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The Fundamental Group of the Real Projective Plane
The goal of today's post is to prove that the fundamental group of the real projective plane, $\pi_1(\mathbb{R}P^2)$, is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. And unlike our proof that $\pi_1(S^1)\cong\mathbb{Z}$, today's proof is fairly short, thanks to the van Kampen theorem!
An important observation
To make our application of van Kampen a little easier, we start with a simple observation:
projective plane - disk = Möbius strip
Below is an excellent animation which captures this quite clearly. Recall that the real projective plane is the set of all lines passing through the origin in $\mathbb{R}^3$. As a quotient space, this is the same as a sphere whose antipodal points are identified. Equivalently, $\mathbb{R}P^2$ is the quotient of a disc whose antipodal points along the boundary are identified. This is illustrated nicely in the video. (I highly recommend you also check out the animator's commentary which you can find here.)
We can also visualize the above by putting a CW-complex structure* on the projective plane and then removing the disc:
Applying van Kampen
Next, keeping the same CW-complex structure on $\mathbb{R}P^2$, we apply van Kampen by writing $\mathbb{R}P^2=A\cup B$ where $A$ is the red disc, $B$ is the blue annulus, and the intersection of $A$ and $B$ is the little, purple-ish annulus as shown below.
Van Kampen tells us that
I'll explain what "amalgamation" means in a little bit, but first let's determine $\pi_1$ of $A$, $B$, and $A\cap B$. We quickly see that $A$ is a (contractible) disc and so $$\pi_1(A)\cong 1=\langle\; 1 \;|\; \varnothing\;\rangle.$$ Also, since $A\cap B$ is an annulus it deformation retracts onto a circle whose fundamental group is generated by, say, $\gamma$, hence $$\pi_1(A\cap B)\cong \mathbb{Z}=\langle\; \gamma \;|\; \varnothing\;\rangle. $$ (Recall that if two spaces $X$ and $Y$ are homotopy equivalent, then $\pi_1(X)\cong \pi_1(Y)$. In particular this holds when that homotopy equivalence comes from a deformation retraction.) To determine $\pi_1(B)$, notice that $B$ is $\mathbb{R}P^2$ minus a disc. As we saw in our Observation above, this space is precisely a Möbius strip! Further, a Möbius strip deformation retracts onto its core circle (by a simple projection) - see the pink circle in the drawing down below. So let's call the generator of the fundamental group of this core circle $b$. Then $$\pi_1(B)\cong \mathbb{Z}=\langle\; b \;|\; \varnothing\;\rangle $$ and so
But what group is $1\ast_{\mathbb{Z}} \mathbb{Z}$?? To answer this question, we wish to find a presentation of $\pi_1(\mathbb{R}P^2)$. To do
that we must amalgamate, i.e. list the generators and relations of $\pi_1(A)$ and $\pi_1(B)$ along with new relations which we find by looking at $\pi_1(A\cap B)$**:
Here, $i_{A_*}:\pi_1(A\cap B)\to \pi_1(A)$ is the homomorphism induced by the injection map $i_A:A\cap B\hookrightarrow A$, and similarly $i_{B_*}:\pi_1(A\cap B)\to \pi_1(B)$ is the homomorphism induced by $i_B:A\cap B\hookrightarrow B.$ So the only tricky part to finding $\pi_1(\mathbb{R}P^2)$ is determining what the maps $i_{A_*}$ and $i_{B_*}$ should be. The first one is easy since $\pi_1(A)=1$ implies that $i_{A_*}$ must be the trivial map, i.e. $$i_{A_*}([\gamma])=1.$$ But to find the image of $\gamma$ under $i_{B_*}$, we need to think a bit about the relationship between $b$ and $\gamma$. (Recall that a homomorphism is completely determined by where it maps generators.) Let's revisit our graphic above, this time paying attention to these two generators:
We see that $b$ is the core circle of the Möbius band while $\gamma$ is its boundary circle. From here, the relationship between $b$ and $\gamma$ is evident: every
one loop around $\gamma$ corresponds to two loops around $b$! (This is shown near the 2:30 mark in the video above.) Hence$$i_{B_*}([\gamma])=b^2.$$ This allows us to write down the final presentation of $\pi_1(\mathbb{R}P^2)$, namely
QED!
Footnotes:
*with one 0-cells, one 1-cell, and one 2-cell which is glued onto the 1-cell according to the arrows indicated in the drawing.
** More generally, if we let $R_A$ denote the set of relations of the generators of $\pi_1(A)$ (and similarly for $R_B$) and if $\pi_1(A)=\langle \alpha_1,\ldots,\alpha_n|R_A\rangle$ and $\pi_1(B)=\langle \beta_1,\ldots,\beta_m|R_B\rangle$ then the amalgamation process tells us that for $X=A\cup B$, $$\pi_1(X)=\langle \alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m\;|\; R_A,\: R_B,\: i_{A_*}([\gamma_1])=i_{B_*}([\gamma_1]), \ldots,i_{A_*}([\gamma_t])=i_{B_*}([\gamma_t]) \rangle$$ where $\gamma_1,\ldots,\gamma_t$ are the generators of $\pi_1(A\cap B)$ and $i_{A_*}:\pi_1(A\cap B)\to\pi_1(A)$ and $i_{B_*}:\pi_1(A\cap B)\to\pi_1(B)$ are the homomorphisms induced by the injection maps $i_A:A\cap B\hookrightarrow A$ and $i_B:A\cap B\hookrightarrow B$, respectively. (Recall, any map $\varphi:X\to Y$ induces a homomorphism $\varphi_*:\pi_1(X)\to\pi_1(Y)$ given by $\varphi_*([f])=[\varphi\circ f]$. ) The only real tricky part here is figuring out what the maps $i_{A_*}$ and $i_{B_*}$ should be.
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De Bruijn-Newman constant
For each real number [math]t[/math], define the entire function [math]H_t: {\mathbf C} \to {\mathbf C}[/math] by the formula
[math]\displaystyle H_t(z) := \int_0^\infty e^{tu^2} \Phi(u) \cos(zu)\ du[/math]
where [math]\Phi[/math] is the super-exponentially decaying function
[math]\displaystyle \Phi(u) := \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3 \pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u}).[/math]
It is known that [math]\Phi[/math] is even, and that [math]H_t[/math] is even, real on the real axis, and obeys the functional equation [math]H_t(\overline{z}) = \overline{H_t(z)}[/math]. In particular, the zeroes of [math]H_t[/math] are symmetric about both the real and imaginary axes. One can also express [math]H_t[/math] in a number of different forms, such as
[math]\displaystyle H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du[/math]
or
[math]\displaystyle H_t(z) = \frac{1}{2} \int_0^\infty e^{t\log^2 x} \Phi(\log x) e^{iz \log x}\ \frac{dx}{x}.[/math]
In the notation of [KKL2009], one has
[math]\displaystyle H_t(z) = \frac{1}{8} \Xi_{t/4}(z/2).[/math]
De Bruijn [B1950] and Newman [N1976] showed that there existed a constant, the
de Bruijn-Newman constant [math]\Lambda[/math], such that [math]H_t[/math] has all zeroes real precisely when [math]t \geq \Lambda[/math]. The Riemann hypothesis is equivalent to the claim that [math]\Lambda \leq 0[/math]. Currently it is known that [math]0 \leq \Lambda \lt 1/2[/math] (lower bound in [RT2018], upper bound in [KKL2009]).
The
Polymath15 project seeks to improve the upper bound on [math]\Lambda[/math]. The current strategy is to combine the following three ingredients: Numerical zero-free regions for [math]H_t(x+iy)[/math] of the form [math]\{ x+iy: 0 \leq x \leq T; y \geq \varepsilon \}[/math] for explicit [math]T, \varepsilon, t \gt 0[/math]. Rigorous asymptotics that show that [math]H_t(x+iy)[/math] whenever [math]y \geq \varepsilon[/math] and [math]x \geq T[/math] for a sufficiently large [math]T[/math]. Dynamics of zeroes results that control [math]\Lambda[/math] in terms of the maximum imaginary part of a zero of [math]H_t[/math]. Contents [math]t=0[/math]
When [math]t=0[/math], one has
[math]\displaystyle H_0(z) = \frac{1}{8} \xi( \frac{1}{2} + \frac{iz}{2} ) [/math]
where
[math]\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{s/2} \Gamma(s/2) \zeta(s)[/math]
is the Riemann xi function. In particular, [math]z[/math] is a zero of [math]H_0[/math] if and only if [math]\frac{1}{2} + \frac{iz}{2}[/math] is a non-trivial zero of the Riemann zeta function. Thus, for instance, the Riemann hypothesis is equivalent to all the zeroes of [math]H_0[/math] being real, and Riemann-von Mangoldt formula (in the explicit form given by Backlund) gives
[math]\displaystyle N_0(T) - (\frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} - \frac{7}{8})| \lt 0.137 \log (T/2) + 0.443 \log\log(T/2) + 4.350 [/math]
for any [math]T \gt 4[/math], where [math]N_0(T)[/math] denotes the number of zeroes of [math]H_0[/math] with real part between 0 and T.
The first [math]10^{13}[/math] zeroes of [math]H_0[/math] (to the right of the origin) are real [G2004]. This numerical computation uses the Odlyzko-Schonhage algorithm. In [P2017] it was independently verified that all zeroes of [math]H_0[/math] between 0 and 61,220,092,000 were real.
[math]t\gt0[/math]
For any [math]t\gt0[/math], it is known that all but finitely many of the zeroes of [math]H_t[/math] are real and simple [KKL2009, Theorem 1.3]. In fact, assuming the Riemann hypothesis,
all of the zeroes of [math]H_t[/math] are real and simple [CSV1994, Corollary 2].
Let [math]\sigma_{max}(t)[/math] denote the largest imaginary part of a zero of [math]H_t[/math], thus [math]\sigma_{max}(t)=0[/math] if and only if [math]t \geq \Lambda[/math]. It is known that the quantity [math]\frac{1}{2} \sigma_{max}(t)^2 + t[/math] is non-decreasing in time whenever [math]\sigma_{max}(t)\gt0[/math] (see [KKL2009, Proposition A]. In particular we have
[math]\displaystyle \Lambda \leq t + \frac{1}{2} \sigma_{max}(t)^2[/math]
for any [math]t[/math].
The zeroes [math]z_j(t)[/math] of [math]H_t[/math] obey the system of ODE
[math]\partial_t z_j(t) = - \sum_{k \neq j} \frac{2}{z_k(t) - z_j(t)}[/math]
where the sum is interpreted in a principal value sense, and excluding those times in which [math]z_j(t)[/math] is a repeated zero. See dynamics of zeros for more details. Writing [math]z_j(t) = x_j(t) + i y_j(t)[/math], we can write the dynamics as
[math] \partial_t x_j = - \sum_{k \neq j} \frac{2 (x_k - x_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math] [math] \partial_t y_j = \sum_{k \neq j} \frac{2 (y_k - y_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math]
where the dependence on [math]t[/math] has been omitted for brevity.
In [KKL2009, Theorem 1.4], it is shown that for any fixed [math]t\gt0[/math], the number [math]N_t(T)[/math] of zeroes of [math]H_t[/math] with real part between 0 and T obeys the asymptotic
[math]N_t(T) = \frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} + \frac{t}{16} \log T + O(1) [/math]
as [math]T \to \infty[/math] (caution: the error term here is not uniform in t). Also, the zeroes behave like an arithmetic progression in the sense that
[math] z_{k+1}(t) - z_k(t) = (1+o(1)) \frac{4\pi}{\log |z_k|(t)} = (1+o(1)) \frac{4\pi}{\log k} [/math]
as [math]k \to +\infty[/math].
See asymptotics of H_t for asymptotics of the function [math]H_t[/math].
Threads Polymath proposal: upper bounding the de Bruijn-Newman constant, Terence Tao, Jan 24, 2018. Polymath15, first thread: computing H_t, asymptotics, and dynamics of zeroes, Terence Tao, Jan 27, 2018. Polymath15, second thread: generalising the Riemann-Siegel approximate functional equation, Terence Tao and Sujit Nair, Feb 2, 2018. Other blog posts and online discussion Heat flow and zeroes of polynomials, Terence Tao, Oct 17, 2017. The de Bruijn-Newman constant is non-negative, Terence Tao, Jan 19, 2018. Lehmer pairs and GUE, Terence Tao, Jan 20, 2018. A new polymath proposal (related to the Riemann hypothesis) over Tao's blog, Gil Kalai, Jan 26, 2018. Code and data Wikipedia and other references Bibliography [B1950] N. C. de Bruijn, The roots of trigonometric integrals, Duke J. Math. 17 (1950), 197–226. [CSV1994] G. Csordas, W. Smith, R. S. Varga, Lehmer pairs of zeros, the de Bruijn-Newman constant Λ, and the Riemann hypothesis, Constr. Approx. 10 (1994), no. 1, 107–129. [G2004] Gourdon, Xavier (2004), The [math]10^{13}[/math] first zeros of the Riemann Zeta function, and zeros computation at very large height [KKL2009] H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathematics, 22 (2009), 281–306. Citeseer [N1976] C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61 (1976), 246–251. [P2017] D. J. Platt, Isolating some non-trivial zeros of zeta, Math. Comp. 86 (2017), 2449-2467. [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. arXiv:1801.05914 [T1986] E. C. Titchmarsh, The theory of the Riemann zeta-function. Second edition. Edited and with a preface by D. R. Heath-Brown. The Clarendon Press, Oxford University Press, New York, 1986. pdf
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As a toy problem, I was looking at a classical second order system of the following form:
\begin{align} \ddot{x}(t) + c \dot{x}(t) + k x(t) = 0 \end{align}
Instead of the basic system above, I want to replace $x(t)$ with a Low-Pass filtered version of $x(t)$. This can change the equation to the following form:
\begin{align} \ddot{x}(t) + c \dot{x}(t) + \frac{k}{\tau} \int_{0}^{t} \exp\left(\frac{-(t-\hat{t})}{\tau}\right)x(\hat{t}) d\hat{t} = 0 \end{align}
Given this form, how might I be able to find numerical stability analytically? In the original system, I could break it into a system of two first order ODEs and analyze the eigenvalues of the matrix relating independent variables to their derivatives. In this latter case, I don't see a strategy like that working.
This problem is motivated by an empirical observation myself and a colleague noted when we were simulating our missile system and found it would develop large errors for Explicit Euler, no matter the step size, but would work fine for 4th Order Runge Kutta. We reproduced it on two separate simulation codebases of our weapon, so it seems to be more than potentially bugs in the code.
Since our system can be approximately viewed as a second order system with a Low-Pass Filter (due to filtering sensors for state estimations), I am interested in trying to understand the second equation from an accuracy and stability perspective. Any hints or insights about how to analyze this scenario would be appreciated.
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A Rational Game
This post is to set forth a little game that attempts to demonstrate something that I find to be intriguing about the real numbers. The game is one that takes place in a theoretical sense only. It starts by assuming we have two pieces of paper. On each is a line segment of length two: [0,2]. Each piece of paper is then stuck to a wall, one just above the other, so that the line segments on each piece of paper are perfectly aligned. This means the line segments on each piece of paper run parallel to each other and drawing a vertical line between each 0 and each 2 would form a rectangle.
On the upper line segment we mark two numbers, $r$ and $r+1$, where:
$$r \in \, (0,1\, ) \text{ and } r \notin \mathbb{Q}$$
On the lower line segment we mark two more numbers, $r’$ and $r’+1$, where:
$$r’ > r,$$
$$r’ \in \, (0,1\, ),$$
$$r’ \notin \mathbb{Q}, \text{ and}$$
$$r’ - r \notin \mathbb{Q}$$
We are now ready to start our game. The object of the game is to slide the top piece of paper horizontally to the right so as to make all of the rational numbers in $\, (r,r+1\, )$ align vertically with all of the rational numbers in $\, (r’,r’+1\, )$. Fortunately, for purposes of this game, we are given the ability to slide ‘perfectly’ given that this game takes place only in a theoretical sense.
Despite our ability to slide the uppermost piece of paper with God-like precision, it is seemingly still impossible to win the game. The length that we would have to slide the upper piece of paper to the right must be a rational number because sliding an irrational length to the right would imply that none of the rational numbers align (ie, a rational plus an irrational will always be irrational). However, if we slide a rational length to the right, then $r$ and $r’$ could not be vertically aligned. In that case, the Archimedean property of the real numbers would ensure that an infinite number of rational numbers on each number line could also not be vertically aligned.
Therein lies what I feel is paradoxical. Seemingly, we should be able to slide the upper piece of paper so as to make all of the rational numbers on each line segment vertically aligned. If we choose to believe that we can in fact do so, then we contradict the standard view that the reals adhere to the Archimedean property which ensures there are an infinite number of rational numbers between any two irrational numbers.
Is there anyone who shares in my view that we ought to be able to get each rational number vertically aligned so as to win the game? Assuming that my view is correct (and in noting that I am not asserting it is), what would this imply?
This seems a lot like the sliding thing we went around on a couple of years ago. Is it basically another variant on the same idea?
ps -- Here it is. http://mymathforum.com/number-theory...rationals.html
Is this the same question in different form?
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Although I understand the standard view as a rigorously proven mathematical concept, a part of me doesn't agree with it on a common sense level. I think we may assert too much based on the notion of what the infinite leads us to believe when perhaps we shouldn't be incorporating the infinite into rigorous mathematical arguments in the first place. There is no irrational that the rationals cannot approach down to an infinitesimal level and Physics could care less, for example, so why the need to distinguish between the two kinds of numbers? My common sense view is that the rationals must be alignable just as Achilles must catch the tortoise (see Zeno's paradoxes), but the math leads us to believe otherwise. I question the Archimedean property and with it many other things.
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Can you please write a short and clear summary of your core issue? Your second post in this thread is handwavy and doesn't say anything. Frankly it's cranky. You know I've suggested before that the reason you come down so hard on Zylo is that you have a bit of the crank in you yourself. You denied it.
So prove me wrong and write something clear and precise and sensible. You can't just say, "Ooh my intuition this and that so the Archimedean property is wrong and besides, physics!" and then bust some OTHER crank's chops. You're just projecting.
You know that I'm not gratuitously attacking you. When you write math, I respond with math. When you get cranky, I call you cranky. That's fair.
How about perhaps a numerical example to show your shifting issue. I'm sure you know that the difference of irrationals is usually irrational, so you can't hope to shift by an irrational and make the rationals line up or make a single point "fall off the end" which is what you were trying to do last time.
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Just post some math. I'll respond with some math. A small numeric example would be helpful. Like, $r_1 = \sqrt 2$ and $r_2 = \pi$ and we shift this by that and now the question is ...
Just like that.
And I think you're the one on drugs. If you insist on making personal attacks I can do that as well as anyone. WTF is wrong with you to say something like that about a member who isn't even here? And I live in the US where the vast majority of the population is whacked out on one substance or another, legal and otherwise. What of it?
By the way I never encouraged Zylo. I do support free speech and a tolerant moderation policy that allows for alternative views. That includes your stuff too. Nobody here has spent more time working with your ideas than I have. "No good deed goes unpunished" as they say.
Get back to the math. If you can explain your idea clearly I'll respond the best I can.
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When it comes to Zylo, there is a slew of research regarding immersing one's self in something like mathematical research and its connection to mental illness. Cantor himself suffered, as did many other great mathematicians. It's not a laughing matter. I am concerned for Zylo and perhaps you should be too. It's not an attack on him.
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Here's my thoughts on the issue, in slightly hand-wavy terms:
Imagine for a second we're not trying to align the rational numbers, but a sequence of randomly spaced dots. Obviously there's no guarantee that we can line up two sets of dots spread over different sections of the line.
Having said that, this situation with the rationals is more complicated. Talking about the spacing of the rationals and whether they are "evenly spaced" makes little to no sense because of their density over the reals. In such cases where we are talking about arbitrarily small (or large) values, intuition is your enemy. Unfortunately, sometimes we have to convince ourselves that something is true because the maths says it is.
As for assuming whether your intuition is correct and what the consequences would be, look up the explosion principle.
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Because I am choosing to question the Archimedean property with this, I might ask what the explosion principle would imply if it were flawed. My hand waivy comparison was to assert that the paper will cross the point where the rationals align just as Achilles will pass the tortoise in Zeno’s paradoxes linked above. Zeno shows how the math suggests Achilles cannot pass the tortoise, but when viewed properly we know he does. Here, we also know the paper will pass the point where the rationals align even though the math implies it isn’t possible because we cannot compute the rational number by which to shift by.
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@Josh: I don't have any precise reference in mind for that, maybe in the Coddington-Levinson? but as far as I remember it is mostly for linear ODE's
Note first that reversing time $t\to -t$ is equivalent to changing $\sigma\to-\sigma$, so you only need to study one side (say $t\to+\infty$). The case $\sigma=0$ is a borderline one that I am not too sure how to deal with. I assume below that $\sigma\neq 0$ (but I guess $\sigma$ is the propagation speed of the wave, so it should be OK to discard stationary waves).
The starting point is to rewrite $w''+\sigma w'+f(w)=0$ as$$(e^{\sigma t}w')'=-e^{\sigma t}f(w).\hspace{2cm}(E)$$
Step 1The first thing you need to show is that $w_{\pm}$ are necessarily steady-states, i-e $f(w_{\pm})=0$ (stationary equilibrium solutions of the ODE, usually one is stable while the other is unstable). In order to see this assume by contradiction that $f(w_+)\neq 0$ (again, it is enough to look at $t\to+\infty$). If $\sigma>0$ then by (E) we have $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ not integrable when $t\to\infty$, so $e^{\sigma t}w'\sim C\int e^{\sigma t}=Ce^{\sigma t}$ hence $w'\sim C\neq 0$. This shows that $w$ blows-up linearly and contradicts $w(\infty)=w_+$. If now $\sigma<0$ then $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ becomes integrable, and thus $e^{\sigma t}w'\to C$ for some limit $C\in \mathbb{R}$. If $C\neq 0$ then $w'\sim Ce^{-\sigma t}$ blows exponentially, which contradicts again $w(\infty)=w_+$. Thus $C=0$, and integrating $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ from $t$ to $\infty$ you get $e^{\sigma t}w'-0\sim C(e^{\sigma t}-0)$, hence again linear blow-up $w'\sim C\neq 0$.
Step 2Once you know that $f(w)\to f(w_+)=0$ (here I am definitely using the continuity of $f$) the heuristic idea is quite simple: the initial ODE $w''+\sigma w+f(w)=0$ roughly becomes a 1st order linear ODE in $v=w'$\begin{equation}v'+\sigma v=-f(w)\approx -f(w_+)= 0,\qquad v=w'.\hspace{2cm}(E')\end{equation}
This linear ODE $v'+\sigma v=0$ gives either the trivial solution $v'=v=0$ (which means precisely $w''=w'=0$), or $v'$ and $v$ proportional to $e^{-\sigma t}$. If $\sigma>0$ you see that both the trivial and exponential cases are admissible and lead to $v',v=w'',w'\to 0$ when $t\to\infty$. If now $\sigma<0$ the exponential blow-up $v=w'\sim e^{-\sigma t}$ is excluded because you assume $w(t)\to w^+=cst$, so the only possibility is again $w'',w'=u',u=0$.
Of course this step 2 is only formal, and rigorously justifying (E') from (E) requires tedious and technical computations similar to those in step 1. For that you may want to use again (E) with now $f(w)\to f(w_+)=0$ hence$$(e^{\sigma t}w')'=o\left(e^{\sigma t}\right),$$and distinguish again integrability or linear/exponential blowup at infinity as in step 1.
