{3}}}=\text{1mol}\]
\[\text{1 L}\times \text{1 }\dfrac{\text{mol}}{\text{L}}=\text{1mol}\]
Problems such as Example \(\PageIndex{2}\) are easier for some persons to solve if the solution concentration is expressed in millimoles per cubic centimeter (mmol cm
–3) or millimoles per ml (1 ml = 1 cm –3) instead of moles per cubic decimeter. Since the SI prefix m means 10 –3, 1 mmol = 10 –3 mol, and
\[\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{1 dm}^{\text{3}}} ~ \times ~ \dfrac{\text{1 dm}^{\text{3}}}{\text{1 L}} ~ = ~ \dfrac{\text{1 mol}}{\text{L}}\]
\[\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{L}} ~ \times ~ \dfrac{\text{10}^{\text{-3}}\text{ L}}{\text{1 ml}} ~ \times ~ \dfrac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~ = ~ \dfrac{\text{1 mmol}}{\text{1 ml}}\]
\[\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{1 dm}^{\text{3}}} \times \dfrac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \times \dfrac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~ = ~ \dfrac{\text{1 mmol}}{\text{1 cm}^{\text{3}}} \]
Thus a concentration of 0.1396 mol dm
–3 (0.1396 M) can also be expressed as 0.1396 mmol cm –3, 0.1396 mol/L or 0.1396 mmol/mL. Expressing the concentration this way is very convenient when dealing with laboratory glassware calibrated in milliliters or cubic centimeters.
Example \(\PageIndex{3}\): Mass of NaOH
Exactly 25.0 ml NaOH solution whose concentration is 0.0974
M was delivered from a pipet. What amount of NaOH was present? What mass of NaOH would remain if all the water evaporated? Solution: a) Since 0.0974 M means 0.0974 mol dm –3, or 0.0974 mmol cm –3, we choose the latter, more convenient quantity as a conversion factor:
\[n_{\text{NaOH}}=\text{25}\text{.0 cm}^{\text{3}}\times \frac{\text{0}\text{.0974 mol}}{\text{1 cm}^{\text{3}}}=\text{2}\text{.44 mmol}=\text{2}\text{.44}\times 10^{\text{-3}}\text{ mol}\]
b) Using molar mass, we obtain
\[m_{\text{NaOH}}=\text{2}\text{.44}\times 10^{\text{-3}}\text{ mol}\times \frac{\text{40}\text{.01 g}}{\text{1 mol}}=9.\text{76}\times 10^{\text{-2}}\text{g}\]
The symbols \(n_{\ce{NaOH}}\) and \(m_{\ce{NaOH}}\) refer to the amount and mass of the
solute \(\ce{NaOH}\), respectively. They do not refer to the solution. If we wanted to specify the mass of aqueous \(\ce{NaOH}\) solution, the symbol \(m_{\ce{NaOH(aq)}}\) could be used.
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Suppose that we have finite extensions $F \subset L \subset M$ and $\sigma \in Gal(M/F)$ and assume that $F \subset L$ is radical. Prove that $F \subset \sigma L$ is also radical.
Since the extension $F \subset L$ is radical, there exist $\alpha_1, \dots , \alpha_n \in \overline{F}$ such that
$i) \ L =F( \alpha_1, \dots , \alpha_n)$
$ii) \ \alpha_1$is a root of the polynomial $t^{l_1} - a_1 \in F[t]$
$iii)$ For $2 \leq i \leq n, \alpha_i$ is a root of $t^{l_i}-a_i \in F(\alpha_1, \dots , \alpha_{i-1})[t]$
Now to prove that $F\subset \sigma L$ is radical, I started by taking $\sigma(\alpha_1),\dots ,\sigma(\alpha_n)$ and showing that they satisfy $i), ii)$ and $iii)$. For $i)$, since $F \subset L$ is radical,
$$\sigma L = \sigma F(\alpha_1, \, \dots, \alpha_n) = F(\sigma(\alpha_1,\dots,\alpha_n))$$
because $\sigma$ leaves $F$ fixed. Now to prove $ii)$, I need to show that $\sigma(\alpha_1)$ is a root of $t^{l_1}-a_1$. But since $\sigma(a_i)=a_i$ because $a_i \in F$, then
$$(\sigma(\alpha_1))^{l_1}-a_1= \sigma(\alpha^{l_1})-\sigma(a_1)=\sigma(\alpha^{l_1}-a_1)=\sigma(0)=0$$
But I'm stuck on proving $iii)$ (considering that the arguments above are tru, which I'm not particularly convinced of). So if anyone could give me any hints or show me the correct way to prove this, I'd be grateful. Thanks in advance!
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The general theme is game dynamics leading to equilibrium concepts.The plan is to deal with the following topics (all concepts will be defined, and proofs / proof outlines will be provided):(1) An integral approach to the construction of calibrated forecasts and their use for Nash equilibrium dynamics.(2) Blackwell's Approachability Theorem and its use for correlated equilibrium dynamics (regret-matching).(3) Communication complexity and its use for the speed of convergence of uncoupled dynamics.
Title:On the evaluation of sums of periodic GaussiansAbstract:Discrete sums of the form$\sum_{k=1}^N q_k \cdot \exp\left( -\frac{t – s_k}{2 \cdot \sigma^2} \right)$where $\sigma>0$ and $q_1, \dots, q_N$ are real numbers and$s_1, \dots, s_N$ and $t$ are vectors in $R^d$,are frequently encountered in numerical computations across a variety of fields.We describe an algorithm for the evaluation of such sums under periodic boundary conditions, provide a rigorous error analysis, and discuss its implications on the computational cost and choice of parameters.
The geodesic flow on a normal cover of a compact hyperbolic surface admits a "random walk" on the group of decks transformations $G$. In this talk, I'll provide some recent results which connect this walk to the geometric properties of the cover and $G$.
In the last 35 years, geometric flows have proven to be a powerful tool in geometry and topology. The Mean Curvature Flow is, in many ways, the most natural flow for surfaces in Euclidean space. In this talk, which will assume no prior knowledge, I will illustrate how mean curvature flow could be used to address geometric questions.
Manchester Building (Hall 2), Hebrew University Jerusalem
Let u be a harmonic function on the plane. The Liouville theorem claims that if |u| is bounded on the whole plane, then u is identically constant. It appears that if u is a harmonic function on the lattice Z^2, and |u| < 1 on 99,99% of Z^2, then u is a constant function. Based on a joint work with A. Logunov, Eu. Malinnikova and M. Sodin.
Speaker: Benny Sudakov, ETH, ZurichTitle: Subgraph statisticsAbstract:Consider integers $k,\ell$ such that $0\le \ell \le \binom{k}2$. Givena large graph $G$, what is the fraction of $k$-vertexsubsets of $G$ which span exactly $\ell$ edges? When $G$ is empty orcomplete, and $\ell$ is zero or $\binom k 2$,this fraction can be exactly 1. On the other hand if $\ell$ is not onethese extreme values, then by Ramsey's theorem, thisfraction is strictly smaller than 1.The systematic study of the above question was recently initiated by
Manchester Building (Hall 2), Hebrew University Jerusalem
The proof of decoupling grew out of an area of Fourier analysis called restriction theory. In this talk, we will describe some of the basic problems and tools of restriction theory, especially wave packets, which are a crucial idea in the proof of decoupling.
Title: Local root numbers for Heisenberg representations
Abstract: On the Langlands program, explicit computation of the local root numbers (or epsilon factors) for Galois representations is an integral part. But for arbitrary Galois representation of higher dimension, we do not have explicit formula for local root numbers. In our recent work (joint with Ernst-Wilhelm Zink) we consider Heisenberg representation (i.e., it represents commutators by scalar matrices) of the Weil
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I am using the usual estimator for kurtosis, $$\hat{K}=\frac{\hat{\mu}_4}{\hat{\sigma}^4}$$, but I notice that even small 'outliers' in my empirical distribution, i.e. small peaks far from the center, affect it tremendously. Is there a kurtosis estimator which is more robust?
There are several. You will find an exhaustive comparison in this link to an ungated version of the paper (proper reference at the bottom of this answer).
Because of the constraints of the problem, the breakdown of the most robust of these algorithms (the L/RMC) is at most 12.5%. An advantage to the the L/RMC is that it is based on quantiles and remain interpretable even when the underlying distribution has no moments. Another advantage is that it does not assume symmetry of the distribution of the uncontaminated part of the data to measure tail weight: in fact, the algorithm returns two numbers: the RMC for the right tail weight and the LMC for the left tail weight.
The robustness of an estimator can be measured by its breakdown point.However, the notion of breakdown point is a complicated one in this context. Intuitively, it means that an adversary would need to control at least 12.5% of your sample to make this estimator take on
arbitrary values (that is to be understood as an arbitrary value within the range of values that the estimator can return, since the measure of tail weight is always in $[0,1]$ by construction: no amount of contamination can for example cause the algorithm to return -1!). In practice, one finds that one can replace about 5% of the sample with even very pathological outliers without causing the most affected of the estimates (there are always two) to depart too much from the value it had on the uncontaminated sample.
The L/RMC is also widely implemented. For example you can find an R implementation here. As explained in the article linked above, to compute the L/RMC, you need to compute the MC (the estimator implemented in the link) separately on the left and right half of you data. Here, (left) right half are the sub-samples formed of the observation (smaller) larger than the median of your original sample.
Brys, Hubert, Struyf. (2006). Robust Measures of Tail Weight.
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I came across John Duffield Quantum Computing SE via this hot question. I was curious to see an account with 1 reputation and a question with hundreds of upvotes.It turned out that the reason why he has so little reputation despite a massively popular question is that he was suspended.May I ...
@Nelimee Do we need to merge? Currently, there's just one question with "phase-estimation" and another question with "quantum-phase-estimation". Might we as well use just one tag? (say just "phase-estimation")
@Blue 'merging', if I'm getting the terms right, is a specific single action that does exactly that and is generally preferable to editing tags on questions. Having said that, if it's just one question, it doesn't really matter although performing a proper merge is still probably preferable
Merging is taking all the questions with a specific tag and replacing that tag with a different one, on all those questions, on a tag level, without permanently changing anything about the underlying tags
@Blue yeah, you could do that. It generally requires votes, so it's probably not worth bothering when only one question has that tag
@glS "Every hermitian matrix satisfy this property: more specifically, all and only Hermitian matrices have this property" ha? I though it was only a subset of the set of valid matrices ^^ Thanks for the precision :)
@Nelimee if you think about it it's quite easy to see. Unitary matrices are the ones with phases as eigenvalues, while Hermitians have real eigenvalues. Therefore, if a matrix is not Hermitian (does not have real eigenvalues), then its exponential will not have eigenvalues of the form $e^{i\phi}$ with $\phi\in\mathbb R$. Although I'm not sure whether there could be exceptions for non diagonalizable matrices (if $A$ is not diagonalizable, then the above argument doesn't work)
This is an elementary question, but a little subtle so I hope it is suitable for MO.Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$.The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda_iI$, and we have the Jordan canonical form:$$ J = \begin...
@Nelimee no! unitarily diagonalizable matrices are all and only the normal ones (satisfying $AA^\dagger =A^\dagger A$). For general diagonalizability if I'm not mistaken onecharacterization is that the sum of the dimensions of the eigenspaces has to match the total dimension
@Blue I actually agree with Nelimee here that it's not that easy. You get $UU^\dagger = e^{iA} e^{-iA^\dagger}$, but if $A$ and $A^\dagger$ do not commute it's not straightforward that this doesn't give you an identity
I'm getting confused. I remember there being some theorem about one-to-one mappings between unitaries and hermitians provided by the exponential, but it was some time ago and may be confusing things in my head
@Nelimee if there is a $0$ there then it becomes the normality condition. Otherwise it means that the matrix is not normal, therefore not unitarily diagonalizable, but still the product of exponentials is relatively easy to write
@Blue you are right indeed. If $U$ is unitary then for sure you can write it as exponential of an Hermitian (time $i$). This is easily proven because $U$ is ensured to be unitarily diagonalizable, so you can simply compute it's logarithm through the eigenvalues. However, logarithms are tricky and multivalued, and there may be logarithms which are not diagonalizable at all.
I've actually recently asked some questions on math.SE on related topics
@Mithrandir24601 indeed, that was also what @Nelimee showed with an example above. I believe my argument holds for unitarily diagonalizable matrices. If a matrix is only generally diagonalizable (so it's not normal) then it's not true
also probably even more generally without $i$ factors
so, in conclusion, it does indeed seem that $e^{iA}$ unitary implies $A$ Hermitian. It therefore also seems that $e^{iA}$ unitary implies $A$ normal, so that also my argument passing through the spectra works (though one has to show that $A$ is ensured to be normal)
Now what we need to look for is 1) The exact set of conditions for which the matrix exponential $e^A$ of a complex matrix $A$, is unitary 2) The exact set of conditions for which the matrix exponential $e^{iA}$ of a real matrix $A$ is unitary
@Blue fair enough - as with @Semiclassical I was thinking about it with the t parameter, as that's what we care about in physics :P I can possibly come up with a number of non-Hermitian matrices that gives unitary evolution for a specific t
Or rather, the exponential of which is unitary for $t+n\tau$, although I'd need to check
If you're afraid of the density of diagonalizable matrices, simply triangularize $A$. You get $$A=P^{-1}UP,$$ with $U$ upper triangular and the eigenvalues $\{\lambda_j\}$ of $A$ on the diagonal.Then$$\mbox{det}\;e^A=\mbox{det}(P^{-1}e^UP)=\mbox{det}\;e^U.$$Now observe that $e^U$ is upper ...
There's 15 hours left on a bountied question, but the person who offered the bounty is suspended and his suspension doesn't expire until about 2 days, meaning he may not be able to award the bounty himself?That's not fair: It's a 300 point bounty. The largest bounty ever offered on QCSE. Let h...
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Analysis of a phase field Navier-Stokes vesicle-fluid interaction model
1.
Department of Mathematics, Pennsylvania State University, University Park, PA 16802, United States
2.
Department of Mathematics, The Pennsylvania State University, University Park, PA 16802
Keywords:Weak Solution, Phase Field, Bending Elastic Energy., Existence, Uniqueness, Navier-Stokes, Fluid Interaction, Vesicle Membrane, Well-posedness. Mathematics Subject Classification:Primary: 58F1. Citation:Qiang Du, Manlin Li, Chun Liu. Analysis of a phase field Navier-Stokes vesicle-fluid interaction model. Discrete & Continuous Dynamical Systems - B, 2007, 8 (3) : 539-556. doi: 10.3934/dcdsb.2007.8.539
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Well-posedness results for the Navier-Stokes equations in the rotational framework.
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Semigroup well-posedness of a linearized, compressible fluid with an elastic boundary.
[8]
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Global well-posedness for inhomogeneous Navier-Stokes equations with logarithmical hyper-dissipation.
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Global well-posedness of axisymmetric Navier-Stokes equations with one slow variable.
[11]
Yoshihiro Shibata.
Local well-posedness of free surface problems for the Navier-Stokes equations in a general domain.
[12] [13]
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On the very weak solution for the Oseen and Navier-Stokes equations.
[14]
Chao Deng, Xiaohua Yao.
Well-posedness and ill-posedness for the 3D generalized Navier-Stokes equations in $\dot{F}^{-\alpha,r}_{\frac{3}{\alpha-1}}$.
[15]
Michele Colturato.
Well-posedness and longtime behavior for a singular phase field system with perturbed phase dynamics.
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Henry Jacobs, Joris Vankerschaver.
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Well-posedness results for phase field systems with memory effects in the order parameter dynamics.
[18]
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Well-posedness and optimal control of a hemivariational inequality for nonstationary Stokes fluid flow.
[19] [20]
2018 Impact Factor: 1.008
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Degree $n$ : $45$ Transitive number $t$ : $12$ Group : $C_5\times C_9:C_3$ Parity: $1$ Primitive: No Nilpotency class: $2$ Generators: (1,8,13,20,26,31,38,45,6,10,18,23,30,35,40,2,9,15,19,25,33,39,44,5,12,16,22,29,34,42,3,7,14,21,27,32,37,43,4,11,17,24,28,36,41), (1,20,38,10,30,2,19,39,12,29,3,21,37,11,28)(4,23,42,13,33)(5,24,40,14,31)(6,22,41,15,32)(7,25,45,17,34,9,26,43,16,35,8,27,44,18,36) $|\Aut(F/K)|$: $15$
|G/N| Galois groups for stem field(s) 3: $C_3$ x 4 5: $C_5$ 9: $C_3^2$ 27: $C_9:C_3$
Resolvents shown for degrees $\leq 10$
There are no siblings with degree $\leq 10$ Data on whether or not a number field with this Galois group has arithmetically equivalent fields has not been computed.
There are 55 conjugacy classes of elements. Data not shown.
Order: $135=3^{3} \cdot 5$ Cyclic: No Abelian: No Solvable: Yes GAP id: [135, 4]
Character table: Data not available.
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Definition:Partial Derivative/Real Analysis Contents Definition
Let $U\subset\R^n$ be an open set.
Let $f : U \to \R$ be a real-valued function.
Let $a = (a_1,\ldots,a_n)^\intercal \in U$.
Let $f$ be differentiable at $a$.
Let $i\in\{1,\ldots, n\}$.
The
partial derivative of $f$ with respect to $x_i$ at $a$ is denoted and defined as: $\dfrac {\partial f}{\partial x_i}(a) := g_i'(a_i)$
where:
$g_i$ is the real function defined as $g \left({x_i}\right) = f \left({a_1, \ldots, x_i, \dots, a_n}\right)$ $g'(a_i)$ is the derivative of $g$ at $a_i$.
The
$i$th partial derivative of $f$ at $a$ is the limit: $\dfrac{\partial f}{\partial x_i}(a) = \displaystyle \lim_{x_i \to a_i} \frac {f\left( a_1,\ldots, x_i, \ldots,a_n\right) - f\left(a\right)}{x_i - a}$ When spoken, $\dfrac {\partial y}{\partial x}$, "the partial derivative of $y$ with respect to $x$" is often shortened to "partial $y$ partial $x$", or "del $y$ del $x$".
Let $U\subset\R^n$ be an open set.
Let $f : U \to \R$ be a real-valued function.
Let $f$ be differentiable in $U$.
$\dfrac {\partial f} {\partial x_i}$ $\dfrac {\partial} {\partial x_i} f$ $f_{x_i} \left({\mathbf x}\right)$ $f_{x_i} \left({x_1, x_2, \cdots, x_n}\right)$ $f_{x_i}$ $\partial_{x_i}f$ $\partial_i f$ $D_i f$ $\dfrac {\partial z} {\partial x_i}$ $z_{x_i}$
where $z = f \left({x_1, x_2, \cdots, x_n}\right)$.
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I'm very confused about some contradicatory statements, and I hope someone can help me clarify this.
Let $\Gamma$ be a congruence subgroup. It is well known that modular forms of weight $k$ for $\Gamma$ can be constructed as global sections of a sheaf $\mathcal{G}_k$ on the modular curve $X(\Gamma)$. If $\Gamma$ contains neither -1 nor elliptic elements, then the sheaf $\mathcal{G}_k$ is actually invertible: there exists a universal elliptic curve $\mathcal{E}$ over $X(\Gamma)$, and $\mathcal{G}_k$ is the $k$th tensor power of the pullback of $\Omega_{\mathcal{E}/X(\Gamma)}$ via the zero section.
If $\Gamma$ contains elliptic elements, then there is no universal family above $X$. However, one can still construct the sheaf $\mathcal{G}_k$, this construction is described in Diamond and Im, Section 12.1. Take a normal congruence subgroup $\Gamma' \subset \Gamma$ not containing $-1$ or an elliptic element. In terms of modular curves, we have the following situation (if I understand correctly). If $G = \Gamma / \Gamma'$, then $X(\Gamma) = X(\Gamma')/G$ and we have a quotient map $\pi : X(\Gamma') \rightarrow X(\Gamma)$. Let $\mathcal{G}'_k$ be the invertible sheaf as described above. Then for any open $V \subset X(\Gamma)$, $G$ acts on $\mathcal{G'}(\pi^{-1}(V))$, so define $\mathcal{G}_k(V) = \mathcal{G}'_k(\pi^{-1}(V))^G$. This gives the sheaf we want.
The article of Diamond and Im says that this sheaf is invertible unless $-1 \in \Gamma$ and $k$ is odd. (without giving any justification). However I fail to see how this can be the case even for $k$ even. For example, say we construct this sheaf $\mathcal{G}_k$ for modular forms of level $1$ of weight $k \geq 6$ such that $k \equiv 2 \pmod{4}$. Suppose $\mathcal{G}_k$ is invertible. We know from the theory of invertible sheaves that if $\mathcal{G}_k$ is an invertible sheaf on a curve $X$ of genus $g$ such that $\deg \mathcal{G}_k \geq 2g$, then $\mathcal{G}_k$ has no base points. But in this case we find that any modular form of weight $k$ vanishes at the elliptic point $SL_2(\mathbb{Z})i$. A similar argument can be made with any group $\Gamma_0(N)$ that has elliptic elements.
Question 1 So is this an error in the article? And if the sheaves $\mathcal{G}_k$ are not in general invertible, can we still say they are coherent? Question 2 Related to this question, I've been trying to look at the situation of a $G$-equivariant invertible sheaf on an affine curve (noetherian, integral, etc). Say the curve is $X = Spec A$, $G$ a finite group of automorphisms of $X$, and $\mathcal{L}$ a $G$-equivariant invertible sheaf on $X$. Let $Y = Spec(A^G)$ be the quotient curve, and consider the $G$-invariant pushforward $\mathcal{F} = \pi_{\ast}(\mathcal{L})^G$. If I understand things correctly, the algebraic description is then: we have a finitely generated $A$-module $M$, and an action of $G$ on $M$ such that:
$g(ax) = g(a)g(x)$ for all $a \in A$, $x \in M$, and $g \in G$.
Then the module $N = M^G$ is a finitely generated $A^G$-module. Now i'm asking, will $N$ be locally free? If the action of $G$ is free, then I know that it will be (as explained in Mumford, Abelian Varieties).
Suppose there is a prime $\mathfrak{q} \in A$ such that $g(\mathfrak{q}) = \mathfrak{q}$ for all $g \in G$, and let $\mathfrak{p} = \mathfrak{q}\cap A^G$ be the prime under it. Then I think $(M_{\mathfrak{q}})^G = N_{\mathfrak{p}}$, which means that $N$ is locally free hence also projective. This seems to say that $\pi_{\ast}(\mathcal{L})^G$ is invertible. So this is contradictory to what I'm asking about in Question 1.
I'm making a mistake (or several!) somewhere. Can someone point them out to me?
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A well-known theorem by Martin asserts that the degrees of maximal sets are precisely the high recursively enumerable degrees, and the same is true with ‘maximal’ replaced by ‘dense simple’, ‘r-maximal’, ‘strongly hypersimple’ or ‘finitely strongly hypersimple’. Many other constructions can also be carried out in any given high r. e. degree, for instance r-maximal or hyperhypersimple sets without maximal supersets . In this paper questions of this type are considered systematically. Ultimately it is shown that every conjunction of simplicity- (...) and non-extensibility properties can be accomplished, unless it is ruled out by well-known, elementary results. Moreover, each construction can be carried out in any given high r. e. degree, as might be expected. For instance, every high r. e. degree contains a dense simple, strongly hypersimple set A which is contained neither in a hyperhypersimple nor in an r-maximal set. The paper also contains some auxiliary results, for instance: every r. e. set B can be transformed into an r. e. set A such that A has no dense simple superset, the transformation preserves simplicity- or non-extensibility properties as far as this is consistent with , and A [TRIPLE BOND]TB if B is high, and A ≥TB otherwise. Several proofs involve refinements of known constructions; relationships to earlier results are discussed in detail. (shrink)
G. E. Moore's ‘A Defence of Common Sense’ has generated the kind of interest and contrariety which often accompany what is new, provocative, and even important in philosophy. Moore himself reportedly agreed with Wittgenstein's estimate that this was his best article, while C. D. Broad has lamented its very great but largely unfortunate influence. Although the essay inspired Wittgenstein to explore the basis of Moore's claim to know many propositions of common sense to be true, A. J. Ayer judges its (...) enduring value to lie in provoking a more sophisticated conception of the very type of metaphysics which disputes any such unqualified claim of certainty. (shrink)
In this article we show that it is possible to completely classify the degrees of r.e. bases of r.e. vector spaces in terms of weak truth table degrees. The ideas extend to classify the degrees of complements and splittings. Several ramifications of the classification are discussed, together with an analysis of the structure of the degrees of pairs of r.e. summands of r.e. spaces.
We construct a nonlow2 r.e. degree d such that every positive extension of embeddings property that holds below every low2 degree holds below d. Indeed, we can also guarantee the converse so that there is a low r.e. degree c such that that the extension of embeddings properties true below c are exactly the ones true belowd.Moreover, we can also guarantee that no b ≤ d is the base of a nonsplitting pair.
In this philosophy classic, which was first published in 1951, E. R. Dodds takes on the traditional view of Greek culture as a triumph of rationalism. Using the analytical tools of modern anthropology and psychology, Dodds asks, "Why should we attribute to the ancient Greeks an immunity from 'primitive' modes of thought which we do not find in any society open to our direct observation?" Praised by reviewers as "an event in modern Greek scholarship" and "a book which it would (...) be difficult to over-praise," _The Greeks and the Irrational _was Volume 25 of the Sather Classical Lectures series. (shrink)
By constructing a maximal incomplete d.r.e. degree, the nondensity of the partial order of the d.r.e. degrees is established. An easy modification yields the nondensity of the n-r.e. degrees and of the ω-r.e. degrees.
By constructing a maximal incomplete d.r.e. degree, the nondensity of the partial order of the d.r.e. degrees is established. An easy modification yields the nondensity of the n-r.e. degrees and of the ω-r.e. degrees.
We investigate dependence of recursively enumerable graphs on the equality relation given by a specific r.e. equivalence relation on ω. In particular we compare r.e. equivalence relations in terms of graphs they permit to represent. This defines partially ordered sets that depend on classes of graphs under consideration. We investigate some algebraic properties of these partially ordered sets. For instance, we show that some of these partial ordered sets possess atoms, minimal and maximal elements. We also fully describe the isomorphism (...) types of some of these partial orders. (shrink)
The lattice of r.e. equivalence relations has not been carefully examined even though r.e. equivalence relations have proved useful in logic. A maximal r.e. equivalence relation has the expected lattice theoretic definition. It is proved that, in every pair of r.e. nonrecursive Turing degrees, there exist maximal r.e. equivalence relations which intersect trivially. This is, so far, unique among r.e. submodel lattices.
We use some simple facts about the wtt-degrees of r.e. sets together with a construction to answer some questions concerning the join and meet operators in the r.e. degrees. The construction is that of an r.e. Turing degree a with just one wtt-degree in a such that a is the join of a minimal pair of r.e. degrees. We hope to illustrate the usefulness of studying the stronger reducibility orderings of r.e. sets for providing information about Turing reducibility.
Important examples of $\Pi^0_1$ classes of functions $f \in {}^\omega\omega$ are the classes of sets (elements of ω 2) which separate a given pair of disjoint r.e. sets: ${\mathsf S}_2(A_0, A_1) := \{f \in{}^\omega2 : (\forall i < 2)(\forall x \in A_i)f(x) \neq i\}$ . A wider class consists of the classes of functions f ∈ ω k which in a generalized sense separate a k-tuple of r.e. sets (not necessarily pairwise disjoint) for each k ∈ ω: ${\mathsf S}_k(A_0,\ldots,A_k-1) := (...) \{f \in {}^\omega k : (\forall i < k) (\forall x \in A_i) f(x) \neq i\}$ . We study the structure of the Medvedev degrees of such classes and show that the set of degrees realized depends strongly on both k and the extent to which the r.e. sets intersect. Let ${\mathcal S}^m_k$ denote the Medvedev degrees of those ${\mathsf S}_k(A_0,\ldots,A_{k-1})$ such that no m + 1 sets among A 0,...,A k-1 have a nonempty intersection. It is shown that each ${\mathcal S}^m_k$ is an upper semi-lattice but not a lattice. The degree of the set of k-ary diagonally nonrecursive functions $\mathsf{DNR}_k$ is the greatest element of ${\mathcal S}^1_k$ . If 2 ≤ l < k, then 0 M is the only degree in ${\mathcal S}^1_l$ which is below a member of ${\mathcal S}^1_k$ . Each ${\mathcal S}^m_k$ is densely ordered and has the splitting property and the same holds for the lattice ${\mathcal L}^m_k$ it generates. The elements of ${\mathcal S}^m_k$ are exactly the joins of elements of ${\mathcal S}^1_i$ for $\lceil{k \over m}\rceil \leq i \leq k$. (shrink)
We show that the Π 4 -theory of the partial order of recursively enumerable weak truth-table degrees is undecidable, and give a new proof of the similar fact for r.e. T-degrees. This is accomplished by introducing a new coding scheme which consists in defining the class of finite bipartite graphs with parameters.
We show that the $\Pi_4$-theory of the partial order of recursively enumerable weak truth-table degrees is undecidable, and give a new proof of the similar fact for r.e. T-degrees. This is accomplished by introducing a new coding scheme which consists in defining the class of finite bipartite graphs with parameters.
This paper analyzes several properties of infima in Dn, the n-r.e. degrees. We first show that, for every n> 1, there are n-r.e. degrees a, b, and c, and an -r.e. degree x such that a < x < b, c and, in Dn, b c = a. We also prove a related result, namely that there are two d.r.e. degrees that form a minimal pair in Dn, for each n < ω, but that do not form a minimal pair (...) in Dω. Next, we show that every low r.e. degree branches in the d.r.e. degrees. This result does not extend to the low2 r.e. degrees. We also construct a non-low r.e. degree a such that every r.e. degree b a branches in the d.r.e. degrees. Finally we prove that the nonbranching degrees are downward dense in the d.r.e. degrees. (shrink)
A recursive algebra is a structure for which A is a recursive set of numbers and the Fi are uniformly recursive operations. We define an r.e. quotient algebra to be the quotient by an r.e. congruence .We say that is recursively stable among r.e. quotient algebras if, for each r.e. quotient algebra and each isomorphism from onto ′, the set {a,baA,bB and =[b]′} is r.e.We shall consider examples of recursive stability. Then, assuming that has a recursive existential diagram, we show (...) that the task of determining its recursive stability among r.e. quotient algebras can be reduced to a more routine consideration of syntactical conditions. To this result, we provide a counter-example which demonstrates the necessity of having a recursive existential diagram. This result and counter-example are on similar lines to ones obtained by Goncharov , for the recursive stability of recursive structures. (shrink)
We study the global properties of [Formula: see text], the Turing degrees of the n-r.e. sets. In Theorem 1.5, we show that the first order of [Formula: see text] is not decidable. In Theorem 1.6, we show that for any two n and m with n < m, [Formula: see text] is not a Σ1-substructure of [Formula: see text].
Let d be a Turing degree containing differences of recursively enumerable sets (d.r.e.sets) and R[d] be the class of less than d r.e. degrees in whichd is relatively enumerable (r.e.). A.H.Lachlan proved that for any non-recursive d.r.e. d R[d] is not empty. We show that the r.e. degree defined by Lachlan for a d.r.e.set $D\in$ d is just the minimum degree in which D is r.e. Then we study for a given d.r.e. degree d class R[d] and show that there (...) exists a d.r.e.d such that R d] has a minimum element $>$ 0. The most striking result of the paper is the existence of d.r.e. degrees for which R[d] consists of one element. Finally we prove that for some d.r.e. d R[d] can be the interval [a,b] for some r.e. degrees a,b, a $<$ b $<$ d. (shrink)
A set of natural numbers is called d.r.e. if it may be obtained from some recursively enumerable set by deleting the numbers belonging to another recursively enumerable set. Sacks showed that for each non-recursive recursively enumerable set A there are disjoint recursively enumerable sets B, C which cover A such that A is recursive in neither A ∩ B nor A ∩ C. In this paper, we construct a counterexample which shows that Sacks's theorem is not in general true when (...) A is d.r.e. rather than r.e. (shrink)
Lachlan observed that the infimum of two r.e. degrees considered in the r.e. degrees coincides with the one considered in the ${\Delta_2^0}$ degrees. It is not true anymore for the d.r.e. degrees. Kaddah proved in (Ann Pure Appl Log 62(3):207–263, 1993) that there are d.r.e. degrees a, b, c and a 3-r.e. degree x such that a is the infimum of b, c in the d.r.e. degrees, but not in the 3-r.e. degrees, as a < x < b, c. In (...) this paper, we extend Kaddah’s result by showing that such a structural difference occurs densely in the r.e. degrees. Our result immediately implies that the isolated 3-r.e. degrees are dense in the r.e. degrees, which was first proved by LaForte. (shrink)
We construct by diagonalization a non-well-founded primitive recursive tree, which is well-founded for co-r.e. sets, provable in Σ1 0. It follows that the supremum of order-types of primitive recursive well-orderings, whose well-foundedness on co-r.e. sets is provable in Σ1 0, equals the limit of all recursive ordinals ω1 ck . RID=""ID="" Mathematics Subject Classification (2000): 03B30, 03F15 RID=""ID="" Supported by the Deutschen Akademie der Naturforscher Leopoldina grant #BMBF-LPD 9801-7 with funds from the Bundesministerium für Bildung, Wissenschaft, Forschung und Technologie. RID=""ID="" (...) I would like to thank A. SETZER for his hospitality during my stay in Uppsala in December 1998 – these investigations are inspired by a discussion with him; S. BUSS for his hospitality during my stay at UCSD and for valuable remarks on a previous version of this paper; and M. MÖLLERFELD for remarks on a previous title. (shrink)
R. E. Ewin has argued that corporations are moral persons, but Ewin describes them as being unable to think or to act in virtuous and vicious ways. Ewin thinks that their impoverished emotional life would not allow them to act in these ways. In this brief essay I want to challenge the idea that corporations cannot act virtuously. I begin by examining deficiencies in Ewin''s notion of corporate personhood. I argue that he effectively reduces corporations to the status of incompetent (...) patients. I shall make use of a richer notion of corporate personhood as I explore the logical relationship between corporate action and the quality of the corporate emotional life. After discussing an alternate methodology for making moral assessments of action I consider briefly two corporate disasters: the crash on Mt. Erebus, the Imperial Foods plant fire. These cases are used to show the inadequacy of Ewin''s thesis that only corporate managers are capable of displaying vice. (shrink)
Weakly semirecursive sets have been introduced by Jockusch and Owings . In the present paper their investigation is pushed forward by utilizing r.e. partial orderings, which turn out to be instrumental for the study of degrees of subclasses of weakly semirecursive sets.
Let β be an arbitrary limit ordinal. A β-r.e. set is l-finite iff all its β-r.e. subsets are β-recursive. The l-finite sets correspond to the ideal of finite sets in the lattice of r.e. sets. We give a characterization of l-finite sets in terms of their ordertype: a β-r.e. set is l-finite iff it has ordertype less than β * , the Σ 1 projectum of β.
We provide three new results about interpolating 2-r.e. or 2-REA degrees between given r.e. degrees: Proposition 1.13. If c h are r.e. , c is low and h is high, then there is an a h which is REA in c but not r.e. Theorem 2.1. For all high r.e. degrees h g there is a properly d-r.e. degree a such that h a g and a is r.e. in h . Theorem 3.1. There is an incomplete nonrecursive r.e. A (...) such that every set REA in A and recursive in 0′ is of r. e. degree. The first proof is a variation on the construction of Soare and Stob . The second combines highness with a modified version of the proof strategy of Cooper et al. . The third theorem is a rather surprising result with a somewhat unusual proof strategy. Its proof is a 0‴ argument that at times moves left in the tree so that the accessible nodes are not linearly ordered at each stage. Thus the construction lacks a true path in the usual sense. Two substitute notions fill this role: The true nodes are the leftmost ones accessible infinitely often; the semitrue nodes are the leftmost ones such that there are infinitely many stages at which some extension is accessible. Another unusual feature of the construction is that it involves using distinct priority orderings to control the interactions of different parts of the construction. (shrink)
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J. PANT
Articles written in Journal of Astrophysics and Astronomy
Volume 40 Issue 2 April 2019 Article ID 0009
We report optical observations of TGSS J1054 $+$ 5832, a candidate high-redshift ($z = 4.8 \pm 2$) steep-spectrum radio galaxy, in $r$ and $i$ bands, using the faint object spectrograph and camera mounted on 3.6-m Devasthal Optical Telescope (DOT). The source previously detected at 150 MHz from Giant Meterwave Radio Telescope (GMRT) and at 1420 MHz from Very Large Array has a known counterpart in near-infrared bands with $K$-band magnitude of AB 22. The source is detected in $i$-band with AB24.3 $\pm$ 0.2 magnitude in theDOT images presented here. The source remains undetected in the $r$-band image at a 2.5$\sigma$ depth of AB 24.4 mag over an $1.2^{\prime\prime}\times 1.2^{\prime\prime}$ aperture. An upper limit to $i−K$ color is estimated to be $\sim$2.3, suggesting youthfulness of the galaxy with active star formation. These observations highlight the importance and potential of the 3.6-mDOT for detections of faint galaxies.