I hope this helps!
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Here's a typical op-amp inverting amplifier:
The input impedance is simply \$R_{in}\$, so for your requirements, \$R_{in} = 10k\Omega\$. \$R_f\$ is then whatever it needs to be to realize the desired gain. You want a gain of -10, so:
$$ -\frac{R_f}{10k\Omega} = -10 \\R_f = 10 \cdot 10k\Omega = 100k\Omega $$
Why does the input impedance depend only on \$R_{in}\$? So long as the op-amp is not saturated, the inverting input is held at the same potential as the non-inverting input. Here, that's just ground, though any DC voltage works. So, you might as well consider \$R_{in}\$ as connected to ground, because the voltage is the same as it was. Once you realize the inverting input is effectively ground, it's easy to see the input impedance is just \$R_{in}\$.
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You're quite right that the other fundamental forces of Nature possess mediator particles, e.g. the photon for the electromagnetic force. For gravity, a
graviton particle has been postulated, and is included in the five standard string theories which are candidates for quantum gravity. From a quantum field theory perspective, the graviton arises as an excitation of the gravitational field. String theory, of course, postulates it arises in the spectrum of a closed string.
Mass certainly gives rise to a gravitational field, but many other quantities do as well, according to the field equations of general relativity. As you're a high school student, I'll present them as,
$$\underbrace{G_{\mu\nu}}_{\text{geometry}}\sim \underbrace{T_{\mu\nu}}_{\text{matter}}$$
Spacetime geometry, and hence the gravitational effects, are equated to the matter present in a system, which may include
energy, pressure and other quantities other than mass.
From a general relativity standpoint, the gravitational field may be viewed, or interpreted, as the curvature of spacetime, which is a manifold, i.e. surface. If we take space to be infinitely large, then the gravitational field must extend indefinitely; otherwise where would we choose to truncate? Even from a Newtonian perspective, we see that given the equation,
$$F_g \sim \frac{1}{r^2}$$
gravity must extend infinitely, as we never reach the point $r=\infty$ where it is truly zero.
As you asked, if the graviton is postulated, what is the need for a field? Well, we know that
particle number is not conserved; we can have virtual particle and anti-particle pair production, and as such the idea that a field propagates throughout space, and the particles are excitations of the field, is a more compatible viewpoint. In addition, the concept of a field arises because of locality. From empirical evidence we know gravitation and electromagnetism do not act instantaneously, at every point.
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I am writing my undergrad thesis on the harmonic oscillator on a lattice. So far I have implemented the Metropolis Monte Carlo algorithm to generate trajectories $x_j$ for $0 \leq j < N$, where $N$ is the number of time lattice divisions.
I get a pretty good histogram for $|\psi(x)|^2$ and it matches the expectation from the theory. Now I have to calculate the energy eigenvalues of the system. The ground energy state can be calculated with $E_0 \propto \langle x^2 \rangle$ which has worked out so far.
The first energy, $E_1$, should be computable using the exponential slope of the correlations, Creutz and Freedman (1980) wrote that this would be doable with this: $$ E_1 = \frac{-1}{\Delta \tau} \log\left( \frac{\langle x(0) x(\tau + \Delta \tau)\rangle}{\langle x(0) x(\tau) \rangle} \right)$$
With the $x_j$ that I have in my computations, I can calculate those correlations to some value of $\tau$, limited by the $N$.
My advisor has given me a paper from Blossier et al(2009) where they introduce a correlation
matrix like so (2.1):
$$ C_{ij} (t) = \sum_{n=1}^\infty \langle 0 | \hat O_i | n \rangle \langle 0 | \hat O_j | n \rangle^* $$
Where $O_i$ are “some interpolating fields $O_i(x_0)$ already projected to a definite momentum and other quantum number such as parity”.
From that, they show how the eigenvalues to the generalized eigenvalue problem (GEVP) consisting of $C(t)$ and $C(t_0)$ will give the energy eigenvalues, which I am interested in.
Lüscher and Wolff (1990) then write on page 245:
In a numerical simulation, the correlation matrix $C(t)$ can be expected to be computable for some range of $t$, and the basic technical problem then is to extract the levels $W_\alpha$. from $C(t)$.
I do not see how I can get the $C_{ij}(t)$ out of my data. What are my fields $O$ for a simple harmonic oscillator to start with? Are those the ladder operators, so that $O_j$ = $(a^\dagger)^j$?
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Consider 2 mathematical problems:
$$ f_1(x) = a - x \\ f_2(x) = e^x -1 $$
The condition number for a function is defined as follows:
$$ k(f) = \left| x \cdot \frac{f'}{f} \right| $$
Lets analyze conditioning first:
$$ k(f_1) = \frac{x}{x - a}, $$
which means that $f_1$ is
ill-conditioned near $x = a$;
$$ k(f_2) = \frac{x \cdot e^x}{e^x - 1}, $$
which is undefined near $x = 0$, so lets use L'Hospital:
$$ k(f_2) = \frac{e^x + x \cdot e^x}{e^x}, $$
which means that $f_2$ is
well-conditioned everywhere (including $x = 0$ proximity).
Now lets analyze stability of these 2 algorithms (if we were to implement them on the computer directly):
$$ \frac{(a - x) \cdot (1 + \epsilon_1) - (a - x)}{a - x} = \epsilon_1, $$
where $\epsilon_1 \leq \epsilon_m$, and which means that no (numerical) amplification of errors occurs and the algorithm is
stable;
$$ \frac{(e^x \cdot (1 + \epsilon_1) - 1) \cdot (1 + \epsilon_2) - (e^x - 1)}{e^x - 1} \approx \{\epsilon_1 \cdot \epsilon_2 \to 0\} \approx \frac{e^x \cdot \epsilon_1 + (e^x - 1) \cdot \epsilon_2}{e^x - 1} = \epsilon_2 + \epsilon_1 \cdot \frac{e^x}{e^x - 1}, $$
where $\epsilon_1, \epsilon_2 \leq \epsilon_m$, and which means that the algorithm is
unstable near $x = 0$.
Although everything is allright from the mathematical point of view, i.e. if we obey the formulas and raw theory when obtaining such results, but I begin to doubt in the validity of these results when I try add some logic and reasoning behind it.
First of all, as far as I understand, when we study conditioning of the mathematical problem we think of it in
exact arithmetics (i.e. we do not think about computers, rounding, floating-point arithmetic, and etc.). Therefore, if I forget for a moment about the result obtained by analyzing $k(f_1)$ and just look on the simple mathematical problem $f_1(x) = a - x$, then I merely don't see how on earth it could be ill-conditioned near $x = a$. What is the physical reasoning behind it? What kind of bad thing can happen in exact arithmetic near $x = a$?
My curriculum pointed out
cancellation error as an explanation. What kind of cancellation error? From my point of view, there is no such thing as cancellation error in exact arithmetic...
So, my reason against it would be straightforward, since
ill-condition implies that the output changes drastically when the input changes slightly, then $f_1$ is clearly linear (moreover with coefficient $1$) and any slight changes of $x$ (regardless of whether near $a$ or not) will always result in the quantitatively equal change of $y = f_1(x)$ (i.e. $\Delta x \equiv \Delta y$). Therefore, I insist that $f_1$ can in no way be ill-conditioned neither at $x = a$ nor anywhere else.
Are there any flaws in my reasoning? Please, clarify this for me as I'm actually stuck on it.
Secondly, why it turns out that $f_2$ is
well-conditioned while it looks the same as $f_1$? I mean if I follow the same logic as for $f_1$ (i.e. that it's ill-conditioned at $x = a$ because of cancellation as my curriculum states), then I could say the same here - ill-conditioned at $x = 0$, just by looking on the definition of $f_2$! However, mathematics show us that it's not true, but rather that the direct implementation of $f_2$ evaluation on computer would result in unstable algorithm. Due to what? I guess now it's cancellation error because we are in the floating-point world now. But why?
And still the question is why these two seemingly similar problems are actually so different? I'd really appreciate an exhaustive breakdown of these problems as I feel that I'm missing something very basic and it might prevent my understanding of more challenging phenomena.
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TL;DR: Do
not just memorise thermodynamics equations! And if you have an issue with the equations $\Delta U = 0$ or $\Delta H = 0$ for an isothermal process, read the answer. The first problem
You said that an exothermic reaction corresponds to $\Delta U < 0$. This is
not true. It is defined by $\Delta H < 0$.
However, there is still room to discuss this issue because $H$ is also a state function and in some cases $H$ is dependent only on temperature.
How to study thermodynamics
This kind of issue in thermodynamics frequently crops up here, and it is a very common mistake amongst students to indiscriminately use equations that they have learnt.
So, you have an equation that says $\Delta U = 0$ for an isothermal process, i.e. one at constant $T$. The important question here is
not "what is the equation" or "what is the answer"! Instead, you should be asking yourself "how do I derive this result", and the answer will naturally follow. This should really apply to everything you do - how can you expect to apply a formula that you do not actually understand?
Furthermore, if you are able to actually understand how an equation comes about, then you don't need to
memorise it - you can derive it. For example, what's the point of memorising $w = -nRT\ln(V_2/V_1)$, along with the conditions, if you can simply obtain the equation from the very definition of work $đw = -\int p \,\mathrm{d}V$ by substituting in $p = nRT/V$ and integrating? What's more, the fact that you actually substituted in the ideal gas law should tell you something about the conditions that accompany the equations - for one, it's only applicable to ideal gases. What about when you reach the equation $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$ and happily memorise it, only to find out later that there are actually three more equations that look exactly like it? Are you going to memorise all four? No - you need to know where they come from. Internal energy
In general, the internal energy $U$ of a substance is a function of at least two variables, for example $T$ and $p$. However, it can be shown using statistical mechanics that for an ideal gas,
$$U = n\left(\frac{3NRT}{2} + U_0\right)$$
Here, $N$ is the number of atoms in one molecule of the gas (for example, $N = 2$ for $\ce{H2}$, and $N = 3$ for $\ce{H2O}$). $U_0$ is the internal energy at absolute zero (this contains terms such as electronic energy), and crucially, this depends on the exact identity of the gas. This result is one example of the equipartition theorem in action (although the theorem itself is more powerful).
We do not need to concern ourselves with its derivation now. The point is that:
this equation can only be applied to a fixed amount of a fixed ideal gas at a time. For example: if I have one mole of $\ce{H2}$ at $298\mathrm{~K}$, it is going to have a different internal energy from one mole of $\ce{Ar}$ at $298\mathrm{~K}$, even though the amount of substance and the temperature are the same.
What does this mean?
Firstly, it means that for anything that is
not an ideal gas, you cannot assume that $U = U(T)$ and therefore you cannot assume that $\Delta T = 0$ implies $\Delta U = 0$. Let's say I compress $10\mathrm{~g}$ of water by reversibly increasing the pressure from $1\mathrm{~atm}$ to $10\mathrm{~atm}$, at a constant temperature of $323\mathrm{~K}$. Is the change in internal energy equal to zero? No, because water is not an ideal gas.
Secondly, it means that if the amount of ideal gas changes in any way, $U$ is going to change. Let's say I have a balloon filled with $1\mathrm{~mol}$ of argon gas, and I add a further \pu{2 mol} argon gas into the balloon, all at a fixed temperature of $298\mathrm{~K}$. Is the change in internal energy equal to zero?
No, in fact it will triple because you are changing $n$ from $1\mathrm{~mol}$ to $3\mathrm{~mol}$!
Lastly, it means that if the chemical composition is changing in any way, you are automatically
not allowed to say that $\Delta T = 0$ implies $\Delta U = 0$. And this is the case even if all reactants and products are ideal gases. Why not? Let's look at this reaction, and let's assume that everything there behaves as an ideal gas, and let's say that the reaction vessel is kept at a constant temperature of $300\mathrm{~K}$.
$$\ce{H2 (g) + Cl2 (g) -> 2HCl (g)}$$
Let's write an expression for the internal energy of the reactants. On the left-hand side, we have:
$$U_\mathrm{reactants} = [3RT + 2U_0(\ce{H2})] + [3RT + 2U_0(\ce{Cl2})]$$
On the right-hand side, we have:
$$U_\mathrm{products} = 6RT + 2U_0(\ce{HCl})$$
So, is $\Delta U$ equal to zero?
No! Even though the multiples of $RT$ cancel out with each other, the values of $U_0$ for the reacting species are different, which makes $\Delta U = 2U_0(\ce{HCl}) - U_0(\ce{H2}) - U_0(\ce{Cl2}) \neq 0$. Enthalpy
Let's go
back to our (closed) system where there is only one ideal gas and it's not undergoing any chemical reaction. Within this system, we can safely say that if $\Delta T = 0$, then $\Delta U = 0$. So:
$$\begin{align}\Delta H &= \Delta (U + pV) \\&= \Delta U + \Delta (pV) \\&= \Delta (pV) \qquad \\&= \Delta (nRT) \\&= nR \Delta T \\&= 0\end{align}$$
We haven't made any assumptions here apart from the ideality of the gas (the fact that $n$ is constant came from the fact that it is a closed system). So, for an ideal gas, $\Delta T = 0$ implies $\Delta H = 0$.
However, we did use the fact that $\Delta U = 0$. Therefore, what I said about the cases above where $\Delta U \neq 0$ also applies here.
Regarding the terms
endothermic and exothermic, they are used to describe chemical reactions. And for chemical reactions, there is absolutely no way you can say that constant temperature implies $\Delta U = 0$ or $\Delta H = 0$. That's why we can actually use the terms endothermic and exothermic. If every single chemical reaction had $\Delta H = 0$, life would be rather plain.
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Update
Since version 12, this functionality in integrated in Mathematica via the Option FitRegularization
Following on @Ajasja's answer in the spirit of this answer one can in fact provide controlled smoothing va an explicit Tichonov like penalty as follows:
ff = Function[{x, y}, basis // Evaluate];
a = ff @@ # & /@ (Most /@ data);
so that
fit[x_, y_] =
basis.LinearSolve[
Transpose[a]. a + 0 IdentityMatrix[Length[basis]],
Transpose[a].( Last /@ data )];
pl0 = PlotPointsAndSurface[data, fit[x, y], "fit and data"];
reproduces exactly @Ajasja's fit, whereas e.g.
fit[x_, y_] =
basis.LinearSolve[
Transpose[a]. a + 10^1 IdentityMatrix[Length[basis]],
Transpose[a].( Last /@ data )];
pl1 = PlotPointsAndSurface[data, fit[x, y], "fit and data"];
would correspond to a smoother solution.
Show[pl1, pl0]
Note the
hyper parameter (here 10^1), which fixes the sought level of smoothness imposed onto the solution, by effectively correlating the coefficients of the basis expansion.
The main advantage is that one need not focus too much on the exact properties of the chosen basis.
For instance we could also use
BSplineBasis
knots = Range[-RANGEX - 2, RANGEX + 2];
basis = Flatten@ Table[BSplineBasis[{3, knots}, i, x]
BSplineBasis[{3, knots}, j, y], {i, 0, 2 RANGEX}
, {j, 0, 2 RANGEX}];
Then, as previously
ff = Function[{x, y}, basis // Evaluate];
a = ff @@ # & /@ (Most /@ data);
Then one could use a penalty function based on second derivatives:
s = SparseArray[{{i_, i_} -> -1, {i_, j_} /; i - j == 1 -> 2,
{i_, j_} /; i - j == 2 -> -1}, {17, 15}] // Transpose;
s1 = ArrayFlatten[TensorProduct[s, s]];
pen = Transpose[s1].s1; pen//ArrayPlot
built so that
s.( Range[17]*0 + 1) and
s.Range[17] are both null (i.e. there is no penalty to have a constant or linear function of
x and
y.
Then, as previously
fit3[x_, y_] =
basis.LinearSolve[Transpose[a]. a + 10^1 pen,
Transpose[a].( Last /@ data )];
pl1 = PlotPointsAndSurface[data, fit3[x, y], "fit and data"]
The main advantage of this second approach is that it is the penalty which sets smoothing, not the sampling of the basis function. Even if the conditioning of
Transpose[a]. a is poor, the inverse will be well conditioned thanks to the regularisation terms
pen.
Note that for the sake of efficiency and memory one could fill the
a matrix using sparse matrices following this answer.
With[{xOrder = Ordering[Join[data[[All, 1]], knots]],
yOrder = Ordering[Join[data[[All, 2]], knots]]},
With[{xPar = xOrder[[# + 1 ;; #2 - 1]] & @@@ Partition[Ordering[xOrder, -Length[knots]], 2, 1],
yPar = yOrder[[# + 1 ;; #2 - 1]] & @@@ Partition[Ordering[yOrder, -Length[knots]], 2, 1]},
nonzero = Join @@ Outer[Intersection, Union @@@ Partition[xPar, 4, 1],
Union @@@ Partition[yPar, 4, 1], 1];]]
colIndex = Range[Length[basis]];
a2 = SparseArray[Join @@ MapThread[Thread[Thread[{#2, #3}] ->
Function[{x, y}, #] @@@ data[[#2, {1, 2}]]] &, {basis, nonzero, colIndex}]]; a == a2
(* True *)
The choice of optimal level of smoothing can be done via generalised cross validation, i.e. by choosing the penalty weight to correspond to the minimum of$$\hat \lambda = {\rm min}_\lambda\left\{\frac{||( \mathbf{1}- \tilde{\mathbf{a}}) \cdot {\mathbf{y}} ||^2}{\left[{\rm trace}( \mathbf{1}- \tilde{\mathbf{a}}) \right]^2} \right\} \,.$$having defined$$ \tilde{\mathbf{a}}(\lambda) =\mathbf{a} \cdot ({\mathbf{a}^{\rm T}} \cdot \mathbf{a} + \lambda\, \mathbf{s}^{\rm T}\cdot \mathbf{s})^{-1} \cdot {\mathbf{a}^{\rm T}}$$
Table[at = a.Inverse[Transpose[a]. a + 10^i pen].Transpose[a];
{i, ((IdentityMatrix[289] - at).(Last /@ data) // #.# &)/
Tr[IdentityMatrix[289] - at]^2}, {i, -3, 3, 1/2}] // ListLinePlot
Update:
Note that if smoothing is not an issue, then in version 10 and above mathematica can deal directly with the data as demonstrated here
PlotPointsAndSurface2[points_, surface_, label_] :=
Module[{},
Show[ListPlot3D[points, PlotLabel -> label, ImageSize -> Medium,
PlotStyle -> Directive[Orange, Opacity[0.5]]],
Plot3D[surface, {x, -RX, RX}, {y, -RY, RY},
PlotStyle -> Directive[Purple, Opacity[0.1]]]]];
pl2 = PlotPointsAndSurface2[data, fit3[x, y], "fit and data"];
Show[pl1,pl2]
as can be seen the regularised and un regularised surfaces look quite similar.
It would be great if mathematica allowed for adding a penalty to the built in function behind ListPlot3D, ListContourPlot or ListInterpolate !
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Difference between revisions of "Gay-Berne model"
m
Line 47: Line 47:
:<math>\frac{\chi \prime }{\alpha \prime^{2}}=1- {\left(\frac{\epsilon_{ee}}{\epsilon_{ss}}\right)} ^{\frac{1}{\mu}}.</math>
:<math>\frac{\chi \prime }{\alpha \prime^{2}}=1- {\left(\frac{\epsilon_{ee}}{\epsilon_{ss}}\right)} ^{\frac{1}{\mu}}.</math>
− + +
==References==
==References==
#[http://dx.doi.org/10.1063/1.441483 J. G. Gay and B. J. Berne "Modification of the overlap potential to mimic a linear site–site potential", Journal of Chemical Physics '''74''' pp. 3316-3319 (1981)]
#[http://dx.doi.org/10.1063/1.441483 J. G. Gay and B. J. Berne "Modification of the overlap potential to mimic a linear site–site potential", Journal of Chemical Physics '''74''' pp. 3316-3319 (1981)]
#[http://dx.doi.org/10.1103/PhysRevE.54.559 Douglas J. Cleaver, Christopher M. Care, Michael P. Allen, and Maureen P. Neal "Extension and generalization of the Gay-Berne potential" Physical Review E '''54''' pp. 559 - 567 (1996)]
#[http://dx.doi.org/10.1103/PhysRevE.54.559 Douglas J. Cleaver, Christopher M. Care, Michael P. Allen, and Maureen P. Neal "Extension and generalization of the Gay-Berne potential" Physical Review E '''54''' pp. 559 - 567 (1996)]
− −
[[category:liquid crystals]]
[[category:liquid crystals]]
[[category:models]]
[[category:models]]
Revision as of 16:14, 19 February 2008
The
Gay-Berne model is used extensively in simulations of liquid crystalline systems. The Gay-Berne modelis an anistropic form of the Lennard-Jones 12:6 potential.where, in the limit of one of the particles being spherical, gives:
and
with
and
Phase diagram Main article: Phase diagram of the Gay-Berne model References J. G. Gay and B. J. Berne "Modification of the overlap potential to mimic a linear site–site potential", Journal of Chemical Physics 74pp. 3316-3319 (1981) Douglas J. Cleaver, Christopher M. Care, Michael P. Allen, and Maureen P. Neal "Extension and generalization of the Gay-Berne potential" Physical Review E 54pp. 559 - 567 (1996)
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$\pi R^2 \Delta p$ (1)
and the surface tension is:
$2\pi R \gamma$ (2)
and a sphere has two surfaces so, roughly:
$\Delta p = \frac{4 \gamma}{R}$
but if we take the limit:
$\lim_{R \to 0} \frac{4 \gamma}{R}$,
then obviously the pressure difference between the outside and inside surface of the bubble is $\infty$! How does that make sense? Is there any way I can get round this? Or have I done something wrong with my algebra?
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This is not really an independent answer; just an explanation of why the answer given by Levieux is optimal. (Of course it's easy to check by computer, but you might prefer to have a more human-comprehensible explanation.)
First of all, obviously
each number's digits must be in descending order because if not, reordering one number's digits to be in descending order increases that number while leaving the other unchanged.
Next, write the two numbers one above the other with most significant digits aligned. (Pad with zeros if you need to.) Then
apart from the first digit, the smaller number's digits must be larger than the corresponding digits of the larger number because otherwise swapping two digits for which this isn't true increases the product. (Because if $\delta$ is the difference between the digits taking place value into account, then we are replacing $ab$ with $(a-\delta)(b+\delta)=ab+\delta(a-b-\delta)=ab+\delta(a'-b')$ where $a',b'$ are the new numbers, and we still have $a'>b'$.)
So we have a situation like this:
$\begin{array} \\9 & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet \\\downarrow & & \uparrow & & \uparrow & & \uparrow & & \uparrow \\\bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet\end{array}$
where arrows mean $\geqslant$ and you should think of the pattern continuing indefinitely to the right where there are infinitely many zeros.
But we could take this diagram and shift the lower number left one place, making it the larger number, and then all but the leftmost vertical arrow would reverse. (Because we would still have a maximal product, so the reasoning above applies "with roles reversed".) In other words, we actually have this (again, continuing infinitely to the right, or at least until whichever number is longer runs out):
$\begin{array} \\9 & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet \\\downarrow & & \uparrow & \searrow & \uparrow & \searrow & \uparrow & \searrow & \uparrow \\\bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet\end{array}$
and now I want to draw attention to the fact that we have a chain connecting all the digits:
$\begin{array} \\9 & & \bullet & & \bullet & & \bullet & & \bullet \\\downarrow & & \uparrow & \searrow & \uparrow & \searrow & \uparrow & \searrow & \uparrow \\\bullet & \rightarrow & \bullet & & \bullet & & \bullet & & \bullet\end{array}$
and therefore only one way to fill them in, which is the one in Levieux's answer.