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Consider a quadrilateral ABCDABCDABCD . Find the necessary and sufficient condition with proof so that there exists a point PPP in the interior of ABCDABCDABCD such that A(PAB)=A(PBC)=A(PCD)=A(PDA)A(PAB)=A(PBC)= A(PCD)= A(PDA)A(PAB)=A(PBC)=A(PCD)=A(PDA).
A( ) represents Area\text{A( ) represents Area}A( ) represents Area
Nice solutions are always welcome!
Note by Nihar Mahajan 4 years, 5 months ago
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Such a point PPP exists if and only if one diagonal bisects the area of the quadrilateral.
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Can you please provide the proof sir ?
Sir, thanks for your answer.But it may be more beneficial for us if you post a nice proof.
Now that you know the right condition, you should try proving it.
Sir, actually this problem is from the homework of our class, and our sir has a different condition-P exists iff one diagonal bisects the other.
Yeah , thats why I posted this as a note.
@Nihar Mahajan – Did you get a proof
@Pranav Kirsur – I got the proof for the condition- P exists iff one diagonal bisects the other.I did not get proof for the condition that Jon Hausmann has posted.
@Nihar Mahajan – yes,same
In a quadrilateral, the conditions "one diagonal bisects the area" and "one diagonal bisects the other diagonal" are equivalent.
@Jon Haussmann – Oh yes, thank you for pointing that out to me
@Calvin Lin @Brian Charlesworth @Trevor Arashiro@Pi Han Goh @Chew-Seong Cheong @Mehul Arora @Sharky Kesa @Sanjeet Raria @Sudeep Salgia @Ronak Agarwal @everyone help us.
Nihar, Tujhe pata hai bhai ki Mujhe Geometry nahi aati :3 Jo Bhi Ho. Thanks For @Mentioning Me :D
I mentioned you so that you will have something interesting in geometry. ;)
@Nihar Mahajan – Yeah, Thanks! :) :D :)
me , @Kalash Verma @Harsh Shrivastava @CH Nikhil are in need of a nice solution.Thanks!
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Given a $2\times2$ matrix, which entries are functions in the complex plane $$\hat{A}(z)=\left(\begin{array}{cc}a(z)&b(z)\\c(z)&d(z)\end{array}\right)$$Where $a(z),b(z),c(z)$ and $d(z)$ are functions in the complex plane. I would like to obtain two new matrices, denoted as $\hat{\phi}_+(z)$ and $\hat{\phi}_-(z)$, such $\hat{A}(z)=\hat{\phi}_+(z)\hat{\phi}_-(z)$. The meaning of the subindexes is related to the discrete Fourier transform. If we take $\hat{\phi}_+(z)$ $$\int dz e^{i k z}\hat{\phi}_+(z)=0,\;\;\forall\;k<0$$ Similar definitions can be given for $\hat{\phi}_-(z)$ with $k>0$. It means that the Fourier transform of the two functions is only non-zero in a half complex plane.
As far as I know, this is called Wiener-Hopf factorization. My questions are about the conditions that ensure the existence of such a factorization, and how this factorization can be determined (references or examples may help at this point).
Thank you for any help!
Note: Suppose that the entries of $A(z)$ are continium and smooth, and the matrix is invertible a.e. ($\mbox{det}[\hat{A}(z)]\neq0,\forall z\in \mathbb{C}$) By the moment I don't care about pathological cases
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Definition:Derivative/Higher Derivatives/Second Derivative Definition
Hence $f'$ is defined on $I$ as the derivative of $f$.
Let $\xi \in I$ be a point in $I$.
Let $f'$ be differentiable at the point $\xi$.
Then the
second derivative $\map {f''} \xi$ is defined as: $\displaystyle f'' := \lim_{x \mathop \to \xi} \dfrac {\map {f'} x - \map {f'} \xi} {x - \xi}$ If $f'$ is differentiable, then it is said that $f$ is doubly differentiable, or twice differentiable. Also defined as
It may also be seen defined as:
$\displaystyle \map {f''} \xi = \lim_{h \mathop \to 0} \frac {\map {f'} {\xi + h} - \map {f'} \xi} h$
The
second derivative is variously denoted as: $\map {f''} \xi$ $D^2 \map f \xi$ $D_{xx} \map f \xi$ $\map {\dfrac {\d^2} {\d x^2} } \xi$ If $y = \map f x$, then it can also expressed as $y''$: $y'' := \map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} }$
and written:
$\dfrac {\d^2 y} {\d x^2}$ Also see Sources 1937: Eric Temple Bell: Men of Mathematics... (previous) ... (next): Chapter $\text{VI}$: On the Seashore 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables... (previous) ... (next): $\S 13$: Higher Derivatives: $13.43$ 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach... (previous) ... (next): $\S 10.2$
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The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say
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Edit. As the OP points out, for his purpose it suffices to take the zero locus of a single (nonzero) partial derivative. So the OP produces a proper closed subset of $V$ containing the singular locus and having degree bounded by $(\text{deg}(V)-1)\text{deg}(V)$. Although this is not what the OP asks, there are cases where we need an upper bound on the degree of the singular locus (or at least the union of all components of the singular locus that have the maximum dimension). This often occurs for bounding the set of "bad characteristic" for some property of schemes over a field of (possibly) finite characteristic. The answer below gives an upper bound on the degree of the singular locus.
I am just rewriting the proof of Lemma 4.2.5 of the following as one answer. I learned of this from Fedor Bogomolov.
Jan Gutt
Hwang–Mok rigidity of cominuscule homogeneous varieties in positive characteristic PhD. thesis, 2013 https://arxiv.org/pdf/1305.5296.pdf
The original statement is for projective varieties, but the result for affine varieties follows by intersecting with affine space (a Zariski open subset of projective space).
Lemma [Jan Gutt, 2013 thesis, Lemma 4.2.5] For a purely $r$-dimensional closed subscheme $V$ of projective space $\mathbb{P}^n_k$ with degree $D>1$, if the zero scheme $S$ of the $r^\text{th}$ Fitting ideal of $\Omega_{V/k}$ has dimension $m$, then the corresponding $m$-cycle of $S$ has degree no greater than $D(D-1)^{r-m}$.
Proof. When $m$ equals $r$, then this just says that the $m$-dimensional cycle of $S$ has degree no greater than the degree of the $m$-dimensional cycle of $V$. Thus, without loss of generality, assume that $r>m$. Also, it suffices to prove the result when $k$ is algebraically closed. The proof uses Theorem 1.1 of the following.
MR0282975 (44 #209)
Mumford, David Varieties defined by quadratic equations. 1970 Questions on Algebraic Varieties (C.I.M.E., III Ciclo, Varenna, 1969) pp. 29–100 Edizioni Cremonese, Rome http://www.dam.brown.edu/people/mumford/alg_geom/papers/1970a--CIME-QuadEqns-DAM.pdf
Mumford proves that the ideal sheaf $I$ of $V$ is generated in degree $d$. More precisely, the linear system $H^0(\mathbb{P}^n_k,I(d))$ of sections $g$ of $\mathcal{O}(d)$ on $\mathbb{P}^n_k$ that vanish on $V$ has base locus that equals $V$ set-theoretically and that equals $V$ scheme-theoretically, at least on the dense open subset $V\setminus S$ of $V$.Thus, the common zero scheme in $V$ of the set of partial derivatives, $\partial g/\partial t$ (for varying homogeneous coordinates $t$) is contained in $S$ set-theoretically, and contains $S$ scheme-theoretically (since the Fitting ideal contains these partial derivatives, locally).
By Bertini’s theorem, for $r-m$ general polynomials $g = (g_1 , \dots , g_{r-m})$ in this linear system, for a general choice of homogeneous coordinates on $P^n_k$ and for a choice $t = (t_1 , \dots , t_{r-m} )$ of $r-m$ of these coordinates, the common zero scheme in $V$ of the $r-m$ partial derivative polynomials $\partial g_i /\partial t_i$ is $m$-dimensional and contains $S$. Since these partial derivatives are global sections of $\mathcal{O}(D − 1)$, the degree bound follows.
QED.
Will Sawin's Examples. Let $V$ be a subvariety that spans a linear subspace $\mathbb{P}^{r+1}_k \subset \mathbb{P}^n_k$ and that equals a degree-$D$ hypersurface in this linear space with defining polynomial $g=t_{m+1}^D + \dots + t_{r+1}^D$. Assume that the integer $D$ is nonzero in $k$. The Fitting ideal is precisely defined by $t_{m+1}^{D-1},\dots,t_{r+1}^{D-1}$ and the linear polynomials $t_{r+2},\dots,t_n$. Thus, the degree equals $(D-1)^{r+1-m}$, which is close to $(D-1)^{r-m}D$.
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The Ramanujan Summation of some infinite sums is consistent with a couple of sets of values of the Riemann zeta function. We have, for instance, $$\zeta(-2n)=\sum_{n=1}^{\infty} n^{2k} = 0 (\mathfrak{R}) $$ (for non-negative integer $k$) and $$\zeta(-(2n+1))=-\frac{B_{2k}}{2k} (\mathfrak{R})$$ (again, $k \in \mathbb{N} $). Here, $B_k$ is the $k$'th Bernoulli number. However, it does not hold when, for example, $$\sum_{n=1}^{\infty} \frac{1}{n}=\gamma (\mathfrak{R})$$ (here $\gamma$ denotes the Euler-Mascheroni Constant) as it is not equal to $$\zeta(1)=\infty$$. Question: Are the first two examples I stated the only instances in which the ramanujan summation of some infinite series coincides with the values of the Riemann zeta function?
The answer can be obtained with the following interpretation of the Ramanujan summation:
More recently, the use of $C(1)$ has been proposed as Ramanujan's summation, since then it can be assured that one series admits one and only one Ramanujan's summation, defined as the value in 1 of the only solution of the difference equation $R(x) − R(x + 1) = f(x)$ that verifies the condition $\int_1^2 R(x)dx=0$.
The function $R(x)=\zeta(z,x)+C\;$ satisfy $R(x) − R(x + 1) = x^{-z}$, $x>0$, $z\in \mathbb C$, $z\ne-1$, where $$ \zeta(z,x)=\sum_{k=0}^\infty\frac1{(k+x)^z} $$ is the Hurwitz zeta function (or its analytic continuation for $\Re z\le 1$.) The value of $C\;$ can be found with the help of the shift formula for the derivative $\frac{\partial}{\partial x} \zeta(z,x)=-z \zeta(z+1,x)\;$: $$ \int_1^2 R(x)dx= \int_1^2(\zeta(z,x)+C)dx= \int_1^2 \left( -\frac1{z-1}\frac{\partial}{\partial x} \zeta(z-1,x)+C\right)dx= $$ $$ C-\frac1{z-1}(\zeta(z-1,2)-\zeta(z-1,1))=C+\frac1{z-1}=0.$$ Hence $C=-\frac1{z-1}$. Also $\zeta(z,1)=\zeta(z)$ and we have $$ \sum_{n=1}^\infty n^{-z}=\zeta(z)-\frac1{z-1}(\mathfrak{R}),\quad z\ne1. $$ So Ramanujan summation transforms the Riemann zeta function into the regularized zeta function. It explains why the value $\gamma$ should be expected for the summation of the harmonic series.
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I am a fan of good documentation—I feel it's the most undervalued component of many projects but I haven't been able to develop a reliable process to:
Collect information Format it Keep it up to date
This is my attempt to come up with that process.
vim
I want to edit my documentation in vim. I like it because it's an extremely lightweight and focused tool for text editing. My
.vimrc file is
Markdown Blogger
I chose Blogger because it is already integrated with my Google Apps Premier account. I needed something basic and reliable. Documentation is divided into two areas—articles hosted by Blogger and articles hosted on my wiki. We'll begin by setting up documentation on Blogger and then use that to host documentation for setting up the wiki. The wiki will, in turn, host documentation the rest of the server stack.
BitBucket
References to BitBucket code are included via BitBucket's embed function. Note—replace the revision number with
default to always pull down the tip version.
GitHub
Code snippets can also be included via gist on GitHub.
PastebinPastebin is also supported: LaTeX
We will occasionally need to display equations. There are several options including using the utility available here. Or you can use MathJax by adding the following to your page template:
When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
I haven't decided which LaTeX method I prefer. Dropbox
I store all of my articles on Dropbox
~/Dropbox/blogger/content. I'm not convinced that my articles belong in revision control--I don't see myself branching and merging articles. Dropbox provides revision histories.
Pandoc
The original Markdown article must be processed by Pandoc before it can be published to Blogger. I run the following command.
pandoc --no-wrap "${docname}" -o "${tempdoc}"
The output file
${tempdoc} contains the final version that is consumed by Blogger.
Objective
My goal is document, start to finish, the steps need to build and deploy our server stack. This includes
Building a new machine with Ubuntu LTS Installing well-known services DNS DHCP LDAP Postfix Samba VirtualBox Installation and configuration of servers Wiki Zenoss
The process for deployment is:
Build services on a local virtual machine. Test services with other local virtual machines. Deploy to staging environment. Run staging tests. Deploy to production. Monitor.
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Degree $n$ : $44$ Transitive number $t$ : $50$ Parity: $1$ Primitive: No Nilpotency class: $-1$ (not nilpotent) Generators: (1,32,15,27,8,23,22,41,14,37,5,34,20,29,12,26,3,44,17,39,10,35)(2,31,16,28,7,24,21,42,13,38,6,33,19,30,11,25,4,43,18,40,9,36), (1,21)(2,22)(3,19)(4,20)(5,18)(6,17)(7,15)(8,16)(9,14)(10,13)(11,12)(23,25)(24,26)(27,43)(28,44)(29,42)(30,41)(31,39)(32,40)(33,37)(34,38)(35,36), (1,25,14,31)(2,26,13,32)(3,38,12,42)(4,37,11,41)(5,28,10,30)(6,27,9,29)(7,39)(8,40)(15,43,22,36)(16,44,21,35)(17,33,20,24)(18,34,19,23) $|\Aut(F/K)|$: $2$
|G/N| Galois groups for stem field(s) 2: $C_2$ x 7 4: $C_2^2$ x 7 8: $D_{4}$ x 2, $C_2^3$ 16: $D_4\times C_2$ 968: 22T10
Resolvents shown for degrees $\leq 29$
There are no siblings with degree $\leq 29$ Data on whether or not a number field with this Galois group has arithmetically equivalent fields has not been computed.
There are 70 conjugacy classes of elements. Data not shown.
Order: $1936=2^{4} \cdot 11^{2}$ Cyclic: No Abelian: No Solvable: Yes GAP id: [1936, 161]
Character table: Data not available.
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I was reading the following book: http://neuralnetworksanddeeplearning.com/chap2.html
and towards the end of equation 29, there is a paragraph that explains this:
However I am unsure how the equation below is derived:
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I was reading the following book: http://neuralnetworksanddeeplearning.com/chap2.html
and towards the end of equation 29, there is a paragraph that explains this:
However I am unsure how the equation below is derived:
I believe he's just saying that:
$$ \frac{\partial C}{\partial z_j^l} \Delta z_j^l \approx \frac{\partial C}{\partial z_j^l} \partial z_j^l \approx \partial C $$
so that the change in cost function can be arrived at simply for a small enough perturbation $\Delta z_j^l$.
Or, taking that line of approximations backwards, the change in the cost function for a given perturbation is just: $$ \partial C \approx \frac{\partial C}{\partial z_j^l} \partial z_j^l \approx \frac{\partial C}{\partial z_j^l} \Delta z_j^l $$
I think that Nielsen just wanted to convey the idea of the back-propagation algorithm using that formula, as you can read from the next paragraph "Now, this demon is a good demon...", so I don't think that that partial derivative is mathematically correct, provided the partial derivative is still with respect to $z_j^l$.
$C$ is the cost (or loss) function. $z_j^l$ is the linear output of neuron $j$ in layer $l$, which is followed by a non-linear function (e.g. sigmoid), denoted by $\sigma$. So, the actual output of neuron $j$ in layer $l$ is $\sigma(z_j^l)$.
The partial derivative of the cost function $C$ with respect to this neuron's linear output, $z_j^l$, is $$\frac{\partial C}{\partial z_j^l} = \frac{\partial C}{\partial z_j^l} 1 = \frac{\partial C}{\partial z_j^l} \frac{\partial z_j^l}{\partial z_j^l}.$$
If the the linear output of node $j$ in layer $l$ is now $z_j^l + \Delta z_j^l$, then the partial derivative with respect to $z_j^l$ becomes
\begin{align} \frac{\partial C}{\partial z_j^l} &= \frac{\partial C}{\partial (z_j^l + \Delta z_j^l)}\frac{\partial (z_j^l + \Delta z_j^l)}{\partial z_j^l} \\ &= \frac{\partial C}{\partial (z_j^l + \Delta z_j^l)} \left( \frac{\partial z_j^l}{\partial z_j^l} + \frac{\partial \Delta z_j^l}{\partial z_j^l} \right) \\ &= \frac{\partial C}{\partial (z_j^l + \Delta z_j^l)} \left( 1 + \frac{\partial \Delta z_j^l}{\partial z_j^l} \right) \\ &= \frac{\partial C}{\partial (z_j^l + \Delta z_j^l)} + \frac{\partial C}{\partial (z_j^l + \Delta z_j^l)}\frac{\partial \Delta z_j^l}{\partial z_j^l} \\ \end{align}
$\Delta z_j^l$ depends on $z_j^l$, but it is not specified how.
The derivative of a function ($f(x_1,x_2..x_n)$) w.r.t to one of the variables ($x_1,x_2..x_n$) gives us the rate of change of the function w.r.t the rate of change of the variable. This roughly means that by how much will the function value change if we change the variable by a "unit amount" or $+1$. (we cannot use the change as $+1$ as the change needs to be infinitesimally small, this is just a rough explanation)
The image shows a tangent line to a curve or function $f(x)$. The slope of this tangent is given by $\frac{df(x)}{dx}$ at that particular $x$. If you move by a very very small amount in the direction of positive $x$ i.e. $x+\delta x$ the change in the value of the $f(x)= y $ will almost be the same as the change in the value of the $y$ of the tangent line.
Now as per the excerpt the cost function $C$ is a function of $z_j^l$. Thus, it can be written as $$C = f(z_j^l, .....).$$ So, $$\frac{\partial C}{\partial z_j^l}$$ indicates how much $C$ will vary w.r.t $z_j^l$, i.e. when $z_j^l$ is changed by an infinitesimally small amount and thus the formulae: $$\frac{\partial C}{\partial z_j^l} \Delta z_j^l$$ where the author assumed $\Delta z_j^l$ to be very small. This gives the infinitesimal change in $C$ or gives us $\Delta C$, for infinitesimally small change in $z_j^l$ or any varible affecting the cost function. This can be derived by series expansions too (given below), but this is an intuitive explanation.
An explanation can be given from
Taylor Series Theorem which states: :
Let $f(x)$ be a function which is analytic at $x = a$. Then we can write $f(x)$ as the following power series, called the Taylor series of $f(x)$ at $x = a$, then we can write $f(x)$ as:
$$f(x) = f(a) + f'(a)(x-a) + f''(a)\frac{(x-a)^2}{2!} + f'''(a)\frac{(x-a)^3}{3!}... $$
Now if we keep other variables constant and make cost function $f$ vary only with $z^l_j$ and if we put $a=z^l_j$ and $x=z^l_j + \Delta z^l_j$ the equation becomes:
$$f(z^l_j + \Delta z^l_j) = f(z^l_j) + f'(z^l_j)(\Delta z^l_j) + f''(z^l_j)\frac{(\Delta z^l_j)^2}{2!} + f'''(a)\frac{(\Delta z^l_j)^3}{3!}... $$
which if we ignore the higher order terms of $\Delta z^l_j$, since terms containing $\Delta z^l_j$ for powers greater than 1 will be negligible compared to $\Delta z^l_j$ with power 1. Thus the equation now effectively is:
$$f(z^l_j + \Delta z^l_j) = f(z^l_j) + f'(z^l_j)(\Delta z^l_j)$$
$$f(z^l_j + \Delta z^l_j) - f(z^l_j)= f'(z^l_j)(\Delta z^l_j)$$
$$f(z^l_j + \Delta z^l_j) - f(z^l_j)= \frac{\partial f(z^l_j)}{\partial z^l_j}(\Delta z^l_j)$$ where $$f(z^l_j + \Delta z^l_j) - f(z^l_j)$$ can be thought of as $\Delta C$ or the change in cost function for small change in $z^l_j$
NOTE: I have glossed over some requirements for a Taylor Series to be convergent.
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Lephenixnoir af424d1baa update documentation after writing the wiki 3ヶ月前 config 4ヶ月前 include/TeX 3ヶ月前 src 3ヶ月前 .gitignore 3ヶ月前 Makefile 4ヶ月前 README.md 3ヶ月前 TODO.md 3ヶ月前 configure 3ヶ月前 font5x7.bmp 4ヶ月前 font8x9.bmp 4ヶ月前 font10x12.bmp 4ヶ月前
This library is a customizable 2D math rendering tool for calculators. It can be used to render 2D formulae, either from an existing structure or TeX syntax.
\frac{x^7 \left[X,Y\right] + 3\left|\frac{A}{B}\right>} {\left\{\frac{a_k+b_k}{k!}\right\}^5}+ \int_a^b \frac{\left(b-t\right)^{n+1}}{n!} dt+ \left(\begin{matrix} \frac{1}{2} & 5 \\ -1 & a+b \end{matrix}\right)
List of currently supported elements:
\frac)
_ and
^)
\left and
\right)
\sum,
\prod and
\int)
\vec) and limits (
\lim)
\sqrt)
\begin{matrix} ... \end{matrix})
Features that are partially implemented (and what is left to finish them):
See the
TODO.md file for more features to come.
First specify the platform you want to use :
cli is for command-line tests, with no visualization (PC)
sdl2 is an SDL interface with visualization (PC)
fx9860g builds the library for fx-9860G targets (calculator)
fxcg50 builds the library for fx-CG 50 targets (calculator)
For calculator platforms, you can use
--toolchain to specify a different toolchain than the default
sh3eb and
sh4eb. The install directory of the library is guessed by asking the compiler, you can override it with
--prefix.
Example for an SDL setup:
% ./configure --platform=sdl2
Then you can make the program, and if it’s a calculator library, install it. You can later delete
Makefile.cfg to reset the configuration, or just reconfigure as needed.
% make% make install # fx9860g and fxcg50 only
Before using the library in a program, a configuration step is needed. The library does not have drawing functions and instead requires that you provide some, namely:
TeX_intf_pixel)
TeX_intf_line)
TeX_intf_size)
TeX_intf_text)
The three rendering functions are available in fxlib; for monospaced fonts the fourth can be implemented trivially. In gint, the four can be defined as wrappers for
dpixel(),
dline(),
dsize() and
dtext().
The type of formulae is
TeX_Env. To parse and compute the size of a formula, use the
TeX_parse() function, which returns a new formula object (or
NULL if a critical error occurs). The second parameter
display is set to non-zero to use display mode (similar to
\[ .. \] in LaTeX) or zero to use inline mode (similar to
$ .. $ in LaTeX).
char *code = "\\frac{x_7}{\\left\\{\\frac{\\frac{2}{3}}{27}\\right\\}^2}";struct TeX_Env *formula = TeX_parse(code, 1);
The size of the formula can be queried through
formula->width and
formula->height. To render, specify the location of the top-left corner and the drawing color (which will be passed to all primitives):
TeX_draw(formula, 0, 0, BLACK);
The same formula can be drawn several times. When it is no longer needed, free it with
TeX_free():
TeX_free(formula);
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Completeness Theorem for Semantic Tableaus Theorem
Let $\mathbf A$ be a WFF of propositional logic.
Let $\mathbf A$ be unsatisfiable for boolean interpretations.
Proof
The proof proceeds by contradiction.
This claim is proved by inductively establishing the following claim for all nodes $t$ of $T$: Inductively suppose that all children of $t$ satisfy the mentioned condition.
By induction hypothesis, $U \left({t'}\right)$ is satisfiable.
Then $U \left({t'}\right)$ is obtained from $U \left({t}\right)$ by replacing $\mathbf B$ by $\mathbf B_1$ and $\mathbf B_2$.
Here, $\mathbf B_1$ and $\mathbf B_2$ are the formulas such that $\mathbf B$ is semantically equivalent to $\mathbf B_1 \land \mathbf B_2$.
It follows that if: $v \models_{\mathrm{BI}} U \left({t'}\right)$
for some boolean interpretation $v$, then also:
$v \models_{\mathrm{BI}} U \left({t}\right)$
by the boolean interpretation of $\land$.
Consequently, $U \left({t}\right)$ is satisfiable.
Next, the case that $\mathbf B$ is a $\beta$-formula.
Then $\mathbf B$ is semantically equivalent to $\mathbf B_1 \lor \mathbf B_2$.
Then, by the Rule of Addition:
$\mathbf B_1 \models_{\mathrm{BI}} \mathbf B$ $\mathbf B_2 \models_{\mathrm{BI}} \mathbf B$
Consequently, by the definitions of $U \left({t'}\right)$ and $U \left({t''}\right)$:
$U \left({t'}\right) \models_{\mathrm{BI}} U \left({t}\right)$ $U \left({t''}\right) \models_{\mathrm{BI}} U \left({t}\right)$
By hypothesis, one of $U \left({t'}\right)$ and $U \left({t''}\right)$ must be satisfiable.
Hence, so must $U \left({t}\right)$ be.
This establishes our claim.
$\blacksquare$
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Let $q$ be a prime power and let $m$ be a positive integer coprime to $q$. Further, let $E_1(\mathbb F_q)$ and $E_2(\mathbb F_q)$ be two isogenous supersingular elliptic curves over a finite field $\mathbb F_q$ such that
$$E_i(\mathbb F_q)[m] \cong \mathbb Z/m\mathbb Z \times \mathbb Z/m\mathbb Z$$
for $i = 1, 2$. Suppose that an isogeny $\phi \colon E_1 \rightarrow E_2$ is separable and has degree coprime to $m$. So in particular $\phi$ is a one-to-one and onto map between $E_1[m]$ and $E_2[m]$.
Now suppose that you are given a pair of generators $P,Q$ of $E_1[m]$ and a pair of generators $R,S$ of $E_2[m]$. In this setting, I have three questions.
For which matrices
$$ \left( \begin{matrix} a & b\\ c & d \end{matrix} \right) \in \operatorname{GL}_2(\mathbb Z) $$
does there exist an endomorphism $\sigma \colon E_1 \rightarrow E_1$ such that $\sigma(P) = aP+bQ$ and $\sigma(Q)=cP+dQ$?
How hard is it to answer the following question:
does there exist an isogeny $\phi \colon E_1 \rightarrow E_2$ such that $\phi(P)=R$ and $\phi(Q) = S$?Of course in general it is difficult to compute $\phi$, but perhaps answering this question of existence is easier?
Given a matrix
$$ \left( \begin{matrix} a & b\\ c & d \end{matrix} \right) \in \operatorname{GL}_2(\mathbb Z) $$
and a fact that $\phi(P)=R$ and $\phi(Q)=S$ for some isogney $\phi$, does there exist an isogeny $\phi'$ such that $\phi'(P) = aR+bS$ and $\phi'(Q)=cR+dS$? For which matrices is this possible?
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1. The Genomes OnLine Database (GOLD) v.4: status of genomic and metagenomic projects and their associated metadata
Nucleic Acids Research, ISSN 0305-1048, 1/2012, Volume 40, Issue D1, pp. D571 - D579
The Genomes OnLine Database (GOLD, http://www.genomesonline.org/) is a comprehensive resource for centralized monitoring of genome and metagenome projects...
SYSTEM | MONITOR | BIOCHEMISTRY & MOLECULAR BIOLOGY | Databases, Genetic | Metagenomics | User-Computer Interface | Phylogeny | Genomics | Gold | Data processing | genomics | Genomes | Databases | Internet
SYSTEM | MONITOR | BIOCHEMISTRY & MOLECULAR BIOLOGY | Databases, Genetic | Metagenomics | User-Computer Interface | Phylogeny | Genomics | Gold | Data processing | genomics | Genomes | Databases | Internet
Journal Article
Standards in Genomic Sciences, ISSN 1944-3277, 10/2015, Volume 10, Issue 1, p. 86
The DOE-JGI Microbial Genome Annotation Pipeline performs structural and functional annotation of microbial genomes that are further included into the...
Microbial Genome Annotation | IMG | SOP | JGI | SYSTEM | DATABASE | RECOGNITION | GENETICS & HEREDITY | CLASSIFICATION | MICROBIOLOGY | IDENTIFICATION | REPEATS | BASIC BIOLOGICAL SCIENCES
Microbial Genome Annotation | IMG | SOP | JGI | SYSTEM | DATABASE | RECOGNITION | GENETICS & HEREDITY | CLASSIFICATION | MICROBIOLOGY | IDENTIFICATION | REPEATS | BASIC BIOLOGICAL SCIENCES
Journal Article
3. Measurement of prompt and nonprompt charmonium suppression in $$\text {PbPb}$$ PbPb collisions at 5.02$$\,\text {Te}\text {V}$$ TeV
The European Physical Journal C, ISSN 1434-6044, 6/2018, Volume 78, Issue 6, pp. 1 - 27
The nuclear modification factors of $${\mathrm {J}/\psi }$$ J/ψ and $$\psi \text {(2S)}$$ ψ(2S) mesons are measured in $$\text {PbPb}$$ PbPb collisions at a...
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Journal Article
4. DIFFERENTIATING PACEMAKER-MEDIATED TACHYCARDIA FROM PACEMAKER ATRIAL TRACKING DURING WIDE COMPLEX TACHYCARDIA: UTILITY OF V-A-A-V VS. V-A-V RESPONSE AFTER PVARP EXTENSION
JACC (Journal of the American College of Cardiology), ISSN 0735-1097, 2011, Volume 57, Issue 14, pp. E169 - E169
Journal Article
5. Differentiating pacemaker-mediated tachycardia from tachycardia due to atrial tracking: Utility of V-A-A-V versus V-A-V response after postventricular atrial refractory period extension
Heart Rhythm, ISSN 1547-5271, 2011, Volume 8, Issue 8, pp. 1185 - 1191
Background Dual-chamber pacemaker systems can lead to two forms of pacemaker-facilitated tachycardia: pacemaker-mediated tachycardia (PMT) and tracking of...
Cardiovascular | Electrophysiology | Tachycardia | Arrhythmia | Pacemaker | CARDIAC & CARDIOVASCULAR SYSTEMS | Algorithms | Prospective Studies | Humans | Middle Aged | Electrocardiography | Male | Cardiac Pacing, Artificial | Heart Atria - physiopathology | Electrophysiologic Techniques, Cardiac | Tachycardia - physiopathology | Pacemaker, Artificial
Cardiovascular | Electrophysiology | Tachycardia | Arrhythmia | Pacemaker | CARDIAC & CARDIOVASCULAR SYSTEMS | Algorithms | Prospective Studies | Humans | Middle Aged | Electrocardiography | Male | Cardiac Pacing, Artificial | Heart Atria - physiopathology | Electrophysiologic Techniques, Cardiac | Tachycardia - physiopathology | Pacemaker, Artificial
Journal Article
6. Search for a heavy resonance decaying into a Z boson and a vector boson in the v(v)over-barq(q)over-bar final state
JOURNAL OF HIGH ENERGY PHYSICS, ISSN 1029-8479, 07/2018, Issue 7
Journal Article
7. Measurement of prompt and nonprompt charmonium suppression in $\text {PbPb}$ collisions at 5.02$\,\text {Te}\text {V}
European Physical Journal. C, Particles and Fields, ISSN 1434-6044, 06/2018, Volume 78, Issue 6
Journal Article
8. Measurement of b hadron lifetimes in pp collisions at $$\sqrt{s} = 8$$ s=8 $$\,\text {Te}\text {V}$$ TeV
The European Physical Journal C, ISSN 1434-6044, 6/2018, Volume 78, Issue 6, pp. 1 - 28
Measurements are presented of the lifetimes of the $${\mathrm {B}^0}$$ B0 , $$\mathrm {B}^0_\mathrm {s}$$ Bs0 , $$\mathrm {\Lambda }_\mathrm {b} ^0$$ Λb0 , and...
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Journal Article
Physical Review Letters, ISSN 0031-9007, 10/2017, Volume 119, Issue 16, p. 161101
On August 17, 2017 at 12:41:04 UTC the Advanced LIGO and Advanced Virgo gravitational-wave detectors made their first observation of a binary neutron star...
GAMMA-RAY BURST | MASSES | EQUATION-OF-STATE | PHYSICS, MULTIDISCIPLINARY | ADVANCED LIGO | PULSAR | MERGERS | RADIATION | DENSE MATTER | General Relativity and Quantum Cosmology | Physics | Astrophysics
GAMMA-RAY BURST | MASSES | EQUATION-OF-STATE | PHYSICS, MULTIDISCIPLINARY | ADVANCED LIGO | PULSAR | MERGERS | RADIATION | DENSE MATTER | General Relativity and Quantum Cosmology | Physics | Astrophysics
Journal Article
10. Search for third-generation scalar leptoquarks decaying to a top quark and a $$\tau $$ τ lepton at $$\sqrt{s}=13\,\text {Te}\text {V} $$ s=13Te
The European Physical Journal C, ISSN 1434-6044, 9/2018, Volume 78, Issue 9, pp. 1 - 26
A search for pair production of heavy scalar leptoquarks (LQs), each decaying into a top quark and a $$\tau $$ τ lepton, is presented. The search considers...
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Journal Article
11. Search for pair production of third-generation scalar leptoquarks and top squarks in proton-proton collisions at v root s=8 TeV
PHYSICS LETTERS B, ISSN 0370-2693, 12/2014, Volume 739, pp. 229 - 249
Journal Article
12. Erratum to: Measurement of b hadron lifetimes in pp collisions at $$\sqrt{s} = 8$$ s=8 $$\,\text {Te}\text {V}$$ Te
The European Physical Journal C, ISSN 1434-6044, 7/2018, Volume 78, Issue 7, pp. 1 - 15
We have corrected Eqs. (17)–(20) from Section 8 of the original paper.
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Journal Article
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Tutor profile: Ikechukwu "Daniel" A. Questions Subject: Pre-Calculus
Find the quotient of $$\frac{2i+3}{7-6i}$$ and simplify. Then, find the exact modulus of the result.
Multiply both the numerator and denominator by the conjugate of the denominator $$\frac{2i+3}{7-6i} *\frac{7+6i}{7+6i} = \frac{(2i+3)(7+6i)}{85} = \frac{(2i+3)(7+6i)}{85} = \frac{9+32i}{85} $$ Now, we can find the modulus of our result, which is the same as simply finding the distance this complex number is from the origin of the complex plane. $$|\frac{9+32i}{85}| = \sqrt{(\frac{9}{85})^2+(\frac{32}{85})^2} =\sqrt{\frac{13}{85}} = \frac{\sqrt{1105}}{85} $$
Subject: Trigonometry
Prove that the $$\frac{\csc{\theta}}{\sin{\theta}} - \frac{\cot{\theta}}{\tan{\theta}} = 1$$
To prove this identity, we'll start with the left-hand side. To make things simpler for us, we'll rewrite everything in terms of $$\sin\theta$$ and $$\cos\theta$$. $$ \frac{\csc{\theta}}{\sin{\theta}} - \frac{\cot{\theta}}{\tan{\theta}} = \frac{\frac{1}{\sin\theta}}{\sin\theta} - \frac{\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}}$$ $$= \frac{1}{{\sin}^2\theta} - \frac{\cos^2\theta}{\sin^2\theta}$$ $$= \frac{1-\cos^2\theta}{\sin^2\theta}=\frac{\sin^2\theta}{\sin^2\theta} = 1$$ Therefore this identity is true.
Subject: Calculus
Determine the indefinite integral of $$\int \frac{x^2-8}{(4x-3)(2x+1)}dx$$.
$$\int \frac{x-8}{(4x-3)(2x+1)}dx$$ We can break up the function that is being integrated into its partial fraction decomposition like so: $$ \frac{x-8}{(4x-3)(2x+1)} = \frac{A}{4x-3} + \frac{B}{2x+1}$$ After doing this, we rewrite this expression so that there are no fractions: $$ x-8 = A(2x+1) + B(4x-3)$$ At this point, we must solve for $$A$$ and $$B$$. If $$x = -\frac{1}{2} $$, then $$ -\frac{1}{2}-8 = A(2(-\frac{1}{2})+1) + B(4(-\frac{1}{2})-3)$$ $$ -\frac{1}{2}-8 = A(-1+1) + B(-2-3)$$ As we can see, $$A$$ will disappear, so we can solve for $$B$$. $$ \frac{-17}{2} = -5B$$ $$ \frac{17}{10} = B$$ Now we can go and solve for $$A$$. If $$x = -\frac{3}{4} $$, then $$ \frac{3}{4}-8 = A(2(\frac{3}{4})+1) + B(4(\frac{3}{4})-3)$$ $$ \frac{3}{4}-8 = A(\frac{6}{4}+1) + B(3-3)$$ This time, $$B$$ disappears, so we can now go about solving for $$A$$. $$ \frac{-29}{4} = \frac{10}{4}A $$ $$ A = \frac{-29}{10} $$ Now that we know both $$A$$ and $$B$$, we can go back to our original integral and solve. $$\int \frac{A}{4x-3} + \frac{B}{2x+1}dx = \int \frac{-29}{(10)(4x-3)}+\frac{17}{(10)(2x+1)}dx $$ $$ = \frac{-29}{40}\ln|4x-3| + \frac{17}{20}\ln|2x+1| + C$$
Contact tutor
needs and Ikechukwu "Daniel" will reply soon.