[EDITED because @mathreshka kindly pointed out a small error, which I have now fixed. Thanks!]
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I'm asking about definite integrals that can effortlessly be found numerically by high schoolers using software. For example,
$$\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \ \mathrm dx$$
This link shows the numerical integration done by software.
This Reddit comment substantiates that the exact solution is knotty, but not why:
The extent of math that this involves (beyond standard integration techniques usually taught in Calc II, just applied on a large scale), is a significant bit of Complex Analysis (the residue theorem, etc.). In general, everything in his derivation should at least
be understandablehad you taken Calc I-III and Complex Analysis.
How can I explain to someone who is just getting started in calculus why it's so hard to find the exact solution by symbolic integration, when it's so trivial to get a numerical answer using software?
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Open Sets Are Everything
In today's post I want to emphasize a simple - but important - idea in topology which I think is helpful for anyone new to the subject, and that is:
Open sets are everything!
What do I mean by that? Well, for a given set $X$,
all the properties* of $X$ are HIGHLY dependent on how you define an "open set."
Suppose for instance, $X$ has a certain topology $\tau$. Then any subset $U\subset X$ which is an element of $\tau$ is what's known as an open set. So since we know what our open sets are, we can explore properties of $X$ and ask questions like, "Is $X$ connected?" "Is it compact?" "Is it a metric space?" "What do continuous functions on $X$ look like?" and so on. BUT as soon as we
change the topology on $X$ from $\tau$ to some other topology $\tau'$, the answers to all those questions may change drastically! The two topological spaces $(X,\tau)$ and $(X,\tau')$ may look completely different even though they both involve the same set $X$! But don't take my word for it!
Let me give you an example. Take $X$ to be $\mathbb{R}$ and $\tau$ to be the usual topology. Here, a set $O\subset\mathbb{R}$ is called open if for each $x\in O$ there is an open interval $(a,b)$ such that $x\in(a,b)\subset O$. In this topology $\mathbb{R}$ is
not compact since, for instance, the collection of open intervals $\mathscr{U}=\{(-n,n)\}_{n=1}^\infty$ covers all of $\mathbb{R}$, yet any finite subcollection of intervals in $\mathscr{U}$ cannot cover $\mathbb{R}$.
Intuitively, to say "$(\mathbb{R},\tau)$ is not compact" means that $\mathbb{R}$ "stretches out to infinity"
and indeed this is consistent with what we know. But $\tau$ is not the
only topology we can place on $\mathbb{R}$! There are lots! In particular, there is a topology $\tau'$ on $\mathbb{R}$ in which $\mathbb{R}$ is compact! It's called the finite complement topology. Here a set $O\subset \mathbb{R}$ is declared to be open if its complement $\mathbb{R}\smallsetminus O$ is finite.** (One can easily check that the collection of these open sets does indeed form a topology.) Let's see why this claim is true: Claim: $\mathbb{R}$ with the finite complement topology is compact. Proof.*** Let $\mathscr{U}=\{U_{\alpha}\}_{\alpha\in A}$ be an open cover of $\mathbb{R}$ (where $A$ is just some indexing set). Choose an arbitrary $U\in\mathscr{U}$. Since $U$ is open, we know that $\mathbb{R}\smallsetminus U$ is finite, say $\mathbb{R}\smallsetminus U=\{x_1,\ldots,x_n\}$ where the $x_i$ are real numbers. Since $\mathscr{U}$ is a cover of $\mathbb{R}$, for each $k=1,\ldots,n$ there exists a $U_k\in\mathscr{U}$ such that $x_k\in U_{k}$. Hence$$\mathbb{R}\smallsetminus U\subset \bigcup_{k=1}^n U_{k}.$$Rewriting $\mathbb{R}$ as $U\cup(\mathbb{R}\smallsetminus U)$, we see that$$\mathbb{R}\subset U\cup\bigcup_{k=1}^n U_{k}.$$Thus $U\cup\{U_{k}\}_{k=1}^n$ is a finite subcollection of $\mathscr{U}$ which covers $\mathbb{R}$, proving that $\mathbb{R}$ is compact.
So what's the takeaway here?
Recall that compactness is an indication of a set's "finiteness." So when we view $\mathbb{R}$ through the eyes of the finite complement topology, $\mathbb{R}$ becomes "finite"! Now don't get me wrong. I'm not claiming that the cardinality of $\mathbb{R}$ suddenly becomes finite. Certainly $\mathbb{R}$ still has uncountably many elements! But what I
am saying is that because of the nature of the open sets in the finite complement topology (which is vastly different than the nature of the open sets in the usual topology), $\mathbb{R}$ is no longer infinitely large. It can be contained.
To put it another way, suppose your friend (a very cruel friend) charges you with the task of gathering all the uncountably many elements of $\mathbb{R}$ into buckets. At first, this might sound horrible because how are you going to find an infinite number of buckets to get the job done?! But you, being a very smart math student, do not panic. You simply endow $\mathbb{R}$ with the finite complement topology and say, "Voila! Now $\mathbb{R}$ is compact. Hence there exists a finite number of buckets which will contain
all the elements of $\mathbb{R}$." And then you simply go and find those buckets and carry them back to your cruel friend.
But all sillyness aside, do you see what's going on here?
Once you change the open sets in your topological space, you change everything! We've just seen that when $\tau$ is the usual topology and $\tau'$ is the finite complement topology, the spaces $(\mathbb{R},\tau)$ and $(\mathbb{R},\tau')$ are very different! Open sets are everything!
Footnotes:
*Topological properties, that is.
** That is if $\mathbb{R}\smallsetminus O$ looks something like $\{r_1,r_2,\ldots,r_n\}$ where the $r_i$ are real numbers.
***The method used in this proof is a standard trick to show a set $X$ with a topology $\tau$ is compact. You let $\mathscr{U}$ be an open cover of $X$ and pick an arbitrary $U\in\mathscr{U}$. Then, given the special properties of the open sets in $\tau$, you show that only finitely many sets $U_1,\ldots,U_n$ in $\mathscr{U}$ are needed to cover the
complement $X\smallsetminus U$. Then $U,U_1,\ldots, U_n$ is your finite subcover for $X$ since you can always write $X$ as $X=U\cup(X\smallsetminus U)$.
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Suppose we have a binary symmetric channel with $p=\frac{1}{3}$; that is, a communications channel in which each bit is flipped with independent probability $\frac{1}{3}$. I know that there is a code such that, in the (highly probable) event that no more than $p$ of the bits are corrupted, we can guarantee recovery of any message we put through... so in particular, we can certainly recover a message that is correct in $\frac{1}{4}$-th of its positions, reducing the error introduced by the channel.
Now, let's ask this for a useless channel - a binary symmetric channel $C$ with $p=\frac{3}{4}$. Obviously, this can't carry any information, but bear with me. Is there a code such that, if we feed an encoded message through the channel $C$, we can recover (with high probability) a version of the message with at most two-thirds of the bits flipped? That is, is it possible to reduce the error (without actually recovering any information) even in a capacity-0 channel? Obviously, we can't get our error down under $\frac{1}{2}$, since that would violate the channel capacity theorem... but can we get as close as we like? Or is this generally impossible?
Stating this final question: For fixed $\epsilon>0$ and $\delta\in(\frac{1}{2},1)$, is there a code that can implement a channel with symmetric error no more than $\frac{1}{2}+\epsilon$ over a channel with symmetric error bounded by $\delta$? (For what it's worth, I don't care about the rate, just the error bounds.)
I welcome any suggestions that would clarify the question as explained above - I haven't done much coding theory, so I'm not used to the terms.
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Consider a beta distribution for a given set of ratings in [0,1]. After having calculated the mean:
$$ \mu = \frac{\alpha}{\alpha+\beta} $$
Is there a way to provide a confidence interval around this mean?
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While there are specific methods for calculating confidence intervals for the parameters in a beta distribution, I’ll describe a few general methods, that can be used for (almost)
all sorts of distributions, including the beta distribution, and are easily implemented in R.
Let’s begin with maximum likelihood estimation with corresponding profile likelihood confidence intervals. First we need some sample data:
# Sample sizen = 10# Parameters of the beta distributionalpha = 10beta = 1.4# Simulate some dataset.seed(1)x = rbeta(n, alpha, beta)# Note that the distribution is not symmetricalcurve(dbeta(x,alpha,beta))
The real/theoretical mean is
> alpha/(alpha+beta)0.877193
Now we have to create a function for calculating the negative log likelihood function for a sample from the beta distribution, with the mean as one of the parameters. We can use the
dbeta() function, but since this doesn’t use a parametrisation involving the mean, we have have to express its parameters (
α and β) as a function of the mean and some other parameter (like the standard deviation):
# Negative log likelihood for the beta distributionnloglikbeta = function(mu, sig) { alpha = mu^2*(1-mu)/sig^2-mu beta = alpha*(1/mu-1) -sum(dbeta(x, alpha, beta, log=TRUE))}
To find the maximum likelihood estimate, we can use the
mle() function in the
stats4 library:
library(stats4)est = mle(nloglikbeta, start=list(mu=mean(x), sig=sd(x)))
Just ignore the warnings for now. They’re caused by the optimisation algorithms trying invalid values for the parameters, giving negative values for
α and/or β. (To avoid the warning, you can add a
lower argument and change the optimisation
method used.)
Now we have both estimates and confidence intervals for our two parameters:
> estCall:mle(minuslogl = nloglikbeta, start = list(mu = mean(x), sig = sd(x)))Coefficients: mu sig 0.87304148 0.07129112> confint(est)Profiling... 2.5 % 97.5 %mu 0.81336555 0.9120350sig 0.04679421 0.1276783
Note that, as expected, the confidence intervals are
not symmetrical:
par(mfrow=c(1,2))plot(profile(est)) # Profile likelihood plot
(The second-outer magenta lines show the 95% confidence interval.)
Also note that even with just 10 observations, we get very good estimates (a narrow confidence interval).
As an alternative to
mle(), you can use the
fitdistr() function from the
MASS package. This too calculates the maximum likelihood estimator, and has the advantage that you only need to supply the density, not the negative log likelihood, but doesn’t give you profile likelihood confidence intervals, only asymptotic (symmetrical) confidence intervals.
A better option is
mle2() (and related functions) from the
bbmle package, which is somewhat more flexible and powerful than
mle(), and gives slightly nicer plots.
Another option is to use the bootstrap. It’s extremely easy to use in R, and you don’t even have to supply a density function:
> library(simpleboot)> x.boot = one.boot(x, mean, R=10^4)> hist(x.boot) # Looks good> boot.ci(x.boot, type="bca") # Confidence intervalBOOTSTRAP CONFIDENCE INTERVAL CALCULATIONSBased on 10000 bootstrap replicatesCALL : boot.ci(boot.out = x.boot, type = "bca")Intervals : Level BCa 95% ( 0.8246, 0.9132 ) Calculations and Intervals on Original Scale
The bootstrap has the added advantage that it works even if your data doesn’t come from a beta distribution.
For confidence intervals on the mean, let’s not forget the good old asymptotic confidence intervals based on the central limit theorem (and the
t-distribution). As long as we have either a large sample size (so the CLT applies and the distribution of the sample mean is approximately normal) or large values of both α and β (so that the beta distribution itself is approximately normal), it works well. Here we have neither, but the confidence interval still isn’t too bad:
> t.test(x)$conf.int[1] 0.8190565 0.9268349
For just slightly larges values of
n (and not too extreme values of the two parameters), the asymptotic confidence interval works exceedingly well.
Check out Beta regression. A good introduction to how to do it using R can be found here:
Another (really easy) way of constructing a confidence interval would be to use a non-parametric boostrap approach. Wikipedia has good info:
Also nice video here:
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Now we are going to build a polynomial of degree $$n$$ that only goes through one given point $$x_0$$, and its $$n$$ derivatives coincide in this point with the $$n$$ derivatives of the original function $$f(x)$$. This polynomial is named Taylor's polynomial of degree $$n$$ around the point $$x_0$$ of the function $$f (x)$$.
This polynomial is:
$$$ \begin{array}{c} T_n(x)=f(x_0)+\dfrac{f'(x_0)}{1!}(x-x_0)+ \dfrac{f''(x_0)}{2!}(x-x_0)^2+ \\ \dfrac{f'''(x_0)}{3!}(x-x_0)^3+ \dots \dots +\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n \end{array} $$$
Therefore, given a $$\mathcal{C}^{n+1}$$ class function, this means derivable $$n +1$$ times with continuity, and a point $$x_0$$ where we know $$f(x_0), \ f'(x_0), \ f''(x_0), \dots, \ f^{(n)}(x_0)$$, we will be able to calculate Taylor's polynomial.
Let's calculate Taylor's polynomial of order $$6$$ of the function $$f(x)=e^x$$ around $$x = 0$$. We need to know the value of the function and its six first derivatives in the point $$x = 0$$. Let's calculate them:
$$$\begin{array}{lcl} f(0)=1 \\ f'(x)=e^x & \Rightarrow & f'(0)=1 \\ f''(x)=e^x & \Rightarrow & f''(0)=1 \\ f'''(x)=e^x & \Rightarrow & f'''(0)=1 \\ f^{iv}(x)=e^x & \Rightarrow & f^{iv}(0)=1 \\ f^{v}(x)=e^x & \Rightarrow & f^{v}(0)=1 \\ f^{vi}(x)=e^x & \Rightarrow & f^{vi}(0)=1 \end{array} $$$
So Taylor's polynomial is:
$$$T_6(x)=1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\dfrac{1}{24}x^4+ \dfrac{1}{120}x^5+ \dfrac{1}{720}x^6$$$
This polynomial can be understood as a polynomial approach of the function $$f(x)$$. Sometimes it is easier to work with a polynomial than with a complicated function and the Taylor's polynomial becomes very useful. In front of this, we wonder what is our error when using this simplification. Well, this error, which measures how correct is our polynomial compared to the given function, is given by the following expression:
$$$\text{error}=\big|f(x)-T_n(x)\big| = \dfrac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-x_0)^{n+1}$$$
where $$\xi(x)$$ is a point between $$x_0$$ and $$x$$.
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Simplifying polynomials
“Polynomials” are expressions with two or more (poly) types of numbers (nomials). An example of a polynomial expression would be:
$(6x^2+5x+3)+(x^2+3x-2)$
To simplify this expression, we need to first consider PEMDAS. Within each set of parentheses, nothing can be reduced or combined. Also, no exponents can be computed since we don’t know what x stands for—it’s already reduced as far as we take reduce it. There is no multiplication or division. There’s only addition and subtraction. So, we must now “combine like terms.” Like terms are the simply the ones “like each other.” You can ask, “How many
x-squareds are there?” Well, there are 6 x-squares plus 1 more x-squared, which makes 7 x-squared. “How many x’s are there?” There are $5x$’s plus $3x$’s, to make $8x$’s. And looking at the numbers only, 3-2 is 1. So, our simplified polynomial is:
$7x^2+8x+1$
Factoring binomials using F.O.I.L.
Factoring binomials means to multiply binomials (expressions with two types of numbers). For example, this type of problem might be:
$(x+3)(x-1)$
This type of problem is solved through the “F.O.I.L.” method. F.O.I.L. stands for “First, Outer, Inside, and Last.” Using the problem above, let’s F.O.I.L.:
First: $x \times x$ yields $x^2$ Outer: $x \times -1$ yields $-1x$ or just $-x$ Inner: $3 \times x$ yields $3x$ Last: $3 \times -1$ yields $-3$
$x^2-x+3x-3$
Lastly, we combine like terms. The only terms able to be combined are the $-x$ and the $3x$ which gives us $2x$. So, our final answer is:
$x^2+2x-3$
Factoring trinomials
This implies doing just the opposite of what we just did. It’s like taking the expression $x^2+2x-3$ and working backwards to find the factors: $(x+3)(x-1)$
Consider this example for factoring trinomials:
$x^2-9x+20$
To find the factors for this trinomial:
Find factors for $x^2$ : $x$ times $x$ or $(x)(x)$ Find factors for 20: $1 \times 20$, $2 \times 10$, and $4 \times 5$ Find the pair with a sum of -9. -4 and -5 have a sum of -9 Complete the factors: $(x-4)(x-5)$ Double check using F.O.I.L. Be sure the numbers are correct and the positives and negatives are correct. Using F.O.I.L., you’ll get back to $x^2-9x+20$ and know your answer is correct. Dividing Polynomials
Dividing polynomials can appear intimidating. Don’t get psyched out! The basic process is the same as regular old long division that you did in grade school. Consider the division problem of…
It’s a binomial (x + 1) dividing into a polynomial (the parts inside the division sign). Look at the graphic following this sentence for an explanation to solve the problem.
Handling square roots
It’s necessary to be able to simplify, multiply and divide square roots.
Simplifying square roots:
Suppose you have an answer of $\sqrt {2}$, you’d need to simplify it on down to its most basic form. To simplify square roots, think of any perfect squares that are factors of the number inside the square root symbol. In other words, do 4, 9, 25, 36 etc. divide evenly into the number inside the symbol? If so, it needs to be simplified. In this case, it could’ve been written $\sqrt{9 \times 2}$ So, we need to “pull out” the 9. The square root of 9 is 3, so our final and most simplified answer would be $3 \sqrt{2}$
Multiplying square roots:
Multiplying square roots is simple—you multiply the numbers under the square root sign, then look to simplify. Consider the problem: $\sqrt{2} \times \sqrt{8}$. Just multiply $2 \times 8$ to get $\sqrt {16}$ The square root of 16 is 4, the answer.
Dividing square roots:
Consider the problem: $\sqrt{27} \over \sqrt{12}$
First simplify by “pulling out” the 9 in the numerator and the 4 in the denominator. The problem then becomes: $3 \sqrt{3} \over 2\sqrt{3}$ Now simplify. The two $\sqrt{3}$ factors cancel each other out, so you could just scratch them and you’re left with your answer: $3/2$
Distance between two points in a plane
Algebra is often called “linear algebra” because the equations can be graphed on a plane, that is to say, on graph paper. This skill requires you to, when given to points on a plane, calculate the distance between those two points.
There are two methods of finding the distance between two points in a plane:
1. Graphically – you can draw the graph to visualize it. 2. Formulaically – you can use the distance formula shown below.
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
We’ll use both methods to illustrate how to compute the distance.
1. Graphically solving the problem:
Consider the example: Find the distance between point A at (-2, 6) and point B at (5,3). Refer to the graphic below to solve this problem.
2. Formulaically solving the problem:
The formula for computing the distance between two points on a graph is built on the same principles that we used in the problem above. The formula is:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Consider the same two points that we used up above: point A (-2,6) and point B (5,3). Placing the x and y coordinates into the formula gives us:
$d=\sqrt{(5- -2)^2+(3-6)^2}$
Which becomes…
$d=\sqrt{7^2+(-3)^2}$
Which becomes…
$d=\sqrt{49+9}$
Which becomes…
$d=\sqrt{58}$ or $d\approx 7.62$ Ratios or proportions
Ratios or proportions are essentially the same thing. They show how numbers relate to one another. Ratios and proportions can be shown a couple of ways:
In numbers such as 3:1. This says “three to one.” As fractions such as $3 \over 1$ This fraction is the same as writing 3:1. It is read as, "A ratio of 3 to 1." Or in words such as there are 3 teaspoons to every 1 tablespoon. It’s still “three to one.”
Often, questions on standardized tests have you solve a simple proportion problem. It may go something like this:
It took 8.5 bags of mulch to cover a playground with 500 square feet of area. How many bags of mulch would it take to cover an 800 square foot playground?
This may appear intimidating, but it’s very easy. It’s really saying 8.5 is to 500, as what is to 800? First, let “what is” be $x$ Therefore, to define the variable, $x$ = the number of bags to cover 800 square feet. Next, we put the problem in the form of a proportion—that is to say, we put it in fraction form and set them equal to one another. So, we now have:
$8.5 \over500$ = $x \over 800$
This now is read as: “8.5 is to x, as 500 is to 800.”
To solve a proportion like this, we cross-multiply and set each “line of the cross” equal to each other. So $8.5(800)=500x$ This gives us $6800=500x$ Divide both sides by 500 to get the answer, 13.6 bags.
As a note, on a question like this, the answer will likely be “14 bags” not 13.6 because you can’t buy a “.6” bag.
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Let $C$ be a $[n,k]$ linear Code over $\mathbb{F}_q$ .
I want to show that each vector of $\mathbb{F}_q^{n-k} $ is written as a linear combination of $m$ columns of $H$ iff $\rho \leq m$.
I have thought the following:
$$ d(C)=\min \{ d \in \mathbb{N} | \text{ there are d linearly dependent columns of H}\} $$
So $H$ has $ d-1 $ linearly independent columns , so each vector of $\mathbb{F}_q^n$ can be written as a linear combination of these columns.
But what can we say about the vectors of $ \mathbb{F}_q^{n-k}$?
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Multiple nonnegative solutions for elliptic boundary value problems involving the $p$-Laplacian
DOI: http://dx.doi.org/10.12775/TMNA.2005.036
Abstract
In this paper we present a result concerning the existence of two
nonzero nonnegative solutions for the following Dirichlet problem involving the $p$-Laplacian $$ \cases -\Delta_p u=\lambda f(x,u) &\text{\rm in\ } \Omega,\\ u=0 &\text{\rm on\ } \partial \Omega, \endcases $$ using variational methods. In particular, we will determine an explicit real interval $\Lambda$ for which these solutions exist for every $\lambda\in \Lambda$. We also point out that our result improves and extends to higher dimension a recent multiplicity result for ordinary differential equations.
nonzero nonnegative solutions for the following Dirichlet problem involving
the $p$-Laplacian
$$
\cases
-\Delta_p u=\lambda f(x,u) &\text{\rm in\ } \Omega,\\
u=0 &\text{\rm on\ } \partial \Omega,
\endcases
$$
using variational methods. In particular, we will determine an
explicit real interval $\Lambda$ for which these solutions exist
for every $\lambda\in \Lambda$. We also point out that our result
improves and extends to higher dimension a recent multiplicity
result for ordinary differential equations.
Keywords
Variational methods; weak solutions; nonnegative solutions; p-Laplacian; Dirichlet problem
Full Text:FULL TEXT Refbacks There are currently no refbacks.
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I am reading about Dupire local volatility model and have a rough idea of the derivation. But I can't reconcile the local volatility surface to pricing using geometric brownian motion process. If I'm not mistaken the proces $dS=rSdt+\sigma(S;t) S dX$ given final condition $(S-K)^{+}$ will produce correct options values given some strike. So in other words each time I change the strike $K$ I should get the options values that are consistent with the market (or really close to them). Is it true? How does my model know that I changed my strike?
EDIT:
2016/11/19:
I'm still not sure if I understand that correctly.
If I have a matrix of option prices by strikes and maturities then I should fit some 3D function to this data. This can be done by some interpolation/extrapolation or just finding some 3rd degree polynomial. I did the latter.