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Let $k$ be a $p$-adic field, $G$ a connected reductive group over $k$ with minimal parabolic $P_0$ containing a maximal split torus $A_0$. Let $W = N_G(A_0)(k)/Z_G(A_0)(k)$ be the Weyl group, and $S \subset W$ the simple reflections from $P_0$.
For $\theta, \Omega \subset S$, we have the standard parabolic subgroups $P_\theta, P_\Omega$.
Each double coset $P_{\theta}w P_{\Omega}$, for $w \in W$ with a $k$-rational representative, is a locally closed subvariety of $G$, with $P_\theta wP_\Omega(k) = P_\theta(k)wP_\Omega(k)$.
How do we know that the quotient $P_\theta \backslash P_\theta wP_\Omega$ is an algebraic variety over $k$? This variety and its dimension are considered in Casselman's notes on representation theory, Chapter 6. I do not know the general theory of quotients of algebraic group actions which would make sense out of something like this.
Once this variety is made sense of, can we say that its $k$-rational points coincides with $P_\theta(k) \backslash P_\theta(k) wP_\Omega(k)$?
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I need to numerically determine the convergence order of Euler's method for various step-sizes. I am unsure how to go about doing this.
Here is the question:
Problem statement: $\frac{dy}{dt}=\alpha t^{\alpha - 1}, y(0)=0$, where $\alpha > 0$. Use Euler's method to solve the initial value problem for $\alpha = 2.5,1,5,1.1$ with stepsize $h=0.2,0.1,0.05$. Determine numerically the convergence orders of the Euler method for these problems.
I have created an Excel spreadsheet with all of the relevant information tabulated (using Excel formulas), with columns $i, t_i=a+ih, a, b, \alpha, N, w_i, y(t_i), |y(t_i)-w_i|$ where:
$i=$ ith step in iteration
$t_i$= time at ith step
$a=$ initial time value
$b=$ final time value
$\alpha=$ constant in problem
$N=$ number of subintervals
$w_i=$ approximation to solution at ith step
$y(t_i)=$ true solution at ith step
$|y(t_i)-w_i|=$ Magnitude of error
Now, since I cannot estimate the error $\kappa$ from just one integration as the error term is approximately of the form $error_{t=b}(h) = Ah^{\kappa}$, one step size is not adequate to estimate $\kappa$. But if I do a log plot of the error and step size, the slope of the best line will give the required estimate of $\kappa$, which should be close to 1.
My question is how can I do the log plot of the error and step size in Excel 2013? I understand the math behind all this, but don't know how to implement it.
Thanks.
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(19 intermediate revisions by 5 users not shown) Line 1: Line 1: − Stoke's Theorem is a generalization of the [[Fundamental Theorem of Calculus]] , which states that if ''M'' is an oriented piece-wise smooth [[manifold]] of [[dimension]] k and <math>\omega</math> is a smooth (''k''−1)-form with compact support on ''M'' . Let ∂''M'' denotes the boundary of ''M'' with its induced orientation, then + +
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's Theorem
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is a generalization of the [[Fundamental Theorem of Calculus]]states that if ''M'' is an oriented smooth [[manifold]] of [[dimension]] k and <math>\omega</math> is a smooth (''k''−1)-formwith compact support on ''M''∂''M'' denotes the boundary of ''M'' with its induced orientation, then
:<math>\int_M \mathrm{d}\omega = \oint_{\partial M} \omega\!\,</math>,
:<math>\int_M \mathrm{d}\omega = \oint_{\partial M} \omega\!\,</math>,
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where ''d'' is the [[exterior derivative]].
where ''d'' is the [[exterior derivative]].
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Latest revision as of 15:18, 29 July 2016
Stokes' Theorem states that the line integral of a closed path is equal to the surface integral of any capping surface for that path, provided that the surface normal vectors point in the same general direction as the right-hand direction for the contour:
Intuitively, imagine a "capping surface" that is nearly flat with the contour. The curl is the microscopic circulation of the function on tiny loops within that surface, and their sum or integral results in canceling out all the internal circulation paths, leaving only the integration over the outer-most path.
[1] This remains true no matter how the capping surface is expanded, provided that the contour remains as its boundary.
Sometimes the circulation (the left side above) is easier to compute; other times the expresses the surface integral of the curl of vector field is easier to computer (particularly when it is zero).
[2]
Stated another way, Stokes' Theorem equates the line integral of a vector fields to a surface integral of the same vector field. For this identity to be true, the
direction of the vector normal n must obey the right-hand rule for the direction of the contour, i.e., when walking along the contour the surface must be on your left.
This is an extension of Green's Theorem to surface integrals, and is also the analog in two dimensions of the Divergence Theorem. The above formulation is also called as the "Curl Theorem," to distinguish it from the more general form of the Stokes' Theorem described below.
Stokes' Theorem is useful in calculating circulation in mechanical engineering. A conservative field has a circulation (line integral on a simple, closed curve) of zero, and application of the Stokes' Theorem to such a field proves that the curl of a conservative field over the enclosed surface must also be zero.
General Form
In its most general form, this theorem is the fundamental theorem of Exterior Calculus, and is a generalization of the Fundamental Theorem of Calculus. It states that if
M is an oriented piecewise smooth manifold of dimension k and is a smooth ( k−1)-form with compact support on M, and ∂ M denotes the boundary of M with its induced orientation, then ,
where
d is the exterior derivative.
There are a number of well-known special cases of Stokes' theorem, including one that is referred to simply as "Stokes' theorem" in less advanced treatments of mathematics, physics, and engineering:
When k=1, and the terms appearing in the theorem are translated into their simpler form, this is just the Fundamental Theorem of Calculus. When k=3, this is often called Gauss' Theorem or the Divergence Theorem and is useful in vector calculus:
Where R is some region of 3-space, S is the boundary surface of R, the triple integral denotes volume integration over R with dV as the volume element, and the double integral denotes surface integration over S with as the oriented normal of the surface element. The on the left side is the divergence operator, and the on the right side is the vector dot product.
When k=2, this is often just called Stokes' Theorem:
Here S is a surface, E is the boundary path of S, and the single integral denotes path integration around E with as the length element. The on the left side is the curl operator.
These last two examples (and Stokes' theorem in general) are the subject of vector calculus. They play important roles in electrodynamics. The divergence and curl operations are cornerstones of Maxwell's Equations.
Stokes' Theorem is a lower-dimension version of the Divergence Theorem, and a higher-dimension version of Green's Theorem. Green’s Theorem relates a line integral to a double integral over a region, while Stokes' Theorem relates a surface integral of the curl of a function to its line integral. Stokes' Theorem originated in 1850.
References ↑ Notes on Gauss' and Stokes' Theorem, see Section 7.6, p. 87 ↑
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Bulletin of the American Physical Society 65th Annual Meeting of the APS Division of Fluid Dynamics Volume 57, Number 17 Sunday–Tuesday, November 18–20, 2012; San Diego, California
Session M27: Boundary-Layer Instability II Hide Abstracts Chair: Jill Klentzman, University of Arizona
Room:
31C
Tuesday, November 20, 2012
8:00AM - 8:13AM
M27.00001: Receptivity of high-speed boundary layers with real gas effects
Jill Klentzman, Anatoli Tumin
The receptivity of high speed boundary layers in chemical nonequilibrium is investigated. A method is developed for the multi-mode decomposition of boundary layer flows including real gas effects, and the receptivity problem with small disturbances introduced at the wall is examined. The solution of the linearized Navier-Stokes equations, within the parallel flow approximation, is expressed in the form of normal modes, and the resulting differential equations for the amplitude functions are discretized using fourth-order finite differences. This discretized system is then in the form of a generalized eigenvalue problem, which yields a straight-forward definition of the associated adjoint system. A biorthogonality condition is formulated based on the adjoint eigenvectors. Assuming that a complete system of eigenfunctions of the discrete and continuous spectra exists, the biorthogonality condition allows for the projection of a solution onto the discrete modes, which can be utilized in the analysis of DNS results. [Preview Abstract]
Tuesday, November 20, 2012
8:13AM - 8:26AM
M27.00002: Statistical inverse analysis of supersonic boundary-layer transition
Gennaro Serino, Olaf Marxen, Fabio Pinna, Paul Constantine, Catherine Gorle, Gianluca Iaccarino
In environments with low boundary-layer disturbance levels representative of free flight in the atmosphere, the laminar-turbulent transition process for vehicles moving at supersonic speeds is typically governed by the convective amplification of high-frequency disturbances. A valid statistical characterization of the disturbance spectrum upstream of the transition location is a pre-requisite for an accurate prediction of transition. Statistical inverse analysis offers the possibility to provide such a characterization of relevant disturbance spectra. Using measured streamwise distributions of heat-transfer coefficients in the transitional zone, an intermittency factor can be defined. The intermittency factor is fed into an inference algorithm based on the Markov chain Monte Carlo method. This approach is applied to infer two characteristic variables, amplitude and frequency, of the disturbance spectrum upstream of the transition location. It relies on the repeated solution of the forward problem, i.e., the computation of intermittency curve given a certain probability density function of the disturbances. The solution of the forward problem employs linear-stability theory in conjunction with a critical threshold amplitude for transition. [Preview Abstract]
Tuesday, November 20, 2012
8:26AM - 8:39AM
M27.00003: Stability of hypersonic compression cones
Helen Reed, Joseph Kuehl, Eduardo Perez, Travis Kocian, Nicholas Oliviero
Our activities focus on the identification and understanding of the second-mode instability for representative configurations in hypersonic flight. These include the Langley 93-10 flared cone and the Purdue compression cone, both at 0 degrees angle of attack at Mach 6. Through application of nonlinear parabolized stability equations (NPSE) and linear parabolized stability equations (PSE) to both geometries, it is concluded that mean-flow distortion tends to amplify frequencies less than the peak frequency and stabilize those greater by modifying the boundary-layer thickness. As initial disturbance amplitude is increased and/or a broad spectrum disturbance is introduced, direct numerical simulations (DNS) or NPSE appear to be the proper choices to model the evolution, and relative evolution, because these computational tools include these nonlinear effects (mean-flow distortion). [Preview Abstract]
Tuesday, November 20, 2012
8:39AM - 8:52AM
M27.00004: Azimuthal hotwire measurements in a transitional boundary layer on a flared cone in a Mach 6 quiet wind tunnel
Jerrod Hofferth, William Saric
Hotwire measurements of second-mode instability waves and the early stages of nonlinear interaction are conducted on a sharp-tipped, 5$^{\circ}$-half-angle flared cone at zero angle of attack in a low-disturbance Mach 6 wind tunnel at $Re$ = 10$\times$10$^6$ m$^{-1}$. Profiles of mean and fluctuating mass flux are acquired at several axial stations along the cone with a bandwidth of over 300 kHz. Frequencies and relative amplitude growth of second-mode instability waves are characterized and compared with nonlinear parabolized stability (NPSE) computations. Additionally, an azimuthal probe-traversing mechanism is used to investigate the character of the nonlinear stages of transition occurring near the base of the cone. Recent Direct Numerical Simulations (DNS) of a sharp cone at Mach 6 have shown that a fundamental resonance (or Klebanoff-type) breakdown mechanism can arise in the late stages of transition, wherein a pair of oblique waves nonlinearly interacts with the dominant two-dimensional wave to create an azimuthal modulation in the form of $\Lambda$-vortex structures and streamwise streaks. The azimuthal measurements will identify periodicity qualitatively consistent with these computations and with ``hot streaks'' observed in temperature sensitive paints at Purdue. [Preview Abstract]
Tuesday, November 20, 2012
8:52AM - 9:05AM
M27.00005: Surface roughness effects on a blunt hypersonic cone
Nicole Sharp, Jerrod Hofferth, Edward White
The mechanisms through which distributed surface roughness produces boundary-layer disturbances in hypersonic flow are poorly understood. Previous work by Reshotko (AIAA 2008-4294) suggests that transient growth, resulting from the superposition of decaying non-orthogonal modes, may be responsible. The present study examines transient growth experimentally using a smooth 5-degree half-angle conic frustum paired with blunted nosetips with and without quasi-random distributed roughness. Hotwire anemometry in the low-disturbance Texas A{\&}M Mach 6 Quiet Tunnel shows a slight growth of fluctuations as well as vertical offset due to surface roughness at a range of unit Reynolds numbers. Spectral measurements indicate that the model is subcritical with respect to second mode growth, and azimuthal measurements are used to examine the high- and low-speed streaks characteristic of transient growth of stationary disturbances. [Preview Abstract]
Tuesday, November 20, 2012
9:05AM - 9:18AM
M27.00006: Characteristics of a streak disturbance induced by an isolated roughness element
Kyle Bade, Ahmed Naguib
A detailed description of a streak disturbance introduced in a Blasius boundary layer by an isolated roughness element will be presented. This work is motivated by the desire to understand the dependence of the evolution/instability of streamwise-oriented streaks (which play a key role in bypass transition) on the method by which they are generated. The proper scaling of the streamwise evolution of the streak disturbance energy is examined. This expands upon established Re$_{k}^{2}$ scaling (White, et al., Physics of Fluids, 2005) of streak disturbances induced by spanwise-periodic roughness element arrays. Examining different roughness heights, k, and employing a method that accounts for the streamwise growth of the streak's wall-normal and spanwise scales, it is found that the streak energy density scales with Re$_{k}^{7/3}$, in the case of an isolated roughness element. The data used in the analysis are acquired using hotwire anemometry throughout a three-dimensional domain located downstream of a single cylindrical roughness element. These measurements are complemented by smokewire visualizations, which capture clearly three distinct disturbance states, dependent upon roughness element height; namely, stable streaks, streaks with intermittent turbulent bursts, and turbulent disturbances. Correspondence is established between these states and the streamwise evolution of the streak energy and the cross-stream disturbance profiles. [Preview Abstract]
M27.00007: ABSTRACT WITHDRAWN Tuesday, November 20, 2012
9:31AM - 9:44AM
M27.00008: Two dimensional roughness effects on hypersonic boundary layer instability
Kahei Danny Fong, Xiaowen Wang, Xiaolin Zhong
Numerical simulations of 2-D roughness effects on modal growth are conducted for a hypersonic boundary layer. Perturbations correspond to pure mode S {\&} mode F at 100 kHz and a wall normal velocity pulse with a frequency spectrum of 1MHz are considered. The evolution of perturbation at different frequency along the streamwise direction with the effect of surface roughness is studied by FFT. Our results show the importance of the relation between roughness location and the synchronization point, where the synchronization point is the point where mode S and mode F have the same phase velocity and synchronizes with each other. Its location can be obtained from the linear stability theory. The results show that if roughness is placed upstream of the synchronization point, perturbation is amplified. The amplification rate depends strongly on roughness height. On the other hand, if roughness is placed close to or downstream of the synchronization point, perturbation is damped. Similar to amplification, the strength of damping depends strongly on roughness height. A tentative explanation is that roughness alters the mean flow profile (ex: sonic line, inflection point). We believe this can be a candidate to explain the roughness-delayed transition as some experiments have shown. [Preview Abstract]
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Hello friends, We have to find the last two digits of 1×3×5×7×9....×991\times3\times5\times7\times9....\times991×3×5×7×9....×99, I did and came out with 25.But I want to verify.
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From the fact that it is divisible by 252525 but not by 222, we can deduce that it ends with either 252525 or 757575. Now, consider the multiplications modulo 444: 1×3×5×7×9×⋯×99≡1×3×1×3×⋯×3≡125×325≡125×(−1)25≡−1≡75≢25 1 \times 3 \times 5 \times 7 \times 9 \times \dots \times 99 \equiv 1 \times 3 \times 1 \times 3 \times \dots \times 3 \equiv 1^{25} \times 3^{25} \equiv 1^{25} \times (-1)^{25} \equiv -1 \equiv 75 \not\equiv 25 1×3×5×7×9×⋯×99≡1×3×1×3×⋯×3≡125×325≡125×(−1)25≡−1≡75≡25, so the answer is 75\boxed{75}75.
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Argh, you beat me to it xD Well done.
how these type of shortcuts can be found? I want to know .if in any website it is plz tell me.
I did not use any website. In fact, I never solved such a problem before. I guess you'll just need to be familiar with prime factors and the modulo operator.
refer elemantary number theory by David M Burton
I think you used Chinese remainder theorem.If not then please tell me the complete theorem.
He did not use Chinese Remainder Theorem. Which part of his solution are you confused with?
@Sotiri Komissopoulos – I am confused over the multiplication modulo 4...
@Krishna Jha – That is simply the individual remainders left when 444 divides the given odd numbers which will obviously be 333 or 111 and you can easily find out the number of such terms which yield 111 or 333 as remainder.
@Aditya Parson – why only with 4, why not other?
@Budha Chaitanya – Because 444 divides 100100100, and looking at the last two digits of a number is essentially considering de remainder when that number is divided by 100100100. As 444 divides 100100100, and we know that the number is one less than a multiple of 444, then the last two digits of the number are also one less than a multiple of 444. Hence, it's a nice trick to use here, as 757575 is one less than a multiple of 444, and 252525 is not.
@Tim Vermeulen – thank you
Its 75... courtesy mathematica ... :-P...Bt how do we solve these sort of problems is still a mystery to me..
There is no theorem that says "oh, if you multiply the first 50 odd digits then the last 2 digits are 75." Rather, it takes intuition along with a knowledge of mathematical theorems. In this case, somebody figured "okay, well, there are more than 1 factors of "5" in this, but no factors of 2 so it must end in either 25 or 75. Then, to figure out if it was a 75 or 25 at the end, he counted the numbers of factors of 3, because any number ending in 25 when multiplied by 3 ends with 75, and any number ending in 75 multiplied by 3 ends with 25.
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It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced.
Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit.
@Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form.
A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts
I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it.
Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis.
Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)?
No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet.
@MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it.
Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow.
@QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary.
@Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer.
@QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits...
@QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right.
OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ...
So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study?
> I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago
In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a...
@MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really.
When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.?
@tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...)
@MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences).
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The cornerstone of the argument is the following:
If the cycle attack works, then you can factor $n$ (see details below). The attacker can choose $e$. I.e., when trying to factor $n$, the attacker is not constrained to use the specific $e$ which you selected for your public key; he can invent his own $e$, since he will do all the computations himself. Therefore, the "cycle attack" is a generic factorization algorithm, against which you cannot protect yourself by selecting a specific $e$; instead, you rely on the cycle attack being hard for any $e$ that the attacker may conceivably choose.
Fortunately, there are only very few values of $e$ with a short cycle length modulo $\phi(n)$. Indeed, $\phi(\phi(n)) = \phi((p-1)(q-1))$; on average, the biggest prime factor of $p-1$ (let's call it $r$) will have a size close to about 30% of the size of $p-1$, i.e. at least 150 bits for a 1024-bit RSA modulus. This implies that:
If the attacker chooses a $e$ with an order multiple of $r$, then the cycle length $k$ will be at least as big as $r$, hence way too long for the attack to be feasible; The chances of a random $e$ having an order which is not a multiple of $r$, are at most $1/r$, i.e. way too small for the attacker hitting one out of pure luck.
There is no known way to efficiently find an $e$ such that the cycle attack is not intolerably expensive; as shown above, a random $e$ is not good enough. In other words, the "cycle attack" does not seem to be an efficient factorization algorithm.
About turning a cycle into a factorization
If you find $e$ and $k$ which yield a cycle, then you know that $x = e^k-1$ is a multiple of $\phi(n)$. Write $x = 2^f·y$ for an integer $f$ and an odd integer $y$ ($e$ is a potential RSA public exponent, hence odd; therefore, $x$ is even, which means that $f \geq 1$).
Now select a random $a$ modulo $n$ such that $\left(\frac{a}{n}\right) = -1$ (that's the Jacobi symbol and it can be efficiently computed for any $a$ and $n$); about half of the integers modulo $n$ have such a property, so you will not have to search long for that. Such an $a$ is then a square modulo $p$ but not modulo $q$, or vice versa. Now do the following:
Compute $b = a^y \pmod n$. If $b = 1$ of $b = n-1$, try again with another $a$. Compute $c = b^2 \pmod n$. If $c = n-1$, then start again this algorithm with another $a$. If $c = 1$, then compute the GCD of $b-1$ and $n$: this will yield a non-trivial factor of $n$. Otherwise, set $b = c$, and start again this step.
The first step will succeed with probability at least $1/2$. The second step will loop at most $f$ times, and will yield a non-trivial factor of $n$ with probability at least $1/2$. Therefore, you will need, on average, no more than four values $a$ to factor $n$, given the cycle length $k$.
Note, though, that $k$ may be quite big. The conditions which we work with imply that it is computationally feasible for the attacker to raise an integer modulo $n$ to the power $e$, and do so $k$ times -- this does not formally imply that the attacker can
store an integer of size $k$ times the size of $e$. Given $k$, finding $f$ is easy by dichotomy (compute $e^k$ modulo $2^f$ for various values of $f$ until you find the biggest for which the result is still $1$). Knowing $f$ and $k$, it should be possible to compute $a^{(e^k-1)/2^f} \pmod n$ with about the same cost than computing $m^{e^k} \pmod n$ took in the first place (it is a bit tricky because it involves using very big integers, and I am too lazy to work out the details this morning, but I instinctively find it plausible).
This leads to a factorization effort of roughly four times the work of the cycle-length-finding effort. That's close enough for my purposes: I want to show that the "cycle attack" cannot be efficient, because otherwise it could be turned into an almost as efficient factorization algorithm.
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Definition:Lower Sum Definition
Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.
Let $P = \left\{{x_0, x_1, x_2, \ldots, x_n}\right\}$ be a finite subdivision of $\left[{a \,.\,.\, b}\right]$. Then: $\displaystyle L^{\left({f}\right)} \left({P}\right) = \sum_{\nu \mathop = 1}^n m_\nu^{\left({f}\right)} \left({x_{\nu} - x_{\nu - 1}}\right)$
is called the
lower sum of $f$ on $\left[{a \,.\,.\, b}\right]$ belonging (or with respect) to (the subdivision) $P$. If there is no ambiguity as to what function is under discussion, $m_\nu$ and $L \left({P}\right)$ are often seen. Also known as
The notation $L \left({P, f}\right)$ can be used in place of $L^{\left({f}\right)} \left({P}\right)$.
The lower sum is also known as the lower Darboux sum or the lower Riemann sum. Also see
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I have been looking for (linear) Extension Operators for Slobodeckii spaces $W^{s,p}(\Omega)$ where $s>1$ and $\Omega \subset\mathbb{R}^N$ is a sufficiently smooth domain, where the influence of $\partial \Omega$ on the constants which define the continuity of this operator are given.
Indeed, by considering $E:W^{s,p}(\Omega)\to W^{s,p}(\mathbb{R}^N)$ such extension, there exists $C=C(N,s,p,\partial \Omega, \Omega)>0$ for which $$ \|E(u)\|_{W^{s,p}(\mathbb{R}^N)}\leq C \|(u)\|_{W^{s,p}(\Omega)}, \forall u \in W^{s,p}(\Omega)$$. I would like to find a reference where the dependence of $\partial \Omega$ is given explicitly.
Despite that the case $s<1$ has some good references, I can't find any good reference for $s>1$. Actually, when $s<1$, there is the nowadays classic paper by Di Nezza, Palatucci, and Valdinoci, Bull. Sci. Math. 136 (2012), no. 5, 521–573, where this kind of operator is build with certain detail. (see Thm. 5.4).
Moreover, there is also the paper of Rychov, JLMS (1999) where a universal extension operator for Besov and Triebel spaces is constructed, however, the dependence of $C$ on $\partial \Omega$ is not given.
Does anyone know any reference for this?
L.
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Are two signals are the same if their
auto-convolution functions are the same?
Almost. Look at the autoconvolution in the frequency domain where the autoconvolution of $x$ (with itself) gives us $(X(f))^2$ in the frequency domain while the autoconvolution of $-x$ (with itself) gives us $(-X(f))^2 = (X(f))^2$. So, given an autoconvolution function, there are two (very related) signals $x$ and $-x$ that have the same autoconvolution function.
Are two signals are the same if their
auto-correlation functions are the same?
Not quite. Now we are given $|X(f)|^2$ in the frequency domain and there are many different factorizations possible. For example, if $y(t)$ is a signal such that values taken in by its Fourier transform always lie on the unit circle in the complex plane (for every $f$, $|Y(f)|=1$) then $|X(f)Y(f)|^2 = |X(f)|^2$ and so $x\star y$ has the same auto-correlation function as $x$. Note that $x(t)$ and $x(t-\tau)$ (which is a
delayed version of $x(t)$) have the same autocorrelation function ($Y(f)$ happens to be $\exp(-j2\pi f \tau)$ here). Another factorization replaces $X(f)$ by $X^*(f)$ which tells us that $x(t)$ and $x(-t)$ (which is just $x(t)$ running backwards in time) have the same autocorrelation function.
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A neural network is a non-linear classifier (separator is not a linear function). It can also be used for regression.
A Shallow neural network is a one hidden layer neural network.
A Vanilla neural network is a regular neural network having layers that do not form cycles.
TensorFlow Playground is an interactive web interface for learning neural networks: http://playground.tensorflow.org.
Computational Graph
Above the computational graph for the function \(f(x) = (x-1)^2\).
Forward propagation
To minimize the function f, we assign a random value to x (e.g. x = 2), then we evaluate y, z, and f (forward propagation).
Backward propagation
Then we compute the partial derivative of f with respect to x step by step (Backward propagation).
\(\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}*\frac{\partial y}{\partial x} + \frac{\partial f}{\partial z}*\frac{\partial z}{\partial x} = 2 \\ \frac{\partial f}{\partial y} = z = 1 \\ \frac{\partial f}{\partial z} = y = 1 \\ \frac{\partial y}{\partial x} = \frac{\partial z}{\partial x} = 1\)
Then we update \(x:= x – ?.\frac{\partial f}{\partial x}\).
We repeat the operation until convergence.
Activation functions
Activation functions introduce nonlinearity into models. The most used activation functions are:
Sigmoid
\(f(x) = \frac{1}{1+exp(-x)}\)
Sigmoid has a positive and non-zero centred output (sigmoid(0) ? 0.5).
When all activation units are positive, then weight update will be in the same direction (all positive or all negative updates) and that will cause a zigzag path during optimization.
\(z=?w_i.a_i+b \\ \frac{dL}{dw_i}=\frac{dL}{dz}.\frac{dz}{dw_i}=\frac{dL}{dz}.ai\)
If all ai>0, then the gradient will have the same sign as \(\frac{dL}{dz}\) (all positive or all negative).
TanH
\(f(x) = \frac{2}{1+exp(-2x)} -1\)
When x is large, the derivative of the sigmoid or Tanh function is around zero (vanishing gradient/saturation).
ReLU (Rectified Linear Unit)
f(x) = max(0, x)
Leaky ReLU
f(x) = max(0.01x, x)
Leaky Relu was introduced to fix the �Dying Relu� problem.
\(z=?w_i.a_i+b \\ f=Relu(z) \\ \frac{dL}{dw_i}=\frac{dL}{df}.\frac{df}{dz}.\frac{dz}{dw_i}\)
When z becomes negative, then the derivative of f becomes equal to zero, and the weights stop being updated.
PRelu (Parametric Rectifier)
f(x) = max(?.x, x)
ELU (Exponential Linear Unit)
f(x) = {x if x>0 otherwise ?.(exp(x)-1)}
Other activation functions: Maxout
Cost function
\(J(?) = \frac{1}{m} \sum_{i=1}^{m} loss(y^{(i)}, f(x^{(i)}; ?))\)
We need to find ? that minimizes the cost function: \(\underset{?}{argmin}\ J(?)\)
Neural Network Regression
Neural Network regression has no activation function at the output layer.
L1 Loss function
\(loss(y,?) = |y – ?|\)
L2 Loss function
\(loss(y,?) = (y – ?)^2\)
Hinge loss function
Hinge loss function is recommended when there are some outliers in the data.
\(loss(y,?) = max(0, |y-?| – m)\)
Two-Class Neural Network
Binary Cross Entropy Loss function
\(loss(y,?) = – y.log(?) – (1-y).log(1 – ?)\)
Multi-Class Neural Network – One-Task
Using Softmax, the output ? is modeled as a probability distribution, therefore we can assign only one label to each example.
Cross Entropy Loss function
\(loss(Y,\widehat{Y}) = -\sum_{j=1}^c Y_{j}.log(\widehat{Y}_{j})\)
Hinge Loss (SVM) function
\(y = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},\ ? = \begin{bmatrix} 2 \\ -5 \\ 3 \end{bmatrix} \\ loss(y,?) = \sum_{c?1} max(0, ?_c – ?_1 + m)\)
For m = 1, the sum will be equal to 2.
Multi-Class Neural Network – Multi-Task
In this version, we assign multiple labels to each example.
Loss function
\(loss(Y,\widehat{Y}) = \sum_{j=1}^c – Y_j.log(\widehat{Y}_j) – (1-Y_j).log(1 – \widehat{Y}_j)\)
Regularization
Regularization is a very important technique to prevent overfitting.
Dropout
For each training example, ignore randomly p activation nodes of each hidden layer. p is called dropout rate (p?[0,1]). When testing, scale activations by the dropout rate p.
Inverted Dropout
With inverted dropout, scaling is applied at the training time, but inversely. First, dropout all activations by dropout factor p, and second, scale them by inverse dropout factor 1/p. Nothing needs to be applied at test time.
Data Augmentation
As a regularization technique, we can apply random transformations on input images when training a model.
Early stopping
Stop when error rates decreases on training data while it increases on dev (cross-validation) data.
L1 regularization
\(J(?) = \frac{1}{m} \sum_{i=1}^{m} loss(y^{(i)}, f(x^{(i)}; ?)) \color{blue} { + ? .\sum_{j} |?_j|} \)
? is called regularization parameter
L2 regularization
\(J(?) = \frac{1}{m} \sum_{i=1}^{m} loss(y^{(i)}, f(x^{(i)}; ?)) \color{blue} { + ? .\sum_{j} ?_j^2} \)
Lp regularization
\(J(?) = \frac{1}{m} \sum_{i=1}^{m} loss(y^{(i)}, f(x^{(i)}; ?)) \color{blue} { + ? .\sum_{j} ?_j^p} \)
For example, if the cost function \(J(?)=(?_1 – 1)^2 + (?_2 – 1)^2\), then the \(L_2\) regularized cost function is \(J(?)=(?_1 – 1)^2 + (?_2 – 1)^2 + ? (?_1^2 + ?_2^2)\)
If ? is large, then the point that minimizes the regularized J(?) will be around (0,0) –> Underfitting.
If ? ~ 0, then the point that minimizes the regularized J(?) will be around (1,1) –> Overfitting.
Elastic net
Combination of L1 and L2 regularizations.
Normalization
Gradient descent converges quickly when data is normalized Xi ? [-1,1]. If features have different scales, then the update of parameters will not be in the same scale (zig-zag).
For example, if the activation function g is the sigmoid function, then when W.x+b is large g(W.x+b) is around 1, but the derivative of the sigmoid function is around zero. For this reason the gradient converges slowly when the W.x+b is large.
Below some normalization functions.
ZScore
\(X:= \frac{X – ?}{?}\)
MinMax
\(X:= \frac{X – min}{max-min}\)
Logistic
\(X:= \frac{1}{1+exp(-X)}\)
LogNormal
\(X:= \frac{1}{?\sqrt{2?}} \int_{0}^{X} \frac{exp(\frac{-(ln(t) – ?)^2}{2?^2})}{t} dt\)
Tanh
\(X:= tanh(X)\)
Weight Initialization
Weight initialization is important because if weights are too big then activations explode. If weights are too small then gradients will be around zero (no learning).
When we normalize input data, we make the mean of the input features equals to zero, and the variance equals to one. To keep the activation units normalized too, we can initialize the weights \( W^{(1)}\) so \(Var(g(W_{j}^{(1)}.x+b_{j}^{(1)}))\) is equals to one.
If we suppose that g is Relu and \(W_{i,j}, b_j, x_i\) are independent, then:
\(Var(g(W_{j}^{(1)}.x+b_{j}^{(1)})) = Var(\sum_{i} W_{i,j}^{(1)}.x_i+b_{j}^{(1)}) =\sum_{i} Var(W_{i,j}^{(1)}.x_i) + 0 \\ = \sum_{i} E(x_i)^2.Var(W_{i,j}^{(1)}) + E(W_{i,j}^{(1)})^2.Var(x_i) + Var(W_{i,j}^{(1)}).Var(x_i) \\ = \sum_{i} E(x_i)^2.Var(W_{i,j}^{(1)}) + E(W_{i,j}^{(1)})^2.Var(x_i) + Var(W_{i,j}^{(1)}).Var(x_i) \\ = \sum_{i} 0 + 0 + Var(W_{i,j}^{(1)}).Var(x_i) = n.Var(W_{i,j}^{(1)}).Var(x_i) \)
Xavier initialization
If we define \(W_{i,j}^{(1)} ? N(0,\frac{1}{\sqrt{n}})\), then the initial variance of activation units will be one (n is number of input units).
We can apply this rule on all weights of the neural network.
Batch Normalization
Batch normalization is a technique to provide any layer in a Neural Network with normalized inputs. Batch Normalization has a regularizing effect.
After training, ? will converge to the standard deviation of the mini-batches and ? will converge to the mean. The ?, ? parameters give more flexibility when shifting or scaling is needed.
Hyperparameters
Neural network hyperparameters are:
Learning rate (?) (e.g. 0.1, 0.01, 0.001,…) Number of hidden units Number of layers Mini-bach size Momentum rate (e.g. 0.9) Adam optimization parameters (e.g. ?1=0.9, ?2=0.999, ?=0.00000001) Learning rate decay Local Minimum
The probability that gradient descent gets stuck in a local minimum in a high dimensional space is extremely low. We could have a saddle point, but it�s rare to have a local minimum.
Transfer Learning
Transfer Learning consists in the use of parameters of a trained model when training new hidden layers of an extended version of that model.
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Counter-question: why should every type be inhabited by a term? You could not have the Curry-Howard correspondence between typing systems and logic if every type was inhabited.
Concrete answer: I don't have Pierce's book handy, but I think you are talking about the system $\lambda\underline{\omega}$ in Barendregt's $\lambda$-cube, a classification of typing systems along three orthogonal axes. (Please correct me if I'm wrong.)
$\lambda\underline{\omega}$ is the simplest extension of the simply typed $\lambda$-calculus allowing
type-level computation. Note that $\lambda\underline{\omega}$ doesn't have parametric polymorphism ortype dependency. One way of thinking about $\lambda\underline{\omega}$ is that type-level computation is carried out by having another $\lambda$-calculus, but this time at the type level.You can think of type-level computation as being run at 'compile time'. Why use a typed language to run type-level computation? Why not use the untyped $\lambda$-calculus at the type level? Because the things that could go wrong at the term level with untyped terms (e.g. ill-formed programs like $3 + hello$) could now go wrong at the type level (e.g. $\mathbb{B}\; Pair$). So $\lambda\underline{\omega}$ needs a way of preventing ill-formed type-level computation.
But how? Well, let's use the simply typed $\lambda$-calculus again, but now at the type-level, to carry out, and constrain type-level computation? In order to avoid terminological confusion, we speak of kinds of for this second simply typed $\lambda$-calculus. In summary:
(As an aside, you can iterate this and have kind-level computation in the same way and so on, but that's not done in $\lambda\underline{\omega}$.)
So far, I've said that kinds classify types. That's another way of saying that well-formed programs for type-level computation inhabit kinds. In $\lambda\underline{\omega}$ kinds are given by the grammar
$$\newcommand{\TY}{\mathsf{Ty}}\kappa\quad ::= \quad \TY\ \ |\ \ \kappa \rightarrow \kappa$$
Here $\kappa \rightarrow \kappa'$ is the kind of type-level functions such as$\lambda t^{\kappa}. \mathbb{N} \rightarrow \kappa$.But what is $\TY$?Answer: the base kind. The kind that is inhabited by all types that can potentially be inhabited by terms, e.g. types like $\mathbb{B}$or $\mathbb{N} \rightarrow \mathbb{B}$. It is the only kind that isinhabited by types. This gives a neat classification of type-level programs intotypes usable to be inhabited by terms, and type-level programs that are only used as components in the computation of such types.
Operators like $Pair$ are not kinded by $\TY$ and so cannot beinhabited by terms. Instead $Pair$ has the kind $\TY$ $\rightarrow$$\TY$ $\rightarrow$ $\TY$, which, by its very shape, cannot beinhabited by types, it can only be used as a program in a type-levelcomputation. Now $\mathbb{N}$ (integers) and $\mathbb{B}$ (Booleans)both are kinded $\TY$, hence the type level program $Pair\;\mathbb{N} \;\mathbb{B}$ also has kind $\TY$ and can therefore beinhabited by a program.
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Dini criterion for the convergence of Fourier series
A criterion first proved by Dini for the convergence of Fourier series in [Di].
TheoremConsider a summable $2\pi$ periodic function $f$ and a point $x\in \mathbb R$. If thereis a number $S$ and a $\delta>0$ such that\begin{equation}\label{e:Dini}\int_0^\delta |f(x+u) + f(x-u)-2S| \frac{du}{u} < \infty\end{equation}then the Fourier series of $f$ converges to $S$ at $x$.
Cp. with Section 38 of Chapter I in volume 1 of [Ba] and Section 6 of Chapter II in volume 1 of [Zy]. Observe that, if \eqref{e:Dini} holds and in addition the limit $\lim_{t\downarrow0}[f(x+t)+f(x-t)]$ exists, then $S=\lim_{t\downarrow0}[f(x+t)+f(x-t)]$. [Di]
From Dini's statement it is possible to conclude several classical corollaries, for instance
the convergence of the Fourier series of $f$ to $f(x)$ at every point where $f$ is differentiable the convergence of the Fourier series of $f$ to $f$ when $f$ is Hölder continuous.
It is also a (sharp) statement in the following sense. If $\omega: ]0, \infty[\to ]0, \infty[$ is a continuous function such that $\frac{\omega (t)}{t}$ is not integrable in a neighborhood of the origin, then there is a continuous $2\pi$-periodic function $f:\mathbb R \to \mathbb R$ such that $|f(t)-f(0)|\leq \omega (|t|)$ for every $t$ and the Fourier series of $f$ diverges at $0$.
References
[Ba] N.K. Bary, "A treatise on trigonometric series" , Pergamon, 1964. [Di] U. Dini, "Serie di Fourier e altre rappresentazioni analitiche delle funzioni di una variabile reale" , Pisa (1880). [Ed] R. E. Edwards, "Fourier series". Vol. 1. Holt, Rineheart and Winston, 1967. [Zy] A. Zygmund, "Trigonometric series" , 1–2 , Cambridge Univ. Press (1988) MR0933759 Zbl 0628.42001 How to Cite This Entry:
Dini criterion.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Dini_criterion&oldid=28457
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also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling
@heather well, there's a spectrum
so, there's things like New Journal of Physics and Physical Review X
which are the open-access branch of existing academic-society publishers
As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di...
Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago
> A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service”
for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty
> for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing)
@0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals.
@BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work...
@BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions.
Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley.
I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea.
@EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results...
Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town...
@EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit.
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Now showing items 1-10 of 26
Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
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Now showing items 1-1 of 1
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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We've seen that classical logic is closely connected to the logic of subsets. For any set \( X \) we get a poset \( P(X) \), the
power set of \(X\), whose elements are subsets of \(X\), with the partial order being \( \subseteq \). If \( X \) is a set of "states" of the world, elements of \( P(X) \) are "propositions" about the world. Less grandiosely, if \( X \) is the set of states of any system, elements of \( P(X) \) are propositions about that system.
This trick turns logical operations on propositions - like "and" and "or" - into operations on subsets, like intersection \(\cap\) and union \(\cup\). And these operations are then special cases of things we can do in
other posets, too, like join \(\vee\) and meet \(\wedge\).
We could march much further in this direction. I won't, but try it yourself!
Puzzle 22. What operation on subsets corresponds to the logical operation "not"? Describe this operation in the language of posets, so it has a chance of generalizing to other posets. Based on your description, find some posets that do have a "not" operation and some that don't.
I want to march in another direction. Suppose we have a function \(f : X \to Y\) between sets. This could describe an
observation, or measurement. For example, \( X \) could be the set of states of your room, and \( Y \) could be the set of states of a thermometer in your room: that is, thermometer readings. Then for any state \( x \) of your room there will be a thermometer reading, the temperature of your room, which we can call \( f(x) \).
This should yield some function between \( P(X) \), the set of propositions about your room, and \( P(Y) \), the set of propositions about your thermometer. It does. But in fact there are
three such functions! And they're related in a beautiful way!
The most fundamental is this:
Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq Y \) define its inverse image under \(f\) to be
$$ f^{\ast}(S) = \{x \in X: \; f(x) \in S\} . $$ The pullback is a subset of \( X \).
The inverse image is also called the
preimage, and it's often written as \(f^{-1}(S)\). That's okay, but I won't do that: I don't want to fool you into thinking \(f\) needs to have an inverse \( f^{-1} \) - it doesn't. Also, I want to match the notation in Example 1.89 of Seven Sketches.
The inverse image gives a monotone function
$$ f^{\ast}: P(Y) \to P(X), $$ since if \(S,T \in P(Y)\) and \(S \subseteq T \) then
$$ f^{\ast}(S) = \{x \in X: \; f(x) \in S\}
\subseteq \{x \in X:\; f(x) \in T\} = f^{\ast}(T) . $$ Why is this so fundamental? Simple: in our example, propositions about the state of your thermometer give propositions about the state of your room! If the thermometer says it's 35°, then your room is 35°, at least near your thermometer. Propositions about the measuring apparatus are useful because they give propositions about the system it's measuring - that's what measurement is all about! This explains the "backwards" nature of the function \(f^{\ast}: P(Y) \to P(X)\), going back from \(P(Y)\) to \(P(X)\).
Propositions about the system being measured also give propositions about the measurement apparatus, but this is more tricky. What does "there's a living cat in my room" tell us about the temperature I read on my thermometer? This is a bit confusing... but there is an answer because a function \(f\) really does also give a "forwards" function from \(P(X) \) to \(P(Y)\). Here it is:
Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq X \) define its image under \(f\) to be
$$ f_{!}(S) = \{y \in Y: \; y = f(x) \textrm{ for some } x \in S\} . $$ The image is a subset of \( Y \).
The image is often written as \(f(S)\), but I'm using the notation of
Seven Sketches, which comes from category theory. People pronounce \(f_{!}\) as "\(f\) lower shriek".
The image gives a monotone function
$$ f_{!}: P(X) \to P(Y) $$ since if \(S,T \in P(X)\) and \(S \subseteq T \) then
$$f_{!}(S) = \{y \in Y: \; y = f(x) \textrm{ for some } x \in S \}
\subseteq \{y \in Y: \; y = f(x) \textrm{ for some } x \in T \} = f_{!}(T) . $$ But here's the cool part: Theorem. \( f_{!}: P(X) \to P(Y) \) is the left adjoint of \( f^{\ast}: P(Y) \to P(X) \). Proof. We need to show that for any \(S \subseteq X\) and \(T \subseteq Y\) we have
$$ f_{!}(S) \subseteq T \textrm{ if and only if } S \subseteq f^{\ast}(T) . $$ David Tanzer gave a quick proof in Puzzle 19. It goes like this: \(f_{!}(S) \subseteq T\) is true if and only if \(f\) maps elements of \(S\) to elements of \(T\), which is true if and only if \( S \subseteq \{x \in X: \; f(x) \in T\} = f^{\ast}(T) \). \(\quad \blacksquare\)
This is great! But there's also
another way to go forwards from \(P(X)\) to \(P(Y)\), which is a right adjoint of \( f^{\ast}: P(Y) \to P(X) \). This is less widely known, and I don't even know a simple name for it. Apparently it's less useful. Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq X \) define
$$ f_{\ast}(S) = \{y \in Y: x \in S \textrm{ for all } x \textrm{ such that } y = f(x)\} . $$ This is a subset of \(Y \).
Puzzle 23. Show that \( f_{\ast}: P(X) \to P(Y) \) is the right adjoint of \( f^{\ast}: P(Y) \to P(X) \).
What's amazing is this. Here's another way of describing our friend \(f_{!}\). For any \(S \subseteq X \) we have
$$ f_{!}(S) = \{y \in Y: x \in S \textrm{ for some } x \textrm{ such that } y = f(x)\} . $$This looks almost exactly like \(f_{\ast}\). The only difference is that while the left adjoint \(f_{!}\) is defined using "for some", the right adjoint \(f_{\ast}\) is defined using "for all". In logic "for some \(x\)" is called the
existential quantifier \(\exists x\), and "for all \(x\)" is called the universal quantifier \(\forall x\). So we are seeing that existential and universal quantifiers arise as left and right adjoints!
This was discovered by Bill Lawvere in this revolutionary paper:
By now this observation is part of a big story that "explains" logic using category theory.
Two more puzzles! Let \( X \) be the set of states of your room, and \( Y \) the set of states of a thermometer in your room: that is, thermometer readings. Let \(f : X \to Y \) map any state of your room to the thermometer reading.
Puzzle 24. What is \(f_{!}(\{\text{there is a living cat in your room}\})\)? How is this an example of the "liberal" or "generous" nature of left adjoints, meaning that they're a "best approximation from above"? Puzzle 25. What is \(f_{\ast}(\{\text{there is a living cat in your room}\})\)? How is this an example of the "conservative" or "cautious" nature of right adjoints, meaning that they're a "best approximation from below"?
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I recently gave a talk on Schubert calculus in the Student Algebraic Geometry Seminar at UC Berkeley. As a combinatorialist, this is a branch of enumerative geometry that particularly strikes my fancy. I also made a handout for the talk called “Combinatorics for Algebraic Geometers,” and I thought I’d post it here in blog format.
Motivation Enumerative geometry
In the late 1800’s, Hermann Schubert investigated problems in what is now called enumerative geometry, or more specifically, Schubert calculus. As some examples, where all projective spaces are assumed to be over the complex numbers:
How many lines in $\newcommand{\PP}{\mathbb{P}} \newcommand{\CC}{\mathbb{C}} \newcommand{\RR}{\mathbb{R}} \newcommand{\ZZ}{\mathbb{Z}} \DeclareMathOperator{\Gr}{Gr} \DeclareMathOperator{\Fl}{Fl} \DeclareMathOperator{\GL}{GL}\PP^n$ pass through two given points? How many planes in $\PP^3$ contain a given line $l$ and a given point $P$? How many lines in $\PP^3$ intersect four given lines $l_1,l_2,l_3,l_4$? How many $(r-1)$-dimensional subspaces of $\PP^{m-1}$ intersect each of $r\cdot (m-r)$ general subspaces of dimension $m-r-1$ nontrivially? Answer: One, as long as the points are distinct. Answer: One, as long as $P\not\in l$. Answer: Two, as long as the lines are in sufficiently “general” position. Answer: $$\frac{(r(m-r))!\cdot (r-1)!\cdot (r-2)!\cdot \cdots\cdot 1!}{(m-1)!\cdot(m-2)!\cdot\cdots\cdot 1!}$$
The first two questions are not hard, but how would we figure out the other two? And what do we mean by “sufficiently general position”?
Schubert’s 19th century solution to problem 3 above would have invoked what he called the “Principle of Conservation of Number,” as follows. Suppose the four lines were arranged so that $l_1$ and $l_2$ intersect at a point $P$, $l_3$ and $l_4$ intersect at $Q$, and none of the other pairs of lines intersect. Then the planes formed by each pair of crossing lines intersect at another line $\alpha$, which necessarily intersects all four lines. The line $\beta$ through $P$ and $Q$ also intersects all four lines, and it is not hard to see that these are the only two in this case.
Schubert would have said that since there are two solutions in this configuration and it is a finite number of solutions, it is true for every configuration of lines for which the number is finite by continuity. Unfortunately, due to degenerate cases involving counting with multiplicity, this led to many errors in computations in harder questions of enumerative geometry. Hilbert’s 15th problem asked to put Schubert’s enumerative methods on a rigorous foundation. This led to the modern-day theory known as Schubert calculus.
Describing moduli spaces
Schubert calculus can also be used to describe intersection properties in simpler ways. As we will see, it will allow us to easily prove statements such as:
The variety of all lines in $\PP^4$ that are both contained in a general $3$-dimensional hyperplane $S$ and intersect a general line $l$ nontrivially is isomorphic to the variety of all lines in $S$ passing through a specific point in that hyperplane.
(Here, the specific point in the hyperplane is the intersection of $S$ and $L$.)
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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I recently gave a talk on Schubert calculus in the Student Algebraic Geometry Seminar at UC Berkeley. As a combinatorialist, this is a branch of enumerative geometry that particularly strikes my fancy. I also made a handout for the talk called “Combinatorics for Algebraic Geometers,” and I thought I’d post it here in blog format.
Motivation Enumerative geometry
In the late 1800’s, Hermann Schubert investigated problems in what is now called enumerative geometry, or more specifically, Schubert calculus. As some examples, where all projective spaces are assumed to be over the complex numbers:
How many lines in $\newcommand{\PP}{\mathbb{P}} \newcommand{\CC}{\mathbb{C}} \newcommand{\RR}{\mathbb{R}} \newcommand{\ZZ}{\mathbb{Z}} \DeclareMathOperator{\Gr}{Gr} \DeclareMathOperator{\Fl}{Fl} \DeclareMathOperator{\GL}{GL}\PP^n$ pass through two given points? How many planes in $\PP^3$ contain a given line $l$ and a given point $P$? How many lines in $\PP^3$ intersect four given lines $l_1,l_2,l_3,l_4$? How many $(r-1)$-dimensional subspaces of $\PP^{m-1}$ intersect each of $r\cdot (m-r)$ general subspaces of dimension $m-r-1$ nontrivially? Answer: One, as long as the points are distinct. Answer: One, as long as $P\not\in l$. Answer: Two, as long as the lines are in sufficiently “general” position. Answer: $$\frac{(r(m-r))!\cdot (r-1)!\cdot (r-2)!\cdot \cdots\cdot 1!}{(m-1)!\cdot(m-2)!\cdot\cdots\cdot 1!}$$
The first two questions are not hard, but how would we figure out the other two? And what do we mean by “sufficiently general position”?
Schubert’s 19th century solution to problem 3 above would have invoked what he called the “Principle of Conservation of Number,” as follows. Suppose the four lines were arranged so that $l_1$ and $l_2$ intersect at a point $P$, $l_3$ and $l_4$ intersect at $Q$, and none of the other pairs of lines intersect. Then the planes formed by each pair of crossing lines intersect at another line $\alpha$, which necessarily intersects all four lines. The line $\beta$ through $P$ and $Q$ also intersects all four lines, and it is not hard to see that these are the only two in this case.
Schubert would have said that since there are two solutions in this configuration and it is a finite number of solutions, it is true for every configuration of lines for which the number is finite by continuity. Unfortunately, due to degenerate cases involving counting with multiplicity, this led to many errors in computations in harder questions of enumerative geometry. Hilbert’s 15th problem asked to put Schubert’s enumerative methods on a rigorous foundation. This led to the modern-day theory known as Schubert calculus.
Describing moduli spaces
Schubert calculus can also be used to describe intersection properties in simpler ways. As we will see, it will allow us to easily prove statements such as:
The variety of all lines in $\PP^4$ that are both contained in a general $3$-dimensional hyperplane $S$ and intersect a general line $l$ nontrivially is isomorphic to the variety of all lines in $S$ passing through a specific point in that hyperplane.
(Here, the specific point in the hyperplane is the intersection of $S$ and $L$.)
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I have a function
fthat I want to restrict to a certain sub-domain of the domain it has been originally defined for. Usually, in mathematics this is illustrated by a vertical bar followed by the restricted set in the lower right of the function symbol. E.g. usually I write
f_{\mid A}
which means
f restricted to the set
A.
Unfortunately, I have a sequence of functions
f_i that has already a subscript. If I write
f_{i_{{\mid A}}}
I do net get an LaTeX error, but the result is semantically wrong.
\bar A is
not a subscript of
i which happens to be a subscript of
f, but
\mid A is a subscript of
f_i. Especially, the font size of
{\mid A} should not be smaller than the size of
i, but have the same size.
However, if a write
{f_i}_{\mid A}
(which btw would be semantically correct) I get a "double subscript error".
Alternatively, I could write
f_{i \; \mid A}
and get the correct font size for
\mid A but it is placed too high.
\mid A should be below
i.
MWE
\documentclass{article}\usepackage[TS1,T1]{fontenc}\usepackage[utf8]{inputenc}\usepackage{lmodern}\begin{document}\begin{equation}{\hat{f}_i}_{\mid A}\end{equation}\end{document}
As I have already found out, the problem occurs only
with the accent above
f. But I need the accent there.
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Journal of Symbolic Logic J. Symbolic Logic Volume 58, Issue 1 (1993), 249-290. On Strong Provability Predicates and the Associated Modal Logics Abstract
PA is Peano Arithmetic. $\mathrm{Pr}(x)$ is the usual $\Sigma_1$-formula representing provability in PA. A strong provability predicate is a formula which has the same properties as $Pr(\cdot)$ but is not $\Sigma_1$. An example: $Q$ is $\omega$-provable if $\mathrm{PA} + \neg Q$ is $\omega$-inconsistent (Boolos [4]). In [5] Dzhaparidze introduced a joint provability logic for iterated $\omega$-provability and obtained its arithmetical completeness. In this paper we prove some further modal properties of Dzhaparidze's logic, e.g., the fixed point property and the Craig interpolation lemma. We also consider other examples of the strong provability predicates and their applications.
Article information Source J. Symbolic Logic, Volume 58, Issue 1 (1993), 249-290. Dates First available in Project Euclid: 6 July 2007 Permanent link to this document https://projecteuclid.org/euclid.jsl/1183744189 Mathematical Reviews number (MathSciNet) MR1217189 Zentralblatt MATH identifier 0795.03082 JSTOR links.jstor.org Citation
Ignatiev, Konstantin N. On Strong Provability Predicates and the Associated Modal Logics. J. Symbolic Logic 58 (1993), no. 1, 249--290. https://projecteuclid.org/euclid.jsl/1183744189
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I’ve been doing a lot of reading on confidence interval theory. Some of the reading is more interesting than others. There is one passage from Neyman’s (1952) book “Lectures and Conferences on Mathematical Statistics and Probability” (available here) that stands above the rest in terms of clarity, style, and humor. I had not read this before the last draft of our confidence interval paper, but for those of you who have read it, you’ll recognize that this is the style I was going for. Maybe you have to be Jerzy Neyman to get away with it.
Neyman gets bonus points for the footnote suggesting the “eminent”, “elderly” boss is so obtuse (a reference to Fisher?) and that the young frequentists should be “remind[ed] of the glory” of being burned at the stake. This is just absolutely fantastic writing. I hope you enjoy it as much as I did.
[begin excerpt, p. 211-215]
[Neyman is discussing using “sampling experiments” (Monte Carlo experiments with tables of random numbers) in order to gain insight into confidence intervals. \(\theta\) is a true parameter of a probability distribution to be estimated.]
The sampling experiments are more easily performed than described in
detail. Therefore, let us make a start with \(\theta_1 = 1\), \(\theta_2 = 2\), \(\theta_3 = 3\) and \(\theta_4 = 4\). We imagine that, perhaps within a week, a practical statistician is faced four times with the problem of estimating \(\theta\), each time from twelve observations, and that the true values of \(\theta\) are as above [ie, \(\theta_1,\ldots,\theta_4\)] although the statistician does not know this. We imagine further that the statistician is an elderly gentleman, greatly attached to the arithmetic mean and that he wishes to use formulae (22). However, the statistician has a young assistant who may have read (and understood) modern literature and prefers formulae (21). Thus, for each of the four instances, we shall give two confidence intervals for \(\theta\), one computed by the elderly Boss, the other by his young Assistant.
[Formula 21 and 22 are simply different 95% confidence procedures. Formula 21 is has better frequentist properties; Formula 22 is inferior, but the Boss likes it because it is intuitive to him.]
Using the first column on the first page of Tippett’s tables of random
numbers and performing the indicated multiplications, we obtain the following four sets of figures.
The last two lines give the assertions regarding the true value of \(\theta\) made by the Boss and by the Assistant, respectively. The purpose of the sampling experiment is to verify the theoretical result that the long run relative frequency of cases in which these assertions will be correct is, approximately, equal to \(\alpha = .95\).
You will notice that in three out of the four cases considered, both assertions (the Boss’ and the Assistant’s) regarding the true value of \(\theta\) are correct and that in the last case both assertions are wrong. In fact, in this last case the true \(\theta\) is 4 while the Boss asserts that it is between 2.026 and 3.993 and the Assistant asserts that it is between 2.996 and 3.846. Although the probability of success in estimating \(\theta\) has been fixed at \(\alpha = .95\), the failure on the fourth trial need not discourage us. In reality, a set of four trials is plainly too short to serve for an estimate of a long run relative frequency. Furthermore, a simple calculation shows that the probability of at least one failure in the course of four independent trials is equal to .1855. Therefore, a group of four consecutive samples like the above, with at least one wrong estimate of \(\theta\), may be expected one time in six or even somewhat oftener. The situation is, more or less, similar to betting on a particular side of a die and seeing it win. However, if you continue the sampling experiment and count the cases in which the assertion regarding the true value of \(\theta\), made by either method, is correct, you will find that the relative frequency of such cases converges gradually to its theoretical value, \(\alpha= .95\).
Let us put this into more precise terms. Suppose you decide on a number \(N\) of samples which you will take and use for estimating the true value of \(\theta\). The true values of the parameter \(\theta\) may be the same in all \(N\) cases or they may vary from one case to another. This is absolutely immaterial as far as the relative frequency of successes in estimation is concerned. In each case the probability that your assertion will be correct is exactly equal to \(\alpha = .95\). Since the samples are taken in a manner insuring independence (this, of course, depends on the goodness of the table of random numbers used), the total number \(Z(N)\) of successes in estimating \(\theta\) is the familiar binomial variable with expectation equal to \(N\alpha\) and with variance equal to \(N\alpha(1 – \alpha)\). Thus, if \(N = 100\), \(\alpha = .95\), it is rather improbable that the relative frequency \(Z(N)/N\) of successes in estimating \(\alpha\) will differ from \(\alpha\) by more than
\(
2\sqrt{\frac{\alpha(1-\alpha)}{N}} = .042
\)
This is the exact meaning of the colloquial description that the long run relative frequency of successes in estimating \(\theta\) is equal to the preassigned \(\alpha\). Your knowledge of the theory of confidence intervals will not be influenced by the sampling experiment described, nor will the experiment prove anything. However, if you perform it, you will get an intuitive feeling of the machinery behind the method which is an excellent complement to the understanding of the theory. This is like learning to drive an automobile: gaining experience by actually driving a car compared with learning the theory by reading a book about driving.
Among other things, the sampling experiment will attract attention to
the frequent difference in the precision of estimating \(\theta\) by means of the two alternative confidence intervals (21) and (22). You will notice, in fact, that the confidence intervals based on \(X\), the greatest observation in the sample, are frequently shorter than those based on the arithmetic mean \(\bar{X}\). If we continue to discuss the sampling experiment in terms of cooperation between the eminent elderly statistician and his young assistant, we shall have occasion to visualize quite amusing scenes of indignation on the one hand and of despair before the impenetrable wall of stiffness of mind and routine of thought on the other. [See footnote] For example, one can imagine the conversation between the two men in connection with the first and third samples reproduced above. You will notice that in both cases the confidence interval of the Assistant is not only shorter than that of the Boss but is completely included in it. Thus, as a result of observing the first sample, the Assistant asserts that
\(
.956 \leq \theta \leq 1.227.
\)
On the other hand, the assertion of the Boss is far more conservative and admits the possibility that \(\theta\) may be as small as .688 and as large as 1.355. And both assertions correspond to the same confidence coefficient, \(\alpha = .95\)! I can just see the face of my eminent colleague redden with indignation and hear the following colloquy.
Boss: “Now, how can this be true? I am to assert that \(\theta\) is between .688 and 1.355 and you tell me that the probability of my being correct is .95. At the same time, you assert that \(\theta\) is between .956 and 1.227 and claim the same probability of success in estimation. We both admit the possibility that \(\theta\) may be some number between .688 and .956 or between 1.227 and 1.355. Thus, the probability of \(\theta\) falling within these intervals is certainly greater than zero. In these circumstances, you have to be a nit-wit to believe that
\(
\begin{eqnarray*}
P\{.688 \leq \theta \leq 1.355\} &=& P\{.688 \leq \theta < .956\} + P\{.956 \leq \theta \leq 1.227\}\\
&& + P\{1.227 \leq \theta \leq 1.355\}\\
&=& P\{.956 \leq \theta \leq 1.227\}.\mbox{”}
\end{eqnarray*}
\) Assistant: “But, Sir, the theory of confidence intervals does not assert anything about the probability that the unknown parameter \(\theta\) will fall within any specified limits. What it does assert is that the probability of success in estimation using either of the two formulae (21) or (22) is equal to \(\alpha\).” Boss: “Stuff and nonsense! I use one of the blessed pair of formulae and come up with the assertion that \(.688 \leq \theta \leq 1.355\). This assertion is a success only if \(\theta\) falls within the limits indicated. Hence, the probability of success is equal to the probability of \(\theta\) falling within these limits —.” Assistant: “No, Sir, it is not. The probability you describe is the a posteriori probability regarding \(\theta\), while we are concerned with something else. Suppose that we continue with the sampling experiment until we have, say, \(N = 100\) samples. You will see, Sir, that the relative frequency of successful estimations using formulae (21) will be about the same as that using formulae (22) and that both will be approximately equal to .95.”
I do hope that the Assistant will not get fired. However, if he does, I would remind him of the glory of Giordano Bruno who was burned at the stake by the Holy Inquisition for believing in the Copernican theory of the solar system. Furthermore, I would advise him to have a talk with a physicist or a biologist or, maybe, with an engineer. They might fail to understand the theory but, if he performs for them the sampling experiment described above, they are likely to be convinced and give him a new job. In due course, the eminent statistical Boss will die or retire and then —.
[footnote] Sad as it is, your mind does become less flexible and less receptive to novel ideas as the years go by. The more mature members of the audience should not take offense. I, myself, am not young and have young assistants. Besides, unreasonable and stubborn individuals are found not only among the elderly but also frequently among young people.
[end excerpt]
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Vague question:Is there a criterion to deduce that a morphism between algebraic stacks is finite based on the local deformation functors?
For sure this is not enough, so let me be more specific.
Suppose that $f: X \to Y$ is a morphism of algebraic stacks. Let us assume in addition that the diagonal $\Delta_f : X \to X \times_Y X$ is finite and unramified.
Now assume that for each point $x \in |X|$ of finite type over $y \in |Y|$ the morphism between the (pro-representing rings of the) deformation functors is finite.
Specific question:Under these conditions, can we assume that $f \colon X \to Y$ is finite? Edit: This too is not enough as Piotr demonstrates in the comments. The key observation is that finiteness is not local on the source and even an open immersion is a counterexample to the question above.
Let me then modify the question like so:
Modified question:What global criteria, such as properness, on the morphism $f:X \to Y$ need to be assumed to make finiteness (formal) local on the source?
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Defining parameters
Level: \( N \) = \( 3600 = 2^{4} \cdot 3^{2} \cdot 5^{2} \) Weight: \( k \) = \( 1 \) Character orbit: \([\chi]\) = 3600.ec (of order \(20\) and degree \(8\)) Character conductor: \(\operatorname{cond}(\chi)\) = \( 400 \) Character field: \(\Q(\zeta_{20})\) Newforms: \( 0 \) Sturm bound: \(720\) Trace bound: \(0\) Dimensions
The following table gives the dimensions of various subspaces of \(M_{1}(3600, [\chi])\).
Total New Old Modular forms 64 16 48 Cusp forms 0 0 0 Eisenstein series 64 16 48
The following table gives the dimensions of subspaces with specified projective image type.
\(D_n\) \(A_4\) \(S_4\) \(A_5\) Dimension 0 0 0 0
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Abstract
We prove two results. (1) There is an absolute constant $D$ such that for any finite quasisimple group $S$, given $2D$ arbitrary automorphisms of $S$, every element of $S$ is equal to a product of $D$ ‘twisted commutators’ defined by the given automorphisms.
(2) Given a natural number $q$, there exist $C=C(q)$ and $M=M(q)$ such that: if $S$ is a finite quasisimple group with $\left| S/\mathrm{Z}(S)\right| > C$, $\beta_{j}$ $ (j=1,\ldots,M)$ are any automorphisms of $S$, and $q_{j}$ $ (j=1,\ldots,M)$ are any divisors of $q$, then there exist inner automorphisms $\alpha_{j}$ of $S$ such that $S=\prod_{1}^{M}[S,(\alpha _{j}\beta_{j})^{q_{j}}]$.
These results, which rely on the classification of finite simple groups, are needed to complete the proofs of the main theorems of Part I.
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How to find the equations of the lines that make up the ruled surface z= 2xy? There are a lot of lines and solving this seems non-obvious. I imagine the first thing to know is the equation of a line in 3d. What is the equation for a 3D line?
The points $\;M=\Bigl(x,\dfrac 2x, 4\Bigr)$ and $\;N=\Bigl(x,-\dfrac 2x, -4\Bigr)$ are on the quadric. One easily check that any point $P=tM+(1-t)N$ also lies on the quadric. Indeed the coordinates of such a point are $$\Bigl(tx+(1-t)x, \dfrac{2t}x-\dfrac{(1-t)2}x, 4t-4(1-t)\Bigr)=\Bigl(x, \dfrac{2(2t-1)}x, 4(2t-1)\Bigr).$$
The lines are given by the planes intersection
$z=xy$ $x=k$
and
$z=xy$ $y=k$
In parametric form for
$z=xy$ $x=k$
we have that a generic point is
$$P=(k,t,tk)=(k,0,0)+t(0,1,k)$$
that is the parametric equation for this family of lines and similarly we can find that for the other family of lines.
A systematic way to find these lines uses the following facts:
If the surface is given by the implicit equation $f(\mathbf x)=0$, then the line $\mathbf p+t\mathbf v$ is a generator iff $f(\mathbf p+t\mathbf v)=0$ for all $t$. For an $n$th-degree surface, this equation is an $n$th-degree polynomial in $t$ and the coordinates of $\mathbf v$.
A generator of the surface lies in a tangent plane to the surface. Therefore, if $\mathbf p$ lies on the surface, then $\nabla f(\mathbf p)\cdot(\mathbf p+t\mathbf v)=0$. If $\nabla f(\mathbf p)\ne0$, this is a linear equation in the coordinates of $\mathbf v$ that can be used to eliminate one of the variables.
The length of $\mathbf v$ is unimportant, but it must be non-zero, so one of its coordinates can be set to $1$. You might have to consider several cases here.
Facts 2 and 3 eliminate two of the variables, leaving you with an equation in $t$ and the remaining variable. Viewed as a polynomial in $t$, all of the coefficients must vanish, which gives you a system of single-variable polynomial equations to solve. You will end up with one or two solutions, depending on whether the surface is singly or doubly ruled.
So, starting with $f(x,y,z)=2xy-z$ we have $$2(p_x+t v_x)(p_y+t v_y)-p_z+t v_z = 0. \tag 1$$ The orthogonality condition of fact 2 gives us $2(p_x v_y+p_yv_x)-v_z=0$, which we can solve for $v_z$ and substitute into (1) to obtain, after simplifying, $$2p_x p_y-p_z+2t^2 v_x v_y = 0. \tag 2$$ Setting $v_x=1$ results in $v_y=0$ and vice-versa, therefore there are two generators through each nonsingular point $\mathbf p$ on the surface: $\mathbf p+t(1,0,2p_y)$ and $\mathbf p+t(0,1,2p_x)$. ${\partial f\over\partial z}=-1$ everywhere, so the gradient never vanishes and there are no singular points to consider.
If you need a directrix curve for this surface, find a plane that’s not parallel to any of these generators and intersect it with the surface. For this one, the plane $x=y$ works, producing a curve that can be parameterized as $(s,s,2s^2)$, which then generates the parameterization $(s,s+t,2s(s+t))$ of the surface.
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The question
I consider the Laplacian $\Delta = \partial_1^2 + \partial_2^2 + \partial_3^2$ in $\mathbb{R}^3$. By the "standard" fundamental solution of the Laplacian, I mean the function
$$ \displaystyle E(x) = C|x|^{-1} $$
where $C$ is some normalization constant.
I would like to know if one can construct a fundamental solution of the Laplacian that decays faster than any rational function at infinity. I have an idea how one could maybe construct this, but I'm not sure if my idea is correct as the idea of a suspicious fundamental solution seems highly suspicious to me. In the following, I sketch my idea: An attempt at an answer
I already know that one can construct a rapidly decaying parametrix $F$ by multiplying $\hat E \sim |x|^{-2}$ with a cutoff function $\chi$ supported in the complement of the unit ball and then taking the inverse Fourier transform. Then, one has $$x^\alpha F(x) = \int D^\alpha(\chi\hat E)(\xi)e^{ix\xi}\text{d}\xi $$ and the right hand side is absolutely convergent for large $\alpha$. This proves the decay at infinity.
I want to modify this construction in order to turn the parametrix $F$ into a proper fundamental solution. To do this, I want to use a simplified version of the construction by Hörmander (see "Distribution" by Duistermaat and Kolk, Theorem 17.11). Letting $\eta$ be a unit vector, I set
$$ G(x) = (2\pi)^{-n} \int \frac{1- \chi(\xi)}{2\pi i}\int_{|z| = 100} \frac{e^{ix\cdot(\xi + z \eta)}}{|\xi + z\eta|^2}\frac{\text{d}z}{z}\text{d}\xi\,.$$
With Cauchy's integral theorem, one can show that $$\tilde E = F + G$$ is indeed a fundamental solution. The idea here is that we avoid the singularity at $0$ by taking a contour integral around it. It seems as if one can use the same trick that one used for $F$ to show the rapid decay of $G$. This suggest that one can indeed find a rapidly deacying fundamental solution.
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How to find multiple roots?
find_root is finding only "one" root, when there are many in a given range? Very simple question. For ex: sin(x)=cos(pi/2+x) may have multiple solutions in the selected range.
find_root is finding only "one" root, when there are many in a given range? Very simple question. For ex: sin(x)=cos(pi/2+x) may have multiple solutions in the selected range.
What's about recursion? See my answer to this question:
Hi,
If you want to find the zeros numerically, my first attempt would be to partition the given interval into smaller subintervals; the precondition for several of these root-finding algorithms (as explained in the docs) is that the function changes sign in the given interval.
If you want to find the roots symbolically, Sage provides the
solve method, you can access the documentation by typing
solve? and the ENTER key. In your example,
sage: solve(sin(x) == cos(pi/2 + x), x)[x == 0]
In this case
solve, which refers to Maxima's solve algorithm, gives only the root at
x=0, although other cases (eg. try with polynomials), it can give all roots. But Sage allows to interface many different mathematical software. We can try SymPy's
solveset:
sage: from sympy import solveset, symbols, sin, cos, pisage: x = symbols('x')sage: solveset(sin(x) - cos(pi/2 + x), x)ImageSet(Lambda(_n, 2*_n*pi), Integers()) U ImageSet(Lambda(_n, 2*_n*pi + pi), Integers())
One interprets this answer as:
$$ S = \{ 2 n \pi : n \in \mathbb{Z}\} \cup \{ 2 n \pi + \pi : n \in \mathbb{Z}\}. $$
The nuisance with this approach is that you have to make many imports by hand.. it is sometimes helpful to use the function
sympy.sympify, as in
sympify(sin(x) - cos(pi/2 + x)), which does the imports automatically.
Asked:
2017-04-23 10:59:42 -0500
Seen:
1,478 times
Last updated:
Apr 24 '17
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Aprahamian, Mary and Higham, Nicholas J. (2013)
The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential. [MIMS Preprint]
PDF
paper.pdf
Download (412kB)
Abstract
A new matrix function corresponding to the scalar unwinding number of Corless, Hare, and Jeffrey is introduced. This matrix unwinding function, $\mathcal{U}$, is shown to be a valuable tool for deriving identities involving the matrix logarithm and fractional matrix powers, revealing, for example, the precise relation between $\log A^\alpha$ and $\alpha \log A$. The unwinding function is also shown to be closely connected with the matrix sign function. An algorithm for computing the unwinding function based on the Schur--Parlett method with a special reordering is proposed. It is shown that matrix argument reduction using the function $\mathrm{mod}(A) = A-2\pi i\, \mathcal{U}(A)$, which has eigenvalues with imaginary parts in the interval $(-\pi,\pi]$ and for which $\e^A = \e^{\mathrm{mod}(A)}$, can give significant computational savings in the evaluation of the exponential by scaling and squaring algorithms.
Item Type: MIMS Preprint Uncontrolled Keywords: matrix unwinding function, unwinding number, matrix logarithm, matrix power, matrix exponential, argument reduction Subjects: MSC 2010, the AMS's Mathematics Subject Classification > 15 Linear and multilinear algebra; matrix theory
MSC 2010, the AMS's Mathematics Subject Classification > 65 Numerical analysis
Depositing User: Nick Higham Date Deposited: 02 Oct 2013 Last Modified: 08 Nov 2017 18:18 URI: http://eprints.maths.manchester.ac.uk/id/eprint/2023 Available Versions of this Item The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential. (deposited 07 May 2013) The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential. (deposited 15 May 2013) The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential. (deposited 02 Oct 2013) [Currently Displayed] The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential. (deposited 02 Oct 2013) The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential. (deposited 15 May 2013) Actions (login required)
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GR8677 #94
Problem
Electromagnetism}LR Circuit
Once the switch S is closed, although the initial current through the inductor is 0, the change in current through it is maximal. Recalling that , one realizes that the voltage across must be non-zero at the start. Thus, plots (C), (D), and (E) are eliminated.