The formula for instantenous volatility is $\sigma(E;T)=\sqrt{\frac{\frac{\partial V}{\partial T}+rE\frac{\partial V}{\partial E}}{\frac{1}{2}E^2\frac{\partial^2 V}{\partial^2 E}}}$
If I want to perform Monte Carlo simulation I should evaluate $\sigma(E;T)$ at each timestep. At each timestep I simulate current stock price $S_c$ and I pretend it is $E$ so I evaulate $\sigma(E;T)$ at $E=S_c; T=t$. Is this correct? If so, then $\frac{\partial V}{\partial T}$ is really a instantenous change of option price which has the strike $S_c$ and it comes from Matrix of option prices. Is this correct? So really any stochastic process $dS=rSdt+\sigma(S;t) S dX$ should have the same diffusion for all Strikes. If they have exactly the same diffusion, the probability density function will be the same and hence the realized volatility will be exactly the same for all options, but market data differentiate volatility between strike and option price. If I have realized volatility different than implied, there is no way I should get the same option prices as the market.
For example for option with strike K=100 realized vol should be 20% (this is implied from quoted option prices), and for option with strike K=110 realized vol should be 15%, but actually with the dupire formula it will be the same for both of them. Could you guys clarify?
Edit 2016/11/21: Ok guys, I think I understand it now. I performed MC simulation and got the correct numbers. In fact the pdf will be tlhe same but it will allow to replicate implied vol surface. Thanks for the explanation, it was helpful.
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Let $m,n \in \mathbb{Z}$ and let $x \in \mathbb{R}$. Let $[x]$ denote the floor function.
We will attempt to prove
$$\Big[\frac{x+n}{m}\Big] = \bigg[\frac{[x]+n}{m}\bigg]$$
Suppose without loss of generality that $m > 0$.
By definition of the floor function on $\frac{x+n}{m}$, we have
$$\Big[\frac{x+n}{m}\Big] \leq \frac{x+n}{m} \tag{1}$$
We observe that $\Big[\frac{x+n}{m}\Big]$ is the closest integer to $\frac{x+n}{m}$ satisfying $(1)$.
If $[x]+n \leq x+n$, then
$$\frac{[x]+n}{m} \leq \frac{x+n}{m} \tag{2}$$
Once more applying the definition of the floor function on $(2)$,
$$\bigg[\frac{[x]+n}{m}\bigg] \leq \frac{[x]+n}{m} \leq \frac{x+n}{m} \tag{3}$$
To prove that there are no integers in the interval $\Big(\frac{[x]+n}{m}, \frac{x+n}{m}\Big)$, specifically $\big[\frac{x+n}{m}\big]$, we suppose to the contrary that such an integer $q$ exists satisfying
$$\frac{[x]+n}{m} < q < \frac{x+n}{m} \tag{4}$$
But $(4)$ implies that $[x] < mq-n < x$ where $(mq-n) \in \mathbb{Z}$, which is a contradiction, since $[x]$ is the greatest intger less than or equal to $x$.
Because there are no integers in such an interval, $\big[\frac{x+n}{m}\big]$ must satisfy
$$\Big[\frac{x+n}{m}\Big] \leq \frac{[x]+n}{m} \leq \frac{x+n}{m} \tag{5}$$
which implies that $\Big[\frac{x+n}{m}\Big]$ is the closest integer less than or equal to $\frac{[x]+n}{m}$; that is, $\Big[\frac{x+n}{m}\Big]$ is the greatest integer less than or equal to $\frac{[x]+n}{m}$.
Therefore, $\Big[\frac{x+n}{m}\Big] = \bigg[\frac{[x]+n}{m}\bigg]$, as desired.
Is there anything wrong with the proof?
Note: A proof can also be achieved using the division algorithm.
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I'm a high school student, so I have no idea what a Jacobian or a manifold is, but as someone who's self-studied linear algebra and abstract algebra, I think it's pretty complex and takes a rather smart/dedicated person to pass these classes, so you're definitely smart enough to understand these concepts.
In my opinion, these fields are kind of intuitive in some ways. Honestly, I have no idea how mathematicians came up with all of this -- especially the more advanced (well, advanced to me) parts of these fields like field arithmetic or Galois theory or all of this spectral theorem and Schur decomposition/lemma stuff that I'm learning about right now. However, I've seen a lot of good online explanations, especially from looking at answers at this site, so I think I have a better intuition than some college students. If you're a college student on a strict schedule, you don't have the free time to explore maths like I do, so my guess is that it's b) and that the education system doesn't focus enough attention on comprehension and instead focuses on getting students passing grades and degrees.
Now, I know the beginning of linear algebra quite well since I've reviewed and cemented those concepts and I can definitely tell you how matrices have helped me. Maybe this and some of the links below will help you gain intuition on some things.
Matrices help us solve problems of systems of equations. For example:$$3x+2y=1$$$$4x+5y=2$$Using Algebra I knowledge, we can solve this using elimination. Divide the first equation by $3$, subtract the second equation by $4$ times that to get rid of $x$, divide the second equation by $\frac 7 3$ to solve for $y$ and subtract the first equation by the $\frac 2 3$ the secone equation to solve for $x$.
This is elimination, but this is the same process used for RREFing the following matrix:$$\left[\begin{matrix}3 & 2 & 1 \\4 & 5 & 2\end{matrix}\right]$$If you RREF that matrix manually, you'll basically end up with the same row operations as we manipulated the equations. Thus, matrices and RREFing them is literally just solving systems of equations with elimination. However, it's hard to see that because there's no variables; we're just manipulating coefficients. It's even harder to see that with 3x3 and 4x4 matrices when this becomes increasingly more time-lengthy and when some people would rather use substitution or guess and check for such systems. However, with this method of matrix and RREFing,
we have an algorithm that makes it much easier to do this without thinking or with a computer. By using this repetitive, boring algorithm, I bet mathematicians were able to solve systems a lot faster since they didn't need to keep track of variables and they didn't need to choose between elimination or substitution. They could just do the algorithm out and there's no thought involved. By using this kind of repetitive/familiar format of matrices, it makes solving systems of equations faster. RREFing basically solves the problem of linear systems, so using this rote method matrices and RREFing allows mathematicians to solve linear systems quickly and focus on more complex, interesting problems.
Now, hopefully, this explanation helped you understand matrices, but there are some more advanced concepts that I think other explanations can help you with best:
Least squares approximation $\rightarrow$ This video taught me why in the world we need things like "left nullspace" and "orthogonal complements." These concepts fit into real world problems like least squares approximations and it's pretty cool how abstract concepts in linear algebra can come together in something useful like this. Eigenbases $\rightarrow$ This taught me how eigenbases can speed computations up and thus how eigenvectors can be useful if someone needs to apply a certain linear transformation repeatedly. Abstract Algebra: Theory and Applications $\rightarrow$ This online textbook shows a lot of the applications behind abstract algebra and gives us a motivation for doing it. It has a lot of practice problems and these proof problems took me hours and hours to solve, but I have a much better intuition about this field because of it.
In short, I've learned my intuition about mathematics because of good online explanations and a lot of practice problems. Explore mathematics, listen to explanations online, ask questions to your professors and possibly this forum and you will gain a better mathematical intuition about mathematical concepts that seem complex to you now, but down the road, will make much more sense. One year ago, all of the concepts I've explained above (except maybe matrices) made no sense to me, but now, I've gained a much better understanding of these concepts, even though I know I have a long way to go and I think that's probably the place you are in. In a few years, the concepts that seem bizarre and odd to us now will hopefully make a lot more sense to us then, at which point we might be struggling with even more complicated mathematical concepts. In any case, good luck and I hope all of this helped!
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We have to evaluate the following integration $$\int_0^{\pi/2}\sin 2x\arctan(\sin x)dx.$$
In this question I thought of using integration by parts .
But stuck in that.
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Substituting $u=\sin x $ gives us $$I =\int_{0}^{1} 2u\arctan u du $$ Integrating by parts, keeping $f=\arctan u $ and $g'=u $, we get $$I =(u^2\arctan u-u + \arctan u)|_{0}^{1} $$ giving us the answer as $$\boxed {\frac {\pi}{2}-1}$$ Hope it helps.
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Given a convex hexagon $ABCDEF$. All its sides are equal (it can be irregular). Furthermore, $AD = BE = CF$. How can I prove that a circle can be inscribed in this hexagon?
Let $O$ be the intersection point of diagonals $AD$ and $BE$ (see diagram below) and set $a=OB$, $b=OD$, $d=AD=AE$, $\theta=\angle AOB=\angle DOE$. Then by the cosine rule we have: $$ a^2+(d-b)^2-2a(d-b)\cos\theta=b^2+(d-a)^2-2b(d-a)\cos\theta, $$ which simplified gives: $$ (b-a)(1-\cos\theta)=0, \quad\hbox{that is:}\quad a=b. $$ It follows that $OBC\cong ODC$ and $OAF\cong OEF$, so that $\angle COB\cong \angle COD\cong \angle AOF\cong (\pi-\theta)/2$ and points $COF$ are aligned.
By repeating the same argument given above for diagonals $AD$ and $CF$ we get $OF=a$, $OC=d-a$. It follows that all six triangles in the diagram are congruent between them and in particular they all have the same altitude from their common vertex $O$. A circle of center $O$ and having that common altitude as radius will therefore touch all six red sides of the hexagon.
Notice that if $a\ne d/2$ the hexagon is NOT regular. All six angles with vertex at $O$ are however congruent and they thus measure 60°.
EDIT.
One could prove that $a=b$ even without trigonometry. Triangles $ABO$ and $EDO$ have congruent between them the angles of vertex $O$ and the sides opposite to them. In addition, the other two sides of each triangle have the same difference.
It is well known that if two triangles have the same base $PQ$ and the angles opposite to $PQ$ both congruent, then the third vertex of both triangles lies on the arc of a well defined circle having $PQ$ as a chord.
Moreover, if the difference of the distances of the third vertex from $P$ and $Q$ is the same for both triangles, then the third vertex must also belong to a well defined hyperbola having $P$ and $Q$ as foci. This third vertex is then one of the two intersections between circle arc and hyperbola, and both possible triangles are congruent between them.
Lemma. Diagonal $BE$ is the angle bisector of both angles $\angle \, ABC$ and $\angle \, DEF$; (1) Diagonal $CF$ is the angle bisector of both angles $\angle \, BCD$ and $\angle \, EFA$; (2) Diagonal $AD$ is the angle bisector of both angles $\angle \, FAB$ and $\angle \, CDE$; (3) Finally, the three diagonals $AD, \, BE$ and $CF$ intersect at a common point I. Proof. Look at quadrilateral $ACDF$. Reflect point $A$ in the line $DF$ and let $A^*$ be the symmetric image of $A$ with respect to $DF$. Then $$ \angle \, DA^*F = \angle \, DAF $$ as well as $FA^* = FA = CD$ and $A^*D = AD = CF$. Since $A^*D = CF$ and $FA^*=CD$ the quad $FCDA^*$ is a parallelogram and thus $$\angle \,DCF = \angle \, DA^*F = \angle \, DAF$$ Therefore quad $ACDF$ is inscribed in a circle and since $FA = CD$, the quad $ACDF$ is in fact an isosceles trapezoid with $AC$ parallel to $DF$. For that reason, the parallel segments $AC$ and $DF$ have a common orthogonal bisector and since $BA = BC$ and $ED =EF$, the points $B$ and $E$ must lie on that orthogonal bisector, i.e. the line $EF$ is the orthogonal bisector of segments $AC$ and $DF$ simultaneously. However, since triangles $ABC$ and $DEF$ are isosceles, the orthogonal bisector $BE$ is at the same time the angle bisector of both angles $\angle \, ABC$ and $\angle \, DEF$. Observe that since the line $BE$ is the orthogonal bisector of the two parallel sides $AC$ and $DF$ of the isosceles trapezoid $ACDF$ its two diagonals $AD$ and $CF$ intersect in a common point which lies on $BE$. The rest of the lemma follows analogously.
$$ $$
Completing the proof: Observe that by the Lemma, the lines $AD, \, BE,$ and $CF$ all meet at a common point, we denote by $I$. Consequently, by the Lemma, $I$ is the common intersection point of all angle bisectors of the angles at the vertices of the hexagon $ABCDEF$. Therefore, there is a circle inscribed in the hexagon with point $I$ the incenter. As a consequence we get that $IA=IC=IE$ and $IB = ID=IF$.
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I have an equation: $\ddot{x}+(\delta+\epsilon\cos{t})x=0$ known as the Mathieu equation.The $\delta-\epsilon$ parameter space of this equation looks something like
The red lines in this diagram indicate that if a point on the red lines is chosen, and the corresponding value of $(\delta,\epsilon)$ is plugged into the Mathieu equation, then it is guaranteed that one would get a periodic solution. I have managed to generate these curves by something known as the arc-length continuation method, thus I have the exact data to plot these curves with me.
On the other hand, I have an analytic relation approximating these red lines which looks something like: $\delta=\frac{1}{4}+f_1(k)\epsilon+f_2(k)\epsilon^2+f_3(k)\epsilon^3+...$
Here, the functions $f_1(k),f_2(k)...$ are known and are highly nonlinear. This equation is for just one of these red lines. This way I have equations for all the red lines. Also, it is only possible to approximate these curves till an upper bound on $\epsilon$, as I have only one degree of freedom (the parameter k) in analytic $\delta-\epsilon$ relation.
First let me choose one particular red line and the corresponding $\delta-\epsilon$ relation for it.Now, what I want to do is two things: 1) I want to find the optimum $\epsilon$ by which I can fit $\delta-\epsilon$ curve. 2) I want to find the optimum $k$ for that optimum $\epsilon$.
I want to find both the above things by some numerical algorithms which converge. This is necessary as my functions $f_1(k),f_2(k)...$ are highly non-linear in nature.
I will tell you where I have reached so far. I have no idea how to even approach point 1). For point 2), I have managed to implement the mean-square errors approach, by which I tried minimizing the error, but that leaves me to solve a polynomial with powers of k like $k^{3500}$. This polynomial is going to have a lot of local minimas and I have no clue of any numerical algorithms which help me in finding a global minima.
Finally, if I manage to implement 1) and 2), I want to do some analysis (like R-squared coefficient) to see how well I have fit my curve.
Please help me out if you know of any numerical procedures that can achieve the above. Thanks.
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Consider a rectangular $(m \times n)$ matrix $\underline E_1$ with $m < n$ that has only $0$ or $1$ entries. It has exactly one $1$ entry in each row and not more than one $1$ entry in each column. Consider it being a selection of $m$ rows out of a $(n \times n)$ permutation matrix $\underline P$.
Given $\underline E_1$ I'm looking for an elegant way to describe the set $\mathcal{P}$ of $((n-m) \times n)$ matrices so that any $\underline E_2 \in \mathcal{P}$ combined with $\underline E_1$, like so
$\underline E = \begin{pmatrix} \underline E_1 \\\ \underline E_2 \end{pmatrix}$
forms a valid $(n \times n)$ permutation matrix $\underline E$, i.e. something like $\mathcal{P} = \operatorname{percomp}_n( \underline E_1 )$ (there are $(n-m)!$ elements in $\mathcal{P}$)
Is there anything like this used in mathematical parlance already?
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Let's say you want to estimate a quantity $\mu$, but you have only access to unbiased estimates of its logarithm, i.e., $\log\mu$. Can you obtain an unbiased estimate of $\mu$?
There used to be a quite nice blog post online for the estimator solving this problem, but it is now in a weird shape, not compiling the Latex code. Recently I was discussing this estimator with friends and realized that, at the moment, there is no quick explanation of it on the web. So I decided to write one.
Let's say, as said in the blurb, you want to estimate a quantity but only have access to unbiased estimates of its $\log$. A typical situation is that you have a quantity of the form: \begin{align*} \mu(x) = \exp{\left(-\frac{1}{n} \sum_{i=1}^n g_i(x)\right)}. \end{align*} If $n$ is large, computing this might be prohibitive, but you can get cheap unbiased estimates of $\log \mu(x)$ by subsampling the functions and computing the average on a mini-batch. However, an unbiased estimate of $\log\mu$ is not enough to obtain an unbiased estimate of $\mu$. Let's denote this unbiased estimate $\widehat{\log\mu}$ and we naturally have $\bE[\widehat{\log\mu}] = \log \mu$. Next, note that \begin{align*} \bE\left[ e^{\widehat{\log\mu}} \right] \geq e^{\bE\left[ \widehat{\log\mu} \right]} = \mu, \end{align*} by Jensen's inequality. Therefore, naively computing the unbiased estimate of $\widehat{\log\mu}$ and exponentiating it will not produce the right result. Is there a way around?
It turns out that there is -- provided that you can obtain many i.i.d unbiased estimates. So returning to the original example, let's say you can obtain i.i.d random variables $\lambda_j$ where $\bE[\lambda_j] = \log \mu$. Then, incredibly, the following estimator is an unbiased estimator of $\mu$: \begin{align*} \widehat{\mu} = e^{\delta} \prod^J_{j=1} \frac{\lambda_j}{\delta}, \quad \text{where} \quad J \sim \textsf{Poisson}(J;\delta) \end{align*} where $\delta > 0$ is a parameter of the estimator. This estimator is called as the Poisson estimator.
Why though? Let's prove a quick and sketchy result. First note $J \sim \textsf{Poisson}(J; \delta)$ has the density: \begin{align*} P(J = k;\delta) = \frac{\delta^k e^{-\delta}}{k!}. \end{align*} Now if we take the expectation of $\widehat{\mu}$: \begin{align*} \widehat{\bE[ \widehat{\mu}]} &= \sum_{k = 0}^\infty P(J = k;\delta) e^{\delta} \prod^k_{j=1} \frac{\lambda_j}{\delta}, \\ &= \sum_{k = 0}^\infty \frac{\delta^k e^{-\delta}}{k!} e^{\delta} \prod^k_{j=1} \frac{\lambda_j}{\delta}, \\ &= 1 + \sum_{k = 1}^\infty \frac{1}{k!} \prod^k_{j=1} \lambda_j. \end{align*} Note that $\widehat{\bE[ \widehat{\mu}]}$ is still a random quantity (hence the hat), since $\lambda_j$ are unbiased and stochastic estimates of $\log\mu$. Therefore, taking another expectation, we now obtain: \begin{align*} \bE[ \widehat{\mu}] = 1 + \sum_{k = 1}^\infty \frac{\lambda^k}{k!}, \end{align*} where $\lambda = \bE[\lambda_j] = \log\mu$, since $\lambda_j$ are i.i.d. Now it suffices to observe that this is indeed the series representation of the exponential function, therefore: \begin{align*} \bE[ \widehat{\mu}] = 1 + \sum_{k = 1}^\infty \frac{\lambda^k}{k!} = \exp(\lambda) = \mu. \end{align*} Although this result is nice and produced some practical applications, this estimator exhibits a huge variance. A partial way around this is to introduce another variable $c \in \bR$ and define the estimator as: \begin{align} \widehat{\mu} = e^{\delta + c} \prod^J_{j=1} \frac{\lambda_j – c}{\delta}, \end{align} where $J \sim \textsf{Poisson}(J;\delta)$. You can check that this estimator is still unbiased. However, as mentioned in this blogpost, this estimator still exhibits a big variance, hence is not solving a lot of the problems we encounter in practice.
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Discretization-invariant Bayesian inversion and Besov space priors
1.
Department of Mathematics and Statistics, University of Helsinki, P.O. Box 68 (Gustaf Hallstromin katu 2b) FI-00014, Finland, Finland
2.
Tampere University of Technology,Institute of Mathematics,, P.O. Box 553, 33101 Tampere
εis considered, where $U$ is a function on a domain of $\R^d$. Here $A$ is a smoothing linear operator and εis Gaussian white noise. The data is a realization $m_k$ of the random variable $M_k = P_kA U+P_k$ ε, where $P_k$ is a linear, finite dimensional operator related to measurement device. To allow computerized inversion, the unknown is discretized as $U_n=T_nU$, where $T_n$ is a finite dimensional projection, leading to the computational measurement model $M_{kn}=P_k A U_n + P_k$ ε. Bayes formula gives then the posterior distribution
$\pi_{kn}(u_n\|\m_{kn})$~ Π
$(u_n)\exp(-\frac{1}{2}$||$\m_{kn} - P_kA u_n$||$\_2^2)$
n
in $\R^d$,
and the mean $\u_{kn}$:$=\int
u_n \ \pi_{kn}(u_n\|\m_k)\ du_n$ is considered as the reconstruction of $U$. We discuss a systematic way of choosing
prior distributions Π
for all $n\geq n_0>0$ by achieving them as projections of a distribution in a infinite-dimensional limit case.
Such choice of prior distributions is n discretization-invariantin the sense that Π represent the same n a prioriinformation for all $n$ and that the mean $\u_{kn}$ converges to a limit estimate as $k,n$→$\infty$. Gaussian smoothness priors and wavelet-based Besov space priors are shown to be discretization invariant. In particular, Bayesian inversion in dimension two with $B^1_11$ prior is related to penalizing the $\l^1$ norm of the wavelet coefficients of $U$. Keywords:wavelet, discretization invariance, Inverse problem, statistical inversion, Besov space., Bayesian inversion, reconstruction. Mathematics Subject Classification:60F17, 65C20, 42C4. Citation:Matti Lassas, Eero Saksman, Samuli Siltanen. Discretization-invariant Bayesian inversion and Besov space priors. Inverse Problems & Imaging, 2009, 3 (1) : 87-122. doi: 10.3934/ipi.2009.3.87
[1]
Lassi Roininen, Janne M. J. Huttunen, Sari Lasanen.
Whittle-Matérn priors for Bayesian statistical inversion with applications in electrical impedance tomography.
[2]
Lassi Roininen, Mark Girolami, Sari Lasanen, Markku Markkanen.
Hyperpriors for Matérn fields with applications in Bayesian inversion.
[3] [4] [5]
Xiaomao Deng, Xiao-Chuan Cai, Jun Zou.
A parallel space-time domain decomposition method for unsteady source inversion problems.
[6]
Tan Bui-Thanh, Quoc P. Nguyen.
FEM-based discretization-invariant MCMC methods for PDE-constrained Bayesian inverse problems.
[7] [8]
Igor E. Shparlinski.
Close values of shifted modular inversions and the decisional modular inversion hidden number problem.
[9] [10]
Hongmei Cao, Hao-Guang Li, Chao-Jiang Xu, Jiang Xu.
Well-posedness of Cauchy problem for Landau equation in critical Besov space.
[11] [12] [13] [14]
Liying Wang, Weiguo Zhao, Dan Zhang, Linming Zhao.
A geometric inversion algorithm for parameters calculation in Francis turbine.
[15] [16] [17] [18]
Vianney Perchet, Marc Quincampoix.
A differential game on Wasserstein space. Application to weak approachability with partial monitoring.
[19]
Josep M. Olm, Xavier Ros-Oton.
Approximate tracking of periodic references in a class of bilinear systems via stable inversion.
[20]
Eric Bedford, Kyounghee Kim.
Degree growth of matrix inversion: Birational maps of symmetric, cyclic matrices.
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Difference between revisions of "Vopenka"
(weakly!)
(→Generic: unless...)
(One intermediate revision by the same user not shown) Line 117: Line 117:
** $gVP(κ, \mathbf{Σ_{n+1}})$ for a proper class of $κ$
** $gVP(κ, \mathbf{Σ_{n+1}})$ for a proper class of $κ$
** There is a proper class of $n$-remarkable cardinals.
** There is a proper class of $n$-remarkable cardinals.
+ + + +
* $κ$ is the least for which $gVP^∗(κ, \mathbf{Σ_{n+1}})$ holds. $\iff κ$ is the least $n$-remarkable cardinal.