Now, one must decide between choices (A) and (B). Once the circuit reaches equilibrium, i.e., the elements reach their asymptotic values. Specifically, the voltage at A goes to 0, since the inductor has no potential difference across it (to wit ). Once the switch is opened, the current suddenly changes, and , thus the inductor has a voltage across it, and the voltage at A becomes nonzero. Because of the diode, this would necessarily have to be a negative voltage. Since there is only to dissipate the voltage, the magnitude of the voltage once the switch is opened should be bigger than that initially, when the switch is closed at . Choose (B).
Comments
casseverhart13 2019-08-13 09:33:49 great problem here. . . tree removal psychonautQQ 2013-12-22 20:23:25 why does having a diode mean that the voltage at point a would have to be negative? first of all it's a zener diode,second of all even if it wasn't i'd still be confused neo55378008 2012-09-06 13:02:52 We can narrow it down to A or B as the other's pointed out, because we know eventually the energy stored in the inductor will oppose the battery.
As for when the switch is opened, just think of an inductive spike. If you've ever had a device with a transformer near powered speakers, you may have heard a pop in the speakers when the device was turned off. The magnetic field around the conductor collapses, and a voltage spike is made.
B shows this large spike FatheadVT 2010-11-09 14:17:57 What's the effect of the Ground in this problem?
Almno10 2010-11-12 20:27:39 The ground there keeps the potential on that side of the inductor at zero. Thus, when the current starts and stops, the potential at A has a large non-zero magnitude.
Almno10 2010-11-12 20:29:36 It also forces the potential at A to go to zero as the circuit reaches a steady state, since the inductor will harbor potential difference when the current is steady. The ground is everything here.
Almno10 2010-11-12 20:30:40 will harbor NO potential difference, that is madfish 2007-11-02 19:12:21 always remember:
Low Frequencies:
inductors are shorts/Capacitors are open circuits
High Frequencies:
Capacitors are shorts/inductors are open
this will help you out greatly to save time
wangjj0120 2008-08-30 13:03:15 What does the comment mean?
Who can explain this~
Poop Loops 2008-11-05 23:40:06 What that means is that Capacitors cannot pass through DC current. It just builds up. Capacitors only pass current if there is an AC signal.
Conversly, Inductors are great for DC currents, since they are just a loop of wire, so it acts like any other wire. But, when you apply some sort of AC signal, the magnetic field around the loops of wire tries to change, inducing another emf from the Inductor to stop it. So Inductors aren't good at passing through AC current.
That's why when you first flip the switch in this problem, dI/dt is huge so the emf induced in the Inductor is big. It's even bigger when you close the switch because now current flow the *other* way, so dI/dt is twice as big as when you first turned on the switch. Jeremy 2007-10-20 12:53:21 After eliminating everything but (A) and (B), it made the most sense to me to consider the time constants () for the two RL circuits. For the first decay, , and for the second, . It is given that , so that , i.e. . This is answer choice (B).
alemsalem 2010-09-21 23:25:02 Yeah i think this is the way to go, because of the diode current first runs through R1 only, then when the switch is open it runs through R2 only which should take less time to decay. wzm 2006-11-03 11:43:29 With switch closed, calling emf V, steady state current is
Open the switch and approximate that the inductor momentarily causes the current through it to remain the same. To do this requires the voltage at the top of the inductor to be
This is approximately what is shown in B.
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Abstract
Consider a system of $N$ bosons in three dimensions interacting via a repulsive short range pair potential $N^2V(N(x_i-x_j))$, where $\mathbf{x}=(x_1, \ldots, x_N)$ denotes the positions of the particles. Let $H_N$ denote the Hamiltonian of the system and let $\psi_{N,t}$ be the solution to the Schrödinger equation. Suppose that the initial data $\psi_{N,0}$ satisfies the energy condition \[ \langle \psi_{N,0}, H_N^k \psi_{N,0} \rangle \leq C^k N^k \; \] for $k=1,2,\ldots\; $. We also assume that the $k$-particle density matrices of the initial state are asymptotically factorized as $N\to\infty$. We prove that the $k$-particle density matrices of $\psi_{N,t}$ are also asymptotically factorized and the one particle orbital wave function solves the Gross-Pitaevskii equation, a cubic nonlinear Schrödinger equation with the coupling constant given by the scattering length of the potential $V$. We also prove the same conclusion if the energy condition holds only for $k=1$ but the factorization of $\psi_{N,0}$ is assumed in a stronger sense.
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Fractions on a Binary Tree II
We have already discussed an instance of a binary tree related to the Stern-Brocot tree. A fraction $\frac{a}{b}$ on the tree served a parent to two other fractions: $\displaystyle\frac{a}{a + b}$ and $\displaystyle\frac{b}{a + b}.$ At the time we only considered the case where $a \lt b.$ With the root equal to $\displaystyle\frac{1}{2},$ the tree has been populated by all positive fractions less than $1.$ Each fraction on the tree appeared in lowest terms. Had we started with the root of $\displaystyle\frac{2}{1},$ we would have obtained a tree (of reciprocals) populated by all fractions greater than $1,$ again all in lowest terms.
By a little change of procedure Calkin and Wilf arrived at a tree with many additional and no less remarkable features. In the modified tree, each fraction $\displaystyle\frac{a}{b}$ has two children, which in order of their appearance on the tree are $\displaystyle\frac{a}{a + b},$ as before, and $\displaystyle\frac{a + b}{b},$ which is the reciprocal of the second term in the previous construct. The diagram below depicts the two trees -- the new and the old -- rooted at the same fraction $\displaystyle\frac{a}{b}:$
(1)
In terms of functions $f_{L}$ and $f_{R},$ defined for the fractions on the Stern-Brocot tree, the new algorithm generates two children for any fraction $\displaystyle\frac{a}{b}:$ $f_{L}(\frac{a}{b})$ and $\displaystyle\frac{1}{f_{R}(\frac{a}{b})}.$ These are exactly the fractions generated on the Stern-Brocot tree, albeit, in an entirely different order. Thus we may claim that row by row the two trees contain exactly same fractions.
Starting with $\displaystyle\frac{a}{b} = \frac{1}{1},$ the new tree appears as
(2)
As the Stern-Brocot tree, it, too, contains all positive rational numbers and each of them appears exactly once. Calkin and Wilf noticed that in each row any two neighboring fraction have an additional property:
(3)
The denominator of the left neighbor coincides with the numerator of the right neighbor.
Also, $1$ being the numerator of the first fraction in each row as well as the denominator of the last one, (3) holds for the sequence of fractions obtained by writing the tree terms row by row:
(4)
$\displaystyle\frac{1}{1},\,\frac{1}{2},\,\frac{2}{1},\,\frac{1}{3},\,\frac{3}{2},\,\frac{2}{3},\,\frac{3}{1},\,\frac{1}{4},\,\frac{4}{3},\,\frac{3}{5},\,\frac{5}{2},\,\frac{2}{5},\,\frac{5}{3},\,\frac{3}{4},\,\frac{4}{1},\,\ldots$
To prove (3), first note that the property is obvious for two immediate descendants of the same parent. Two neighbors that are not siblings but share a grandparent also have the same property, see (1). In general, two neighboring fractions that are not siblings, descend from two neighboring fractions in the row above. The situation calls for an inductive argument. Assume, for the sake of induction, that the two parents are $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{b}{c}.$ Their four children then are
$\displaystyle\frac{a}{a + b},\,\frac{a + b}{b},\,\frac{b}{b + c},\,\frac{b + c}{c}$,
and the inductive step follows. A look at the first rows of the tree (2) supplies a base for induction. The proof is complete. (An inductive argument based on $\displaystyle f_{R}(\frac{b}{a}) = \frac{1}{f_{L}(\frac{a}{b})}$ shows that each row of the tree has central symmetry.)
Based on (3), the fractions in (4) can be written as $\displaystyle\frac{q(n)}{q(n + 1)},$ $n \ge 0,$ where $q(n)$ is the sequence of numerators
(5)
$1$, $1$, $2$, $1$, $3$, $2$, $3$, $1$, $4$, $3$, $5$, $2$, $5$, $3$, $4,$ $\ldots,$
with $q(0) = q(1) = 1,$ $q(2) = 2,$ $q(3) = 1,$ etc. (The sequence (5) is listed as #A002487 in the On-Line Encyclopedia of Integer Sequences, where it's called
Stern's diatomic series. The sequence is remarkable in its own right.)
Function $q(n)$ provides an effective way for enumeration of rational numbers. For $n = 0, 1, 2, \ldots,$ fraction $\#n$ equals $\displaystyle\frac{q(n)}{q(n+1)}.$ That simple.
The sequence (5) has been discovered several times over. E. W. Dijkstra described it in a letter to a friend in 1976. Then later same year he learned that function q(n) has been investigated by G. de Rham yet in 1947. Calkin and Wilf proved that $q(n)$ gives a number of
hyperbinary representations of the number $n.$ (In other words, $q(n)$ is the number of ways of writing the integer n as a sum of powers of $2,$ each power being used at most twice (i.e., once more than the legal limit for binary expansions).) Berlekamp et al (1981) identified $q(n)$ with the number of distinct nim-sums of two numbers that add up in the ordinary sense to the number n. Hence, the sequence (5) can be read from an XOR table: $q(n)$ is the number of distinct terms on the $n^{th}$ SW-NE diagonal of the XOR table:
The proof of the latter two assertions is based on the fact that function $q(n)$ satisfies a recurrence relation:
(6)
$\begin{align} q(2n + 1) &= q(n),\\ q(2n + 2) &= q(n) + q(n + 1),\,n = 0, 1, \ldots \end{align}$
(This is because the children of $\displaystyle\frac{q(n)}{q(n + 1)}$ are $\displaystyle\frac{q(2n + 1)}{q(2n + 2)}$ and $\displaystyle\frac{q(2n + 2)}{q(2n + 3)}.$ As it happens, we have the following
Proposition
The number $b(n)$ of hyperbinary representations of $n$ satisfies the recurrence (6) with the same initial condition, $b(0) = 1.$
The number $x(n)$ of distinct nim-sums $k \oplus m$, where $k + m = n,$ satisfies the recurrence (6) with the same initial condition, $x(0) = 1.$
Proof
Now, the binary and hence any hyperbinary expansion of an odd number $2n + 1$ ends in $1.$ It follows by first subtracting $1$ and then dividing by $2$ that there is a 1-1 correspondence between the hyperbinary expansions of $2n + 1$ and $n.$ Thus $b(2n + 1) = b(n).$ Furthermore, $b(2n + 2) = b(n) + b(n + 1),$ because a hyperbinary expansion of $2n + 2$ might have either two $1\mbox{'s}$ or no $1\mbox{'s}$ in it. If it has two $1\mbox{'s}$, then by deleting them and dividing by $2$ we'll get an expansion of $n.$ If it has no $1\mbox{'s}$, then by just dividing by $2$ we get an expansion of $n+1.$
If $k + m = 2n + 1,$ then the binary expansion of one of $k$ or $m$ ends in $1,$ the other in $0.$ By removing this one and dividing by $2,$ we'll get $k_{1} + m_{1} = n$, for some $k_{1}$ and $m_{1}$ such that $k_{1} \oplus m_{1} = k \oplus m.$ If $k + m = 2n + 2,$ then k and m have binary expansions that either both end in 1 or both end in 0. In the first case, asas before, we find $k_{1} + m_{1} = n$ and also $k_{1} \oplus m_{1} = k \oplus m.$ In the second case, $k_{1} + m_{1} = n + 1$ and still $k_{1} \oplus m_{1} = k \oplus m.$
"Playing with figures" on Ascension day 1976 E. W. Dijkstra came across a shifted function (which he called $fusc$):
(6')
$\begin{align} fusc(1) &= 1,\\ fusc(2n) &= fusc(n),\\ fusc(2n + 1) &= fusc(n) + fusc(n + 1),\,n = 1, \ldots \end{align}$
so that $fusc(k) = q(k - 1),$ $k = 1, 2, 3, \ldots$ Compatible with (6'), Dijkstra also defines $fusc(0) = 0.$ He derived an iterative algorithm to compute $fusc(n)$ and, based on it, has established several properties of the function. The algorithm is cute:
k = n
a = 1 b = 0 while (k > 0 ) { if (k is even) { k = k/2 a = a + b } else { k = (k-1)/2 b = a + b } } fusc(n) = b
Function fusc has the following properties:
For $n$ odd, let $m$ be obtained from n by inverting the internal (all, but the first and the last) digits of the binary representation of $n$ $(0$ to $1,$ $1$ to $0).$ Then $fusc(m) = fusc(n).$
Let the binary representation of $m$ be the reverse (written in the reverse order) of the binary representation of $n.$ Then $fusc(m) = fusc(n).$
$fusc(n)$ is even iff $3|n.$ (Proof)
References E. R. Berlekamp, J. H. Conway, R. K. Guy, Winning Ways for Your Mathematical Plays, Volume 1, A K Peters, 2001, pp. 115-116. E. W. Dijkstra, An exercise for Dr. R. M. Burstall, published in Edsger W. Dijkstra, Selected Writings on Computing: A Personal Perspective, Springer-Verlag, 1982, pp. 215-216. E. W. Dijkstra, More about the function "fusc", published in Edsger W. Dijkstra, Selected Writings on Computing: A Personal Perspective, Springer-Verlag, 1982, pp. 230-232. N. Calkin, H. S. Wilf, Recounting the Rationals, Am Math Monthly, Vol. 107 (2000), pp. 360-363. Fine features Binary Encoding Binary Encoding, a Second Interpretation Continued Fractions on the Stern-Brocot Tree Fractions on a Binary Tree Fractions on a Binary Tree II Farey series Pick's Theorem Countability of Rational Numbers Counting Ordered Pairs Countable Times Countable Is Countable Countability of Rational Numbers Countability of Rational Numbers: PWW Countability Principle Countability of Rational Numbers via Conitnued Fractions Fractions on a Binary Tree II Countability of Rational Numbers as a Union of Finite Sets A New Proof That the Rationals Are Countable
Copyright © 1966-2016 Alexander Bogomolny
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We continued discussing recursion. We also discussed memoization and demonstrated it.
from IPython.core.interactiveshell import InteractiveShellInteractiveShell.ast_node_interactivity = "all"
A bus driver needs to give an exact change and she has coins of limited types. She has infinite coins of each type. Given the amount of change ($amount$) and the coin types (the list $coins$), how many ways are there?
def change(amount, coins): if amount == 0: return 1 elif amount < 0 or coins == []: return 0 return change(amount, coins[:-1]) +\ change(amount - coins[-1], coins) change(5, [1,2,3])
5
Why is it counting unique solutions? For example, why is [2,2,1] counted once and not [1,2,2]?
Consider the case where $amount = n^2, coins = [1,2,\ldots, n]$. When we call $change(amount, coins)$ the first level of the recursion calls $change([1,\ldots, n-1], n^2)$ and $change([1,\ldots, n], n^2 - n)$. This means that the list size in the recusrive calls is at least $n-1$ and $amount \geq n^2 -n$.
One can show (using induction) that in the first $k \leq n$ levels of recursion the list size is at least $n - k$ and $amount \geq n^2-n k$, thus there are two recursive calls at each of these layers.
This gives a running time of at least $2^n$ by the same argument as that we applied in the subset sum problem.
It is interesting to note that while the two problems share some similarities, one major difference is that in the subset sum problem we are asking whether a
solution exists while in the change problem we are trying to count the number of valid solutions to a problem.
This is a recurring theme in CS that we will encounter many times in the future.
Question 2(a) from the 2015 fall semester exam (Moed B).
Given a list $L$ of non-negative integers with $len(L) = d$ we consider $L$ as a point in a $d$-dimensional space.
For example: $L = [0, 0, 0]$ represents the origin in a $3$-dimensional space.
Our goal is to find how many ways we have of getting from $[0, \ldots, 0]$ to $L$ by advancing
only forward.
For example, if $L=[1, 2]$ then we have three such paths:
Again, we first think of the base case, and then reduce big problems to smaller ones.
This gives rise to a simple recursive algorithm:
def cnt_paths(L): if all_zeros(L): return 1 result = 0 for i in range(len(L)): if L[i] != 0: L[i] -= 1 result += cnt_paths(L) L[i] += 1 return resultdef all_zeros(L): for i in L: if i != 0: return False return Trueprint(cnt_paths([3,4,5]))
27720
Note that the leaves in the tree are exactly the "legal paths" which we count.
Let $cnt$ be the returned value of $cnt\_paths(L)$. Since we increment $cnt$ by $1$ for each legal path, this means that the running time is
at least $cnt$.
Using the following combinatorial analysis, which is
for reference only, we show that$cnt$ is exponential in $d,n$, implying that the running time of $cnt\_paths(L)$ is at least exponential in $d,n$.Can we do better than the recursive solution in terms of running time? ...yes!
Let's take a simple case where $L = [n, n, \ldots, n]$ and $|L| = d$. I.e. - we are in a $d$-dimensional space and we want to get to $[n, \ldots, n]$.
Can you think of a combinatorial solution for cnt_paths? Let's take the case above where $L = [n, n, \ldots, n]$ and $|L| = d$.
In each step we subtract $1$ from one of the $d$ coordinates (which is currently positive) and in exactly $nd$ steps we need to get to the all-zero vector.
Think about the first coordinate - there are precisely $n$ steps along our path where we change this coordinate, thus we have $\binom{nd}{n}$ options to choose where the moves for the first coordinate are located.
What about the second coordainte? Well now we are left with $nd - n = n(d-1)$ places out of which we again pick $n$ places to advance the second coordinate. By indcution, we get: $$cnt = \prod_{i=1}^d \tbinom{n(d-i+1)}{n} = \prod_{i=1}^d \tbinom{ni}{n} $$
How long does it take to compute this number? We need to multiply $d$ elements, and for each of those we need to compute factorials and divide numbers in the range of $1,\ldots,n$. This is clearly done in time polynomial in $d,n$.
And how big is $cnt$ exactly? Recall that $cnt$ is a bound on the running time of cnt_paths.
We now claim that $cnt = exp(n,d)$. To see this, we write it explicitly:$$cnt = \tbinom{n}{n}\cdot \tbinom{2n}{n} \cdots \tbinom{dn}{n} = \frac{n!}{(n-n)! n!} \cdot \frac{(2n)!}{(2n-n)! n!} \cdot \frac{(3n)!}{(3n-n)! n!}\cdots \frac{(nd)!}{(nd-n)! n!} $$
This product has a telescopic property - thus we can cancel out elements and get:$$cnt = \frac{(nd)!}{(n!)^d} = \frac{1\cdot 2 \cdot 3 \cdots \cdot nd}{(1 \cdot 2 \cdots n) \cdots(1 \cdot 2 \cdots n)} $$
Break this product into two terms:$$cnt = \frac{1 \cdots 2n}{n! \cdot n!} \cdot \frac{(2n + 1) \cdots nd }{(n!)^{d - 2}}$$
The first multiplicand is clearly larger than $1$, as for the second, each term in the numerator is
at least $2n$ and each term in the denominator is at most $n$ thus clearly: $$cnt \gg \frac{2n \cdot 2n \cdots 2n}{n \cdot n \cdot n} = 2^{dn - 2n} = exp(n,d)$$
Conclusion? Using the combinatorial computation we reduce a running time which is exponential in $n,d$ into a running time polynomial in $n,d$.
The number of sets of size $k$ selected from a set of $n$ elements (with no repetitions) Recursive formula (Pascal): $\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$ where the stopping criterion is $\binom{n}{0} = \binom{n}{n} = 1$
The time complexity of binom is exponential in $n$ (worst case behaviour is when $k=\frac{n}{2}$)
def binom(n,k): if n < 0 or k < 0 or n < k: return 0 elif (k==0 or n==k): return 1 return binom(n-1,k-1) + binom(n-1,k)binom(4,2)
6
def binom_trace(n,k): result = binom_trace(n,k) return resultdef binom_trace(n,k,indent=1): #indent = how much to indent the printouts if (k<0 or n<0 or n<k): # safety checks return 0 elif (k==0 or k==n): # halting conditions print(">>"*indent + "({},{})".format(n,k)) print("<<"*indent + "({},{})".format(n,k)) return 1 print(">>"*indent + "({},{})".format(n,k)) indent+=1 val = binom_trace(n-1,k,indent) + binom_trace(n-1,k-1,indent) indent-=1 print("<<"*indent + "({},{})".format(n,k)) return valbinom_trace(4,2)
>>(4,2) >>>>(3,2) >>>>>>(2,2) <<<<<<(2,2) >>>>>>(2,1) >>>>>>>>(1,1) <<<<<<<<(1,1) >>>>>>>>(1,0) <<<<<<<<(1,0) <<<<<<(2,1) <<<<(3,2) >>>>(3,1) >>>>>>(2,1) >>>>>>>>(1,1) <<<<<<<<(1,1) >>>>>>>>(1,0) <<<<<<<<(1,0) <<<<<<(2,1) >>>>>>(2,0) <<<<<<(2,0) <<<<(3,1) <<(4,2) 6
Now with memoization:
def binom_fast(n,k): d = {} return binom_mem(n,k,d)def binom_mem(n,k,mem): if n < 0 or k < 0 or n < k: return 0 elif (k==0 or n==k): return 1 if (n,k) not in mem: mem[(n,k)] = binom_mem(n-1,k, mem) + \ binom_mem(n-1,k-1, mem) return mem[(n,k)]binom_fast(4,2)binom_fast(50,25)
6 126410606437752
Printing the recursive calls, with memoization:
def binom_fast_trace(n,k): mem = dict() result = binom_mem_trace(n,k,mem) return resultdef binom_mem_trace(n,k,mem,indent=1): #indent = how much to indent the printouts if (k<0 or n<0 or n<k): # safety checks return 0 elif (k==0 or k==n): # halting conditions print(">>"*indent + "({},{})".format(n,k)) print("<<"*indent + "({},{})".format(n,k)) return 1 print(">>"*indent + "({},{})".format(n,k)) indent+=1 if (n,k) not in mem: mem[(n,k)] = binom_mem_trace(n-1,k,mem,indent) + binom_mem_trace(n-1,k-1,mem,indent) indent-=1 print("<<"*indent + "({},{})".format(n,k)) return mem[(n,k)]binom_fast_trace(4,2)
>>(4,2) >>>>(3,2) >>>>>>(2,2) <<<<<<(2,2) >>>>>>(2,1) >>>>>>>>(1,1) <<<<<<<<(1,1) >>>>>>>>(1,0) <<<<<<<<(1,0) <<<<<<(2,1) <<<<(3,2) >>>>(3,1) >>>>>>(2,1) <<<<<<(2,1) >>>>>>(2,0) <<<<<<(2,0) <<<<(3,1) <<(4,2) 6
To analyze the time complexity of the function, we will construct an $ (n+1) \times (k+1)$ table, where the cell in position $(i,j)$ will denote a call to compute $\binom{i}{j}$.
In this method, the running time can be computed by $$\text{number of visited cells} \times \text{number of visits per cell} \times \text{time per cell (without recursive calls)}$$
Consider a cell in position $(i,j)$. By our recursive formula, this cell will be called exactly in the cases where we need to compute either $(i + 1, j)$ or $(i + 1, j + 1)$.
Now, assume $(i+1, j)$ was the first cell to call $(i,j)$, then:
It follows that each cell will be accessed at most twice, and thus the running time is clearly $O(nk)$. In the diagram below we see that we can even give a more precise running time based on the fact that many cells in the table are base cases.
A bus driver needs to give an exact change and she has coins of limited types. She has infinite coins of each type. Given the amount of change ($amount$) and the coin types (the list $coins$), how many ways are there?
def change(amount, coins): if amount == 0: return 1 elif amount < 0 or coins == []: return 0 return change(amount, coins[:-1]) +\ change(amount - coins[-1], coins) change(5, [1,2,3])
5
Now with memoization:
def change_fast(amount, coins): d = {} return change_mem(amount, coins, d)def change_mem(amount, coins, d): if amount == 0: return 1 elif amount < 0 or coins == []: return 0 #if (amount, tuple(coins)) not in d: if (amount, len(coins)) not in d: #d[(amount, tuple(coins))] = \ d[(amount, len(coins))] = \ change_mem(amount, coins[:-1], d) +\ change_mem(amount - coins[-1], coins, d) #return d[(amount, tuple(coins))] return d[(amount, len(coins))]change_fast(500, [1,3,2])
21084
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GR8677 #96
Comments
casseverhart13 2019-08-12 07:04:10 I’d been just browsing every now and then and also got to learn to read this problem. power washing GRE0717 2015-10-20 21:28:08 I couldn\'t figure why the odd terms would give zero coef cause n=1 ,3, 5,... for square well are even functions. The question had n=0 as the ground state instead, so everything shifted by -1. neo55378008 2012-09-06 13:11:22 All the odd n's should have a node at the middle of the well. If there is a voltage at zero, the wavefunctions will be altered.
Did I over simplify this too much? Donofnothing 2010-10-05 07:53:43 one thing i've always had trouble with is guessing if ETS is starting with n=0 or n=1 for ground states in these kinda problems. in this one, look at the index on the summation. it's starting at 0, so that means 0 is your ground state!rn
wavicle 2011-11-08 14:37:14 Yes it's so silly. Technically you don't have to guess, just read carefully. But seriously. The energy for infinite square well goes as n^2 with the ground state being nonzero. Hence if n=0 corresponds to ground state, to do an energy calculation, in code for example, you would have to offset by 1. Then again, if you live in an ivory tower and only do math on paper, what does it matter to you? mangree 2008-10-18 08:19:55 To sum it up,because there is a lot of misunderstanding about this:
V and V' are both even,so the perturbed potential V+V' is even.
Therefore the perturbed problem has even AND odd eigenfunctions.
The (new) groundstate however is even,thus it is a mixture of even states only =0 for n=1,3,...
Saint_Oliver 2013-09-14 14:05:41 This is an important point. Eigenfunctions of both the unperturbed and the perturbed potential will have both symmetric and antisymmetric members. The key in this problem is that it is asking about the ground state, which will be composed of only the symmetric eigenfunctions. magneton 2008-09-28 21:06:14 The unperturbed ground state wave function is even.
The perturbed wave function must remain even in ALL orders of the perturbation theory. Explanation:
the matrix elements of the perturbation potential V' (it is an even function of x, depending on |x| only) are only between the states of the SAME parity: even-even or odd-odd. Since the unperturbed ground state wave function is even, only the states with even n =0, 2, 4, ... are admixed to it. Therefore, all a_0n with odd n must be zero, as in C. kolahalb 2007-11-21 17:43:42 Friends,the symmetry in the potential must reflect in the form of the wave function.
Here,the potential is even.=>The wave-functions are even=>odd terms are zero.
Poop Loops 2008-09-23 20:49:00 Okay, but the non-perturbed potential is also even, yet you get wave functions that are odd. I am not understanding why even potential = even wave function.
mangree 2008-10-18 08:11:15 Even potential the eigenfunctions of the problem are even AND odd.
p40515 2010-10-01 14:59:07 symmetric perturbation potential all even and odd eigenfunctions will keep their geometric characteristic after perturbation ground state,that was even, will be even after per....rnbut remember to ignore unperturbed ground state from series in the answer;this is at most true(because of the first-order perturbation formula showing perturbed eigenstate)
lelandr 2011-04-26 11:07:17 I would like to try and clarify some things:
1) An even potential does NOT mean that all of the eigenfunctions will be even.
2) An even potential DOES mean that the square of the wave function will be even.
3) An even potential DOES mean that the GROUND STATE wave function will be even.
so in fact - "Here,the potential is even.=>The wave-functions are even=>odd terms are zero." is not correct.
I believe this could more accurately read:
Here the potential and the perturbing potential are both even -> the perturbed GROUND STATE wave function will also be even -> If we write this as a linear combination of our non-perturbed basis of wave-functions, we can write this using only the even-wave functions (which here are the EVEN terms - thus the ODD coefficients are 0) Richard 2007-11-01 13:16:41 I like Yosun's solution.
Jeremy 2007-11-02 18:04:10 I think the first paragraph is great. My trouble is with the second paragraph. In my opinion, it's almost assuming the reader knows how to solve the problem. If I've looked up the right answer, I already know that the sum does not contain odd terms, and that I must work my way towards this or another equivalent statement. Now, let's say I represent someone who took the test, got the problem wrong, and then came to this page for an explanation (which is, in fact, true). Could I not solve the problem because I failed to realize that or are even? Maybe, but it's more likely that I couldn't figure out why this eliminated the odd terms in the sum. However, this is not explained. And there's no mention of the strange convention being used in the indices. I guess the short way of putting it is "This solution didn't help me." I've learned a lot of physics using this site, but the explanations here fell short of the typical quality and clarity. tin2019 2007-10-29 09:11:05 Well let's TeX out the derivation in case somebody needs it. We are supposing that our Hamiltonian is a sum of an unperturbed hamiltonian which is and a small pertubation . Then our stationary Schrodinger equation becomes
.
As the solutions of the unperturbed system make a complete set of functions our function of the perturbed system must be a superposition of the functions , i.e. solutions of the equation . In other words
.
Replacing this into our initial Schrodinger equation we get
or
,
where are energy levels of unperturbed system. We now multiply both sides by and integrate to get
or
.
where It would be beneficial to memorize this result for the future as from it is easier to proceed with almost any problem conserning perturbation. Note that had the problem been the one with degenerate eigenstates the result would be somewhat different, as our development would have to include eigenfunctions with the same eigenvalue of energy. Now we expand and into series, i.e. and . Note that our perturbation can be expanded into series where the zeroth term would be zero, as the zeroth term corresponds to no perturbation. For the first order perturbation we have:
.
The level of interest is the ground level state. For a ground state we then have that the zeroth order coefficients are zero for every except , i.e. that our unperturbed ground state wavefunction has only one term in the expansion, namely itself. Now it follows that for we have
or that
.
As the unperturbed ground state is an even function and the perturbation is an even function we have that are zero for such that is odd. This is true for , therefore the coefficients vanish for odd .
kb 2007-10-29 14:44:57 yes i agree u just showed that the coefficients vanish when m = odd for the first order correction.
but the old unperturbed wavefunction does not change so the total wavefunction (psi0 + correction) must still contain both even and odd terms.
but my problem is i think the question its very ambiguous when it askes for psi' , is that the total wavefunction or only the correction part? kb 2007-10-24 19:16:27 I am confused, if this question is asking for the TOTAL (psi0 + psi1) perturbed wavefunction, not simply the first order CORRECTION... then all even and odd wfs should remain because the perturbation is so small...
only the odd terms in the CORRECTION should go away, not the actual wavefunction
Jeremy 2007-11-02 17:16:21 You bring up an important point that I hadn't thought of. I have already argued for why the first order correction for the ground state, psi1 as you put it, is even. Your psi0 we already know to be even*. Therefore, psi0 + psi1 is even, as expressed in answer choice (B).
* The ground state wave function for a well spanning to (I would use , but keeps the same well width as the test problem) is: . You can get the ground state for this potential (the unperturbed one) using , so that , which is indeed an even function. Jeremy 2007-10-10 12:23:25 This problem is EVIL! I picked (C) on this problem. I spent quite a while trying to figure out why (B) is right; and it was only when I got around to writing some equations down that I realized I had fallen for a subtle trap: THIS PROBLEM USES N=0, NOT N=1, FOR THE GROUND STATE!!! So of course I get the wrong answer because I'm thinking even produces an odd wave function, but just the opposite is true - you get an even wave function.
Also, the explanations on this page are weak at best, unless perturbation theory is a fresh topic in your mind. Not that I'm about to throw down a derivation, but here is the pertinent formula:
,
where (See Griffiths 6.1, problem 6.1 (b) might give you deja vu). This is the first-order correction to the wave function - not what the question is asking for, but close enough I think because the perturbation is "very small." The bracket represents the integral of the product of the three functions shown. is even (about the center of the well). We're after the new ground state (), so is the unperturbed ground state - another even function. Thus, the parity of determines the parity of the entire integrand; if it is odd, the integrand is odd, and the integral is zero. With 0 representing the ground state, odd values (i.e. m, n = 1, 3, 5, ...) produce odd wave functions, making the coefficients zero. This is answer choice (B).
94709 2007-10-14 21:49:38 I agree it's evil in the sense they used n=0 for the ground state,,,
Two fundamental knowledges...Q.M. and Perturbation.
Very fundamental knowledge about QM; inifinite solution form a complete basis set to span hilbert space.
Perturbation is not only in quantum mechanics, but everywhere else in general. Better be fresh...
1. Sudden approximation tells that all even functions about the origine is likely to be gone.
2. Using odd functions in original summation, one must be able to complete the basis set since there are infinite many.
3. Since the original potential is an infinite well, the ground state is, of course, even. So 1st excited states, 3rd, 5th, will remain.
Only 3 steps in head, and no calculation but fundamental knowledge about Q.M. Easier than those problems that require memorization of equations.
Jeremy 2007-10-20 12:26:48 94709,
There are a couple things you say that I disagree with:
1. "Even functions... likely to be gone." It's just the opposite actually. Answer choice (B) says the new ground state is a sum of even functions - remember, odd n corresponds to odd unperturbed wave functions.
2. "Using odd functions in original summation, one must be able to complete the basis set since there are infinite many." Not true. For example, you can't write an even function as a sum of odd functions (the sum of odd functions is an odd function).
94709 2007-10-21 21:06:18 Jeremy,
Thank you for your reply.
About the first correction:
Given that the potential is infinite-well, I think the ground state is an even function with one node at the origin. New potential is an even function which surpress all even functions. Defining the ground state with n=0, all even n terms express even functions. Hence, odd n remains, meaning that odd functions remain. This is an argument of "Is the wavefunction of the ground state in infinite potential well even or odd function?" With one node at the center, x=0, the ground state is even. Hence, even n express even functions, and odd n expresse odd functions.
About the second correction:
I am sorry that I did not explicitly mention my assumption in every step. As a consequence from previous steps, what I mean by "function can be spanned by odd terms" is that odd functions can be spanned by odd terms. I stupidly assumed this is very clear, and did not mention at this stage.
Thanks for your comment. freeform 2006-11-30 18:29:07 Rather than a square well, the potential is a well with a dent in the center so that it slightly looks like a W. Only eigenfunctions with nodes at center of well will be admitted---there are the .
Jeremy 2007-10-10 12:39:59 freeform, you made the same mistake I did. Because the counting starts at , even corresponds to even wave functions. The new ground state does NOT have a node at the center of the well as it is an even function. dirichlet 2006-11-21 11:27:46 I think all the even term will go out because the even functions are of odd parity in this case.
dirichlet 2006-11-21 11:22:42
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Higher harmonic flow coefficients of identified hadrons in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Springer, 2016-09)
The elliptic, triangular, quadrangular and pentagonal anisotropic flow coefficients for $\pi^{\pm}$, $\mathrm{K}^{\pm}$ and p+$\overline{\mathrm{p}}$ in Pb-Pb collisions at $\sqrt{s_\mathrm{{NN}}} = 2.76$ TeV were measured ...
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In Higher Algebra Proposition 5.4.4.10
1, Lurie proves that for a coCartesian fibration of $\infty$-operads $q:\mathscr{C}^\otimes\to\mathscr{O}^\otimes$, where $\mathscr{O}^\otimes$, when viewed as an $\infty$-category, is pointed and $\mathscr{C}$ is a stable $\mathscr{O}$-monoidal $\infty$-category under the coCartesian fibration $q$. There is an equivalence of $\infty$-categories $\mathrm{Alg^{nu}}_\mathscr{O}(\mathscr{C})\to\mathrm{Alg^{aug}_{\mathscr{O}}(\mathscr{C})}$.
Let $\mathscr{O}^\otimes=\mathrm{N}(\mathscr{F}\mathrm{in}_{*})$, where $\mathscr{F}\mathrm{in}_{*}$ is Segal's category of pointed finite sets (the $n$lab denotes it as $\Gamma$, if I'm not wrong). A $\mathbb{E}_\infty$-ring is a commutative monoidal object, hence $\mathbb{E}_\infty$-monoidal object, of $\mathrm{Sp}$, and the sphere spectrum $S$ is naturally a $\mathbb{E}_\infty$-ring, so we can let $\mathscr{C}$ to be $\mathrm{Mod}_S$. Then, the $\infty$-category of nonunital $\mathbb{E}_\infty$-rings is equivalent to the $\infty$-category of augmented $\mathbb{E}_\infty$-rings over the sphere spectrum.
1
This is an expansion of Haugseng's comment
above.
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Linear Algebra
A matrix is a 2-D array of numbers.
A tensor is a n-D array of numbers.
Matrices are associative but not commutative.
Not all square matrices have inverse. A matrix that is not invertible is called Singular or Degenerative.
A matrix is “singular” if any of the following are true:
-Any row or column contains all zeros.
-Any two rows or columns are identical.
-Any row or column is a linear combination of other rows or columns.
A square matrix A is singular, if and only if the determinant(A) = 0
Inner product of two vectors \(\overrightarrow{u}\) and \(\overrightarrow{v}\):
\(\overrightarrow{u} * \overrightarrow{v} = p * ||\overrightarrow{u}|| = p * \sqrt{\sum_{i=1}^{n}u_{i}^{2}} = u^{T} * v = \sum_{i=1}^{n}u_{i} * v_{i}\)
p is the projection of \(\overrightarrow{v}\) on \(\overrightarrow{u}\) and \(||\overrightarrow{u}||\) is the Euclidean norm of \(\overrightarrow{u}\).
\(A = \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}\)
Determinant(A) = |A| = a(ei-hf) – b(di-gf) + c(dh – ge)
If \(f: R^{n*m} \mapsto R\)
\(\frac{\partial}{\partial A} f(A) = \begin{bmatrix} \frac{\partial}{\partial a_{11}}f(A) & … & \frac{\partial}{\partial a_{1m}} f(A)\\ … & … & …\\ \frac{\partial}{\partial a_{n1}}f(A) & … & \frac{\partial}{\partial a_{nm}}A\\ \end{bmatrix}\)
Example:
\( f(A) = a_{11} + … + a_{nm} \)\(\frac{\partial}{\partial A} f(A) = \begin{bmatrix} 1 & … & 1\\ … & … & …\\ 1 & … & 1 \\ \end{bmatrix}\)
If A is a squared matrix:
trace(A) = \(\sum_{i=1}^n A_{ii}\)
trace(AB) = trace(BA)
trace(ABC) = trace(CAB) = trace(BCA)
trace(B) = trace(\(B^T\))
trace(a) = a
\(\frac{\partial}{\partial A} trace(AB) = B^T\)\(\frac{\partial}{\partial A} trace(ABA^TC) = CAB + C^TAB^T\)
Eigenvector
Given a matrix A, if a vector μ satisfies the equation A*μ = λ*μ then μ is called an eigenvector for the matrix A, and λ is called the eigenvalue for the matrix A. The principal eigenvector for the matrix A is the eigenvalue with the largest eigenvalue.
Example:
The normalized eigenvectors for \(\begin{bmatrix}0 & 1 \\1 & 0 \end{bmatrix}\) are \(\begin{bmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}\) and \(\begin{bmatrix}-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}\), the eigenvalues are 1 and -1.
Eigendecomposition
Given a squared matrix A∈\(R^{n*n}\), ∃ Q∈\(R^{n*n}\), Λ∈\(R^{n*n}\) and Λ diagonal, such as \(A=QΛQ^T\).
Q’s columns are the eigenvectors of \(A\)
Λ is the diagonal matrix whose diagonal elements are the eigenvalues
Example:
The eigendecomposition of \(\begin{bmatrix}0 & 1 \\1 & 0 \end{bmatrix}\) is Q=\(\begin{bmatrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{bmatrix}\), Λ=\(\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{bmatrix}\)
Eigenvectors are vect
Single Value Decomposition
Given a matrix A∈\(R^{n*m}\), ∃ U∈\(R^{n*m}\), D∈\(R^{m*m}\) and D diagonal, V∈\(R^{m*m}\) such as \(A=UDV^T\).
U’s columns are the eigenvectors of \(AA^T\)
V’s columns are the eigenvectors of \(A^TA\)
Example:
The SVD decomposition of \(\begin{bmatrix} 0 & 1 \\1 & 0 \end{bmatrix}\) is U=\(\begin{bmatrix}0 & -1 \\ -1 & 0\end{bmatrix}\), D=\(\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\), V=\(\begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}\).
More details about SVD can be found here: http://www.youtube.com/watch?v=9YtmGy-wfE4
The Moore-Penrose Pseudoinverse
The Moore-Penrose pseudoinverse is a matrix that can act as a partial replacement for the matrix inverse in cases where it does not exist (e.g. non-square matrices).
The pseudoinverse matrix is defined as: \(pinv(A) = \lim_{α \rightarrow 0} (A^TA + αI)^{-1} A^T\)
Analysis
0! = 1
exp(1) = 2.718
exp(0) = 1
ln(1) = 0
\(ln(x) = log_e(x) \\ log_b (b^a) = a\)
exp(a + b) = exp(a) * exp(b)
ln(a * b) = ln(a) + ln(b)
\(cos(x)^2 + sin(x)^2 = 1\)
Euler’s formula
exp(iθ) = cos(θ) + i sin(θ)
Complex numbers
Rectangular form
z = a + ib (real part + imaginary part and i an imaginary unit satisfying \(i^2 = −1\)).
Polar form
z = r (cos(θ) + i sin(θ))
Exponential form
z = r.exp(iθ)
Multivariate equations
The solution set of a system of linear equations with 3 variables is the intersection of hyperplanes defined by each linear equation.
Derivatives
\(\frac{\partial f(x)}{\partial x} = \lim_{h \rightarrow 0} \frac{f(x+h) – f(x)}{h}\)
Function
Derivative
x^n
n * x^(n-1)
exp(x)
exp(x)
f o g (x)
g’(x) * f’ o g(x)
ln(x)
1/x
sin(x)
cos(x)
cos(x)
-sin(x)
Integration by parts
\(\int_{a}^{b} (f(x) g(x))’ dx = \int_{a}^{b} f'(x) g(x) dx+ \int_{a}^{b} f(x) g'(x) dx\)
Binomial theorem
\((x + y)^n = \sum_{k=0}^{n} C_n^k x^k y^{n-k}\)
Chain rule
Z = f(x(u,v), y(u,v))
\(\frac{\partial Z}{\partial u} = \frac{\partial Z}{\partial x} * \frac{\partial x}{\partial u} + \frac{\partial Z}{\partial y} * \frac{\partial y}{\partial u}\)
Entropy
Entropy measures the uncertainty associated with a random variable.
\(H(X) = -\sum_{i=1}^n p(x^{(i)}) log(p(x^{(i)}))\)
Example:
Entropy({1,1,1,1}) = 0
Entropy({0,1,0,1}) = -½ (4*log(½))
Hessian
\(H = \begin{bmatrix}\frac{\partial^2 f(θ)}{\partial θ_1\partial θ_1} & \frac{\partial^2 f(θ)}{\partial θ_1 \partial θ_2} \\ \frac{\partial^2 f(θ)}{\partial θ_2\partial θ_1} & \frac{\partial^2 f(θ)}{\partial θ_2\partial θ_2} \end{bmatrix}\)
Example:
\(f(θ) = θ_1^2 + θ_2^2 \\ H(f) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\)
A function f(θ) is convex if its Hessian matrix is positive semidefinite (\(x^T.H(θ).x >= 0\), for every \(x∈R^2\)).
\(x^T.H(θ).x = \begin{bmatrix} x_1 & x_2 \end{bmatrix} . \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} . \begin{bmatrix} x_1\\ x_2 \end{bmatrix} = 2 x_1^2 + 2 x_2^2 >= 0\)
Method of Lagrange Multipliers
To maximize/minimize f(x) with the constraints \(h_r(x) = 0\) for r in {1,..,l},
We need to define the Lagrangian: \(L(x, α) = f(x) – \sum_{r=1}^l α_r h_r(x)\) and find x, α by solving the following equations:
\(\frac{\partial L}{\partial x} = 0\)
\(\frac{\partial L}{\partial α_r} = 0\) for all r
\(h_r(x) = 0\) for all r
Calculate the Hessian matrix (f”(x)) to know if the solution is a minimum or maximum.
Method of Lagrange Multipliers with inequality constraints
To minimize f(x) with the constraints \(g_i(x) \geq 0\) for i in {1,..,k} and \(h_r(x) = 0\) for r in {1,..,l},
We need to define the Lagrangian: \(L(x, α, β) = f(w) – \sum_{i=1}^k α_i g_i(x) – \sum_{r=1}^l β_r h_r(x)\) and find x, α, β by solving the following equations:
\(\frac{\partial L}{\partial x} = 0\)
\(\frac{\partial L}{\partial α_i} = 0\) for all i
\(\frac{\partial L}{\partial β_r} = 0\) for all r
\(h_r(x) = 0\) for all r
\(g_i(x) \geq 0\) for all i
\(α_i * g_i(x) = 0\) for all i (Karush–Kuhn–Tucker conditions)
\(α_i >= 0\) for all i (KTT conditions)
Lagrange strong duality – hard to understand 🙁
Lagrange dual function \(d(α, β) = \underset{x}{min} L(x, α, β)\), and x satisfies equality and inequality constraints.
We define \(d^* = \underset{α \geq 0, β}{max}\ d(α, β)\)
We define \(p^* = \underset{w}{min}\ f(x) \) (x satisfies equality and inequality constraints)
Under certain conditions (Slater conditions: f convex,…), \(p^* = d^*\)
Jensen’s inequality
If f a convex function, and X a random variable, then f(E[X]) <= E[f(X)].
If f a concave function, and X a random variable, then f(E[X]) >= E[f(X)].
If f is strictly convex (f”(x) > 0), then f(E[X]) = E[f(X)] holds true only if X = E[X] (X is a constant).
Probability
Below the main probability theorems.
Law of total probability
If A is an arbitrary event, and B are mutually exclusive events such as \(\sum_{i=1}^{n} P(B_{i}) = 1\), then:
\(P(A) = \sum_{i=1}^{n} P(A|B_{i}) P(B_{i}) = \sum_{i=1}^{n} P(A,B_{i})\)
Example:
Suppose that 15% of the population of your country was exposed to a dangerous chemical Z. If exposure to Z quadruples the risk of lung cancer from .0001 to .0004. What’s the probability that you will get lung cancer.
P(cancer) = .15 * .0004 + .85 * .0001 = .000145
Bayes’ rule
P(A|B) = P(B|A) * P(A) / P(B)
Where A and B are events.
Example:
Suppose that 15% of the population of your country was exposed to a dangerous chemical Z. If exposure to Z quadruples the risk of lung cancer from .0001 to .0004. If you have lung cancer, what’s the probability that you were exposed to Z?
P(Z|Cancer) = P(Cancer|Z) * P(Z) / P(Cancer)
We can calculate the P(Cancer) using the law of total probability: P(Cancer) = P(Cancer|Z) * P(Z) + P(Cancer|~Z) * P(~Z)
P(Z|Cancer) = .0004 * 0.15 / (.0004 * 0.15 + .0001 * .85) = 0.41
Chain rule
\(P(A_1,A_2,…,A_n) = P(A_1) P(A_2|A_1) ….P(A_n|A_{n-1},…,A_1)\)
P(Y,X1,X2,X3,X4) = P(Y,X4,X3,X2,X1)
= P(Y|X4,X3,X2,X1) * P(X4|X3,X2,X1) * P(X3|X2,X1) * P(X2|X1) * P(X1)
= P(Y|X4,X3) * P(X4) * P(X3|X2,X1) * P(X2) * P(X1)
The Union Bound
\(P(A_1 \cup A_2 … \cup A_n) \leq P(A_1) + P(A_2) + … + P(A_n) \)
Nb of permutations with replacement
Nb of permutations with replacement = \({n^r}\), r the number of events, n the number of elements.
Probability = \(\frac{1}{n^r}\)
Example:
Probability of getting 3 six when rolling a dice = 1/6 * 1/6 * 1/6
Probability of getting 3 heads when flipping a coin = 1/2 * 1/2 * 1/2
Nb of permutations without replacement and with ordering
Nb of permutations without replacement = \(\frac{n!}{(n-r)!}\), r the number of events, n the number of elements.
Probability = \(\frac{(n-r)!}{n!}\)
Example:
Probability of getting 1 red ball and then 1 green ball from an urn that contains 4 balls (1 red, 1 green, 1 black and 1 blue) = 1/4 * 1/3
Nb of combinations without replacement and without ordering
Nb of combinations \(\frac{n!}{(n-r)! \, r!}\), r the number of events, n the number of elements.
Probability = \(\frac{(n-r)! \, r!}{n!}\)
Example:
Probability of getting 1 red ball and 1 green ball from an urn that contains 4 balls (1 red, 1 green, 1 black and 1 blue) = 1/4 * 1/3 + 1/4 * ⅓
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My setup is similar to Emrakul's, but since the OP asked for patterns, I will explore what Emrakul left as an exercise. And while I'll mostly discuss the same tips as in Xynariz's answer, his approach does not consider carry digits generally, which can miss possible solutions.
Like in Sudoku, there are quite a few patterns, some of which are very situational and convoluted and it'd be a hard exercise to not forget some case. However, there are some patterns that are more direct and frequent than others.
Generally, the themes involve pinpointing some letter to the values 0, 1 or 9, or deducing that a letter must be odd, even or within some range. The intersection of such clues should narrow down the possible values for a letter. For example, if you have $A = even$ and $A \lt 4$, then you know $A \in \{0, 2\}$ (this notation means A can be one of the following values).
Before starting looking for clues, I'd advise explicitly writing all equations including a carry digit, $c_n$, where $c_n$ is the overflow from the previous column, $n-1$. Since the starting column is 1, $c_0 = 0$. For the example in the OP, that would be:
$D + E = Y \pmod{10}$
$N + R + c_1 = E \pmod{10}$ $E + O + c_2 = N \pmod{10}$ etc...
If you're dealing with only two addends, $c_n \in \{0, 1\}$. For 3 addends, $c_n \in \{0, 1, 2\}$. That's because the highest value for the first column can be $9 + 9 + 9 = 27$ and since $c_1 \le 2$, any subsequent columns can be at most $29$. In general, for $k$ addends, $c_n \le k - 1$.
Look for more digits in the result than in the addends.
For $A + B = CD$, we know that $A + B$ must overflow since $C$ is a leading digit and can't be $0$. This forces us to conclude $C = 1$. Explicitly, that would be written as $0 + 0 + c_1 = C \Leftrightarrow c_1 = C$, with $C \not= 0$ and $c_1 \in \{0, 1\}$. Therefore, $C = 1$.
A special case of this is $A + BC = DEF$. Similar to above we know $D = 1$ and $c_2 = 1$, which means $0 + B + c_1 \ge 10$. But since $B \le 9$, the only possibility is for $B = 9$ and $c_1 = 1$. By extension, $E = 0$. As you can see, by plugging in equations newly deduced values, we can create a chain effect.
Look for the same letter in one of the addends and the sum.
The pattern $A + B + c_n = A \pmod{10}$ means we add a multiple of 10 to $A$. For two addends this means either
$B = 0, c_n = 0, c_{n+1} = 0$, or $B = 9, c_1 = 1, c_{n+1} = 1$
Look for multiple occurances of a letter in the addends.
Here we try to establish the parity of a letter. $A + A + c_n = 2*A + c_n = B \pmod{10}$ can mean either
$c_n = 0, B = even$, or $c_n = 1, B = odd$
Consider the example $xAAx + xAAx = xCBx$, where we don't care what $x$ is or whether it exists; it only serves to show we focus on two columns in local context. Since the result in both sums is different, a carry digit is involved in only one of them. This means either
$c_1 = 1, B = odd, A \le 4, c_2 = 0, C = even$, or $c_1 = 0, B = even, A \ge 5, c_2 = 1, C = odd$
Carry digits matter, too!
Any restrictions you can deduce is progress. Consider the case $xABA + xACA = xADA$. Obviously from the first column we get $A = 0, c_1 = 0$. However, we also notice $A + A + c_2 = A$, which means $c_2 = 0$. Now we can write $B + C = D \le 9$. Imagine we later conclude $B \ge 7$. This would instantly mean $B = 7, C = 1, D = 9$. This is because we require all 3 letters to be different and $C = 0$ would result to $B = D$.
Think of the consequences when you derive a relation.
Here's a sneaky one I left out intentionally. Let's go back to $A + A + c_n = B \pmod{10}$. For the second scenario we assumed $c_n = 1$. However, if we were to mentally list the result for all possible values of $A$, we would also notice that $A \ne 9$. This is because $9 + 9 + 1 = 9 \pmod{10}$ would mean $A = B$. For the same reason, if $c_n = 0$ then $A \not= 0$.
On a similar note, from $ABC + DEF = GHI$ we can instantly say $G \ge 3$. Why? Because we require $X \ge 1$, where $X$ is any of the three leading digits. But since they're all different, the smallest sum for $G$ would be $1 + 2$. And since we have $G \le 9$, this means $A \le 8$.
With practice such patterns will become second nature. But explicitly writing every little clue, no matter how unimportant, may bring your attention to restrictions you didn't see. More importantly, as mentioned in the introduction abut the intersection of clues, you never know when or how a clue might come handy. For example, assume we have narrowed down two letters to even and greater than 5, effectively making one 6 and the other 8. If at some point you conclude a third letter must be 6, then you've reached a contradiction and can stop exploring that branch. Which leads me to...
Brute forcing is fine, too.
It's very likely you will narrow down a letter to a bunch of possible values but you can't continue. By assuming every possible value, you continue your derivation and either reach a valid solution or contradiction. The point here is to leave brute forcing for very end. Squeeze every last constraint you can from all the clues to minimise your trial and error load.
If your puzzle has 9 unique letters and you're down to the last 3, you have at most 4 possible values to choose from for each letter. In this case it's trivial to brute force one and see whether you can reach a solution for the rest.
And don't forget, if you're brute forcing for $A$ and $B$, but you already have the equation $A + B + c_3 = C$, with $C = 4, c_3 = 1, c_4 = 1$, you can rearrange it as $B = 13 - A$ and use it.
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We've seen that classical logic is closely connected to the logic of subsets. For any set \( X \) we get a poset \( P(X) \), the
power set of \(X\), whose elements are subsets of \(X\), with the partial order being \( \subseteq \). If \( X \) is a set of "states" of the world, elements of \( P(X) \) are "propositions" about the world. Less grandiosely, if \( X \) is the set of states of any system, elements of \( P(X) \) are propositions about that system.
This trick turns logical operations on propositions - like "and" and "or" - into operations on subsets, like intersection \(\cap\) and union \(\cup\). And these operations are then special cases of things we can do in
other posets, too, like join \(\vee\) and meet \(\wedge\).
We could march much further in this direction. I won't, but try it yourself!
Puzzle 22. What operation on subsets corresponds to the logical operation "not"? Describe this operation in the language of posets, so it has a chance of generalizing to other posets. Based on your description, find some posets that do have a "not" operation and some that don't.
I want to march in another direction. Suppose we have a function \(f : X \to Y\) between sets. This could describe an
observation, or measurement. For example, \( X \) could be the set of states of your room, and \( Y \) could be the set of states of a thermometer in your room: that is, thermometer readings. Then for any state \( x \) of your room there will be a thermometer reading, the temperature of your room, which we can call \( f(x) \).
This should yield some function between \( P(X) \), the set of propositions about your room, and \( P(Y) \), the set of propositions about your thermometer. It does. But in fact there are
three such functions! And they're related in a beautiful way!
The most fundamental is this:
Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq Y \) define its inverse image under \(f\) to be
$$ f^{\ast}(S) = \{x \in X: \; f(x) \in S\} . $$ The pullback is a subset of \( X \).
The inverse image is also called the
preimage, and it's often written as \(f^{-1}(S)\). That's okay, but I won't do that: I don't want to fool you into thinking \(f\) needs to have an inverse \( f^{-1} \) - it doesn't. Also, I want to match the notation in Example 1.89 of Seven Sketches.
The inverse image gives a monotone function
$$ f^{\ast}: P(Y) \to P(X), $$ since if \(S,T \in P(Y)\) and \(S \subseteq T \) then
$$ f^{\ast}(S) = \{x \in X: \; f(x) \in S\}
\subseteq \{x \in X:\; f(x) \in T\} = f^{\ast}(T) . $$ Why is this so fundamental? Simple: in our example, propositions about the state of your thermometer give propositions about the state of your room! If the thermometer says it's 35°, then your room is 35°, at least near your thermometer. Propositions about the measuring apparatus are useful because they give propositions about the system it's measuring - that's what measurement is all about! This explains the "backwards" nature of the function \(f^{\ast}: P(Y) \to P(X)\), going back from \(P(Y)\) to \(P(X)\).
Propositions about the system being measured also give propositions about the measurement apparatus, but this is more tricky. What does "there's a living cat in my room" tell us about the temperature I read on my thermometer? This is a bit confusing... but there is an answer because a function \(f\) really does also give a "forwards" function from \(P(X) \) to \(P(Y)\). Here it is:
Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq X \) define its image under \(f\) to be
$$ f_{!}(S) = \{y \in Y: \; y = f(x) \textrm{ for some } x \in S\} . $$ The image is a subset of \( Y \).
The image is often written as \(f(S)\), but I'm using the notation of
Seven Sketches, which comes from category theory. People pronounce \(f_{!}\) as "\(f\) lower shriek".
The image gives a monotone function
$$ f_{!}: P(X) \to P(Y) $$ since if \(S,T \in P(X)\) and \(S \subseteq T \) then
$$f_{!}(S) = \{y \in Y: \; y = f(x) \textrm{ for some } x \in S \}
\subseteq \{y \in Y: \; y = f(x) \textrm{ for some } x \in T \} = f_{!}(T) . $$ But here's the cool part: Theorem. \( f_{!}: P(X) \to P(Y) \) is the left adjoint of \( f^{\ast}: P(Y) \to P(X) \). Proof. We need to show that for any \(S \subseteq X\) and \(T \subseteq Y\) we have
$$ f_{!}(S) \subseteq T \textrm{ if and only if } S \subseteq f^{\ast}(T) . $$ David Tanzer gave a quick proof in Puzzle 19. It goes like this: \(f_{!}(S) \subseteq T\) is true if and only if \(f\) maps elements of \(S\) to elements of \(T\), which is true if and only if \( S \subseteq \{x \in X: \; f(x) \in T\} = f^{\ast}(T) \). \(\quad \blacksquare\)
This is great! But there's also
another way to go forwards from \(P(X)\) to \(P(Y)\), which is a right adjoint of \( f^{\ast}: P(Y) \to P(X) \). This is less widely known, and I don't even know a simple name for it. Apparently it's less useful. Definition. Suppose \(f : X \to Y \) is a function between sets. For any \( S \subseteq X \) define
$$ f_{\ast}(S) = \{y \in Y: x \in S \textrm{ for all } x \textrm{ such that } y = f(x)\} . $$ This is a subset of \(Y \).
Puzzle 23. Show that \( f_{\ast}: P(X) \to P(Y) \) is the right adjoint of \( f^{\ast}: P(Y) \to P(X) \).
What's amazing is this. Here's another way of describing our friend \(f_{!}\). For any \(S \subseteq X \) we have
$$ f_{!}(S) = \{y \in Y: x \in S \textrm{ for some } x \textrm{ such that } y = f(x)\} . $$This looks almost exactly like \(f_{\ast}\). The only difference is that while the left adjoint \(f_{!}\) is defined using "for some", the right adjoint \(f_{\ast}\) is defined using "for all". In logic "for some \(x\)" is called the
existential quantifier \(\exists x\), and "for all \(x\)" is called the universal quantifier \(\forall x\). So we are seeing that existential and universal quantifiers arise as left and right adjoints!
This was discovered by Bill Lawvere in this revolutionary paper:
By now this observation is part of a big story that "explains" logic using category theory.
Two more puzzles! Let \( X \) be the set of states of your room, and \( Y \) the set of states of a thermometer in your room: that is, thermometer readings. Let \(f : X \to Y \) map any state of your room to the thermometer reading.
Puzzle 24. What is \(f_{!}(\{\text{there is a living cat in your room}\})\)? How is this an example of the "liberal" or "generous" nature of left adjoints, meaning that they're a "best approximation from above"? Puzzle 25. What is \(f_{\ast}(\{\text{there is a living cat in your room}\})\)? How is this an example of the "conservative" or "cautious" nature of right adjoints, meaning that they're a "best approximation from below"?
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Probabilist often work on Polish spaces. Does somebody know an ("non-exotic") example, for which it is not possible to work on a Polish space, but instead one has to work on a general measurable space? By non-exotic example I mean something like a stochastic process, which is really used in applications, and cannot be defined on a Polish space...(I posted the this question also here).
There are a number of constructions that do not work for Polish spaces, but a certain class of probability spaces, variously known as super-atomless, saturated, nowhere countably generated and a number of other names. A nice overview can be found here.
A probability space $(\Omega,\Sigma,\mu)$ is
saturated if for every two Poilsh spaces $X$ and $Y$, every probability measure $\nu$ on $X\times Y$ and every random variable $f:\Omega\to X$ such that its distribution $\mu f^{-1}$ equals the marginal of $\nu$on $X$, there is a random variable $g:\Omega\to Y$ such that the joint distribution of $(f,g)$ is $\nu$.
The following definition is conceptually different, but can be shown to be equivalent:
A probability space $(\Omega,\Sigma,\mu)$ is
super-atomless if there is no $A\in\Sigma$ satisfying $\mu(A)>0$, such that the pseudo-metric space obtained byendowing the trace $\sigma$-algebra on $A$ with the pseudo-metric $d(A,B)=\mu(A\triangle B)$ is separable.
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First some setup:
Definition 1.9A graph$\Gamma$ consists of a set $V(\Gamma)$ of vertices and a set $E(\Gamma)$ of edges, each edge being associated to an unordered pair of vertices by a function "Ends": $\text{ENDS}(e) = \{v,w\}$ where $v,w \in V$. In this case we call $v$ and $w$ the ends of the edge $e$ and we also say $v$ and $w$ are adjacent.
We also allow the possibility that there are multiple edges with the same associated pair of vertices. Thus for two distinct edges $e$ and $e'$ it can be the case that $\text{ENDS}(e) = \text{ENDS}(e')$. We also allow loops, that is, edges whose associated vertices are the same.
Definition 1.14A symmetryof a graph $\Gamma$ is a bijection $\alpha$ taking vertices to vertices and edges to edges such that if $Ends(e) = \{v,w\}$, then $Ends(\alpha(e)) = \{\alpha (v),\alpha(v)\}$. The symmetry group of $\Gamma$ is the collection of all its symmetries. We note this group by $Sym(\Gamma)$
I take it that a symmetry is a map from $V(\Gamma) \cup E(\Gamma) \to V(\Gamma) \cup E(\Gamma)$.
Let $\overline{f} : V \to V$ be some bijection. In any natural way, does $\overline{f}$ induce a symmetry on $\Gamma$? I.e., does there exists a symmetry $f$ such that $f \bigg|_{V} = \overline{f}$? Here's what I thought. If $v \in V$, simply define $f(v) = \overline{f}(v)$. Now for the edges. Let $e \in E$. Then $\text{ENDS}(e) =\{v,w\}$ for some $v,w \in V$. Define $f(e)$ such that $\text{ENDS}(f(e)) = \{f(v),f(w)\}$.
But this doesn't seem rigorous, particularly that last sentence; and I'm worried about the possibility of multiple edges (which of the multiple edges, if there are any, is it being sent to?). I think it's clear that it preserves adjacency, but how do I verify that it's bijective?
Injectivity: Let $x,y \in V \cup E$ be such that $f(x) =f(y)$. Obviously $x$ and $y$ can't belong to different sets. If $x,y \in V$, injectivity of $\overline{f}$ implies $x=y$. If $x,y \in E$, then $Ends(x) = \{v_x,w_x\}$ and $Ends(y) = \{v_y,w_y\}$, so $Ends(f(x)) = Ends(f(y))$ or $\{f(v_x),f(w_x)\} = \{f(v_y),f(w_y)\}$... ? Surjectivity: Let $x \in E$ (if $x \in V$, there's nothing to prove). Then there are $v,w \in V$ such that $Ends(x) = \{v,w\}$. By surjectivity of $\overline{f}$, there are $v',w' \in V$ such that $f(v') = \overline{f}(v') = v$ and $f(w') = \overline{f}(w') =w$...But the problem is that $v'$ and $w'$ might not be adjacent, right... ?
The reason for my interest is that nearly every example in my book defines a map just on the vertices or just on the edges and simply asserts that it is a graph symmetry. My thought was that the author was presupposing this result I'm asking about.
EDIT
Ah, I think I came up with a counterexample: a square with a line segment connecting the bottom right vertex to a vertex not on the square (so the graph has five vertices). Define $f$ to permute the bottom left vertex and the vertex not on the square, fixing everything else. This defines a bijection on the vertices. But if the follow the above procedure, you get something that maps an edge to a non-edge.
I think the above procedure works for complete graphs. Any other graphs?
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GR0177 #50
Problem This problem is still being typed.
Wave Phenomena}Sound
Since the wavelength of the wave does not change, as the pipe presumably stays approximately the same length, only the frequency varies. If the speed of sound changes, then the frequency changes. If the speed of sound is lower than usual, then the frequency is lower. Thus, choices (A), (B) and (C) remain. Calculate to get choice (B).
Alternate Solutions
uhurulol 2014-10-20 18:24:42 If you're looking for a more mathematical answer, use the simple relation
=
Note that is just , and as such
= .
cancels on both sides and , so plugging everything in and cancelling we have
Cross multiply to obtain , as in choice (B). Hope this helps. dham 2010-10-06 23:07:59 For a quick guess: Cold instruments go flat. (Frequency goes down), leaving choices A-C.
Directly: c=f, so speed of sound is directly proportional to wavelength.
Find the change in wavelength by finding (440)=3*4.4.
That's little over 12.
440-12=428, which is a little over 427 (B).
B is the right answer.
Comments
star19 2019-08-13 10:33:05 Nice blog, the article you have shared is good.\r\ntest\r\n star19 2019-08-13 10:30:46 Nice blog, the article you have shared is good.This article is very useful. My friend suggest me to use this blog.\r\ntest star19 2019-08-13 10:27:32 Our academic pursuits, along with a range of extracurricular activities, help in honing a child\'s skills and ensuring that he/she grows to be a mature and responsible citizen. \r\nbest " target="_blank">http://www.gdgoenkanoida.edu.in\">best school in greater noida\r\n\r\n yummyhat 2017-10-26 12:40:36 use the fact that wavelength*frequency=velocity. notice that the wavelength is determined by the pipe and is constant. use a ratio because the ratio of velocities is already given, and solve for the final frequency. kevintah 2015-10-05 10:05:11 Hey guys, . . . . Kevin Tah Njokom here :)\r\n\r\n Currently studying for the Physics GRE. This is my first response here. I think it is how to attack the problem, although you have to be able to do it fast to account for time. So here goes.\r\n First a few observations. I noticed that the solution was given by @uhurulol using some very quick method, which is probably faster. My solution is an alternative way of arriving at the answer.\r\n\r\n Notice that the pipe is open at one end. This makes the following relation valid. . Here L is the length of the pipe . This is a good thing because we know that the length is not suddenly going to change. \r\n We also know the relation that . \r\n We solve for from this, and then substitute the value into the equation for . This gives us . Since is constant it mus apply for the new and old frequency and velocity.\r\n\r\n It is mentioned that on a cold day, the speed of sound is 3% lower than it would be at 20 degrees.\r\n\r\n This means \r\n \r\nThis makes \r\n\r\nThus:\r\n\r\n \r\n\r\nWhen you crunch things out, you get 426.8 close to 427. This makes B, the answer.\r\n\r\n Good Luck on the Physics GRE everyone. Study hard. Let me know what you think of my answer.
kevintah 2015-10-05 23:39:49 Some how the Latex got distorted, and I can\'t edit it . Oh well uhurulol 2014-10-20 18:24:42 If you're looking for a more mathematical answer, use the simple relation
=
Note that is just , and as such
= .
cancels on both sides and , so plugging everything in and cancelling we have
Cross multiply to obtain , as in choice (B). Hope this helps. jwbrooks0 2014-09-22 21:21:30 I'm confused. The speed of sound should be proportional to the sqrt(T). http://en.wikipedia.org/wiki/Speed_of_sound#Speed_in_ideal_gases_and_in_air
Therefore, why isn't 440*sqrt(0.97) = 433 correct?
Stardust 2015-10-22 15:37:18 The question says the \'SPEED OF SOUND is 3% lower than...\' not the temperature. dham 2010-10-06 23:07:59 For a quick guess: Cold instruments go flat. (Frequency goes down), leaving choices A-C.
Directly: c=f, so speed of sound is directly proportional to wavelength.
Find the change in wavelength by finding (440)=3*4.4.
That's little over 12.
440-12=428, which is a little over 427 (B).
B is the right answer. f4hy 2009-11-06 23:27:57 They give you the 20 degrees as a trick then?
dham 2010-10-06 23:09:25 Not really, it's just an initial condition. michealmas 2006-12-30 11:48:37 use the equation relating freq., velocity, and wavelength. nahmad 2006-03-30 22:21:27 One quick way of doing this calculation in your head is to consider it as a 3% loss. Thus -3 for every 100, which gives a loss of -12. 440-12 = 428.
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This is claimed in a Wikipedia Article:
If two representations (of a $C^*$-algebra $A$) $\rho$ and $\sigma$, on Hilbert spaces $H$ and $G$ respectively, are each unitarily equivalent to a subrepresentation of the other, then they are unitarily equivalent.
I don't really follow the proof given. Indeed, what bothers me is the following: if we let $I,J$ be sets and $f:I\rightarrow J, g:J\rightarrow I$ be injections inducing isometries $\ell^2(I)\rightarrow\ell^2(J)$ and $\ell^2(J)\rightarrow\ell^2(I)$ and we just let $A$ be the complex numbers, then the claim is that there is a unitary $U:\ell^2(I)\rightarrow\ell^2(J)$. But furthermore, the proof indicates that $U$ is constructed from the original isometries, so $U$ maps basis vectors to basis vectors and hence immediately induces a bijection $I\rightarrow J$. That is, we seem to have reproved Schröder–Bernstein using nothing but induction...
Indeed, I am fairly sure I can prove this result using Schröder–Bernstein for projections, working with $\lambda=\pi\oplus\rho$ and using comparison of projections in $\lambda(A)'$. Notice that in the projection case, using the notation of the Wikipedia article, you need to consider $R$.
Am I correct to be a little worried?
Furthermore, Schröder–Bernstein for projections is very standard, and in lots of books. Perhaps I have just not looked hard enough, but I cannot find a textbook proof of the result for the representations.
Is there a reference to this result in a textbook or paper?
Edit: Let me re-write the Wikipedia proof, in the special setting I outlined about. It boils down to observing that we can partition $I = I' \sqcup g(J)$ and $J = J' \sqcup f(I)$ and then repeatedly applying $f$ and $g$, so$$ J = J' \sqcup f(I) = J' \sqcup f(I') \sqcup fg(J)= J' \sqcup f(I') \sqcup fg(J') \sqcup fgf(I) = \cdots $$and$$ f(I) = f(I') \sqcup fg(J) = f(I') \sqcup fg(J') \sqcup fgf(I) = \cdots $$But now we come to Ruy's objection: you cannot conclude that e.g.$$ J = \bigsqcup_{n\geq0} (fg)^n(J') \sqcup \bigsqcup_{n\geq 0} (fg)^nf(I') $$The argument (for example) completely missing the possibility of "cycles", in Andreas's language.
So, I think I've answer my first question. I know how to prove the result. I am still interested in an actual reference in e.g. a textbook.
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This shows you the differences between two versions of the page.
Both sides previous revision Previous revision numerical-shadow [2018/05/22 14:43]
plewandowska [Other names]
numerical-shadow [2018/10/08 08:50] (current)
plewandowska [Definition]
Line 3: Line 3: ===== Definition ===== ===== Definition ===== - For any square matrix $A$ of size $ N$, one defines its //numerical shadow// as a probability distribution $P_A(z)$ on the complex plane, supported in the numerical range $W(A)$, + For any square matrix $A$ of dimension $ d$, one defines its //numerical shadow// as a probability distribution $P_A(z)$ on the complex plane, supported in the numerical range $W(A)$, $$ $$ - P_A(z) := \int_{\ Omega_N} {\rm d} \mu(\psi) \delta\Bigl( z-\langle \psi|A|\psi\rangle\Bigr) . + P_A(z) := \int_{\ Omega_d} {\rm d} \mu(\psi) \delta\Bigl( z-\langle \psi|A|\psi\rangle\Bigr) . $$ $$ - Here $\mu(\psi)$ denotes the unique unitarily invariant (Fubini-Study) measure on the set $\ Omega_N$ of $N$-dimensional pure quantum states. In other words the shadow $P$ of matrix $A$ at a given point $z$ characterizes the likelihood that the expectation value of $A$ among a random pure state is equal to $z$. + Here $\mu(\psi)$ denotes the unique unitarily invariant (Fubini-Study) measure on the set $\ Omega_d$ of $N$-dimensional pure quantum states. In other words the shadow $P$ of matrix $A$ at a given point $z$ characterizes the likelihood that the expectation value of $A$ among a random pure state is equal to $z$. ===== Other names ===== ===== Other names =====
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It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced.
Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit.
@Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form.
A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts
I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it.
Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis.
Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)?
No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet.
@MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it.
Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow.
@QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary.
@Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer.
@QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits...
@QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right.
OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ...
So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study?
> I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago
In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a...
@MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really.
When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.?
@tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...)
@MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences).
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Set between Connected Set and Closure is Connected Theorem
Let $T$ be a topological space.
Let $H$ be a connected set of $T$.