* $κ$ is the least for which $gVP^∗(κ, \mathbf{Σ_{n+1}})$ holds. $\iff κ$ is the least $n$-remarkable cardinal.
−
* If $gVP^∗(Π_n)$, then there is an $n$-remarkable cardinal.
+
* If $gVP^∗(Π_n)$ , then there is an $n$-remarkable cardinal.
* If $gVP^∗(\mathbf{Π_n})$ holds, then there is a proper class of $n$-remarkable cardinals.
* If $gVP^∗(\mathbf{Π_n})$ holds, then there is a proper class of $n$-remarkable cardinals.
* If there is a proper class of $n$-remarkable cardinals, then $gVP(Σ_{n+1})$ holds.<cite>GitmanHamkins2018:GenericVopenkaPrincipleNotMahlo</cite>
* If there is a proper class of $n$-remarkable cardinals, then $gVP(Σ_{n+1})$ holds.<cite>GitmanHamkins2018:GenericVopenkaPrincipleNotMahlo</cite>
Line 127: Line 131:
Open problems:
Open problems:
* Must there be an $n$-remarkable cardinal
* Must there be an $n$-remarkable cardinal
−
** if $gVP(κ, \mathbf{Σ_{n+1}})$ holds for some $κ$
+
** if $gVP(κ, \mathbf{Σ_{n+1}})$ holds for some $κ$
−
** if $gVP(Π_n)$ holds
+
** if $gVP(Π_n)$ holds
==External links==
==External links==
Revision as of 10:44, 9 October 2019
Vopěnka's principle is a large cardinal axiom at the upper end of the large cardinal hierarchy that is particularly notable for its applications to category theory. In a set theoretic setting, the most common definition is the following:
For any language $\mathcal{L}$ and any proper class $C$ of $\mathcal{L}$-structures, there are distinct structures $M, N\in C$ and an elementary embedding $j:M\to N$.
For example, taking $\mathcal{L}$ to be the language with one unary and one binary predicate, we can consider for any ordinal $\eta$ the class of structures $\langle V_{\alpha+\eta},\{\alpha\},\in\rangle$, and conclude from Vopěnka's principle that a cardinal that is at least $\eta$-extendible exists. In fact if Vopěnka's principle holds then there is a stationary proper class of extendible cardinals; bounding the strength of the axiom from above, we have that if $\kappa$ is almost huge, or even almost-high-jump, then $V_\kappa$ satisfies Vopěnka's principle.
Contents Formalizations
As stated above and from the point of view of ZFC, this is actually an axiom schema, as we quantify over proper classes, which from a purely ZFC perspective means definable proper classes. A somewhat stronger alternative is to view Vopěnka's principle as an axiom in second-order set theory capable to dealing with proper classes, such as von Neumann-Gödel-Bernays set theory. This is a strictly stronger assertion. [1] Finally, one may relativize the principle to a particular cardinal, leading to the concept of a Vopěnka cardinal.
Vopěnka's principle can be formalized in first-order set theory as a schema, where for each natural number $n$ in the meta-theory there is a formula expressing that Vopěnka’s Principle holds for all $Σ_n$-definable (with parameters) classes.[1]
Vopěnka principle VP and the Vopěnka scheme VS are not equivalent, but they are equiconsistent and have the same first-order consequences (GBC+VP is conservative over GBC+VS and ZFC+VS, VP makes no sense in the context of ZFC):[2]
If ZFC and the Vopěnka scheme holds, then there is a class forcing extension, adding classes but no sets, in which GBC and the Vopěnka scheme holds, but the Vopěnka principle fails. If ZFC and the Vopěnka scheme holds, then there is a class forcing extension, adding classes but no sets, in which GBC and the Vopěnka principle holds.
Vopěnka cardinal is an inaccessible cardinal $δ$ such that $\langle V_δ , ∈, V_{δ+1} \rangle$ is a model of VP (and the Morse–Kelley set theory). Vopěnka-scheme cardinal is a cardinal $δ$ such that $\langle V_δ , ∈ \rangle$ is a model of ZFC+VS.[2]
Vopěnka cardinals
An inaccessible cardinal $\kappa$ is a
Vopěnka cardinal if and only if $V_\kappa$ satisfies Vopěnka's principle, that is, where we interpret the proper classes of $V_\kappa$ as the subsets of $V_\kappa$ of cardinality $\kappa$. Because of a characterization of Vopěnka's principle in terms of graphs, a cardinal $\kappa$ is Vopěnka if and only if $\kappa$ is inaccessible and any set $\kappa$-sized set $G$ of $<\kappa$-sized nonisomorphic graphs has some $g_0$ and $g_1$ with $g_0$ a proper subgraph of $g_1$. (Need to cite sources)
As we mentioned above, every almost huge cardinal is a Vopěnka cardinal.
Equivalent statements Extendible cardinals
The schema form of Vopěnka's principle is equivalent to the existence of a proper class of $C^{(n)}$-extendible cardinals for every $n$; indeed there is a level-by-level stratification of Vopěnka's principle, with Vopěnka's principle for a $\Sigma_{n+2}$-definable class corresponds to the existence of a $C^{(n)}$-extendible cardinal greater than the ranks of the parameters (see section "Variants”). [4]
The Vopěnka principle is equivalent over GBC to both following statements:[2]
For every class $A$, there is an $A$-extendible cardinal. For every class $A$, there is a stationary proper class of $A$-extendible cardinals. Strong Compactness of Logics
Vopěnka's principle is equivalent to the following statement about logics as well:
For every logic $\mathcal{L}$, there is a cardinal $\mu_{\mathcal{L}}$ such that for any language $\tau$ and any $\mathcal{L}(\tau)$-theory $T$, $T$ is satisfiable if and only if every $t\subseteq T$ such that $|t|<\mu_{\mathcal{L}}$ is satisfiable. [5]
This $\mu_{\mathcal{L}}$ is called the strong compactness cardinal of $\mathcal{L}$. Vopěnka's principle therefore is equivalent to every logic having a strong compactness cardinal. This is very similar in definition to the Löwenheim–Skolem number of $\mathcal{L}$, although it is not guaranteed to exist.
Here are some examples of strong compactness cardinals of specific logics:
If $\kappa\leq\lambda$ and $\lambda$ is strongly compact or $\aleph_0$, then the strong compactness cardinal of $\mathcal{L}_{\kappa,\kappa}$ is at most $\lambda$. Similarly, if $\kappa\leq\lambda$ and $\lambda$ is extendible, then for any natural number $n$, the strong compactness cardinal of $\mathcal{L}^n_{\kappa,\kappa}$ ($\mathcal{L}_{\kappa,\kappa}$ with $n+1$-th order logic) is at most $\lambda$. Therefore for any natural number $n$, the strong compactness cardinal of $n+1$-th order finitary logic is at most the least extendible cardinal. Locally Presentable Categories
Vopěnka's principle is equivalent to the axiom stating "no large full subcategory $C$ of any locally presentable category is discrete." (Sources needed). Equivalently, no large full subcategory of Graph (the category of all graphs) is discrete; that is, for any proper class of simple directed graphs, there is at least one pair of nonequal graphs $G$ and $H$ in the class such that $G$ is a subgraph of $H$. This is a $\Pi^1_1$ statement, so the least Vopěnka cardinals are not even weakly compact (although the least weakly compact cardinal is much, much, much smaller than the least Vopěnka cardinal, if it exists).
Intuitively, a "category" is just a class of mathematical objects with some notion of "morphism", "homomorphism", "isomorphism", (etc.). For example, in Set, the category of all sets, homomorphisms are just injections, and isomorphisms are bijections. In categories of groups and models, homomorphisms and isomorphisms share their actual names.
A "locally small category" $C$ is one with only set-many morphisms between any two objects of $C$. This is one where the objects of $C$ behave "set-like" in the sense that, usually, the number of morphisms between two set-sized objects is at most the number of functions between their universes (like in groups and in graphs). A "locally presentable category" is a locally small category with a couple more really nice properties; you can "generate" all of the objects from set-many objects in the category.
Vopěnka's principle intuitively states that if you have a locally presentable category $C$, then any proper class of objects of $C$ has some nonisomorphic objects $c$ and $d$ where $c$ has a morphism into $d$.
Woodin cardinals
There is a strange connection between the Woodin cardinals and the Vopěnka cardinals. In particular, Vopěnkaness is equivalent to two strengthening variants of Woodinness, namely the Woodin for Supercompactness cardinals and the $2$-fold Woodin cardinals. As a result, every Vopěnka cardinal is Woodin.
Elementary Embeddings Between Ranks
An equivalent statement to Vopěnka's principle is that for any proper class $C\subseteq ORD$, there are $\alpha\in C$, $\beta\in C$, and a nontrivial elementary embedding $j:\langle V_\alpha;\in,P\rangle\rightarrow\langle V_\beta;\in,P\rangle$. Vopěnka's principle quite obviously implies this. The reason the converse holds is because every elementary embedding can be "encoded" (in a sense) into one of these. For more information, see [6].
Other points to note
Whilst Vopěnka cardinals are very strong in terms of consistency strength, a Vopěnka cardinal need not even be weakly compact. Indeed, the definition of a Vopěnka cardinal is a $\Pi^1_1$ statement over $V_\kappa$ (Vopěnka's principle itself is $\Pi^1_1$), and $\Pi^1_1$-indescribability is one of the equivalent definitions of weak compactness. Thus, the least weakly compact Vopěnka cardinal must have (many) other Vopěnka cardinals less than it.
Variants
(Boldface) $VP(\mathbf{Σ_n})$ denotes the fragment of Vopěnka’s Principle for $Σ_n$-definable classes and (lightface) $VP(Σ_n)$ is the weaker principle, where parameters are not allowed in the definition of the class (with analogous definitions for $Π_n$ and $∆_n$).
Vopěnka-like principles $VP(κ, \mathbf{Σ_n})$ for cardinal $κ$ state that for every proper class $\mathcal{C}$ of structures of the same type that is $Σ_n$-definable with parameters in $H_κ$ (the collection of all sets of hereditary size less than $κ$), $\mathcal{C}$ reflects below $κ$, namely for every $A ∈ C$ there is $B ∈ H_κ ∩ C$ that elementarily embeds into $A$.
Results:
For every $Γ$, $VP(κ, Γ)$ for some $κ$ implies $VP(Γ)$. $VP(κ, \mathbf{Σ_1})$ holds for every uncountable cardinal $κ$. $VP(Π_1) \iff VP(κ, Σ_2)$ for some $κ \iff$ There is a supercompact cardinal. $VP(\mathbf{Π_1}) \iff VP(κ, \mathbf{Σ_2})$ for a proper class of cardinals $κ \iff$ There is a proper class of supercompact cardinals. For $n ≥ 1$, the following are equivalent: $VP(Π_{n+1})$ $VP(κ, \mathbf{Σ_{n+2}})$ for some $κ$ There is a $C(n)$-extendible cardinal. The following are equivalent: $VP(Π_n)$ for every n. $VP(κ, \mathbf{Σ_n})$ for a proper class of cardinals $κ$ and for every $n$. $VP$ For every $n$, there is a $C(n)$-extendible cardinal. Generic
(Information in this section from [7] unless noted otherwise)
Definitions:
The Generic Vopěnka’s Principlestates that for every proper class $\mathcal{C}$ of structures of the same type there are $B ≠ A$, both in $\mathcal{C}$, such that $B$ elementarily embeds into $A$ in some set-forcing extension. (Boldface) $gVP(\mathbf{Σ_n})$ and (lightface) $gVP(Σ_n)$ (with analogous definitions for $Π_n$ and $∆_n$) as well as $gVP(κ, \mathbf{Σ_n})$ are generic analogues of corresponding weakenings of Vopěnka's principle. For transitive $∈$-structures $B$ and $A$ and elementary embedding $j : B → A$, we say that $j$ is overspillingif it has a critical point and $j(crit(j)) > rank(B)$. The principle $gVP^∗(Σ_n)$ states that for every $Σ_n$-definable (without parameters) proper class $\mathcal{C}$ of transitive $∈$-structures, there are $B ≠ A$ in $\mathcal{C}$ such that there is an overspilling elementary embedding $j : B → A$ in some set-forcing extension. ($gVP^∗(Π_n)$, $gVP^∗(\mathbf{Π_n})$, and $gVP^∗(κ, \mathbf{Σ_n})$ are defined analogously.)
Results:
The following are equiconsistent: $gVP(Π_n)$ $gVP(κ, \mathbf{Σ_{n+1}})$ for some $κ$ There is an $n$-remarkable cardinal. The following are equiconsistent: $gVP(\mathbf{Π_n})$ $gVP(κ, \mathbf{Σ_{n+1}})$ for a proper class of $κ$ There is a proper class of $n$-remarkable cardinals. Unless there is a transitive model of ZFC with a proper class of $n$-remarkable cardinals, if for some cardinal $κ$, $gVP(κ, \mathbf{Σ_{n+1}})$ holds, then there is an $n$-remarkable cardinal. if $gVP(Π_n)$ holds, then there is an $n$-remarkable cardinal. if $gVP(\mathbf{Π_n})$ holds, then there is a proper class of $n$-remarkable cardinals. $κ$ is the least for which $gVP^∗(κ, \mathbf{Σ_{n+1}})$ holds. $\iff κ$ is the least $n$-remarkable cardinal. If $gVP^∗(Π_n)$ holds, then there is an $n$-remarkable cardinal. If $gVP^∗(\mathbf{Π_n})$ holds, then there is a proper class of $n$-remarkable cardinals. If there is a proper class of $n$-remarkable cardinals, then $gVP(Σ_{n+1})$ holds.[8] If $gVP(Σ_{n+1})$ holds, then either there is a proper class of $n$-remarkable cardinals or there is a proper class of virtually rank-into-rank cardinals.[8] If $0^\#$ exists, then $L$, equipped with only its definable classes, is a model of $gVP$. (By elementary-embedding absoluteness results. The hypothesis can be weakened, because one can chop at off the universe at any Silver indiscernible and use reflection.)[8] The generic Vopěnka scheme is equivalent over ZFC to the scheme asserting of every definable class $A$ that there is a proper class of weakly virtually $A$-extendible cardinals.[8]
Open problems:
Must there be an $n$-remarkable cardinal if $gVP(κ, \mathbf{Σ_{n+1}})$ holds for some $κ$? if $gVP(Π_n)$ holds?
References Bagaria, Joan. $C^{(n)}$-cardinals.Archive for Mathematical Logic 51(3--4):213--240, 2012. www DOI bibtex Hamkins, Joel David. The Vopěnka principle is inequivalent to but conservative over the Vopěnka scheme., 2016. www arχiv bibtex Perlmutter, Norman. The large cardinals between supercompact and almost-huge., 2010. arχiv bibtex Bagaria, Joan and Casacuberta, Carles and Mathias, A R D and Rosický, Jiří. Definable orthogonality classes in accessible categories are small.Journal of the European Mathematical Society 17(3):549--589. arχiv bibtex Makowsky, Johann. Vopěnka's Principle and Compact Logics.J Symbol Logic www bibtex Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Bagaria, Joan and Gitman, Victoria and Schindler, Ralf. Generic {V}opěnka's {P}rinciple, remarkable cardinals, and the weak {P}roper {F}orcing {A}xiom.Arch Math Logic 56(1-2):1--20, 2017. www DOI MR bibtex Gitman, Victoria and Hamkins, Joel David. A model of the generic Vopěnka principle in which the ordinals are not Mahlo., 2018. arχiv bibtex
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My problem Deflection needs a solution and rating.
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@Pranav Arora , @jatin yadav , @Sudeep Salgia , @Maharnab Mitra , @David Mattingly , @Anish Puthuraya Care to read this post...
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I have posted the solution. Kindly, check it.
It is quite an easy question . I want to write a solution but I don't know how to upload a diagram can you please tell me how to upload a diagram.Then I will put up the solution
! [alt text] (the link)...delete the spaces..
The link of what "an uploaded image on facebook" or what
@Ronak Agarwal – upload your diagram on imgur or any other similar website. Then copy the link of that image from that website onto here.
@Anish Puthuraya – Ok
@Ronak Agarwal – Can you please give an example of an image upload
You can use these sites: postimage.org, imgur.com.
I have posted the solution.You can check it out
Also try this challenging question Challenges in mechanics by Ronak Agarwal(Part1)
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In this question it is shown that being able to compute reciprocals (together with sums and differences) is enough to do do multiplication in a field of characteristic $\ne 2$. That made me wonder:
Can we formulate a set of field axioms that are based on the reciprocal function rather than a multiplication operation? Something like
A
fieldis an abelian group $G$ (written additively) together with a designated element $1\ne 0$ and an involution $x\mapsto\frac{1}{x}$ of $G\setminus\{0\}$ such that $\frac 11=1$ ???
Obviously we could derive a sufficient set of axioms by translating the usual field axioms using the algorithm from the earlier question, but that would end up being very ugly (not least because there are several cases that one must handle specially in order not to divide by zero). Is there a nicer way to characterize involutions that arise as the reciprocal of some field?
Partial answers (for example, ones that work only for characteristic 0) would also be interesting.
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Suppose $f$ is continuous on $[a, b]$, if for every continuous function $g$ on $[a, b]$ with $g(a) = g(b) = 0, \int_{a}^{b}f(x)g(x) dx = 0$, Show $f(x) = 0, \forall x \in [a, b]$,
I want to prove by contradiction, and then find a continuous $g$ such that $g(a) = g(b) = 0$ but $\int_{a}^{b}f(x)g(x) dx \neq 0$
Proof: Suppose by contradiction that $f(x) > 0 $ for some $x_0 \in [a, b]$. Since $f$ is continuous, $\exists \delta$ such that $f(x) > 0, \forall x \in [x_0 - \delta, x_0 + \delta]$. Take $g(x) = \begin{cases} 0 & \text{ if } x \in (a, x_0 - \delta) \cup (x_0+\delta, b) \\ -(x - x_0 - \delta)(x - x_0 + \delta) & \text{ if } x \in (x_0 - \delta, x_0 + \delta) \end{cases}$
Now I want to show that since $g$ is continuous on $[a, b]$, then it is integrable. Thus $\int_{a}^{b} g(x) dx = sup L(g, p)$. However, since the lower sum is $> 0$, it follows that the supremum is also $> 0$. Therefore $\int_{a}^{b} > 0$. A contradiction. However, I do not know how to show that $L(g, p) > 0$
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Given a finite group $G$, and a non-identity representative $g$ in a conjugacy class of prime order $p$, I'm trying to show that some nontrivial irreducible character of $G$ must have $\chi(g) \neq 0$ and $\chi(1) \neq 0 \text{ mod } p$.
The suggestion was to show that the nonexistence of such a character would imply that $1/p$ is an algebraic integer.
So far, I've used the fact that $\chi(g) \cdot \frac{p}{\chi(1)} \in \overline{\mathbb{Z}}$ for any character. With some manipulation (and closure of algebraic integers over sums and products), I can get this into a form that allows me to sum over all irreducible characters $$p \sum \chi_i(g)^2 \in \overline{\mathbb{Z}}.$$ My hope was to then use the orthogonality relations on the columns of a character table, but doing so leaves a pesky factor of the order of the group, i.e. $|G|/p \in \overline{\mathbb{Z}}$ instead of $1/p$.
Am I going about this the wrong way? Is there something else I am overlooking?
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I am trying to determine the parameters for the Nelson Siegel Svensson model and am solving a Non-Linear Optimization problem to do this.
I am trying to solve:
$$ \min_\theta{\sum{(p_i - \hat p_i)^2}}. $$
where $p_i$ are the observed dirty prices of the bonds and $\hat p_i$ are the prices that have been calculated using the NSS parameters, $\theta$
I am using the procedure presented in this paper. But I've also read that the Optimization is highly sensitive to the input set of parameters ($\theta$) as mentioned in Page 2 of this paper. Hence, if I don't have data on these parameters how should I look to set the input. I am currently trying to do this for GBP Government Bonds, but am unable to find any published parameters. I was also unable to find how people circumvent this problem.
Currently, I am using the $\theta$ values presented here as I thought they may be similar for the GBP Government Bonds. However, the Optimization proves to be unsolvable.
This is part of the code that I am using in Python to solve the optimization problem.
func just returns the sum of the squared difference in prices (Objective function) and
params refer to $\theta$. These are the input
params I am currently using.
params = [3.15698855, -2.98240445, -3.37586632, -1.67713694, 0.88538977, 3.84324841] #Thetaoptimize.minimize(func, params, method='COBYLA', constraints = cons, options={'disp': True})
Thank You
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Firstly, in connexion with optics, Fermat's principle is always and approximation: it
defines an approximation, namely the first term in the WKB Approximation to the solution of either the quasi-time-harmonic Maxwell's equations or, in a more general setting, the Helmholtz equation. Here the WKB scale parameter is the wavelength, which is taken to be small compared to the features of a medium in question. As explained beautifully in CR Drost's Answer, the intuition behind this is that Fermat's principle defines paths of stationary phase around which diffraction effects are added to get the complete solution to the problem and is extremely widely applicable without approximation, i.e. Fermat's principle will give, without approximation, the first term in the relevant WKB expansion in all the situations discussed below. These include all special relativistic (flat spacetime) problems and static general relativistic ones too.
The extremized quantity, or the optical Lagrangian, is the
optical path length expressed as a phase difference between the ends of the ray path - in waves or radians, for example. A careful reading and mulling over CR Drost's answer clearly shows this fact. So it is not probably not helpful to think of it as least time principle since, as in Michael Siefert's answer this is can be problematic in relativity. In contrast, the phase field of a steady state optical excitation in a medium is a scalar field i.e. it transforms as such.
In
Special Relativity, one can see how the principle plays out in two different, relatively boosted inertial frames by looking at the steady-state optical field in question from the two different frames at the instant when the origins of their co-ordinate systems co-incide. Let's put one of the frames at rest relative to the material mediums that the field is established in. Fermat's principle plays out in the wonted way in this frame.
The relatively moving observer sees the medium in Lorentz-transformed co-ordinates. Maxwell's equations are still Lorentz covariant with the medium present, but the medium properties and the constitutive relationships transform radically. Intuitively you can see this is so; the Lorentz Fitzgerald contraction changes the medium's optical density anisotropically. In fact, if we have a simple, anisotropic medium in the rest frame with electric and magnetic constants $p_e$ and $p_m$ (One shuns epsilons and mus in these kinds of calculations to avoid confusion with Greek indices on tensors), the relatively moving observer sees an anisotropic
magnetoelectric constant such that:
$$\vec{D} = p_e\,\vec{E} + c^{-1} \vec{v}\times \vec{H}$$$$\vec{B} = p_m\,\vec{H} - c^{-1} \vec{v}\times \vec{E}$$
where, naturally, $\vec{v}$ is the relative velocity.
The upshot of all of this is that both observers calculate the same scalar phase field from their version of Maxwell equations and so a ray path is an extremal optical path length path in one frame if and only if it is an extremal path in the other. So we see that the Fermat principle gives us the same rays in both cases.
A quirk here is that, in the anisotropic medium as seen by the relatively moving observer, Snell’s law does not apply to rays at interfaces, although it does apply to wave vectors. Phase fronts are not needfully normal to the Poynting vectors. This is the same situation as in an anisotropic crystal. But Fermat’s principle still applies.