Let $H \subseteq K \subseteq H^-$, where $H^-$ denotes the closure of $H$. Then $K$ is connected.
Let $T$ be a topological space.
Let $H$ be a connected set of $T$.
Let $H^-$ denote the closure of $H$ in $T$.
Then $H^-$ is connected in $T$.
Let $T = \struct{S, \tau}$ be a topological space.
Let $D$ be the discrete space $\left\{{0, 1}\right\}$.
Let $f: K \to D$ be an arbitrary continuous mapping.
We have that:
Thus by definition of connected set:
$f \left({H}\right) = \left\{{0}\right\}$ or $f \left({H}\right) = \left\{{1}\right\}$ Without loss of generality, let $f \left({H}\right) = \left\{{0}\right\}$.
Aiming for a contradiction, suppose $\exists k \in K: f \left({k}\right) = 1$.
Hence by definition of continuous mapping:
$f^{-1} \left({\left\{{1}\right\}}\right)$ is open in $K$.
Let $K$ be given the subspace topology.
Then for some $U$ open in $T$:
$f^{-1} \left({\left\{{1}\right\}}\right) = K \cap U$
We have that:
$k \in f^{-1} \left({\left\{{1}\right\}}\right) \subseteq U$
and:
$k \in H^-$
By definition of topology:
$\exists x \in H \cap U$
As $x \in H$, we have that:
$f \left({x}\right) = 0$
But because $x \in H \cap U \subseteq K \cap U = f^{-1} \left({\left\{{1}\right\}}\right)$:
$f \left({x}\right) = 1$
Thus $K$ is connected.
$\blacksquare$
Let $\operatorname{cl}_K \left({H}\right)$ denote the closure of $H$ in $K$.
$\operatorname{cl}_K \left({H}\right) = K \cap H^-$ $K \subseteq H^-$
and so by Intersection with Subset is Subset:
$\operatorname{cl}_K \left({H}\right) = K$ Let $D$ be the discrete space $\left\{{0, 1}\right\}$.
Let $f: K \to D$ be any continuous mapping.
SWe have that:
Thus by definition of connected set:
$f \left({H}\right) = \left\{{0}\right\}$ or $f \left({H}\right) = \left\{{1}\right\}$ Without loss of generality, let $f \left({H}\right) = \left\{{0}\right\}$. $f \left({\operatorname{cl}_K \left({H}\right)}\right) \subseteq \operatorname{cl}_K \left({f \left({H}\right)}\right) = \left\{{0}\right\}^-$
where $\left\{{0}\right\}^-$ is the closure of $\left\{{0}\right\}$ in $D$.
$\left\{{0}\right\}^- = \left\{{0}\right\}$
That is, $f \left({K}\right) = \left\{{0}\right\}$.
Thus $K$ is connected by definition.
$\blacksquare$
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Gamma-function $\Gamma$-function
2010 Mathematics Subject Classification:
Primary: 33B15 Secondary: 33B2033D05 [MSN][ZBL]$\newcommand{\abs}[1]{\left|#1\right|}\newcommand{\Re}{\mathop{\mathrm{Re}}}\newcommand{\Im}{\mathop{\mathrm{Im}}}$
A transcendental function $\Gamma(z)$ that extends the values of the factorial $z!$ to any complex number $z$. It was introduced in 1729 by L. Euler in a letter to Ch. Goldbach, using the infinite product $$ \Gamma(z) = \lim_{n\rightarrow\infty}\frac{n!n^z}{z(z+1)\ldots(z+n)} = \lim_{n\rightarrow\infty}\frac{n^z}{z(1+z/2)\ldots(1+z/n)}, $$ which was used by L. Euler to obtain the integral representation (Euler integral of the second kind, cf. Euler integrals) $$ \Gamma(z) = \int_0^\infty x^{z-1}e^{-x} \rd x, $$ which is valid for $\Re z > 0$. The multi-valuedness of the function $x^{z-1}$ is eliminated by the formula $x^{z-1}=e^{(z-1)\ln x}$ with a real $\ln x$. The symbol $\Gamma(z)$ and the name gamma-function were proposed in 1814 by A.M. Legendre.
If $\Re z < 0$ and $-k-1 < \Re z < -k$, $k=0,1,\ldots$, the gamma-function may be represented by the Cauchy–Saalschütz integral: $$ \Gamma(z) = \int_0^\infty x^{z-1} \left( e^{-x} - \sum_{m=0}^k (-1)^m \frac{x^m}{m!} \right) \rd x. $$ In the entire plane punctured at the points $z=0,-1,\ldots $, the gamma-function satisfies a Hankel integral representation: $$ \Gamma(z) = \frac{1}{e^{2\pi iz} - 1} \int_C s^{z-1}e^{-s} \rd s, $$ where $s^{z-1} = e^{(z-1)\ln s}$ and $\ln s$ is the branch of the logarithm for which $0 < \arg\ln s < 2\pi$; the contour $C$ is represented in Fig. a. [FIXME] It is seen from the Hankel representation that $\Gamma(z)$ is a meromorphic function. At the points $z_n = -n$, $n=0,1,\ldots$ it has simple poles with residues $(-1)^n/n!$.
Figure: g043310a
Contents Fundamental relations and properties of the gamma-function.
1) Euler's functional equation: $$ z\Gamma(z) = \Gamma(z+1), $$ or $$ \Gamma(z) = \frac{1}{z\ldots(z+n)}\Gamma(z+n+1); $$ $\Gamma(1)=1$, $\Gamma(n+1) = n!$ if $n$ is an integer; it is assumed that $0! = \Gamma(1) = 1$.
2) Euler's completion formula: $$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}. $$ In particular, $\Gamma(1/2)=\sqrt{\pi}$; $$ \Gamma\left(n+\frac{1}{2}\right) = \frac{1.3\ldots(2n-1)}{2^n}\sqrt{\pi} $$ if $n>0$ is an integer; $$ \abs{\Gamma\left(\frac{1}{2} + iy\right)}^2 = \frac{\pi}{\cosh y\pi}, $$ where $y$ is real.
3) Gauss' multiplication formula: $$ \prod_{k=0}^{m-1} \Gamma\left( z + \frac{k}{m} \right) = (2\pi)^{(m-1)/2}m^{(1/2)-mz}\Gamma(mz), \quad m = 2,3,\ldots $$ If $m=2$, this is the Legendre duplication formula.
4) If $\Re z \geq \delta > 0$ or $\abs{\Im z} \geq \delta > 0$, then $\ln\Gamma(z)$ can be asymptotically expanded into the Stirling series: $$ \ln\Gamma(z) = \left(z-\frac{1}{2}\right)\ln z - z + \frac{1}{2}\ln 2\pi + \sum_{n=1}^m \frac{B_{2n}}{2n(2n-1)z^{2n-1}} + O\bigl(z^{-2m-1}\bigr), \quad m = 1,2,\ldots, $$ where $B_{2n}$ are the Bernoulli numbers. It implies the equality $$ \Gamma(z) = \sqrt{2\pi} z^{z-1/2} z^{-z} \left( 1 + \frac{1}{12}z^{-1} + \frac{1}{288}z^{-2} - \frac{139}{51840}z^{-3} - \frac{571}{2488320}z^{-4} + O\bigl(z^{-5}\bigr) \right). $$ In particular, $$ \Gamma(1+x) = \sqrt{2\pi} x^{x+1/2} e^{-x + \theta/12x}, \quad 0 < \theta < 1. $$ More accurate is Sonin's formula [So]: $$ \Gamma(1+x) = \sqrt{2\pi} x^{x+1/2} e^{-x + 1/12(x+\theta)}, \quad 0 < \theta < 1/2. $$
5) In the real domain, $\Gamma(x) > 0$ for $x > 0$ and it assumes the sign $(-1)^{k+1}$ on the segments $-k-1 < x < -k$, $k = 0,1,\ldots$ (Fig. b).
Figure: g043310b
The graph of the function $ $.
For all real $x$ the inequality $$ \Gamma\Gamma^{\prime\prime} > \bigl(\Gamma^\prime\bigr)^2 \geq 0 $$ is valid, i.e. all branches of both $\abs{\Gamma(x)}$ and $\ln\abs{\Gamma(x)}$ are convex functions. The property of logarithmic convexity defines the gamma-function among all solutions of the functional equation $$ \Gamma(1+x) = x\Gamma(x) $$ up to a constant factor (see also the Bohr–Mollerup theorem).
For positive values of $x$ the gamma-function has a unique minimum at $x=1.4616321\ldots$ equal to $0.885603\ldots$. The local minima of the function $\abs{\Gamma(x)}$ form a sequence tending to zero as $x\rightarrow -\infty$.
Figure: g043310c
The graph of the function $ $.
6) In the complex domain, if $\Re z > 0$, the gamma-function rapidly decreases as $\abs{\Im z} \rightarrow \infty$, $$ \lim_{\abs{\Im z} \rightarrow \infty} \abs{\Gamma(z)}\abs{\Im z}^{(1/2)-\Re z}e^{\pi\abs{\Im z}/2} = \sqrt{2\pi}. $$
7) The function $1/\Gamma(z)$ (Fig. c) is an entire function of order one and of maximal type; asymptotically, as $r \rightarrow \infty$, $$ \ln M(r) \sim r \ln r, $$ where $$ M(r) = \max_{\abs{z} = r} \frac{1}{\abs{\Gamma(z)}}. $$ It can be represented by the infinite Weierstrass product: $$ \frac{1}{\Gamma(z)} = z e^{\gamma z} \prod_{n=1}^\infty \left(\left( 1 + \frac{z}{n} \right) e^{-z/n} \right), $$ which converges absolutely and uniformly on any compact set in the complex plane ($\gamma$ is the Euler constant). A Hankel integral representation is valid: $$ \frac{1}{\Gamma(z)} = \frac{1}{2\pi i} \int_{C'} e^s s^{-z} \rd s, $$ where the contour $C'$ is shown in Fig. d.
Figure: g043310d
$ $
G.F. Voronoi [Vo] obtained integral representations for powers of the gamma-function.
In applications, the so-called poly gamma-functions — $k$th derivatives of $\ln\Gamma(z)$ — are of importance. The function (Gauss' $\psi$-function) $$ \psi(z) = \frac{\mathrm{d}}{\mathrm{d}z}\ln\Gamma(z) = \frac{\Gamma'(z)}{\Gamma(z)} = -\gamma + \sum_{n=0}^\infty \frac{z-1}{(n+1)(z+n)} = -\gamma + \int_0^1 \frac{1 - (1-t)^{z-1}}{t} \rd t $$ is meromorphic, has simple poles at the points $z=0,-1,\ldots$ and satisfies the functional equation $$ \psi(z+1) - \psi(z) = \frac{1}{z}. $$ The representation of $\psi(z)$ for $\abs{z}<1$ yields the formula $$ \ln\Gamma(1+z) = -\gamma z + \sum_{k=2}^\infty \frac{(-1)^k S_k}{k} z^k, $$ where $$ S_k = \sum_{n=1}^\infty n^{-k}. $$ This formula may be used to compute $\Gamma(z)$ in a neighbourhood of the point $z=1$.
For other poly gamma-functions see [BaEr]. The incomplete gamma-function is defined by the equation $$ I(x,y) = \int_0^y e^{-t}t^{x-1} \rd t. $$ The functions $\Gamma(z)$ and $\psi(z)$ are transcendental functions which do not satisfy any linear differential equation with rational coefficients (Hölder's theorem).
The exceptional importance of the gamma-function in mathematical analysis is due to the fact that it can be used to express a large number of definite integrals, infinite products and sums of series (see, for example, Beta-function). In addition, it is widely used in the theory of special functions (the hypergeometric function, of which the gamma-function is a limit case, cylinder functions, etc.), in analytic number theory, etc.
References
[An] A. Angot, "Compléments de mathématiques. A l'usage des ingénieurs de l'electrotechnique et des télécommunications", C.N.E.T. (1957) [BaEr] H. Bateman (ed.) A. Erdélyi (ed.), Higher transcendental functions, 1. The gamma function. The hypergeometric functions. Legendre functions, McGraw-Hill (1953) [Bo] N. Bourbaki, "Elements of mathematics. Functions of a real variable", Addison-Wesley (1976) (Translated from French) [JaEm] E. Jahnke, F. Emde, "Tables of functions with formulae and curves", Dover, reprint (1945) (Translated from German) [Ni] N. Nielsen, "Handbuch der Theorie der Gammafunktion", Chelsea, reprint (1965) [So] N.Ya. Sonin, "Studies on cylinder functions and special polynomials", Moscow (1954) (In Russian) [Vo] G.F. Voronoi, "Studies of primitive parallelotopes", Collected works, 2, Kiev (1952) pp. 239–368 (In Russian) [WhWa] E.T. Whittaker, G.N. Watson, "A course of modern analysis", Cambridge Univ. Press (1952) Comments
The $q$-analogue of the gamma-function is given by $$ \Gamma_q(z) = (1-q)^{1-z} \prod_{k=1}^\infty \frac{1-q^{k+1}}{1-q^{k+z}}, \quad z \neq 0,-1,-2,\ldots;\quad 0<q<1, $$ cf. [As]. Its origin goes back to E. Heine (1847) and D. Jackson (1904).
References
[Ar] E. Artin, "The gamma function", Holt, Rinehart & Winston (1964) [As] R. Askey, "The $q$-Gamma and $q$-Beta functions" Appl. Anal., 8 (1978) pp. 125–141 How to Cite This Entry:
Gamma-function.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Gamma-function&oldid=25607
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An Experimental Study of the Decay $$D^0 \to K^- K^- K^+ \pi^+$$ Abstract
Using data from the E791 experiment at Fermi National Accelerator Laboratory (Fermilab), we have studied the Cabibbo favored, but phase space suppressed decay $$D^0 \to K^-K^- K^+ \pi^+$$ with the normalization channel $$D^0 \to K^- \pi^- \pi^+ \pi^+$$. We report the branching ratio of $$D^0 \to K^- K^- K^+ \pi^+$$ relative to the branching ratio of $$D^0 \to K^- \pi^- \pi^+ \pi^+$$. This value is (0.54 $$\pm$$ 0.13 $$\pm$$ 0.07)%. We see a clear signal of $K^-K^+$ resonance in the decay $$D^0 \to K^-K^-K^+\pi^+$$ from which we conclude that about (60 $$\pm$$ 30)% of $$KKK\pi$$ comes from $$D^0 \to \phi K^-\pi^+; \phi \to K^-K^+$$. We also set the range (0.30% - 0.90%) for the ratio $$P_{q\overline{q}} = P_{NoPop}$$ where $$P_{q\bar{q}}$$ is the contribution from either $$D^0 \to K^-K^-K^+\pi^+$$ terms that pop an $$s\bar{s}$$ or corresponding $$D^0 \to K^- \pi^- \pi^+ \pi^+$$ terms that pop either $$u\bar{u}$$ or $$d\bar{d}$$ pair and $$P_{NoPop}$$ is the contributions from the $$D^0 \to K^- \pi^- \pi^+ \pi^+$$ terms that do not have such corresponding popping.
Authors: Cincinnati U. Publication Date: Research Org.: Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Sponsoring Org.: USDOE Office of Science (SC), High Energy Physics (HEP) (SC-25) OSTI Identifier: 1421467 Report Number(s): FERMILAB-THESIS-2000-05; UMI-99-79957 538472 DOE Contract Number: AC02-07CH11359 Resource Type: Thesis/Dissertation Country of Publication: United States Language: English Subject: 73 NUCLEAR PHYSICS AND RADIATION PHYSICS; 72 PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Citation Formats
Devmal, Shiral Cleophas.
An Experimental Study of the Decay $D^0 \to K^- K^- K^+ \pi^+$. United States: N. p., 2000. Web. doi:10.2172/1421467.
Devmal, Shiral Cleophas.
An Experimental Study of the Decay $D^0 \to K^- K^- K^+ \pi^+$. United States. doi:10.2172/1421467.
Devmal, Shiral Cleophas. Mon . "An Experimental Study of the Decay $D^0 \to K^- K^- K^+ \pi^+$". United States. doi:10.2172/1421467. https://www.osti.gov/servlets/purl/1421467.
@article{osti_1421467,
title = {An Experimental Study of the Decay $D^0 \to K^- K^- K^+ \pi^+$}, author = {Devmal, Shiral Cleophas}, abstractNote = {Using data from the E791 experiment at Fermi National Accelerator Laboratory (Fermilab), we have studied the Cabibbo favored, but phase space suppressed decay $D^0 \to K^-K^- K^+ \pi^+$ with the normalization channel $D^0 \to K^- \pi^- \pi^+ \pi^+$. We report the branching ratio of $D^0 \to K^- K^- K^+ \pi^+$ relative to the branching ratio of $D^0 \to K^- \pi^- \pi^+ \pi^+$. This value is (0.54 $\pm$ 0.13 $\pm$ 0.07)%. We see a clear signal of $K^-K^+$ resonance in the decay $D^0 \to K^-K^-K^+\pi^+$ from which we conclude that about (60 $\pm$ 30)% of $KKK\pi$ comes from $D^0 \to \phi K^-\pi^+; \phi \to K^-K^+$. We also set the range (0.30% - 0.90%) for the ratio $P_{q\overline{q}} = P_{NoPop}$ where $P_{q\bar{q}}$ is the contribution from either $D^0 \to K^-K^-K^+\pi^+$ terms that pop an $s\bar{s}$ or corresponding $D^0 \to K^- \pi^- \pi^+ \pi^+$ terms that pop either $u\bar{u}$ or $d\bar{d}$ pair and $P_{NoPop}$ is the contributions from the $D^0 \to K^- \pi^- \pi^+ \pi^+$ terms that do not have such corresponding popping.}, doi = {10.2172/1421467}, journal = {}, number = , volume = , place = {United States}, year = {2000}, month = {5} }
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I’ll be flying to Nebraska tomorrow to teach at the Math Olympiad Summer Program. One of my more advanced classes will be on projective geometry and how it can help us understand Euclidean geometry. I thought I would share the introduction from my handout, since it is an attempt to demonstrate one of the many reasons projective geometry is such a gemstone:
General projective planes
Let’s start with some general theory. A
projective plane is any set $P$, which we call the set of points, and a nonempty collection $L$ of subsets of $P$ called lines, such that the following four axioms hold: P1: Every pair of distinct points are contained in exactly one line. P2: Every pair of distinct lines intersect in exactly one point. P3: There are four distinct points $a,b,c,d\in P$ no three of which lie on a line. P4: Every line contains at least three points.
Axiom
P2 raises some alarms. In Euclidean geometry, we assume that there are parallel lines that never meet, but here, we’re requiring that every pair of lines intersect. So the usual Euclidean plane is not a projective plane.
However, there are many interesting projective planes that do come in handy. The simplest and smallest example of a projective plane is known as the
Fano plane, or $F_7$, consisting of seven points and seven lines as shown below.
Notice the terms “point” and “line” are interchangeable in the axioms above. We can in fact interchange the roles of points and lines in any projective plane to obtain another projective plane, called the “dual” projective plane. Some projective planes, like the Fano plane above, are even
self-dual! (In the diagram above, we can associate each point $A, B, C,\ldots$ with its corresponding lowercase-letter line $a, b, c,\ldots$, and incidence is preserved: points $A$ lies on lines $b$, $c$, and $d$, whereas line $a$ contains points $B$, $C$, and $D$. And so on.)
The projective plane we’ll be dealing with today is the
real projective plane, which is an extension of the Euclidean plane that gets rid of those nasty parallel lines by adding an extra line out at infinity. Let’s take a look. The real projective plane
The real projective plane was first defined by Desargues as an extension of the Euclidean plane $\mathbb{R}^2$ that forces parallel lines to intersect “at infinity”, hence creating a plane that satisfies axiom
P2 rather than the parallel postulate. When you think about it, this is a rather natural model of things we see in reality. Suppose you are standing on parallel train tracks and looking out towards the horizon along the direction of the tracks. The tracks are parallel, but they appear to converge to a point on the horizon. In this sense, Euclidean geometry is simply a local version of projective geometry; parallel lines are only truly “parallel” near your feet!
To make this rigorous, define a
pencil of parallel lines in $\mathbb{R}^2$ to be the set of all lines parallel to a given line in the plane. Then the real projective plane is the set $\mathbb{R}^2\cup l^\infty$ where $l^\infty$ is a set consisting of one point for each pencil of parallel lines in $\mathbb{R}^2$. A line in this plane is then one of: $l^\infty$, Any subset of $\mathbb{R}^2\cup l^\infty$ consisting of a line in $\mathbb{R}^2$ along with the associated point in $l^\infty$.
It’s not hard to see that this satisfies axioms P1-P4 above. Any two lines that were parallel in $\mathbb{R}^2$ now meet at the point at infinity corresponding to their pencil.
Homogeneous coordinates
There is another, more symmetric way to define the real projective plane. We can think of a point in the real projective plane as a ratio $(a:b:c)$ of three real numbers, not all zero, called its
homogeneous coordinates. (Since we only care about the ratio between the numbers, $(2:4:5)$ and $(6:12:15)$ describe the same point.)
A
line is then defined as the set of solutions $(a:b:c)$ to a linear equation $$\alpha x +\beta y + \gamma z=0$$ for some fixed $\alpha$, $\beta$, $\gamma$.
It’s not hard to see why this is equivalent to our first definition of the real projective plane. Each point either has $z=0$ or $z\neq 0$. Those with $z\neq 0$, we can normalize so that $z=1$, and then take $(x:y:1)$ to be the point $(x,y)$ in $\mathbb{R}^2$. The points with $z=0$ then form the new line $l^\infty$, where $(x:y:0)$ corresponds to the direction with “slope” $y/x$ (or the vertical direction, if $x$ is $0$.)
Exercise. Show that the unit circle, expressed in homogeneous coordinates, is described by the equation $$x^2+y^2=z^2.$$ Why doesn’t the equation $x^2+y^2=1$ make sense in the projective plane?
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GR0177 #56
Problem This problem is still being typed.
Mechanics}Fluid Statics
The physical equation required is the buoyancy equation or Archimedes' Principal. The buoyant force is given by, . refers to the density of the fluid that buoy's the object. In this case, is just the density of air (not Helium).
The buoyant force for this problem is . This force must balance the load carried by the balloon. One now has (approximately, to simplify the calculations, since one is neglecting the weight of Helium) a simple sum of the forces problem, .
Now, solving for the volume of Helium required, one has . Since , this is about . However, since the balloon has to lift the weight of the Helium as well, the actual volume should be slightly larger. The closest choice is (D).
Alternate Solutions
tim123456789 2019-09-29 04:06:36 Archimedes Principle: The buoyant force is equal to the weight of the displaced fluid.\r\n\r\nAir is being displaced by the balloon. So:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n1.29 is slightly larger than one, so the quotient should be slightly smaller than 300, which is (D) greenfruit 2008-11-05 08:41:46 It's really simple, conceptually. For the balloon to float:
The mass of the block + the mass of the He in the balloon < the mass of the air displaced by the balloon
300kg + (x * .18 kg/m^3) <= x * 1.29 kg/m^3
with x being the volume of the balloon.
Solve for x, x=270 m^3.
Comments
tim123456789 2019-09-29 04:06:36 Archimedes Principle: The buoyant force is equal to the weight of the displaced fluid.\r\n\r\nAir is being displaced by the balloon. So:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n1.29 is slightly larger than one, so the quotient should be slightly smaller than 300, which is (D) NSF Fellow 2013-10-07 11:12:44 Why don't you just do
300 / (1.29-0.18)=270 Dodobird 2010-11-01 08:14:52 The easiest way to solve this problem is to use dimensional analysis. We are given a mass and two densities.rnThen to get volume just invert the densities and multiply by the mass. The difference between the two densities is nearly one.rnrn300/1.11~300 D)270 anmuhich 2009-03-15 12:22:06 Right I agree with NaijaProfessor. It's just a sum of the forces acting upon the balloon. The bouyant force is the weight displaced by the balloon due to Archimedes principle. This bouyant force equals the weight of the helium plus the weight of the mass. Solving this equation for the volume gives you 270. This amounts to the same thing as what Joe is saying but this is why it works. It's good that many people had the intuition to subtract the densities. If all else fails on this test it's best to go with your intuition. greenfruit 2008-11-05 08:41:46 It's really simple, conceptually. For the balloon to float:
The mass of the block + the mass of the He in the balloon < the mass of the air displaced by the balloon
300kg + (x * .18 kg/m^3) <= x * 1.29 kg/m^3
with x being the volume of the balloon.
Solve for x, x=270 m^3. duckduck_85 2008-10-31 17:58:16 I don't get why the buoyant force uses the volume of He while using the density of air. Isn't it supposed to be equal to the mass of the fluid displaced? Naturally, air by itself can't lift the balloon, but i just don't get the reasoning. Please help. grae313 2007-11-01 13:44:05 It looks like there is some confusion because the problem says, "neglect the mass of the balloon," but by this it means to neglect the mass of whatever material encloses the helium, not the helium itself as Yosun has assumed.
Thank you for this site, Yosun! joe 2006-10-21 21:03:19 One could also find the difference in the densities of air and helium (1.29-0.18=1.11) Then use that number to find the answer (300/1.11=270).
jesford 2008-04-05 10:13:47 I agree with joe... the correct way to do this problem would be to use the differences in densities for finding the bouyant force.
I really appreciate this site! Thanks Yosun, and everyone who contributes!
Poop Loops 2008-11-01 19:49:22 That's the way I did it. I thought it was dubious reasoning, but if others did it the same way then it must be alright to do it that way.
QM320 2008-11-05 20:59:29 That's what I did as well. I figured that the buoyancy force (up) of the He must equal the sum of weight + the downward force exerted by the air. I took the downward force exerted by the air to be where V is the volume of helium (assumed to be approximately equal to the volume of air exerting the force).
carl_the_sagan 2008-11-07 19:42:52 Also did it this way, if you can see a quick trick like this and arrived at a supplied answer, it's almost always a good thing.
jmason86 2009-07-12 15:47:58 I also did it like this.
It didn't seem like there should be any fundamental constants coming into play for this problem, so I just did dimensional analysis to figure out what combination of the values given would yield units. It worked out and only took about 30 seconds to complete the entire problem.. important time saved on this test :) joe 2006-10-21 21:01:53 NaijaProfessor 2005-11-24 06:18:47 If u equate the buoyant force to the weigth of the mass and that of helium, ur answer will be very approximate to 270m3. The helium gas does have weight u know acting downwards. Thus,
0.189Vg + 300g = Bouyant force
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Answer
(a) The simplest way to find 10% of a number is to move the decimal one place to the left. (b) 10% of 50 is 5. 10% x 50 = $\frac{10}{100}\times50 = \frac{1}{10}\times50 = 50\div10 = 5$
Work Step by Step
(a) The simplest way to find 10% of a number is to move the decimal one place to the left. (b) 10% of 50 is 5. 10% x 50 = $\frac{10}{100}\times50 = \frac{1}{10}\times50 = 50\div10 = 5$
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202 35 Summary What happens to invariant mass of an object when it gets closer or further from a gravitational body?
In Special Relativity, you learn that invariant mass is computed by taking the difference between energy squared and momentum squared. (For simplicity, I'm saying c = 1).
[tex] m^2 = E^2 - \vec{p}^2 [/tex] This can also be written with the Minkowski metric as: [tex] m^2 = \eta_{\mu\nu} p^\mu p^\nu [/tex] More generally, if there is a different metric (for example Schwartzchild), you would write it as: [tex] m^2 = g_{\mu\nu} p^\mu p^\nu [/tex] Now the question is, if invariant mass does not change from one metric to the other, you get the equation: [tex] 0 = (g_{\mu\nu} - \eta_{\mu\nu})p^\mu p^\nu [/tex] This seems to give unphysical results. I solved for a photon in the Schwartzchild metric, and the only physical solution available is if the Schwartzchild radius is 0. So this seems to imply that invariant mass (or lack thereof) is not invariant under gravitational fields. Any help here would be much appreciated. Thank you.
[tex] m^2 = E^2 - \vec{p}^2 [/tex]
This can also be written with the Minkowski metric as:
[tex] m^2 = \eta_{\mu\nu} p^\mu p^\nu [/tex]
More generally, if there is a different metric (for example Schwartzchild), you would write it as:
[tex] m^2 = g_{\mu\nu} p^\mu p^\nu [/tex]
Now the question is, if invariant mass does not change from one metric to the other, you get the equation:
[tex] 0 = (g_{\mu\nu} - \eta_{\mu\nu})p^\mu p^\nu [/tex]
This seems to give unphysical results.
I solved for a photon in the Schwartzchild metric, and the only physical solution available is if the Schwartzchild radius is 0. So this seems to imply that invariant mass (or lack thereof) is not invariant under gravitational fields.
Any help here would be much appreciated. Thank you.
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I've been reading Chen's original works on iterated integrals and in order to consider differential forms on the path space $PM$ of a smooth manifold $M$ he gives $PM$ the following "differentiable space" structure:
Let $N$ be a smooth manifold. A continuous map $\alpha: N \to PM$ (where $PM$ has the compact open topology) is said to be
smooth if the adjoint map $\tilde{\alpha}: N \times I \to M$ defined by $(n,t)\mapsto \alpha(n)(t)$ is smooth in the usual sense. The smooth map $\alpha$ is a plot if $N$ is an open convex subset of $\mathbb{R}^n$ for some $n$. Thus, we are modeling $PM$ locally with plots of varying dimension.
Chen defines a differential n-form $\omega$ on $PM$ as a rule which assigns to every plot $\alpha: U \to PM$ a differential form $\omega_{\alpha} \in \Omega^n(U)$. We define $(d\omega)_{\alpha}=d\omega_{\alpha}$. It turns out that for Chen's purposes one does not need to develop more calculus tools on $PM$. He shows a De Rham type result: the cohomology of the complex $\Omega^*(LM)$ (where $LM$ is the free loop space) is isomorphic to the real singular cohomology of $LM$.
However, for other purposes it is useful to consider forms as alternating tensors and to do this in this context we need a notion of vector fields on $PM$. I've always thought of a tangent vector at $\gamma \in PM$ as a vector field $T_\gamma$ along $\gamma$ on $M$, so a vector field on $PM$ assings each point $\gamma \in PM$ a vector field along $\gamma$. However, following Chen, the natural way to define vector fields to make it compatible with his notion of differential forms is as follows: a vector field $T$ on $PM$ is a rule which assings to each plot $\alpha: U \to PM$ a vector field $T_{\alpha}$ on $U$.
How do we reconcile these two notions of vector fields on $PM$? Are they equivalent?
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I have solution that can guarantee the end goal being reached in a maximum of
four five steps. (Many thanks for @dmg and @Taemyr for their comments to fix this solution.) The trick is to:
Reduce the cups into the following configuration:
$\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $
where turning any two cups along the diagonal will satisfy the end goal
For a constructive proof, we first consider the following
two three CASES:
CASE 1) the orientation of a single cup is different from the others, e.g. $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix} $ or $\begin{bmatrix} \bullet & \circ \\ \bullet & \bullet \end{bmatrix} $
CASE 2a) two cups are in each orientation along the diagonal, e.g. $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $ CASE 2b) two cups are in each orientation along two sides, e.g. $\begin{bmatrix} \circ & \bullet \\ \circ & \bullet \end{bmatrix} $
Step 1
Take two cups along the diagonal.
- If they are different, flip $\circ$ to $\bullet$. If this doesn't sound the bell, we either have $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix}$ or $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$, with the majority orientation being $\circ$. In either case, proceed to step 2.
- If they are the same, flip both. If this doesn't sound the bell, we either have $\begin{bmatrix} \bullet & \circ \\ \bullet & \bullet \end{bmatrix}$ or $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Either way, the majority orientation is known, and we can proceed to step 3.
Step 2
Take two cups along the diagonal.
- If they are different, we now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Flip $\bullet$ to $\circ$ to win.
- If they are the same, flip both. If this doesn't sound the bell, we now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Proceed to Step 3.
Step 3
At this point, we have already figured out what the majority orientation is, so we assume we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix} $ WLOG. Then:
Take two cups along the diagonal.
- If the cups are different, flip $\bullet$ to $\circ$ to win.
- If the cups are both $\circ$, flip one of them. We now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \bullet \end{bmatrix}$.
Step 4
Take two adjacent cups along an edge and flip both.
- If the cups chosen are the same, we win.
- If they are different, we now know we have $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $
Step 5
Take two cups along a diagonal and flip both. We win!
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Three-body closed chain of interactive (an)harmonic oscillators and the algebra $sl(4)$
Alexander TURBINER, Willard MILLER JR, Adrian ESCOBAR-RUIZ
- 2019-10-07 (P/19/13)
In this work we study 2- and 3-body oscillators with quadratic and sextic pairwise potentials which depend on relative {\it distances}, $|{\bf r}_i - {\bf r}_j |$, between particles. The two-body harmonic oscillator is two-parametric and can be reduced to a one-dimensional radial Jacobi oscillator, while in the 3-body case such a reduction is not possible in general. Our study is restricted to solutions in the space of relative motion which are functions of mutual (relative) distances only ($S$-states). In general, three-body harmonic oscillator is 7-parametric depending on 3 masses and 3 spring constants, and frequency. It is shown that for certain relations involving masses and spring constants the system becomes maximally (minimally) superintegrable in the case of two (one) relations.
Solutions of loop equations are random matrices
Bertrand EYNARD
- 2019-09-23 (M/19/12)
On volume subregion complexity in Vaidya spacetime
Roberto AUZZI, Giuseppe NARDELLI, Fidel Ivan SCHAPOSNIK MASSOLO, Gianni TALLARITA, Nicolo ZENONI
- 2019-08-29 (P/19/11)
We study holographic subregion volume complexity for a line segment in the AdS3 Vaidya geometry. On the field theory side, this gravity background corresponds to a sudden quench which leads to the thermalization of the strongly-coupled dual conformal field theory. We find the time-dependent extremal volume surface by numerically solving a partial differential equation with boundary condition given by the Hubeny-Rangamani-Takayanagi surface, and we use this solution to compute holographic subregion complexity as a function of time. Approximate analytical expressions valid at early and at late times are derived.
Action potential solitons and waves in axons
Gaspar CANO, Rui DILãO
- 2019-08-07 (P/19/10)
We show that the action potential signals generated inside axons propagate as reaction-diffusion solitons or as reaction-diffusion waves, refuting the Hodgkin and Huxley (HH) hypothesis that action potentials propagate along axons with an elastic wave mechanism. Action potential signals are solitary propagating spikes along the axon, occurring in a type I intermittency regime of the HH model. Reaction-diffusion action potential wave fronts annihilate at collision and at the boundaries of axons with zero flux, in contrast with elastic waves, where amplitudes add up and reflect at boundaries. We calculate numerically the values of the speed of the action potential spikes, as well as the dispersion relations.
These findings suggest several experiments as validating and falsifying tests for the HH model.
Super McShane Identity
Yi HUANG, Robert PENNER, Anton ZEITLIN
- 2019-08-06 (M/19/08)
The McShane identity for the once-punctured
super torus is derived following Bowditch's proof in the
bosonic case using techniques in super Teichmueller
theory developed by the two latter-named authors.
An electrophysiology model for cells and tissues
Rui DILãO
- 2019-08-06 (P/19/09)
We introduce a kinetic model to study the dynamics of ions in aggregates of cells and tissues. Different types of communication channels between adjacent cells and between cells and intracellular space are considered (ion channels, pumps and gap junctions). We shows that stable transmembrane ionic Nernst potentials are due to the coexistence of both specialised ion pumps and channels. Ion pumps or channels alone do not contribute to an equilibrium transmembrane potential drop. The kinetic parameters of the model straightforwardly calibrate with the Nernst potentials and ion concentrations. The model is based on the ATPase enzymatic mechanism for the ions $\hbox{Na}^+$, $\hbox{K}^+$, and it can be generalised for other ion pumps.
We extend the model to account for electrochemical effects, where transmembrane gating mechanism are introduced.
In this framework, axons can be seen as the evolutionary result of the aggregation of cells through gap junctions, which can be identified as the Ranvier nodes. In this kinetic framework, the injection of current in an axon induces the modification of the potassium equilibrium potential along the axon.
McShane identities for Higher Teichmuller theory and the Goncharov-Shen potential
Yi HUANG, Zhe SUN
- 2019-06-25 (M/19/07)
In [GS15], Goncharov and Shen introduce a family of mapping class group invariant regular functions on their A-moduli space to explicitly formulate a particular homological mirror symmetry conjecture. Using these regular functions, we obtain McShane identities for general rank positive surface group representations with loxodromic boundary monodromy and (non-strict) McShane-type inequalities for general rank positive representations with unipotent boundary monodromy. Our identities are expressed in terms of projective invariants, and we study these invariants: we establish boundedness and Fuchsian rigidity results for triple ratios. Moreover, we obtain McShane identities for finite-area cusped convex real projective surfaces by generalizing the Birman--Series geodesic scarcity theorem. We apply our identities to derive the simple spectral discreteness of unipotent bordered positive representations, collar lemmas, and generalizations of the Thurston metric.