In
General Relativity we must be a little careful. The optical Fermat principle applies time-invariant mediums. Therefore, it cannot be applied (at least I am not aware of any extension) to nonstatic spacetime - or at least one without a timelike Killing field - with or without material mediums. This is because, in the first instance, Fermat's principle applies to time-harmonic electromagnetic fields, with pulses and the like being described by Fourier superpositions of time-harmonic solutions;
But for a static, curved spacetime, the situation is similar to the special relativistic one. Different observers see a medium’s material properties and constitutive differently, but such that they would all agree on the scalar phase field for a given steady state optical excitation of the medium, and they would all calculate the same ray paths from the Fermat principle.
In fact, an empty, medium-less curved spacetime has the constitutive relationships (Plebanski's constitutive equations, see J. Plebanski
Phys. Rev. 118 (1960), p1396:
$$D^i = -\epsilon_0\,\frac{\sqrt{-g}}{g_{0 0}}\,g^{i j}\,E_j + c^{-1} \varepsilon^{ij}_{\,\,k}\,\frac{g_{0j}}{g_{00}}H^k$$$$B^i = -\mu_0\,\frac{\sqrt{-g}}{g_{0 0}}\,g^{i j}\,H_j + c^{-1} \varepsilon^{ij}_{\,\,k}\,\frac{g_{0j}}{g_{00}}E^k$$
where we have summed over spatial indices $1,\,2,\,3$ only (note the Roman, rather than Greek indices) and the $\varepsilon$ is the
three dimensional, spatial Levi-Civita symbol. This observation is the starting point for the field of transformational optics: the use of metamaterial mediums to mimic propagation in the spatially curved part of static curved spacetime. These ideas show great promise for the realization of optical cloaking devices, for example: material mediums whose electric, magnetic and magnetooptic constants match the above propagate light, as does of course empty curved space, without scattering. Light can be bent around objects by such mediums without scattering and it's not too hard to see that the object to be cloaked can be hidden inside regions not reachable from an observer. See, for example:
Ulf Leonhardt and Thomas G. Philbin, "Transformation Optics and the Geometry of Light",
Prog. Opt. 53, pp69-152 (2009
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So first, represent your system as the following in your case:
\begin{align}\frac{\partial q_k}{\partial t} - q_k \nabla \cdot F_k(\boldsymbol{q}) &= 0 & \forall k \in \lbrace 1, 2 \rbrace\end{align}
where $\boldsymbol{q} = [q_1, q_2]^T = [f, g]^T$, $F_1(\boldsymbol{q}) = q_2 \hat{e}_1$, and $F_2(\boldsymbol{q}) = q_1 \hat{e}_1$. Then you just perform the normal Galerkin for each equation using a set of weight functions for that equation. Let us assume we have some space of functions defined for each function/equation, call them $\mathcal{V}_k$. Then we can proceed with the Galerkin projection like so:
\begin{align}\int_{\Omega} w_j \left(\frac{\partial q_k}{\partial t} - q_k \nabla \cdot F_k(\boldsymbol{q})\right) d\Omega &= 0 &\forall w_j \in \mathcal{V}_k \\%\int_{\Omega} w_j \frac{\partial q_k}{\partial t} d\Omega &= \int_{\Omega} w_j q_k \nabla \cdot F_k(\boldsymbol{q}) d\Omega &\forall w_j \in \mathcal{V}_k \\%&= \int_{\Omega} \nabla \cdot \left( w_j q_k F_k(\boldsymbol{q}) \right) - F_k(\boldsymbol{q}) \cdot \nabla \left( w_j q_k \right)d\Omega &\forall w_j \in \mathcal{V}_k \\%&= \int_{\Gamma} w_j q_k F_k(\boldsymbol{q}) \cdot \hat{n}\; d\Gamma - \int_{\Omega} F_k(\boldsymbol{q}) \cdot \nabla \left( w_j q_k \right)d\Omega &\forall w_j \in \mathcal{V}_k \\&= G_k(\boldsymbol{q}, w_j) &\forall w_j \in \mathcal{V}_k\end{align}
From here you can discretize things further with your approximate representation for the functions $q_k(x)$ using your space of functions $\mathcal{V}_k$ and come up with the equations you need in terms of any coefficients you are solving for.
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In my previous two posts I showed worked solutions to problems 2.5 and 11.7 in Bulmer’s
Principles of Statistics, both of which involve the characteristics of self-fertilizing hybrid sweet peas. It turns out that problem 11.8 also involves this same topic, so why not work it as well for completeness. The problem asks us to assume that we were unable to find an explicit solution for the maximum likelihood equation in problem 11.7 and to solve it by using the following iterative method:
\( \theta_{1} = \theta_{0} + \frac{S(\theta_{0})}{I(\theta_{0})} \)
where \( S(\theta_{0}) \) is the value of \( \frac{d \log L}{d\theta}\) evaluated at \( \theta_{0}\) and \( I(\theta_{0})\) is the value of \( -E(\frac{d^{2}\log L}{d\theta^{2}})\) evaluated at \( \theta_{0}\).
So we begin with \( \theta_{0}\) and the iterative method returns \( \theta_{1}\). Now we run the iterative method again starting with \( \theta_{1}\) and get \( \theta_{2}\):
\( \theta_{2} = \theta_{1} + \frac{S(\theta_{1})}{I(\theta_{1})} \)
We repeat this process until we converge upon a value. This is called the Newton-Raphson method. Naturally this is something we would like to have the computer do for us.
First, recall our formulas from problem 11.7:
\( \frac{d \log L}{d\theta} = \frac{1528}{2 + \theta} – \frac{223}{1 – \theta} + \frac{381}{\theta} \)
\( \frac{d^{2}\log L}{d \theta^{2}} = -\frac{1528}{(2 + \theta)^{2}} -\frac{223}{(1 – \theta)^{2}} -\frac{381}{\theta^{2}} \)
Let’s write functions for those in R:
# maximum likelihood score mls <- function(x) { 1528/(2 + x) - 223/(1 - x) + 381/x } # the information inf <- function(x) { -1528/((2 + x)^2) - 223/((1 - x)^2) - 381/(x^2) }
Now we can use those functions in another function that will run the iterative method starting at a trial value:
# newton-raphson using expected information matrix nr <- function(th) { prev <- th repeat { new <- prev + mls(prev)/-inf(prev) if(abs(prev - new)/abs(new) <0.0001) break prev <- new } new }
This function first takes its argument and names it "prev". Then it starts a repeating loop. The first thing the loop does it calculate the new value using the iterative formula. It then checks to see if the difference between the new and previous value - divided by the new value - is less than 0.0001. If it is, the loop breaks and the "new" value is printed to the console. If not, the loop repeats. Notice that each iteration is hopefully converging on a value. As it converges, the difference between the "prev" and "new" value will get smaller and smaller. So small that dividing the difference by the "new" value (or "prev" value for that matter) will begin to approach 0.
To run this function, we simply call it from the console. Let's start with a value of \( \theta_{0} = \frac{1}{4}\), as the problem suggests:
nr(1/4) [1] 0.7844304
There you go! We could make the function tell us a little more by outputting the iterative values and number of iterations. Here's a super quick and dirty way to do that:
# newton-raphson using expected information matrix nr <- function(th) { k <- 1 # number of iterations v <- c() # iterative values prev <- th repeat { new <- prev + mls(prev)/-inf(prev) v[k] <- new if(abs(prev - new)/abs(new) <0.0001) break prev <- new k <- k + 1 } print(new) # the value we converged on print(v) # the iterative values print(k) # number of iterations }
Now when we run the function we get this:
nr(1/4) [1] 0.7844304 [1] 0.5304977 0.8557780 0.8062570 0.7863259 0.7844441 0.7844304 [1] 6
We see it took 6 iterations to converge. And with that I think I've had my fill of heredity problems for a while.
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In the previous STT5100 course, last week, we’ve seen how to use monte carlo simulations. The idea is that we do observe in statistics a sample \{y_1,\cdots,y_n\}, and more generally, in econometrics \{(y_1,\mathbf{x}_1),\cdots,(y_n,\mathbf{x}_n)\}. But let’s get back to statistics (without covariates) to illustrate. We assume that observations y_i are realizations of an underlying random variable Y_i. We assume that Y_i are i.id. random variables, with (unkown) distribution F_{\theta}. Consider here some estimator \widehat{\theta} – which is just a function of our sample \widehat{\theta}=h(y_1,\cdots,y_n). So \widehat{\theta} is a real-valued number like . Then, in mathematical statistics, in order to derive properties of the estimator \widehat{\theta}, like a confidence interval, we must define \widehat{\theta}=h(Y_1,\cdots,Y_n), so that now, \widehat{\theta} is a real-valued random variable. What is puzzling for students, is that we use the same notation, and I have to agree, that’s not very clever. So now, \widehat{\theta} is .
There are two strategies here. In classical statistics, we use probability theorem, to derive properties of \widehat{\theta} (the random variable) : at least the first two moments, but if possible the distribution. An alternative is to go for computational statistics. We have only one sample, \{y_1,\cdots,y_n\}, and that’s a pity. But maybe we can create another one \{y_1^{(1)},\cdots,y_n^{(1)}\}, as realizations of F_{\theta}, and another one \{y_1^{(2)},\cdots,y_n^{(2)}\}, anoter one \{y_1^{(3)},\cdots,y_n^{(3)}\}, etc. From those counterfactuals, we can now get a collection of estimators, \widehat{\theta}^{(1)},\widehat{\theta}^{(2)}, \widehat{\theta}^{(3)}, etc. Instead of using mathematical tricks to calculate \mathbb{E}(\widehat{\theta}), compute \frac{1}{k}\sum_{s=1}^k\widehat{\theta}^{(s)}That’s what we’ve seen last friday.
I did also mention briefly that looking at densities is lovely, but not very useful to assess goodness of fit, to test for normality, for instance. In this post, I just wanted to illustrate this point. And actually, creating counterfactuals can we a good way to see it. Consider here the height of male students,
Davis=read.table( "http://socserv.socsci.mcmaster.ca/jfox/Books/Applied-Regression-2E/datasets/Davis.txt") Davis[12,c(2,3)]=Davis[12,c(3,2)] X=Davis$height[Davis$sex=="M"]
We can visualize its distribution (density and cumulative distribution)
u=seq(155,205,by=.5) par(mfrow=c(1,2)) hist(X,col=rgb(0,0,1,.3)) lines(density(X),col="blue",lwd=2) lines(u,dnorm(u,178,6.5),col="black") Xs=sort(X) n=length(X) p=(1:n)/(n+1) plot(Xs,p,type="s",col="blue") lines(u,pnorm(u,178,6.5),col="black")
Since it looks like a normal distribution, we can add the density a Gaussian distribution on the left, and the cdf on the right. Why not test it properly. To be a little bit more specific, I do not want to test if it’s a Gaussian distribution, but if it’s a \mathcal{N}(178,6.5^2). In order to see if this distribution is relevant, one can use monte carlo simulations to create conterfactuals
hist(X,col=rgb(0,0,1,.3)) lines(density(X),col="blue",lwd=2) Y=rnorm(n,178,6.5) hist(Y,col=rgb(1,0,0,.3)) lines(density(Y),col="red",lwd=2) Ys=sort(Y) plot(Xs,p,type="s",col="white",lwd=2,axes=FALSE,xlab="",ylab="",xlim=c(155,205)) polygon(c(Xs,rev(Ys)),c(p,rev(p)),col="yellow",border=NA) lines(Xs,p,type="s",col="blue",lwd=2) lines(Ys,p,type="s",col="red",lwd=2)
We can see on the left that it is hard to assess normality from the density (histogram and also kernel based density estimator). One can hardly think of a valid distance, between two densities. But if we look at graph on the right, we can compare the empirical distribution cumulative distribution \widehat{F} obtained from \{y_1,\cdots,y_n\} (the blue curve), and some conterfactual, \widehat{F}^{(s)} obtained from \{y_1^{(s)},\cdots,y_n^{(s)}\} generated from F_{\theta_0} – where \theta_0 is the value we want to test. As suggested above, we can compute the yellow area, as suggest in Cramer-von Mises test, or the Kolmogorov-Smirnov distance.
d=rep(NA,1e5) for(s in 1:1e5){ d[s]=ks.test(rnorm(n,178,6.5),"pnorm",178,6.5)$statistic } ds=density(d) plot(ds,xlab="",ylab="") dks=ks.test(X,"pnorm",178,6.5)$statistic id=which(ds$x>dks) polygon(c(ds$x[id],rev(ds$x[id])),c(ds$y[id],rep(0,length(id))),col=rgb(1,0,0,.4),border=NA) abline(v=dks,col="red")
If we draw 10,000 counterfactual samples, we can visualize the distribution (here the density) of the distance used a test statistic \widehat{d}^{(1)}, \widehat{d}^{(2)}, etc, and compare it with the one observe on our sample \widehat{d}. The proportion of samples where the test-statistics exceeded the one observed
mean(d>dks) [1] 0.78248
is the computational version of the p-value
ks.test(X,"pnorm",178,6.5) One-sample Kolmogorov-Smirnov test data: X D = 0.068182, p-value = 0.8079 alternative hypothesis: two-sided
I thought about all that a couple of days ago, since I got invited for a panel discussion on “coding”, and why “coding” helped me as professor. And this is precisely why I like coding : in statistics, either manipulate abstract objects, like random variables, or you actually use some lines of code to create counterfactuals, and generate fake samples, to quantify uncertainty. The later is interesting, because it helps to visualize complex quantifies. I do not claim that maths is useless, but coding is really nice, as a starting point, to understand what we talk about (which can be very usefull when there is a lot of confusion on notations).
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Soliton solutions for quasilinear Schrödinger equations involving supercritical exponent in $\mathbb R^N$
1.
Department of Mathematics, University of British Columbia, Vancouver, BC, Canada
$-\epsilon \Delta u+V(x)u-\epsilon k(\Delta(|u|^2))u=g(u), \quad u>0,x \in \mathbb R^N,$
where g has superlinear growth at infinity without any restriction from above on its growth. Mountain pass in a suitable Orlicz space is employed to establish this result. These equations contain strongly singular nonlinearities which include derivatives of the second order which make the situation more complicated. Such equations arise when one seeks for standing wave solutions for the corresponding quasilinear Schrödinger equations. Schrödinger equations of this type have been studied as models of several physical phenomena. The nonlinearity here corresponds to the superfluid film equation in plasma physics.
Mathematics Subject Classification:Primary: 35J10, 35J20; Secondary: 35J2. Citation:Abbas Moameni. Soliton solutions for quasilinear Schrödinger equations involving supercritical exponent in $\mathbb R^N$. Communications on Pure & Applied Analysis, 2008, 7 (1) : 89-105. doi: 10.3934/cpaa.2008.7.89
[1]
Christopher Grumiau, Marco Squassina, Christophe Troestler.
On the
Mountain-Pass algorithm for the quasi-linear Schrödinger equation.
[2]
Dorota Bors.
Application of Mountain Pass Theorem to superlinear equations with fractional Laplacian controlled by distributed parameters and boundary data.
[3]
Masahito Ohta.
Strong instability of standing waves for nonlinear Schrödinger equations with a partial confinement.
[4]
Xiaoyu Zeng.
Asymptotic properties of standing waves for mass subcritical nonlinear Schrödinger equations.
[5]
François Genoud, Charles A. Stuart.
Schrödinger equations with a spatially decaying nonlinearity: Existence and stability of standing waves.
[6]
Soohyun Bae, Jaeyoung Byeon.
Standing waves
of nonlinear Schrödinger equations with optimal conditions for
potential and nonlinearity.
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Daniele Cassani, João Marcos do Ó, Abbas Moameni.
Existence and concentration of solitary waves for a class of quasilinear Schrödinger equations.
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Jaeyoung Byeon, Louis Jeanjean.
Multi-peak standing waves for nonlinear Schrödinger equations with a general nonlinearity.
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Zaihui Gan, Jian Zhang.
Blow-up, global existence and standing waves for the magnetic nonlinear Schrödinger equations.
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Reika Fukuizumi.
Stability and instability of standing waves for the nonlinear Schrödinger equation with harmonic potential.
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François Genoud.
Existence and stability of high frequency standing waves for a nonlinear
Schrödinger equation.
[14]
Zhi Chen, Xianhua Tang, Ning Zhang, Jian Zhang.
Standing waves for Schrödinger-Poisson system with general nonlinearity.
[15]
Yi He, Gongbao Li.
Concentrating solitary waves for a class of singularly perturbed quasilinear Schrödinger equations with a general nonlinearity.
[16]
Jaeyoung Byeon, Ohsang Kwon, Yoshihito Oshita.
Standing wave concentrating on compact manifolds for nonlinear
Schrödinger equations.
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Fábio Natali, Ademir Pastor.
Stability properties of periodic standing waves for the
Klein-Gordon-Schrödinger system.
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Stability of standing waves for a nonlinear SchrÖdinger equation under an external magnetic field.
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Constructing the Tensor Product of Modules The Basic Idea
Today we talk tensor products. Specifically this post covers the construction of the tensor product between two modules over a ring. But before jumping in, I think now's a good time to ask, "What are tensor products good for?" Here's a simple example where such a question might arise:
Suppose you have a vector space $V$ over a field $F$. For concreteness, let's consider the case when $V$ is the set of all $2\times 2$ matrices with entries in $\mathbb{R}$ and let $F=\mathbb{R}$. In this case we know what "$F$-scalar multiplication" means: if $M\in V$ is a matrix and $c\in \mathbb{R}$, then the new matrix $cM$ makes perfect sense. But what if we want to multiply $M$ by
complex scalars too? How can we make sense of something like $(3+4i)M$? That's precisely what the tensor product is for! We need to create a set of elements of the form $$\text{(complex number) "times" (matrix)}$$ so that the mathematics still makes sense. With a little massaging, this set will turn out to be $\mathbb{C}\otimes_{\mathbb{R}}V$.
So in general, if $F$ is an arbitrary field and $V$ an $F$-vector space, the tensor product answers the question "How can I define scalar multiplication by some
larger field which contains $F$?" And of course this holds if we replace the word "field" by "ring" and consider the same scenario with modules.
Now this isn't the only thing tensor products are good for (
far from it!), but I think it's the most intuitive one since it is readily seen from the definition (which is given below). You can read about another motivation for the tensor product here. And for all my physics friends, this blurb by K. Conrad sheds light on the relationship between the physicist's tensor and the mathematician's tensor (see p. 46).
So with this motivation in mind,
let's go!
From English to Math
Let $R$ be a ring with 1 and let $M$ be a right $R$-module and $N$ a left $R$-module and suppose $A$ is any abelian group. Our goal is to create an abelian group $M\otimes_R N$, called the
tensor product of $M$ and $N$, such that if there is an $R$-balanced map $i\colon M\times N\to M\otimes_R N$ and any $R$-balanced map $\varphi\colon M\times N\to A$, then there is a unique abelian group homomorphism $\Phi\colon M \otimes_R N\to A$ such that $\varphi=\Phi\circ i$, i.e. so the diagram below commutes.
Notice that the statement above has the same flavor as the universal mapping property of free groups!
Definition: Let $X$ be a set. A group $F$ is said to be a free group on $X$ if there is a function $i\colon X\to F$ such that for any group $G$ and any set map $\varphi\colon X\to G$, there exists a unique group homomorphism $\Phi\colon F\to G$ such that the following diagram commutes: (i.e. $\varphi=\Phi\circ i$)
set map, so in particular we just want our's to be $R$-balanced:
: Let $R$ be a ring with 1. Let $M$ be a right $R$-module, $N$ a left $R$-module, and $A$ an abelian group. A map $\varphi:M\times N\to R$ is called $\mathbf{R}$-balanced if for all $m,m_1,m_2\in M$, all $n,n_1,n_2\in N$ and all $r\in R$,
$\varphi(m_1+m_2,n)=\varphi(m_1,n)+\varphi(m_2,n)$$\varphi(m,n_1+n_2)=\varphi(m,n_1)+\varphi(m,n_2)$$\varphi(mr,n)=\varphi(m,rn)$
By "replacing" F by a certain quotient group $F/H$! (We'll define $H$ precisely below.)
These observations give us a road map to construct the tensor product. And so we begin:
Step 1
Let $F$ be a free abelian group generated by $M\times N$ and let $A$ be an abelian group. Then by definition (of free groups), if $\varphi:M\times N\to A$ is any set map, and $M\times N \hookrightarrow F$ by inclusion, then there is a unique abelian group homomorphism $\Phi:F\to A$ so that the following diagram commutes.
Step 2
that the inclusion map $M\times N\hookrightarrow F$ is not $R$-balanced! To fix this, we must "modify" the target space $F$ by replacing it with the quotient $F/H$ where $H\leq F$ is the subgroup of $F$ generated by elements of the form
$(m_1+m_2,n)-(m_1,n)-(m_2,n)$ $(m,n_1+n_2)-(m,n_1)-(m,n_2)$ $(mr,n)-(m,rn)$ where $m_1,m_2,m\in M$, $n_1,n_2,n\in N$ and $r\in R$. Why elements of this form? Because if we define the map $i:M\times N\to F/H$ by $$i(m,n)=(m,n)+H,$$ we'll see that $i$ is indeed $R$-balanced! Let's check:
So, are we done now? Can we really just replace $F$ with $F/H$ and replace the inclusion map with the map $i$, and
still retain the existence of a unique homomorphism $\Phi:F/H\to A$? No! Of course not. $F/H$ is not a free group generated by $M\times N$, so the diagram below is bogus, right?
Not
totally. We haven't actually disturbed any structure!
How can we relate the pink and blue lines? We'd really like them to be the same. But we're in luck because they basically are!
Step 3
$H\subseteq \ker(f)$, that is as long as $f(h)=0$ for all $h\in H$. And notice that this condition, $f(H)=0$, forces $f$ to be $R$-balanced!
Let's check: Sooooo... homomorphisms $f:F\to A$ such that $H\subseteq \ker(f)$ are the same as $R$-balanced maps from $M\times N$ to $A$! (Technically, I should say homomorphisms $f$ restricted to $M\times N$.) In other words, we have
In conclusion, to say "abelian group homomorphisms from $F/H$ to $A$ are the same as (isomorphic to) $R$-balanced maps from $M\times N$ to $A$" is the simply the
hand-wavy way of saying Whenever $i:M\times N\to F$ is an $R$-balanced map and $\varphi:M\times N\to A$ is an $R$-balanced map where $A$ is an abelian group, there exists a unique abelian group homomorphism $\Phi:F/H\to A$ such that the following diagram commutes:
And this is just want we want! The last step is merely the final touch:
Step 4
the abelian quotient group $F/H$ to be the tensor product of $M$ and $N$,
whose elements are cosets,
where $m\otimes n$ for $m\in M$ and $n\in N$ is referred to as a
simple tensor. And there you have it! The tensor product, constructed.
Reference:
D. Dummit and R. Foote,
Abstract Algebra, 3rd ed., Wiley, 2004. (ch. 10.4)
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Mass, Weight and Density
Category : Railways
Mass, Weight and Density
Mass
The mass (m) of a body of matter is quantitative measure of its inertia i.e., its resistance to a change in the state of rest or motion of the body, when a force is applied.
Inertia is the property of a mass which resists change from its states of rest or motion. Volume (V) is defined as the amount of space occupied by a three-dimensional object as measured in cubic units. Weight
The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg. Since the weight is a force, its SI unit is the
Newton.
For an object in free fall, so that gravity is the only force acting on it,
Then the expression for weight follows from Newton's second law.
The value of g allows us to determine the net gravity force if it were in freefall and that net gravity force is the weight. Another approach is to consider "g" to be the measure of the intensity of
Ac gravity field in Newtons/kg at our location. We can view the weight as a measure of the mass in kg times the intensity of the gravity field, 9.8 Newton's/kg under standard conditions.
Density
Density (p) is defined as the mass of a substance per unit volume.
There are two kinds of density, "weight density" and "mass density". We will only use mass density and when we say: "density", its means "mass density".
The metric system was designed so that water will have a density of one gram per cubic centimeter or 1000 kilograms per cubic meter. Lead is about 10 times as dense as water and Styrofoam is about one tenth as dense as water.
Fluid Density
Mass per unit volume is defined as density. So density at a point of a fluid is represented as
\[\rho =\underset{\Delta V\to 0}{\mathop{\lim }}\,\frac{\Delta m}{\Delta V}=\frac{dm}{dV}\]
where m is the mass and v is the volume of the fluid. Density is a positive scalar quantity.
SI unit: \[kg/{{m}^{3}}\] Dimensions: \[[M{{L}^{-3}}{{T}^{0}}]\]
i.e.\[{{M}_{body}}={{M}_{sub}}\]then\[{{V}_{body}}={{V}_{sub}}\]and\[{{\rho }_{body}}={{\rho }_{sub}}.\]
But for a hollow body or body with air gaps
\[{{M}_{body}}={{M}_{sub}}\]and\[{{V}_{body}}>{{V}_{sub}}\]then\[{{\rho }_{body}}<{{\rho }_{sub}}\]
\[{{M}_{MIX}}={{m}_{1}}+{{m}_{2}}\]and\[{{V}_{MIX}}={{V}_{1}}+{{V}_{2}}=\]\[\frac{{{m}_{1}}}{{{\rho }_{1}}}+\frac{{{m}_{2}}}{{{\rho }_{2}}}\]
\[\therefore \] \[\rho \,\,mix=\frac{M\,\,mix}{V\,\,mix}=\frac{{{m}_{1}}+{{m}_{2}}}{\underset{\rho 1}{\mathop{\underline{{{m}_{1}}}}}\,+\underset{\rho 2}{\mathop{\underline{{{m}_{2}}}}}\,}\]
\[\rho \,\,mix=\frac{2{{\rho }_{1}}{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}}\](harmonic mean of individual densities)
\[{{V}_{MIX}}={{V}_{1}}+{{V}_{2}}\] and,
\[{{M}_{MIX}}={{m}_{1}}+{{m}_{2}}={{\rho }_{1}}{{V}_{1}}+{{\rho }_{2}}{{V}_{2}}\]
\[\therefore \] \[\rho \,\,mix=\frac{M\,\,mix}{V\,\,mix}=\frac{{{\rho }_{1}}{{V}_{1}}+{{\rho }_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\]
\[\rho \,\,mix=\frac{{{\rho }_{1}}+{{\rho }_{2}}}{2}\](arithmetic mean of individual densities)
Specific Weight or Weight Density (W)
It is defined as the ratio of the weight of the fluid to its volume or the weight acting per unit volume of the fluid.
Specific weight, \[W=\frac{Weight}{Volume}\]
\[W=\frac{mg}{V}=\left[ \frac{m}{V} \right]g=\rho g\]
S I Unit:\[N/{{m}^{3}}\] Dimensions:\[[M{{L}^{-2}}{{T}^{-2}}]\]
Specific weight of pure water at \[4{}^\circ C\] is \[9.81\,\,kN/{{m}^{3}}\]
Relative Density
It is defined as the ratio of the density of the given fluid to the density of pure water at \[4{}^\circ C\]
Density of given liquid
Relative density (R.D).\[=\frac{Density\,\,of\,\,given\,\,liquid}{Density\,\,of\,\,pure\,\,water\,\,at\,\,4{}^\circ C}\]
The density of water is maximum at \[4{}^\circ C\] and is equal to \[1.0\times {{10}^{3}}kg{{m}^{-3}}\]
Relative density or specific gravity is a
unitless and dimensionless positive scalar physical quantity.
Being a dimensionless/unitless quantity R.D. of a substance is same in SI and CGS system.
Specific Gravity
It is defined as the ratio of the specific weight of the given fluid to the specific weight of pure water at\[4{}^\circ C\],
Specific gravity
\[=\frac{Specific\,\,weight\,\,of\,\,given\,liquid}{Specific\,\,weight\,\,of\,\,pure\,\,water\,\,at\,\,4{}^\circ C(9.81kN/{{m}^{3}})}\]
\[=\frac{{{\rho }_{\gamma }}\times g}{{{\rho }_{w}}\times g}=\frac{{{\rho }_{\gamma }}}{{{\rho }_{w}}}\]=R.D. of the liquid
Thus specific gravity of a liquid is numerically equal to the relative density of that liquid and for calculation purposes they are used interchangeably.
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Difference between revisions of "State Feedback"
(→Chapter Summary)
(→Chapter Summary: updated LQR gain notation)
Line 111: Line 111:
<li><p>A ''linear quadratic regulator'' minimizes the cost function
<li><p>A ''linear quadratic regulator'' minimizes the cost function
<center><math>
<center><math>
−
\tilde J = \int_0^\infty \left(x^T
+
\tilde J = \int_0^\infty \left(x^T x + u^T u \right)\,dt.
</math></center>
</math></center>
The solution to the LQR problem is given by a linear control law of
The solution to the LQR problem is given by a linear control law of
the form
the form
<center><math>
<center><math>
−
\left. u = -
+
\left. u = -^{-1} B^T P x \right.
</math></center>
</math></center>
where <math>P \in R^{n \times n}</math> is a positive definite, symmetric
where <math>P \in R^{n \times n}</math> is a positive definite, symmetric
matrix that satisfies the equation
matrix that satisfies the equation
<center><math>
<center><math>
−
\left. P A + A^T P - P B
+
\left. P A + A^T P - P B ^{-1} B^T P + = 0. \right.
</math></center>
</math></center>
This equation is called the''algebraic
This equation is called the''algebraic
Revision as of 23:04, 18 May 2006
Prev: Linear Systems Chapter 6 - State Feedback Next: Output Feedback
This chapter describes how feedback can be used shape the local behavior of a system. The concept of reachability is introduced and used to investigate how to "design" the dynamics of a system through placement of its eigenvalues. In particular, it will be shown that under certain conditions it is possible to assign the system eigenvalues to arbitrary values by appropriate feedback of the system state.
Textbook Contents Lecture Materials Supplemental Information Summary Chapter Summary
This chapter describes how state feedback can be used to design the (closed loop) dynamics of the system:
A linear system with dynamics
is said to be
reachableif we can find an input defined on the interval that can steer the system from a given final point to a desired final point .
The
reachability matrixfor a linear system is given by
A linear system is reachability if and only if the reachability matrix is invertible (assuming a single intput/single output system). Systems that are not reachable have states that are constrained to have a fixed relationship with each other.
A linear system of the form
is said to be in
reachable canonical form. A system in this form is always reachable and has a characteristic polynomial given by
A reachable linear system can be transformed into reachable canonical form through the use of a coordinate transformation .
A state feedback law has the form
where is the reference value for the output. The closed loop dynamics for the system are given by
The stability of the system is determined by the stability of the matrix . The equilibrium point and steady state output (assuming the systems is stable) are given by
Choosing as
gives .
If a system is reachable, then there exists a feedback law of the form
the gives a closed loop system with an arbitrary characteristic polynomial. Hence the eigenvalues of a reachable linear system can be placed arbitrarily through the use of an appropriate feedback control law.
Integral feedbackcan be used to provide zero steady state error instead of careful calibration of the gain . An integral feedback controller has the form
where
is the integral error. The gains , and can be found by designing a stabilizing state feedback for the system dynamics augmented by the integrator dynamics.
A
linear quadratic regulatorminimizes the cost function
The solution to the LQR problem is given by a linear control law of the form
where is a positive definite, symmetric matrix that satisfies the equation
This equation is called the
algebraicRiccati equation and can be solved numerically.
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Link to arXiv Preprint.
Abstract
Batch normalization is often used in an attempt to stabilize and accelerate training in deep neural networks. In many cases it indeed decreases the number of parameter updates required to reduce the training error. However, it also reduces robustness to small input perturbations and noise by double-digit percentages, as we show on five standard datasets. Furthermore, substituting weight decay for batch norm is sufficient to nullify the relationship between adversarial vulnerability and the input dimension. Our work is consistent with a mean-field analysis that found that batch norm causes exploding gradients.
Batch Norm
We briefly review how batch norm modifies the hidden layers’ pre-activations $h$ of a neural network. We use the notation of (Yang, Pennington, Rao, Sohl-Dickstein, & Schoenholz, 2019), where $\alpha$ is the index for a neuron, $l$ for the layer, and $i$ for a mini-batch of $B$ samples from the dataset; $N_l$ denotes the number of neurons in layer $l$, $W^l$ is the matrix of weights and $b^l$ is the vector of biases that parametrize layer $l$. The batch mean is defined as $\mu_\alpha = \frac{1}{B} \sum_i h_{\alpha i}$, and the variance is $\sigma_\alpha^2 = \sqrt{\frac{1}{B} \sum_i {(h_{\alpha i} - \mu_\alpha)}^2 + c}$, where $c$ is a small constant to prevent division by zero. In the batch norm procedure, the mean $\mu_\alpha$ is subtracted from the pre-activation of each neuron $h_{\alpha i}^l$ – consistent with (Ioffe & Szegedy, 2015) –, the result is divided by the standard deviation $\sigma_\alpha$, then scaled and shifted by the learned parameters $\gamma_\alpha$ and $\beta_\alpha$, respectively. This is described in Eqs. \eqref{eq:bn1} and \eqref{eq:bn2}, where a per-unit nonlinearity $\phi$, e.g., ReLU, is applied after the normalization.
\begin{equation} \label{eq:bn1} h_{\alpha i}^l = \gamma_{\alpha} \frac{h_{\alpha i} - \mu_{\alpha}}{\sigma_{\alpha}} + \beta_{\alpha} \end{equation}
\begin{equation} \label{eq:bn2} h_i^l = W^l \phi (h_i^{l - 1}) + b^l \end{equation}
Note that this procedure fixes the first and second moments of all neurons $\alpha$equally. This suppresses the information contained in these moments.Batch norm induces a nonlocal batch-wise nonlinearity, such thattwo mini-batches that differ by only a
single example will havedifferent representations for each example (Yang, Pennington, Rao, Sohl-Dickstein, & Schoenholz, 2019).This difference is further amplified by stacking batch norm layers.We argue that this information loss and inability to maintain relativedistances in the input space reduces adversarial as well as general robustness.
We illustrate this effect by reproducing Fig. 6 of (Yang, Pennington, Rao, Sohl-Dickstein, & Schoenholz, 2019) in the above animation. Two mini-batches that contain the same data points except for one are shown at Layer 0, or input. We propagate the mini-batches through a deep batch-normalized linear network, i.e. with $\phi=id$, and of any practical width. The activations are then projected to their two principal components. This figure reminded me of the Adversarial Spheres dataset for binary classification of concentric spheres on the basis of their differing radii (Gilmer et al., 2018). It turns out that this simple task poses a challenge to the conventional wisdom that batch norm accelerates training and improves generalization; batch norm does the exact opposite in this case, prolonging training by $\approx 50 \times$, increasing sensitivity to the learning rate, and reduces robustness. This is likely why they used the Adam optimizer instead of plain SGD in the original work on Adversarial Spheres, and trained in an online manner for one million steps. The next illustration makes it clear why this is so.
Concentric circles and their representations in a deep linear network with batch norm at initialization. Mini-batch membership is indicated by marker fill and class membership by colour. Each layer is projected to its two principal components. Some samples overlap at Layer 2, and classes are mixed at Layer 14.
In the next visualization, we repeat the experiment of (Yang, Pennington, Rao, Sohl-Dickstein, & Schoenholz, 2019) by training fully-connected nets of depth $L$ and constant-width ReLU layers for ten epochs by SGD, and learning rate $\eta = 10^{-5} B$ for batch size $B$ on MNIST. The batch norm parameters $\gamma$ and $\beta$ were left as default, momentum was disabled, and $c = 10^{-3}$. Trials were averaged over three random seeds.
It turns out that one can predict the theoretical maximum depth solely as afunction of the batch size, due to an – almost paradoxically reliable –gradient explosion due to batch norm.The following function computes this, up to a correctionfactor since the general form works for
any dataset, learning rate, oroptimizer. The dashed line shows the theoretical maximum trainable depthin the context of our experiment.
# Thm. 3.10 Yang et al., (2019)def J_1(c): return (1 / np.pi) * (np.sqrt(1 - c**2) + (np.pi - np.arccos(c))*c)# first derivative of J_1def J_1_prime(c): return - (c / (np.pi * np.sqrt(1 - c**2)))lambda_G = (1 / (B - 3)) * ((B - 1 + J_1_prime(-1 / (B - 1))) / (1 - J_1(-1 / (B - 1))) - 1)print(16 * (1 / np.log(lambda_G))) # 16 is a correction factor
References A Mean Field Theory of Batch Normalization In International Conference on Learning Representations2019 Adversarial Spheres In International Conference on Learning Representations Workshop Track2018 Batch Normalization: Accelerating Deep Network Training by Reducing Internal Covariate Shift In International Conference on Machine Learning2015
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In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
That seems like what I need to do, but I don't know how to actually implement it... how wide of a time window is needed for the Y_{t+\tau}? And how on earth do I load all that data at once without it taking forever?
And is there a better or other way to see if shear strain does cause temperature increase, potentially delayed in time
Link to the question: Learning roadmap for picking up enough mathematical know-how in order to model "shape", "form" and "material properties"?Alternatively, where could I go in order to have such a question answered?
@tpg2114 For reducing data point for calculating time correlation, you can run two exactly the simulation in parallel separated by the time lag dt. Then there is no need to store all snapshot and spatial points.
@DavidZ I wasn't trying to justify it's existence here, just merely pointing out that because there were some numerics questions posted here, some people might think it okay to post more. I still think marking it as a duplicate is a good idea, then probably an historical lock on the others (maybe with a warning that questions like these belong on Comp Sci?)
The x axis is the index in the array -- so I have 200 time series
Each one is equally spaced, 1e-9 seconds apart
The black line is \frac{d T}{d t} and doesn't have an axis -- I don't care what the values are
The solid blue line is the abs(shear strain) and is valued on the right axis
The dashed blue line is the result from scipy.signal.correlate
And is valued on the left axis
So what I don't understand: 1) Why is the correlation value negative when they look pretty positively correlated to me? 2) Why is the result from the correlation function 400 time steps long? 3) How do I find the lead/lag between the signals? Wikipedia says the argmin or argmax of the result will tell me that, but I don't know how
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
Because I don't know how the result is indexed in time
Related:Why don't we just ban homework altogether?Banning homework: vote and documentationWe're having some more recent discussions on the homework tag. A month ago, there was a flurry of activity involving a tightening up of the policy. Unfortunately, I was really busy after th...
So, things we need to decide (but not necessarily today): (1) do we implement John Rennie's suggestion of having the mods not close homework questions for a month (2) do we reword the homework policy, and how (3) do we get rid of the tag
I think (1) would be a decent option if we had >5 3k+ voters online at any one time to do the small-time moderating. Between the HW being posted and (finally) being closed, there's usually some <1k poster who answers the question
It'd be better if we could do it quick enough that no answers get posted until the question is clarified to satisfy the current HW policy
For the SHO, our teacher told us to scale$$p\rightarrow \sqrt{m\omega\hbar} ~p$$$$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$And then define the following$$K_1=\frac 14 (p^2-q^2)$$$$K_2=\frac 14 (pq+qp)$$$$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$The first part is to show that$$Q \...
Okay. I guess we'll have to see what people say but my guess is the unclear part is what constitutes homework itself. We've had discussions where some people equate it to the level of the question and not the content, or where "where is my mistake in the math" is okay if it's advanced topics but not for mechanics
Part of my motivation for wanting to write a revised homework policy is to make explicit that any question asking "Where did I go wrong?" or "Is this the right equation to use?" (without further clarification) or "Any feedback would be appreciated" is not okay
@jinawee oh, that I don't think will happen.
In any case that would be an indication that homework is a meta tag, i.e. a tag that we shouldn't have.
So anyway, I think suggestions for things that need to be clarified -- what is homework and what is "conceptual." Ie. is it conceptual to be stuck when deriving the distribution of microstates cause somebody doesn't know what Stirling's Approximation is
Some have argued that is on topic even though there's nothing really physical about it just because it's 'graduate level'
Others would argue it's not on topic because it's not conceptual
How can one prove that$$ \operatorname{Tr} \log \cal{A} =\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr}e^{-s \mathcal{A}},$$for a sufficiently well-behaved operator $\cal{A}?$How (mathematically) rigorous is the expression?I'm looking at the $d=2$ Euclidean case, as discuss...
I've noticed that there is a remarkable difference between me in a selfie and me in the mirror. Left-right reversal might be part of it, but I wonder what is the r-e-a-l reason. Too bad the question got closed.
And what about selfies in the mirror? (I didn't try yet.)
@KyleKanos @jinawee @DavidZ @tpg2114 So my take is that we should probably do the "mods only 5th vote"-- I've already been doing that for a while, except for that occasional time when I just wipe the queue clean.
Additionally, what we can do instead is go through the closed questions and delete the homework ones as quickly as possible, as mods.
Or maybe that can be a second step.
If we can reduce visibility of HW, then the tag becomes less of a bone of contention
@jinawee I think if someone asks, "How do I do Jackson 11.26," it certainly should be marked as homework. But if someone asks, say, "How is source theory different from qft?" it certainly shouldn't be marked as Homework
@Dilaton because that's talking about the tag. And like I said, everyone has a different meaning for the tag, so we'll have to phase it out. There's no need for it if we are able to swiftly handle the main page closeable homework clutter.
@Dilaton also, have a look at the topvoted answers on both.
Afternoon folks. I tend to ask questions about perturbation methods and asymptotic expansions that arise in my work over on Math.SE, but most of those folks aren't too interested in these kinds of approximate questions. Would posts like this be on topic at Physics.SE? (my initial feeling is no because its really a math question, but I figured I'd ask anyway)
@DavidZ Ya I figured as much. Thanks for the typo catch. Do you know of any other place for questions like this? I spend a lot of time at math.SE and they're really mostly interested in either high-level pure math or recreational math (limits, series, integrals, etc). There doesn't seem to be a good place for the approximate and applied techniques I tend to rely on.
hm... I guess you could check at Computational Science. I wouldn't necessarily expect it to be on topic there either, since that's mostly numerical methods and stuff about scientific software, but it's worth looking into at least.
Or... to be honest, if you were to rephrase your question in a way that makes clear how it's about physics, it might actually be okay on this site. There's a fine line between math and theoretical physics sometimes.
MO is for research-level mathematics, not "how do I compute X"
user54412
@KevinDriscoll You could maybe reword to push that question in the direction of another site, but imo as worded it falls squarely in the domain of math.SE - it's just a shame they don't give that kind of question as much attention as, say, explaining why 7 is the only prime followed by a cube
@ChrisWhite As I understand it, KITP wants big names in the field who will promote crazy ideas with the intent of getting someone else to develop their idea into a reasonable solution (c.f., Hawking's recent paper)
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This note addresses the typical applied problem of estimating from data how a target “conversion rate” function varies with some available scalar score function — e.g., estimating conversion rates from some marketing campaign as a function of a targeting model score. The idea centers around estimating the integral of the rate function; differentiating this gives the rate function. The method is a variation on a standard technique for estimating pdfs via fits to empirical cdfs.
We review the math and code needed to fit a Gaussian Process (GP) regressor to data. We conclude with a demo of a popular application, fast function minimization through GP-guided search. The gif below illustrates this approach in action — the red points are samples from the hidden red curve. Using these samples, we attempt to leverage GPs to find the curve’s minimum as fast as possible.
Appendices contain quick reviews on (i) the GP regressor posterior derivation, (ii) SKLearn’s GP implementation, and (iii) GP classifiers.
We review binary logistic regression. In particular, we derive a) the equations needed to fit the algorithm via gradient descent, b) the maximum likelihood fit’s asymptotic coefficient covariance matrix, and c) expressions for model test point class membership probability confidence intervals. We also provide python code implementing a minimal “LogisticRegressionWithError” class whose “predict_proba” method returns prediction confidence intervals alongside its point estimates.
Our python code can be downloaded from our github page, here. Its use requires the jupyter, numpy, sklearn, and matplotlib packages.
Here, we briefly review a subtlety associated with machine-learning model selection: the fact that the optimal hyperparameters for a model can vary with training set size, $N.$ To illustrate this point, we derive expressions for the optimal strength for both $L_1$ and $L_2$ regularization in single-variable models. We find that the optimal $L_2$ approaches a finite constant as $N$ increases, but that the optimal $L_1$ decays exponentially fast with $N.$ Sensitive dependence on $N$ such as this should be carefully extrapolated out when optimizing mission-critical models.
Queries ping a certain computer server at random times, on average $\lambda$ arriving per second. The server can respond to one per second and those that can’t be serviced immediately are queued up. What is the average wait time per query? Clearly if $\lambda \ll 1$, the average wait time is zero. But if $\lambda > 1$, the queue grows indefinitely and the answer is infinity! Here, we give a simple derivation of the general result — (9) below.
A common task in applied statistics is the pairwise comparison of the responses of $N$ treatment groups in some statistical test — the goal being to decide which pairs exhibit differences that are statistically significant. Now, because there is one comparison being made for each pairing, a naive application of the Bonferroni correction analysis suggests that one should set the individual pairwise test sizes to $\alpha_i \to \alpha_f/{N \choose 2}$ in order to obtain a desired family-wise type 1 error rate of $\alpha_f$. Indeed, this solution is suggested by many texts. However, implicit in the Bonferroni analysis is the assumption that the comparisons being made are each mutually independent. This is not the case here, and we show that as a consequence the naive approach often returns type 1 error rates far from those desired. We provide adjusted formulas that allow for error-free Bonferroni-like corrections to be made.
[edit (7/4/2016): After posting this article, I’ve since found that the method we suggest here is related to / is a generalization of Tukey’s range test — see here.]
[edit (6/11/2018): I’ve added the notebook used below to our Github, here]
In this post, we review two facts about maximum-likelihood estimators: 1) They are consistent, meaning that they converge to the correct values given a large number of samples, $N$, and 2) They satisfy the Cramer-Rao lower bound for unbiased parameter estimates in this same limit — that is, they have the lowest possible variance of any unbiased estimator, in the $N\gg 1$ limit.
Here, we review parameter regularization, which is a method for improving regression models through the penalization of non-zero parameter estimates. Why is this effective? Biasing parameters towards zero will (of course!) unfavorably bias a model, but it will also reduce its variance. At times the latter effect can win out, resulting in a net reduction in generalization error. We also review Bayesian regressions — in effect, these generalize the regularization approach, biasing model parameters to any specified prior estimates, not necessarily zero.
Let $a_1, a_2, \ldots$ be an infinite set of non-negative samples taken from a distribution $P_0(a)$, and write
$$\tag{1} \label{problem} S = 1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \ldots. $$ Notice that if the $a_i$ were all the same, $S$ would be a regular geometric series, with value $S = \frac{1}{1-a}$. How will the introduction of $a_i$ randomness change this sum? Will $S$ necessarily converge? How is $S$ distributed? In this post, we discuss some simple techniques to answer these questions.
Note: This post covers work done in collaboration with my aged p, S. Landy.