Crystal Volumes and Monopole Dynamics
Sergey CHERKIS, Rebekah CROSS
- 2019-06-08 (P/19/06)
Third Kind Elliptic Integrals and 1-Motives
Cristiana BERTOLIN
- 2019-05-21 (M/19/05)
In [5] we have showed that the Generalized Grothendieck's Conjecture of Periods applied to 1-motives, whose underlying semi-abelian variety is a product of elliptic curves and of tori, is equivalent to a transcendental conjecture involving elliptic integrals of the first and second kind, and logarithms of complex numbers.
In this paper we investigate the Generalized Grothendieck's Conjecture of Periods in the case of 1-motives whose underlying semi-abelian variety is a non trivial extension of a product of elliptic curves by a torus. This will imply the introduction of elliptic integrals of the third kind for the computation of the period matrix of M and therefore the Generalized Grothendieck's Conjecture of Periods applied to M will be equivalent to a transcendental conjecture involving elliptic integrals of the first, second and third kind.
Large genus behavior of topological recursion
Bertrand EYNARD
- 2019-05-13 (M/19/04)
We show that for a rather generic set of regular spectral curves, the {\it Topological--Recursion} invariants $F_g$ grow at most like $O((\beta g)! r^{-g}) $ with some $r>0$ and $\beta\leq 5$
Standard conjectures in model theory, and categoricity of comparison isomorphisms. A model theory perspective.
Misha GAVRILOVICH
- 2019-03-22 (M/19/03)
On Phases Of Melonic Quantum Mechanics
Frank FERRARI, Fidel I. SCHAPOSNIK MASSOLO
- 2019-03-18 (P/19/02)
We explore in detail the properties of two melonic quantum mechanical theories which can be formulated either as fermionic matrix quantum mechanics in the new large D limit, or as disordered models. Both models have a mass parameter m and the transition from the perturbative large m region to the strongly coupled "black-hole" small m region is associated with several interesting phenomena. One model, with U(n)^2 symmetry and equivalent to complex SYK, has a line of first-order phase transitions terminating, for a strictly positive temperature, at a critical point having non-trivial, non-mean-field critical exponents for standard thermodynamical quantities. Quasi-normal frequencies, as well as Lyapunov exponents associated with out-of-time-ordered four-point functions, are also singular at the critical point, leading to interesting new critical exponents. The other model, with reduced U(n) symmetry, has a quantum critical point at strictly zero temperature and positive critical mass m∗. For 0
The lure of conformal symmetry
Ivan TODOROV
- 2019-01-22 (P/19/01)
The Clifford algebra ${\rm Cl} (4,1) \simeq {\mathbb C} [4]$, generated by the real (Majorana) $\gamma$-matrices and by a hermitian $\gamma_5$, gives room to the reductive Lie algebra $u(2,2)$ of the conformal group extended by the $u(1)$ helicity operator. Its unitary positive energy ladder representations, constructed by Gerhard Mack and the author 50 years ago, opened the way to a better understanding of zero-mass particles and fields and their relation to the space of bound states of the hydrogen atom. They became a prototypical example of a minimal representation of a non-compact reductive group introduced during the subsequent decade by Joseph.
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Some means inequalities for positive operators in Hilbert spaces 1k Downloads Abstract
In this paper, we obtain two refinements of the ordering relations among Heinz means with different parameters via the Taylor series of some hyperbolic functions and by the way, we derive new generalizations of Heinz operator inequalities. Moreover, we establish a matrix version of Heinz inequality for the Hilbert-Schmidt norm. Finally, we introduce a weighted multivariate geometric mean and show that the weighted multivariate operator geometric mean possess several attractive properties and means inequalities.
Keywordsmeans inequalities positive linear operators Hilbert space MSC47A30 47A63 26D10 26D20 26E60 1 Introduction
Since Heinz proved a series of useful norm inequalities, which are closely related to the Cordes inequality and the Furuta inequality, in 1951, many researchers have devoted themselves to sharping the Heinz inequalities and extending the Heinz norm inequalities to more general cases with the help of a Bernstein type inequality for nonselfadjoint operators, the convexity of norm functions, the Jensen functional and its properties, the Hermite-Hadamard inequality, and so on. With this kind of research, the study of various means inequalities, such as the geometric mean, the arithmetic mean, the Heinz mean, arithmetic-geometric means, and Arithmetic-Geometric-Harmonic (A-G-H) weighted means, has received much attention and development too. For recent interesting work in this area, we refer the reader to [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and references therein.
Based on [1, 2, 3, 4, 5, 6, 7, 8], in this paper, we are concerned with the further refinements of the geometric mean and the Heinz mean for operators in Hilbert spaces. Our purpose is to derive some new generalizations of Heinz operator inequalities by refining the ordering relations among Heinz means with different parameters, and of the geometric mean by investigating geometric means of several operator variables in a weighted setting. Moreover, we will obtain a matrix version of the Heinz inequality for the Hilbert-Schmidt norm.
Tand S. When \(\nu=1/2\) we write \(T\nabla S\) and \(T\sharp S\) in short, respectively. We refer the reader to Kubo and Ando [6] for more information on the means of positive linear operators. aand b. It is clear that Tand S. Clearly, 2 Improved Heinz means inequalities
In this section, we improve the result and give two theorems as follows.
Theorem 2.1 Suppose\(T,S\in\mathfrak{B}^{++}(\mathcal{H})\), and let\(s,t\in[0,1]\) satisfy Then Proof x, we deduce that x, we have x, we have Theorem 2.2 Suppose\(T,S\in\mathfrak{B}^{++}(\mathcal{H})\) and let\(s,t\in[0,1]\) satisfy Then Proof xthat x, we know that, for any \(s,t\in [0,1]\) satisfying \(\vert s-\frac{1}{2}\vert \geq \vert t-\frac{1}{2}\vert \), \(s, t \neq\frac{1}{2}\), x, we have 3 Heinz inequality for the Hilbert-Schmidt norm Tand Sbeing positive semidefinite. For \(T=[a_{ij}]\in M_{n}\), the Hilbert-Schmidt norm of Tis defined by
Next, we prove the following matrix version of Heinz inequality for the Hilbert-Schmidt norm.
Theorem 3.1 Let\(s,t\in[0,1]\) satisfy Then Proof Tand Sare positive semidefinite, we know by the spectral theorem that there exist unitary matrices \(U, V \in M_{n}\) such that 4 The inductive weighted geometric means and means inequalities
Let \(F: \mathcal{D}\rightarrow\mathfrak{B}(\mathcal{H})\) be a mapping of
k variables defined in a convex domain \(\mathcal{D}\subseteq \mathfrak{B}(\mathcal{H})^{k}\). Recall from Hansen [7] that F is regular if: Uon \(\mathcal{H}\). Pand Qbe mutually orthogonal projections acting on \(\mathcal{H}\) and take arbitrary k-tuples \((T_{1},\ldots ,T_{k})\) and \((S_{1},\ldots,S_{k})\) of operators in \(\mathfrak{B}(\mathcal{H})\) such that the compressed tuples \((PT_{1}P,\ldots,PT_{k}P)\) and \((QS_{1}Q, \ldots, QS_{k}Q)\) are in the domain \(\mathcal{D}\). Then the k-tuple of diagonal block matrices Theorem 4.1 Suppose that \(F: \mathcal{D}^{k}_{+}\rightarrow\mathfrak {B}(\mathcal{H})_{\mathrm{sa}}\) is regular, concave, and continuous. Then the perspective function \(\mathcal{P}_{F}\) is monotone. Proof Theorem 4.2 Suppose that\(F: \mathcal{D}^{k}_{+}\rightarrow\mathfrak {B}(\mathcal{H})_{\mathrm{sa}}\) is a regular, concave, and positively homogeneous. Then the perspective function\(\mathcal{P}_{F}\) satisfies the property of congruence invariance: for any invertible operator W on\(\mathcal{H}\). Proof
It follows from Theorem 3.2 of [7] that the perspective function \(\mathcal{P}_{F}\) is concave. Moreover, since
F is positively homogeneous, it is easy to prove that \(\mathcal{P}_{F}\) is also positively homogeneous. Hence, by Proposition 2.3 in [7], we get the conclusion. □
(i) Let \(G^{\alpha}_{1}(T)=T\) for each positive definite invertible operator
T. kvariables we associate an auxiliary mapping \(A_{k}: \mathcal {D}^{k}_{+}\rightarrow\mathfrak{B}(\mathcal{H})\) such that \(A_{k}\) is regular and concave, and βis the weight associated to \(T_{1},\ldots, T_{k}\). Theorem 4.3 The means\(G_{k}^{\alpha}: \mathcal{D}^{k}_{+}\rightarrow \mathfrak{B}(\mathcal{H})^{+}\) constructed as above are regular, positively homogeneous, concave, and they satisfy for\(\mathbb{T}=(T_{1},\ldots T_{k})\in \mathcal{D}^{k}_{+}\). Proof
By the definition of \(G_{k}^{\alpha}\), we know that \(G_{k}^{\alpha}\) for each \(k=2,3,\ldots\) is the perspective of a regular positively homogeneous map. Therefore, \(G_{k}^{\alpha}\) are regular and positively homogeneous. Moreover, since \(G_{k+1}^{\alpha}\) is the perspective of \((G_{k}^{\beta})^{1-\alpha_{k+1}}\), we see that (4.2) holds.
kand the corresponding weight β. For \(\alpha_{k+1}\in[0,1]\), the map \(x\rightarrow x^{1-\alpha_{k+1}}\) is operator monotone (increasing) and operator concave. Then we have Remark 4.4
A similar analysis to Theorem 4.3 in [7] shows that the above conditions uniquely determine the Geometric means \(G_{k}^{\alpha}\) for \(k=1,2,\ldots\) by setting \(G_{1}^{\alpha}(T)=T\).
Theorem 4.5 Set\(\mathbb{T}=(T_{1},\ldots, T_{k})\in\mathcal{D}^{k}_{+}\). The means\(G_{k}^{\alpha}\) constructed as above have the following properties: (P1)
(consistency with scalars) \(G_{k}^{\alpha}(\mathbb {T})=T_{1}^{\alpha_{1}}\cdots T_{k}^{\alpha_{k}}\)
if the\(T_{i}\) ’s commute; (P2)
(joint homogeneity) \(G_{k}^{\alpha}(t_{1} T_{1},\ldots ,t_{k} T_{k}) =t_{1}^{\alpha_{1}}\cdots t_{k}^{\alpha_{k}}G_{k}^{\alpha}(\mathbb {T})\)
for\(t_{i}>0 \); (P3)
(monotonicity)
if\(B_{i}\leq T_{i}\) for all\(1\leq i\leq k\), then\(G_{k}^{\alpha}(\mathbb{B}) \leq G_{k}^{\alpha}(\mathbb{T})\); (P4)
(congruence invariance) \(G_{k}^{\alpha}(W^{*} T_{1} W,\ldots, W^{*} T_{k} W)=W^{*}G_{k}^{\alpha}(\mathbb{T}) W\)
for any invertible operator W on\(\mathcal{H}\); (P5)
(self-duality) \(G_{k}^{\alpha}(\mathbb {T}^{-1})=G_{k}^{\alpha}(\mathbb{T})^{-1}\);
(P6)
(A-G-H weighted mean inequalities) \((\sum_{i=1}^{k} \alpha_{i} T_{i}^{-1})^{-1}\leq G_{k}^{\alpha}(\mathbb{T})\leq\sum_{i=1}^{k} \alpha_{i} T_{i}\);
(P7)
(determinant identity) \(\det G_{k}^{\alpha }(\mathbb{T})=\Pi_{i=1}^{k}(\det T_{i})^{\alpha_{i}}\).
Proof k. Then, by virtue of Notes Acknowledgements
The authors acknowledge support from NSFC (No. 11571229). The authors also would like to thank the reviewers very much for their useful suggestions.
References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Copyright information Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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You are here News
Recent result from PandaX-II was published on Physical Review Letter (Phys. Rev. Lett. 119, 181302, "Editor's Suggestion") on October 30, 2017, back-to-back with the first result from XENON1T experiment. The papers are highlighted by a Physics "Viewpoint" commentary by Dan Hooper from FNAL and Univ. of Chicago, commenting that “...
The PandaX-II collaboration released the official WIMP search results using 54 ton-day exposure on Aug. 23, 2017. No excess events were found above the expected background. The most stringent upper limit on spin-independent WIMP-nucleon cross section was set for a WIMP mass greater than $100 GeV/c^{2}$, with the lowest exclusion at $8.6\times10^{-47} cm^{2}$ at $40 GeV/c^{2}$. The result reported here is more conservative than the preliminary result shown during the TeVPA2017 conference, due to the adoption of updated photon/electron detection efficiencies.
Prof. Xiangdong Ji of Shanghai Jiao Tong University and University of Maryland, spokesperson of the PandaX Collaboration, announced new results on the dark matter (DM) search from the PandaX-II experiment on Monday Aug. 7, during the TeV Particle Astrophysics 2017 Conference at Columbus, Ohio, the United States. No DM candidate was identified within the data from an exposure of 54 ton-day, the largest reported DM direct detection data set to date.
The PandaX observatory uses xenon as target and detector to search for WIMP particles as well as neutrinoless double beta decay ($0\nu\beta\beta$) in ${}^{136}Xe$. At present, the PandaX-II experiment is in operation in CJPL-I. The future PandaX program will pursue the following three main directions:
Xiangdong Ji, Spokesperson and Project Leader of the PandaX experiment located in the China Jin-Ping underground Laboratory (CJPL), announced the first dark matter search results from the PandaX-II 500 kg liquid xenon detector in the 2016 International Identification of Dark Matter conference at Sheffield, UK in the evening of July 21, 2016 (Beijing Time). He reported that no trace of dark matter was observed with an exposure of 33,000 kg·day of liquid xenon, providing the newest constraints on the existence of dark matter.
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Since it was getting boring to always do OOP Languages I decided to dabble in the functional realm of programming.
For that I chose the narrowly missed community-challenge by Edward: Resistor Mania, since it really is a quick-to solve using recursive and functional style.
Challenge description:
In electronics, two resistors in series have a combined resistance \$R_1+R_2\$, and two resistors in parallel have combined resistance of \$\displaystyle \frac{R_1 R_2}{R_1 + R_2}\$. Given an infinite supply of \$270\Omega\$ resistors with \$5\%\$ tolerance, write a program to describe how to combine them into any arbitrary resistance value.
To make things a little easier for me, I decided to ditch the tolerance for this first prototype. Instead of \$270\Omega\$ I also ask the user to provide the values for our resistors.
The code works as expected for rather simple inputs, and I didn't check for more complicated stuff yet.
The ouptut comes as a wiring help that uses simple "equations" to describe parallel and combining wiring.
Output of some manual tests:
*Main> resistor_mania 60 120 "60.0+60.0" *Main> resistor_mania 120 60 "120.0||(120.0)" *Main> resistor_mania 120 120 "120.0"
Code:
module Main whereimport System.IOmain = do hSetBuffering stdin LineBuffering; putStrLn "Please enter the value of resistors we use"; res_input <- getLine; putStrLn "Please enter the value you want to model with these resistors"; target_value <- getLine; putStrLn (show (resistor_mania (read res_input) (read target_value)));{-Calculate the remaining resistor-strenght necessary to get to a given target when going in parallel to the given existing resistor.-}p_resistor :: Fractional a => a -> a -> ap_resistor = (\existing target -> (1 / ((1 / target) - (1 / existing))));{-Resistor_Mania: Create a wiring schema for a given target resistor valuewith one single available kind of resistor-}resistor_mania :: (Show a, Ord a, Fractional a) => a -> a -> Stringresistor_mania resistor target = if resistor == target; then (show resistor); else if resistor < target; then (show resistor) ++ "+" ++ (resistor_mania resistor (target - resistor)); else (show resistor) ++ "||(" ++ (resistor_mania resistor (p_resistor resistor target)) ++ ")";
There's quite a few things that I hope can be improved here. For one I don't like how
p_resistor is declared as a lambda, and
resistor_mania isn't.
Another thing I don't like is the concatenation of Strings in the control structure and the type-signature of
resistor_mania makes me uneasy :/
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I have a question that seems to not been answered directly, i tried using the
empheq package to make boxes around equation (in the align environment ) and it works, but i want a wide box which occupies the hole textwidth and includes the equation number.
Here is an example code:
\documentclass[]{article} \usepackage{amsmath} \usepackage{empheq} \usepackage{xcolor} \empheqset{box=\fcolorbox{white}[gray]{0.85}} \begin{document} \begin{empheq}{align} U_e(r_A)=&-\frac{\hbar\mu_0}{8\pi^2c}\int\limits_0^\infty d\xi\xi^3\alpha(i\xi)\sum\limits_{l=1}^{\infty}(2l+1)\bigg\{r_{lTE}[h^{(1)}_l(kr_A)]^2\nonumber\\ &+r_{lTM}\bigg[l(l+1)\frac{[h^{(1)}_l(kr_A)]^2}{(kr_A)^2}+\frac{[kr_Ah^{(1)}_l(kr_A)]'^2}{(kr_A)^2}\bigg] \bigg\} \end{empheq} \end{document}
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Kimi and Jordan Task
Kimi and Jordan are each working during the summer to earn money in addition to their weekly allowance. Kimi earns \$9 per hour at her job, and her allowance is \$8 per week. Jordan earns \$7.50 per hour, and his allowance is \$16 per week.
Jordan wonders who will have more income in a week if they both work the same number of hours. Kimi says, "It depends." Explain what she means. Is there a number of hours worked for which they will have the same income? If so, find that number of hours. If not, why not? What would happen to your answer to part (b) if Kimi were to get a raise in her hourly rate? Explain. What would happen to your answer to part (b) if Jordan were no longer to get an allowance? Explain. IM Commentary
In the middle grades, students have lots of experience analyzing and comparing linear functions using graphs, tables, symbolic expressions, and verbal descriptions. In this task, students may choose a representation that suits them and then reason from within that representation.
To find the point of intersection in part (b), tabular approaches must extend beyond integer domain values. Halves, quarters, and tenths of hours can help students gradually hone in on the intersection, but more efficient methods use general equation solving techniques or proportional reasoning from values in the table.
Parts (c) and (d) require that students imagine changes in one of the two linear graphs (foreshadowing F-BF.3) and then predict what will happen to the point of intersection.
When used in instruction, this task provides opportunities to compare representations and to make connections among them.
While this problem involves linear functions, other tasks in this set illustrate F.BF.1a in the context of quadratic (Skeleton Tower), exponential (Rumors), and rational (Summer Intern) functions.
Solutions
Solution: Kimi and Jordan
The weekly total of Kimi's allowance and job earnings, $\$ K$, is the sum of her \$8 allowance and $\$9 \cdot h$, where $h$ is the number of hours she works. In other words, $K = 8+9h$. Similarly, the weekly total of Jordan's allowance and job earnings, $J$, is given by $J = 16+7.5h$. The graphs of these two equations are shown below.
The graphs show that if they work less than $5\frac13$ hours, Jordan has a greater total. But if they work more than $5\frac13$ hours, Kimi has a greater total.
Their totals are the same at the point of intersection of the graphs. We find the $h$-coordinate of this point of intersection by finding the number of hours for which Kimi and Jordan will have the same total, assuming they both work with same number of hours in a week. We do this by setting the expressions for Kimi's total and Jordan's total equal and solving for $h$:
\begin{align} 9h+8&=7.5h+16\ 1.5h &= 8\ h&=8/1.5=5\frac13. \end{align} The graphs show that Kimi and Jordan's totals are the same if they both work for $5\frac13$ hours. If Kimi's hourly rate were to increase, the dotted line above would have an increased slope and intersect the vertical axis at the same point, which would move the point of intersection to the left. So Kimi and Jordan's total would be the same if they both worked a number of hours less than $5\frac13$. If Jordan were to have no allowance, the solid line would intersect the vertical axis at $0$ and have the same slope. Because the slope of the solid line is less than that of the dotted line, the lines would not intersect in the first quadrant. So Jordan and Kimi's totals would never be the same.
Solution: Reasoning from the context
Kimi's allowance is less than Jordan's, but her hourly rate is higher. So for only a few hours worked, she will have less job earnings per week than Jordan, but if she works a lot of hours, her higher hourly rate will more than make up for the lower allowance. We can find the number of hours at which this change occurs by calculating how many hours, at $\$1.50$ more per hour, it will take to make up the $\$8$ difference in their allowances: $\$8/(\$1.50/\mbox{hour}) = 5\frac13$ hours, or $5$ hours and $20$ minutes. See part (a). If Kimi's hourly rate were higher she would be able to make up the allowance difference more quickly. In the above calculation, the denominator would increase and the resulting number of hours would decrease. If Jordan were without an allowance, it would be impossible for him to make up for Kimi's allowance because his hourly rate is lower.
Solution: Tabular solution
Based on their allowances and hourly rates, the following table compares their weekly income from 0 to 7 hours. Hours worked, $h$ 0 1 2 3 4 5 6 7 Kimi's weekly total, $K$ 8 17 26 35 44 53 62 71 Jordan's weekly total, $J$ 16 23.5 31 38.5 46 53.5 61 68.5
From the table above, it looks like their incomes will be the same somewhere between 5 and 6 hours. An income table for quarter hours shows that the their incomes might be equal between $5\frac14$ and $5\frac12$ hours. Similarly, a table for tenths of hours shows that the incomes might be equal between 5.3 and 5.4 hours.
A more efficient strategy is to notice from the table that the difference between their totals decreases by \$1.50 each hour, which makes sense as this is the difference between their hourly rates. At 5 hours, the difference is \$0.50, which is $\frac13$ of \$1.50. So it makes sense to try $\frac13$ hour more. During this 20 minutes, Kimi will make \$3.00 and Jordan will make \$2.50. So in $5\frac13$ hours they both will make \$56.
If Kimi's hourly rate were to increase, then the numbers in the middle row of the table (after the 8) would increase, exceeding the bottom row of the table sooner in the row. So Kimi and Jordan's totals be the same at less than $5\frac13$ hours. If Jordan were to have no allowance, all of the entries in the bottom row of the table would decrease by \$16. Because Jordan's total increases more slowly than Kimi's, it would fall further and further behind Kimi's, and their totals would never be the same.
Solution: Purely algebraic solution
The weekly total of Kimi's allowance and job earnings, $K$, is the sum of her \$8 allowance and (\$9 * $h$), where $h$ is the number of hours she works. In other words, $K = 8+9h$. Similarly, the weekly total of Jordan's allowance and job earnings, $J$, is given by $J = 16+7.5h$. Both $K$ and $J$ depend on $h$, so there is no obvious answer to Jordan's question without further investigation. First we find the number of hours for which Kimi and Jordan will have the same total, assuming they both work with same number of hours in a week. We do this by setting the expressions for Kimi's total and Jordan's total equal and solving for $h$: \begin{align} 9h+8&=7.5h+16\\ 1.5h &= 8\\ h&=8/1.5=5\frac13 \end{align} Because Jordan's allowance is greater, this also tells us that Jordan will have a greater total than Kimi if they work less than $5\frac13$ hours. If however, they work more than $5\frac13$ hours, then Kimi's total will be greater. If Kimi's hourly rate were greater, the 1.5 in the above solution would be a greater number, and the result of the division would be less. So they would have the same income at less than $5\frac13$ hours. Without an allowance Jordan's total would be given by $J=7.5h$. And the solution process above would become as follows: \begin{align} 9h+8&=7.5h\\ 1.5h &= -8\\ h&=-8/1.5=-5\frac13 \end{align} This means that their total would be the same if they both worked $-5\frac13$ hours. This solution to the algebraic equation does not make sense in terms of the context, since they both work a positive number of hours. So their totals would never be equal. Kimi and Jordan
Kimi and Jordan are each working during the summer to earn money in addition to their weekly allowance. Kimi earns \$9 per hour at her job, and her allowance is \$8 per week. Jordan earns \$7.50 per hour, and his allowance is \$16 per week.
Jordan wonders who will have more income in a week if they both work the same number of hours. Kimi says, "It depends." Explain what she means. Is there a number of hours worked for which they will have the same income? If so, find that number of hours. If not, why not? What would happen to your answer to part (b) if Kimi were to get a raise in her hourly rate? Explain. What would happen to your answer to part (b) if Jordan were no longer to get an allowance? Explain.
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Here is a closely related pair of examples from operator theory, von Neumann's inequality and the theory of unitary dilations of contractions on Hilbert space, where things work for 1 or 2 variables but not for 3 or more.
In one variable, von Neumann's inequality says that if $T$ is an operator on a (complex) Hilbert space $H$ with $\|T\|\leq1$ and $p$ is in $\mathbb{C}[z]$, then $\|p(T)\|\leq\sup\{|p(z)|:|z|=1\}$. Szőkefalvi-Nagy's dilation theorem says that (with the same assumptions on $T$) there is a unitary operator $U$ on a Hilbert space $K$ containing $H$ such that if $P:K\to H$ denotes orthogonal projection of $K$ onto $H$, then $T^n=PU^n|_H$ for each positive integer $n$.
These results extend to two commuting variables, as Ando proved in 1963. If $T_1$ and $T_2$ are commuting contractions on $H$, Ando's theorem says that there are commuting unitary operators $U_1$ and $U_2$ on a Hilbert space $K$ containing $H$ such that if $P:K\to H$ denotes orthogonal projection of $K$ onto $H$, then $T_1^{n_1}T_2^{n_2}=PU_1^{n_1}U_2^{n_2}|_H$ for each pair of nonnegative integers $n_1$ and $n_2$. This extension of Sz.-Nagy's theorem has the extension of von Neumann's inequality as a corollary: If $T_1$ and $T_2$ are commuting contractions on a Hilbert space and $p$ is in $\mathbb{C}[z_1,z_2]$, then $\|p(T_1,T_2)\|\leq\sup\{|p(z_1,z_2)|:|z_1|=|z_2|=1\}$.
Things aren't so nice in 3 (or more) variables. Parrott showed in 1970 that 3 or more commuting contractions need not have commuting unitary dilations. Even worse, the analogues of von Neumann's inequality don't hold for $n$-tuples of commuting contractions when $n\geq3$. Some have considered the problem of quantifying how badly the inequalities can fail. Let $K_n$ denote the infimum of the set of those positive constants $K$ such that if $T_1,\ldots,T_n$ are commuting contractions and $p$ is in $\mathbb{C}[z_1,\ldots,z_n]$, then $\|p(T_1,\ldots,T_n)\|\leq K\cdot\sup\{|p(z_1,\ldots,z_n)|:|z_1|=\cdots=|z_n|=1\}$. So von Neumann's inequality says that $K_1=1$, and Ando's Theorem yields $K_2=1$. It is known in general that $K_n\geq\frac{\sqrt{n}}{11}$. When $n>2$, it is not known whether $K_n\lt\infty$.
See Paulsen's book (2002) for more. On page 69 he writes:
The fact that von Neumann’s inequality holds for two commuting contractions
but not three or more is still the source of many surprising results and
intriguing questions. Many deep results about analytic functions come
from this dichotomy. For example, Agler [used] Ando’s theorem to deduce an
analogue of the classical Nevanlinna–Pick interpolation formula
for analytic functions on the bidisk. Because of the failure of a von
Neumann inequality for three or more commuting contractions, the analogous
formula for the tridisk is known to be false, and the problem of finding the
correct analogue of the Nevanlinna–Pick formula for polydisks
in three or more variables remains open.
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Abstract
Let $G={\bf SL}l(n,\Bbb{R})$ with $n\geq 6$. We construct examples of lattices $\Gamma \subset G$, subgroups $A$ of the diagonal group $D$ and points $x\in G/\Gamma$ such that the closure of the orbit $Ax$ is not homogeneous and such that the action of $A$ does not factor through the action of a one-parameter nonunipotent group. This contradicts a conjecture of Margulis.
[ber] D. Berend, "Minimal sets on tori," Ergodic Theory Dynam. Systems, vol. 4, iss. 4, pp. 499-507, 1984.
@article {ber, MRKEY = {779709},
AUTHOR = {Berend, Daniel}, TITLE = {Minimal sets on tori}, JOURNAL = {Ergodic Theory Dynam. Systems}, FJOURNAL = {Ergodic Theory Dynam. Systems}, VOLUME = {4}, YEAR = {1984}, NUMBER = {4}, PAGES = {499--507}, ISSN = {0143-3857}, MRCLASS = {28D15 (58F11 58F99)}, MRNUMBER = {86i:28028}, MRREVIEWER = {M. P. Heble}, ZBLNUMBER = {0563.58020}, } [B1] A. Borel, Introduction aux Groupes Arithmétiques, Paris: Hermann, 1969.
@book {B1, MRKEY = {0244260},
AUTHOR = {Borel, Armand}, TITLE = {Introduction aux Groupes Arithmétiques}, SERIES = {Publications de l'Institut de Mathématique de l'Université de Strasbourg, {\rm XV}}, PUBLISHER = {Hermann}, ADDRESS = {Paris}, YEAR = {1969}, PAGES = {125}, MRCLASS = {14.50 (20.00)}, MRNUMBER = {39 \#5577}, MRREVIEWER = {James Humphreys}, ZBLNUMBER = {0186.33202}, } [ekl] M. Einsiedler, A. Katok, and E. Lindenstrauss, "Invariant measures and the set of exceptions to Littlewood’s conjecture," Ann. of Math., vol. 164, iss. 2, pp. 513-560, 2006.
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AUTHOR = {Furstenberg, Harry}, TITLE = {Disjointness in ergodic theory, minimal sets, and a problem in {D}iophantine approximation}, JOURNAL = {Math. Systems Theory}, FJOURNAL = {Mathematical Systems Theory. An International Journal on Mathematical Computing Theory}, VOLUME = {1}, YEAR = {1967}, PAGES = {1--49}, ISSN = {0025-5661}, MRCLASS = {28.70 (10.00)}, MRNUMBER = {35 \#4369}, MRREVIEWER = {W. Parry}, DOI = {10.1007/BF01692494}, ZBLNUMBER = {0146.28502}, } [kss] D. Kleinbock, N. Shah, and A. Starkov, "Dynamics of subgroup actions on homogeneous spaces of Lie groups and applications to number theory," in Handbook of Dynamical Systems, Vol. 1 A, Amsterdam: North-Holland, 2002, pp. 813-930.
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AUTHOR = {Kleinbock, Dmitry and Shah, Nimish and Starkov, Alexander}, TITLE = {Dynamics of subgroup actions on homogeneous spaces of {L}ie groups and applications to number theory}, BOOKTITLE = {Handbook of Dynamical Systems, {V}ol. 1{\rm A}}, PAGES = {813--930}, PUBLISHER = {North-Holland}, ADDRESS = {Amsterdam}, YEAR = {2002}, MRCLASS = {22F30 (11E57 22E30 37A45 37C85)}, MRNUMBER = {2004b:22021}, MRREVIEWER = {Mikhail S. Kulikov}, ZBLNUMBER = {1050.22026}, } [lw] E. Lindenstrauss and B. Weiss, "On sets invariant under the action of the diagonal group," Ergodic Theory Dynam. Systems, vol. 21, iss. 5, pp. 1481-1500, 2001.
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AUTHOR = {Margulis, Gregory A. and Qian, Nantian}, TITLE = {Rigidity of weakly hyperbolic actions of higher real rank semisimple {L}ie groups and their lattices}, JOURNAL = {Ergodic Theory Dynam. Systems}, FJOURNAL = {Ergodic Theory and Dynamical Systems}, VOLUME = {21}, YEAR = {2001}, NUMBER = {1}, PAGES = {121--164}, ISSN = {0143-3857}, MRCLASS = {22F05 (22E46 37C85 37D20)}, MRNUMBER = {2003a:22019}, DOI = {10.1017/S0143385701001109}, ZBLNUMBER = {0976.22009}, } [MP] D. Meiri and Y. Peres, "Bi-invariant sets and measures have integer Hausdorff dimension," Ergodic Theory Dynam. Systems, vol. 19, iss. 2, pp. 523-534, 1999.
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AUTHOR = {Ratner, Marina}, TITLE = {Raghunathan's topological conjecture and distributions of unipotent flows}, JOURNAL = {Duke Math. J.}, FJOURNAL = {Duke Mathematical Journal}, VOLUME = {63}, YEAR = {1991}, NUMBER = {1}, PAGES = {235--280}, ISSN = {0012-7094}, CODEN = {DUMJAO}, MRCLASS = {22E40 (22D40 28D10)}, MRNUMBER = {93f:22012}, MRREVIEWER = {Gopal Prasad}, DOI = {10.1215/S0012-7094-91-06311-8}, ZBLNUMBER = {0733.22007}, } [S] N. A. Shah, "Uniformly distributed orbits of certain flows on homogeneous spaces," Math. Ann., vol. 289, iss. 2, pp. 315-334, 1991.
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AUTHOR = {Shah, Nimish A.}, TITLE = {Uniformly distributed orbits of certain flows on homogeneous spaces}, JOURNAL = {Math. Ann.}, FJOURNAL = {Mathematische Annalen}, VOLUME = {289}, YEAR = {1991}, NUMBER = {2}, PAGES = {315--334}, ISSN = {0025-5831}, CODEN = {MAANA}, MRCLASS = {22E40 (58F11)}, MRNUMBER = {93d:22010}, MRREVIEWER = {T. N. Venkataramana}, DOI = {10.1007/BF01446574}, ZBLNUMBER = {0702.22014}, } [tom] G. Tomanov, "Actions of maximal tori on homogeneous spaces," in Rigidity in Dynamics and Geometry (Cambridge, 2000), New York: Springer-Verlag, 2002, pp. 407-424.
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Best Compromise
There is a saying “a good compromise makes no one happy”. As distressing as that may be, I guess most people still think it is a good idea to seek the best agreement. The problem though, is to define the measure in which the word “best” gets a meaning. Let us consider $n$ people, each with a firm belief on $m$ different yes/no issues. We may think of the people’s beliefs as binary strings $b_ i$ of length $m$ for $0<i\leq n$, i.e. one string per person, with entries reflecting a person’s belief in each of the $m$ issues. An
agreement is also a binary string $a$ of length $m$, indicating the agreed outcome of each of the $m$ issues. Some would argue that the agreement $a$ that minimises the maximum of $H(b_ i-a)$ over all $i$, where $H(x)$ is the Hamming function counting the number of non-zero entries of the vector $x$, is the best compromise. Unfortunately, it is widely believed that finding this agreement is computationally infeasible for $n$ and $m$ large. Another reasonable suggestion is the agreement $a$ that minimises $\left(\sum _{i=1}^ n H(b_ i-a)^ p\right)^{1/p}$ for some positive integer $p$. Note in particular that when $p\rightarrow \infty $, this measure coincides with the former. I for one am most content with the choice of $p=1$ though, don’t you agree? Input
On the first line of input is a single positive integer $t$, telling the number of test scenarios to follow. Each test scenario is described by two positive integers $n\leq 100$ and $m\leq 100$, on a line of their own. Then follow $n$ lines, each containing a binary string $b_ i$ on $m$ ‘0’ and ‘1’ characters.
Output
For each test scenario, output the assignment $a$ that minimises the measure above with $p = 1$, on a row of its own. If there are several solutions, output anyone.
Sample Input 1 Sample Output 1 2 5 5 00000 00100 01001 01101 00101 1 7 0110010 00101 0110010
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Banach Journal of Mathematical Analysis Banach J. Math. Anal. Volume 8, Number 1 (2014), 55-63. On the essential spectrum of the sum of self-adjoint operators and the closedness of the sum of operator ranges Abstract
Let $\mathcal{H}$ be a complex Hilbert space, and $A_1,\ldots,A_N$ be bounded self-adjoint operators in $\mathcal{H}$ such that $A_i A_j$ is compact for any $i\neq j$. It is well-known that $\sigma_e(\sum_{i=1}^N A_i)\setminus\{0\}=(\cup_{i=1}^N\sigma_e(A_i))\setminus\{0\}$, where $\sigma_e(B)$ stands for the essential spectrum of a bounded self-adjoint operator $B$. In this paper we get necessary and sufficient conditions for $0\in\sigma_e(\sum_{i=1}^N A_i)$. This conditions are formulated in terms of the projection valued spectral measures of $A_i$, $i=1,\ldots,N$. Using this result, we obtain necessary and sufficient conditions for the sum of ranges of $A_i$, $i=1,\ldots,N$ to be closed.
Article information Source Banach J. Math. Anal., Volume 8, Number 1 (2014), 55-63. Dates First available in Project Euclid: 14 October 2013 Permanent link to this document https://projecteuclid.org/euclid.bjma/1381782087 Digital Object Identifier doi:10.15352/bjma/1381782087 Mathematical Reviews number (MathSciNet) MR3161682 Zentralblatt MATH identifier 1310.47003 Subjects Primary: 47B15: Hermitian and normal operators (spectral measures, functional calculus, etc.) Secondary: 46C07: Hilbert subspaces (= operator ranges); complementation (Aronszajn, de Branges, etc.) [See also 46B70, 46M35] Citation
Feshchenko, Ivan S. On the essential spectrum of the sum of self-adjoint operators and the closedness of the sum of operator ranges. Banach J. Math. Anal. 8 (2014), no. 1, 55--63. doi:10.15352/bjma/1381782087. https://projecteuclid.org/euclid.bjma/1381782087
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