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A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content.
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
triple poster
Code: Select all
#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
someone who likes this
Code: Select all
#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
someone who probably broke the record for the most posts in a row
Code: Select all
#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
someone who is still doing this
Code: Select all
#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
bidibangboom Posts: 34 Joined: May 10th, 2019, 6:38 pm
someone who will stop now
Code: Select all
#C A period 8 oscillator that was found in 1972.
#C http://www.conwaylife.com/wiki/index.php?title=Roteightor
x = 14, y = 14, rule = 23/3
bo12b$b3o8b2o$4bo7bob$3b2o5bobob$10b2o2b2$6b2o6b$5b2obo5b$6b3o5b$2b2o
3b3o4b$bobo5b2o3b$bo7bo4b$2o8b3ob$12bo!
fun
fluffykitty
Posts: 638 Joined: June 14th, 2014, 5:03 pm
Someone who posted 7 times in a row
I like making rules
A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact:
Someone who responded to someone who was wrong about breaking the record for most posts in a row (see here
).
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
Moosey Posts: 2479 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
A person who is being responded to by a person who is writing what this link
goes to
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
testitemqlstudop Posts: 1182 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
Somebody who used a self-reference instead of a previous-person-reference when the reference relations are relaxed.
A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact:
Somebody who I replied to.
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
testitemqlstudop Posts: 1182 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
Somebody who made the same aforementioned error that I attempted to correct by the second-previous post.
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
this post before reading the post itself.
Last edited by PkmnQ
on May 19th, 2019, 5:42 am, edited 1 time in total.
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who can't use the new paste rle feature of LifeViewer.
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who likes to size stack every now and then.
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who has posted 4 times in a row, including this one
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is going for 12 posts (5)
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is wondering why he did this (6)
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is already past the halfway mark (7)
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is running out of descriptions for himself (8)
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Alternating rule (9)
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who wants to have a profile picture (10)
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who is almost done (11)
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
PkmnQ
Posts: 665 Joined: September 24th, 2018, 6:35 am Location: Server antipode
Someone who has posted
12
times in a row and is now done
Code: Select all
x = 12, y = 12, rule = AnimatedPixelArt
4.P.qREqWE$4.2tL3vSvX$4.qREqREqREP$4.vS4vXvS2tQ$2.qWE2.qREqWEK$2.2vX
2.vXvSvXvStQtL$qWE2.qWE2.P.K$2vX2.2vX2.tQ2tLtQ$qWE4.qWE$2vX4.2vX$2.qW
EqWE$2.4vX!
i like loaf
testitemqlstudop Posts: 1182 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
PkmnQ wrote:this post before reading the post itself.
Someone who either doesn't have or chooses to ignore the rules of English
Saka
Posts: 3138 Joined: June 19th, 2015, 8:50 pm Location: In the kingdom of Sultan Hamengkubuwono X
A person that is trick work to.
Airy Clave White It Nay
Code: Select all
x = 17, y = 10, rule = B3/S23
b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5b
o2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo!
(Check gen 2)
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This is essentially an addition to the list of @4tnemele
I'd like to add some earlier work to this list, namely Discrete Gauge Theory.
Discrete gauge theory in 2+1 dimensions arises by breaking a gauge symmetry with gauge group $G$ to some lower
discrete subgroup $H$, via a Higgs mechanism. The force carriers ('photons') become massive which makes the gauge force ultra-short ranged. However, as the gauge group is not completely broken we still have the the Aharanov-Bohm effect. If H is Abelian this AB effect is essentially a 'topological force'. It gives rise to a phase change when one particle loops around another particle. This is the idea of fractional statistics of Abelian anyons.
The particle types that we can construct in such a theory (i.e. the one that are "color neutral") are completely determined by the residual, discrete gauge group $H$. To be more precise: a particle is said to be charged if it carries
a representation of the group H. The number of different particle types that carry a charge is then equal to the number of irreducible representations of the group H. This is similar to ordinary Yang-Mills theory where charged particles (quarks) carry the fundamental representation of the gauge group (SU(3). In a discrete gauge theory we can label all possible charged particle types using the representation theory of the discrete gauge group H.
But there are also other types of particles that can exist, namely those that carry flux. These flux carrying particles are also known as magnetic monopoles. In a discrete gauge theory the flux-carrying particles are labeled by the
conjugacy classes of the group H. Why conjugacy classes? Well, we can label flux-carrying particles by elements of the group H. A gauge transformation is performed through conjugacy, where $ |g_i\rangle \rightarrow |hg_ih^{-1}\rangle $ for all particle states $|g_i\rangle$ (suppressing the coordinate label). Since states related by gauge transformations are physically indistinguishable the only unique flux-carrying particles we have are labeled by conjugacy classes.
Is that all then? Nope. We can also have particles which carry both charge and flux -- these are known as dyons. They are labeled by both an irrep and a conjugacy class of the group $H$. But, for reasons which I wont go into, we cannot take all possible combinations of possible charges and fluxes.
(It has to do with the distinguishability of the particle types. Essentially, a dyon is labeled by $|\alpha, \Pi(g)\rangle$ where $\alpha$ is a conjugacy class and $\Pi(N(g))$ is a representation of the associated normalizer $N(\alpha)$ of the conjugacy class $\alpha$.)
The downside of this approach is the rather unequal setting of flux carrying particles (which are labeled by conjugacy classes), charged particles (labeled by representations) and dyons (flux+compatible charge). A unifying picture is provided by making use of the (quasitriangular) Hopf algebra $D(H)$ also known as a quantum double of the group $H$.
In this language
all particles are (irreducible) representations of the Hopf algebra $D(H)$. A Hopf Algebra is endowed with certain structures which have very physical counterparts. For instance, the existence of a tensor product allows for the existence of multiple particle configurations (each particle labeled by their own representation of the Hopf algebra). The co-multiplication then defines how the algebra acts on this tensored space. the existence of an antipode (which is a certain mapping from the algebra to itself) ensures the existence of an anti-particle. The existence of a unit labels the vacuum (=trivial particle).
We can also go beyond the structure of a Hopf algebra and include the notion of an R-matrix. In fact, the quasitriangular Hopf Algebra (i.e. the quantum double) does precisely this: add the R-matrix mapping. This R-matrix describes what happens when one particle loops around another particle (braiding). For non-Abelian groups $H$ this leads to non-Abelian statistics. These quasitriangular Hopf algebras are also known as quantum groups.
Nowadays the language of discrete gauge theory has been replaced by more general structures, referred to by topological field theories, anyon models or even modular tensor categories. The subject is huge, very rich, very physical and a lot of fun (if you're a bit nerdy ;)).
Sources:
http://arxiv.org/abs/hep-th/9511201 (discrete gauge theory)
http://www.theory.caltech.edu/people/preskill/ph229/ (lecture notes: check out chapter 9. Quite accessible!)
http://arxiv.org/abs/quant-ph/9707021 (a simple lattice model with anyons. There are definitely more accessible review articles of this model out there though.)
http://arxiv.org/abs/0707.1889 (review article, which includes potential physical realizations)This post imported from StackExchange Physics at 2015-11-01 19:23 (UTC), posted by SE-user Olaf
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This is essentially an addition to the list of @4tnemele
I'd like to add some earlier work to this list, namely Discrete Gauge Theory.
Discrete gauge theory in 2+1 dimensions arises by breaking a gauge symmetry with gauge group $G$ to some lower
discrete subgroup $H$, via a Higgs mechanism. The force carriers ('photons') become massive which makes the gauge force ultra-short ranged. However, as the gauge group is not completely broken we still have the the Aharanov-Bohm effect. If H is Abelian this AB effect is essentially a 'topological force'. It gives rise to a phase change when one particle loops around another particle. This is the idea of fractional statistics of Abelian anyons.
The particle types that we can construct in such a theory (i.e. the one that are "color neutral") are completely determined by the residual, discrete gauge group $H$. To be more precise: a particle is said to be charged if it carries
a representation of the group H. The number of different particle types that carry a charge is then equal to the number of irreducible representations of the group H. This is similar to ordinary Yang-Mills theory where charged particles (quarks) carry the fundamental representation of the gauge group (SU(3). In a discrete gauge theory we can label all possible charged particle types using the representation theory of the discrete gauge group H.
But there are also other types of particles that can exist, namely those that carry flux. These flux carrying particles are also known as magnetic monopoles. In a discrete gauge theory the flux-carrying particles are labeled by the
conjugacy classes of the group H. Why conjugacy classes? Well, we can label flux-carrying particles by elements of the group H. A gauge transformation is performed through conjugacy, where $ |g_i\rangle \rightarrow |hg_ih^{-1}\rangle $ for all particle states $|g_i\rangle$ (suppressing the coordinate label). Since states related by gauge transformations are physically indistinguishable the only unique flux-carrying particles we have are labeled by conjugacy classes.
Is that all then? Nope. We can also have particles which carry both charge and flux -- these are known as dyons. They are labeled by both an irrep and a conjugacy class of the group $H$. But, for reasons which I wont go into, we cannot take all possible combinations of possible charges and fluxes.
(It has to do with the distinguishability of the particle types. Essentially, a dyon is labeled by $|\alpha, \Pi(g)\rangle$ where $\alpha$ is a conjugacy class and $\Pi(N(g))$ is a representation of the associated normalizer $N(\alpha)$ of the conjugacy class $\alpha$.)
The downside of this approach is the rather unequal setting of flux carrying particles (which are labeled by conjugacy classes), charged particles (labeled by representations) and dyons (flux+compatible charge). A unifying picture is provided by making use of the (quasitriangular) Hopf algebra $D(H)$ also known as a quantum double of the group $H$.
In this language
all particles are (irreducible) representations of the Hopf algebra $D(H)$. A Hopf Algebra is endowed with certain structures which have very physical counterparts. For instance, the existence of a tensor product allows for the existence of multiple particle configurations (each particle labeled by their own representation of the Hopf algebra). The co-multiplication then defines how the algebra acts on this tensored space. the existence of an antipode (which is a certain mapping from the algebra to itself) ensures the existence of an anti-particle. The existence of a unit labels the vacuum (=trivial particle).
We can also go beyond the structure of a Hopf algebra and include the notion of an R-matrix. In fact, the quasitriangular Hopf Algebra (i.e. the quantum double) does precisely this: add the R-matrix mapping. This R-matrix describes what happens when one particle loops around another particle (braiding). For non-Abelian groups $H$ this leads to non-Abelian statistics. These quasitriangular Hopf algebras are also known as quantum groups.
Nowadays the language of discrete gauge theory has been replaced by more general structures, referred to by topological field theories, anyon models or even modular tensor categories. The subject is huge, very rich, very physical and a lot of fun (if you're a bit nerdy ;)).
Sources:
http://arxiv.org/abs/hep-th/9511201 (discrete gauge theory)
http://www.theory.caltech.edu/people/preskill/ph229/ (lecture notes: check out chapter 9. Quite accessible!)
http://arxiv.org/abs/quant-ph/9707021 (a simple lattice model with anyons. There are definitely more accessible review articles of this model out there though.)
http://arxiv.org/abs/0707.1889 (review article, which includes potential physical realizations)This post imported from StackExchange Physics at 2015-11-01 19:23 (UTC), posted by SE-user Olaf
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I'm trying to prove the non-existance of three positive integers $x,y,z$ with $x\geq z$ such that\begin{align} (x-z)^2+y^2 &\text{ is a perfect square,}\\ x^2+y^2 &\text{ is a perfect square,}\\ (x+z)^2+y^2 &\text{ is a perfect square.} \end{align} I failed tying to find such triple numerically. I tried to look at the differences, but that didn't give me much. Also, I tried the general solution of $a^2+b^2=c^2$, that is, $a=2pq$, $b=q^2-p^2$, and $c=q^2+p^2$ for some $p<q$, but this got ugly quite fast. I am not sure whether or not such $x,y,z$ exist. If it exists, one counterexample suffices to disprove my conjecture. Thanks in advance!
$$(1904-1040)^2+990^2=1314^2\\ 1904^2+990^2=2146^2\\ (1904+1040)^2+990^2=3106^2$$
Found using a computer search.
It is impossible that non-existance because there are infinitely many counterexamples. In fact, we must have by the Pythagorean triples $$x-z=t^2-s^2;\space y=2ts\qquad (*)$$ $$x=t_1^2-s_1^2; \space y=2t_1s_1$$ $$x+z=t_2^2-s_2^2;\space \space y=2t_2s_2\qquad (**)$$ so we have from $(*)$ and $(**)$ $$x=\frac{t^2-s^2+t_2^2-s_2^2}{2}=t_1^2-s_1^2\iff t^2+t_2^2-2t_1^2=s^2+s_2^2-2s_1^2$$ On the other hand the identity $$(m^2-n^2)^2+(2mn)^2=2(m^2+n^2)^2$$ shows infinitely many solutions to the equation $X^2+Y^2=2Z^2$.
Thus we can get an infinite set of possible $t_1,t_2,t_3$
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Let $X$ be an algebraic stack. In Perfect complexes on algebraic stacks (4.1) a
perfect complex $P \in \mathsf{D}_{\mathrm qc}(X)$ is defined to be a complex such that, for any smooth morphism $\operatorname{Spec}(A) \to X$ (where $A$ is a commutative ring), the complex $\mathbf{R}\Gamma(X,P_{|\operatorname{Spec}(A)})$ is a perfect complex of $A$-modules.
On the other hand, if $X$ is just a scheme, then a complex $P \in \mathsf{D}_{\mathrm qc}(X)$ is called perfect if it is quasi-isomorphic to a bounded complex of locally free sheaves of finite rank.
Question: if $X$ is a scheme, how can I prove that the two definitions agree? Is there a reference with a proof of this fact?
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The possibility of recursive self-improvement is often brought up as a reason to expect that an intelligence explosion is likely to result in a singleton - a single dominant agent controlling everything. Once a sufficiently general artificial intelligence can make improvements to itself, it begins to acquire a compounding advantage over rivals, because as it increases its own intelligence, it increases its ability to increase its own intelligence. If returns to intelligence are not substantially diminishing, this process could be quite rapid. It could also be difficult to detect in its early stages because it might not require a lot of exogenous inputs.
However, this argument only holds if self-improvement is not only a rapid route to AI progress, but the fastest route. If an AI participating in the broader economy could make advantageous trades to improve itself faster than a recursively self-improving AI could manage, then AI progress would be coupled to progress in the broader economy.
If algorithmic progress (and anything else that might seem more naturally a trade secret than a commodity component) is shared or openly licensed for a fee, then a cutting-edge AI can immediately be assembled whenever profitable, making a single winner unlikely. However, if leading projects keep their algorithmic progress secret, then the foremost project could at some time have a substantial intelligence advantage over its nearest rival. If an AI project attempting to maximize intelligence growth would devote most of its efforts towards such private improvements, then the underlying dynamic begins to resemble the recursive self-improvement scenario.
This post reviews a prior mathematization of the recursive self-improvement model of AI takeoff, and then generalizes it to the case where AIs can allocate their effort between direct self-improvement and trade.
Contents
1 A recalcitrance model of AI takeoff 2 Deciding between trade and self-improvement A recalcitrance model of AI takeoff
In
Superintelligence, Nick Bostrom describes a simple model of how fast an intelligent system can become more intelligent over time by working on itself. This exposition loosely follows the one in the book.
We can model the intelligence of the system as a scalar quantity , and the work, or optimization power, applied to system in order to make it more intelligent, as another quantity . Finally, at any given point in the process, it takes some amount of work to augment the system's intelligence by one unit. Call the marginal cost of intelligence in terms of work recalcitrance, , which may take different values at different points in the progress. So, at the beginning of the process, the rate at which the system's intelligence increases is determined by the equation .
We then add two refinements to this model. First, assume that intelligence is nothing but a type of optimization power, so and can be expressed in the same units. Second, if the intelligence of the system keeps increasing without limit, eventually the amount of work it will be able to put into things will far exceed that of the team working on it, so that . is now the marginal cost of intelligence in terms of applied intelligence, so we can write .
Constant recalcitrance
The simplest model assumes that recalcitrance is constant, . Then , or . This implies exponential growth.
Declining recalcitrance Superintelligence also considers a case where work put into the system yields increasing returns. Prior to takeoff, where , this would look like a fixed team of researchers with a constant budget working on a system that always takes the same interval of time to double in capacity. In this case we can model recalcitrance as , so that , so that for some constant , which implies that the rate of progress approaches infinity as approaches ; a singularity.
How plausible is this scenario? In a footnote, Bostrom brings up Moore's Law as an example of increasing returns to input, although (as he mentions) in practice it seems like increasing resources are being put into microchip development and manufacturing technology, so the case for increasing returns is far from clear-cut. Moore's law is predicted by the experience curve effect, or Wright's Law, where marginal costs decline as cumulative production increases; the experience curve effect produces exponentially declining costs under conditions of exponentially accelerating production.
1 This suggests that in fact accelerating progress is due to an increased amount of effort put into making improvements. Nagy et al. 2013 show that for a variety of industries with exponentially declining costs, it takes less time for production to double than for costs to halve.
Since declining costs also reflect broader technological progress outside the computing hardware industry, the case for declining recalcitrance as a function of input is ambiguous.
Increasing recalcitrance
In many cases where work is done to optimize a system, returns diminish as cumulative effort increases. We might imagine that high intelligence requires high complexity, and more intelligent systems require more intelligence to understand well enough to improve at all. If we model diminishing returns to intelligence as , then . In other words, progress is a linear function of time and there is no acceleration at all.
Generalized expression
The recalcitrance model can be restated as a more generalized self-improvement process with the functional form :
Increasing recalcitrance Constant progress Increasing recalcitrance Polynomial progress Constant recalcitrance Exponential progress 1" /> Declining recalcitrance Singularity Deciding between trade and self-improvement
Some inputs to an AI might be more efficiently obtained if the AI project participates in the broader economy, for the same reason that humans often trade instead of making everything for themselves. This section lays out a simple two-factor model of takeoff dynamics, where an AI project chooses how much to engage in trade.
Suppose that there are only two inputs into each AI: computational hardware available for purchase, and algorithmic software that the AI can best design for itself. Each AI project is working on a single AI running on a single hardware base. The intelligence of this AI depends both on hardware progress and software progress, and holding either constant, the other has diminishing returns. (This is broadly consistent with trends described by Grace 2013.) We can model this as .
At each moment in time, the AI can choose whether to allocate all its optimization power to making money in order to buy hardware, improving its own algorithms, or some linear combination of these. Let the share of optimization power devoted to algorithmic improvement be .
Assume further that hardware earned and improvements to software are both linear functions of the optimization power invested, so , and .
What is the intelligence-maximizing allocation of resources ?
This problem can be generalized to finding that maximizes for any monotonic function . This is maximized whenever is maximized. (Note that this is no longer limited to the case of diminishing returns.)
This generalization is identical to the Cobb-Douglas production function in economics. If then this model predicts exponential growth, if 1" /> it predicts a singularity, and if then it predicts polynomial growth. The intelligence-maximizing value of is .
2
In our initial toy model , where , that implies that no matter what the price of hardware, as long as it remains fixed and the indifference curves are shaped the same, the AI will always spend exactly half its optimizing power working for money to buy hardware, and half improving its own algorithms.
Changing economic conditions
The above model makes two simplifying assumptions: that the application of a given amount of intelligence always yields the same amount in wages, and that the price of hardware stays constant. This section relaxes these assumptions.
Increasing productivity of intelligence
We might expect the productivity of a given AI to increase as the economy expands (e.g. if it discovers a new drug, that drug is more valuable in a world with more or richer people to pay for it). We can add a term exponentially increasing over time to the amount of hardware the application of intelligence can buy: .
This does not change the intelligence-maximizing allocation of intelligence between trading for hardware and self-improving.
3 Declining hardware costs
We might also expect the long-run trend in the cost of computing hardware to continue. This can again be modeled as an exponential process over time, . The new expression for the growth of hardware is , identical in functional form to the expression representing wage growth, so again we can conclude that .
Maximizing profits rather than intelligence
AI projects might not reinvest all available resources in increasing the intelligence of the AI. They might want to return some of their revenue to investors if operated on a for-profit basis. (Or, if autonomous, they might invest in non-AI assets where the rate of return on those exceeded the rate of return on additional investments in intelligence.) On the other hand, they might borrow if additional money could be profitably invested in hardware for their AI.
If the profit-maximizing strategy involves less than 100% reinvestment, then whatever fraction of the AI's optimization power is reinvested should still follow the intelligence-maximizing allocation rule , where is now the share of reinvested optimization power devoted to algorithmic improvements.
If the profit-maximizing strategy involves a reinvestment rate of slightly greater than 100%, then at each moment the AI project will borrow some amount (net of interest expense on existing debts) , so that the total optimization power available is . Again, whatever fraction of the AI's optimization power is reinvested should still follow the intelligence-maximizing allocation rule , where is now the share of economically augmented optimization power devoted to algorithmic improvements.
This strategy is no longer feasible, however, once \frac{\alpha}{\alpha+\beta}" />. Since by assumption hardware can be bought but algorithmic improvements cannot, at this point additional monetary investments will shift the balance of investment towards hardware, while 100% of the AI's own work is dedicated to self-improvement.
References [ + ]
1. ↑ Moore's Law describes costs diminishing exponentially as a function of time : . Wright's law describes log costs diminishing as a linear function of log cumulative production : . If cumulative production increases exponentially over time, then Wright's law simplifies to Moore's law. 2. ↑
is maximized by maximizing
We can ignore the term here, since at any moment the value of materially affects only the rates of change and , not the constants or current endowments of hardware, software, or intelligence. Therefore we merely need to find the value of that maximizes :
, or
In the initial case we considered where , this constraint just means that the marginal product of intelligence applied to hardware purchases or software improvements must be the same, . There's no constraint on yet. In the generalized case, we also have scaling factors to account for differing marginal benefit curves for hardware and software.
To find the intelligence-maximizing value of we can consider a linear approximation to our initial set of equations:
Since initial endowments and are assumed to have been produced through the intelligence-maximizing process, we can substitute in the identity :
3. ↑ We again need to find such that . Using the new equation for , we get:
, or .
Thus, we should expect that the proportion of each AI's hardware endowment to its software endowment grows proportionally to the wage returns of intelligence.
To find the intelligence-maximizing value of we can again use a linear approximation, this time including the exponential growth of wages in our approximation of hardware growth:
Since initial endowments and are assumed to have been produced through the intelligence-maximizing process, we can apply the relation :
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I’m attempting to write a C program to gather values from a coupled spring system: There is a wall, connected to a mass $m_1$ by a spring, then this mass is connected to a second mass $m_2$ by another spring.
The values I require are the positions of the two masses and their velocities.
Using initial values for position and velocity I intend to calculate the movement of the two masses using the 4th-order Runge–Kutta method. I have successfully done this for positions, but I can’t figure out how to do it for the velocities. The differential equations for the velocities are:
$$ \begin{alignat}{2} \dot v_1 &= -\omega_1^2(x_1-R_1) &+ \omega_2^2(x_2-x_1-w-R_2)\\ \dot v_2 &= &- \omega_2^2(x_2-x_1-w-R_2) \end{alignat} $$
Where $R$ is the rest length of the springs, $x$ is the position, $w$ is the width. The equations are correct; that isn’t the problem.
The problem I have is that I just don’t know how to express the Runge–Kutta algorithm for the velocities when
both $x$ positions are changing, do I increase them both at the same time in the algorithm? Do I increase one and then take the average?
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I am doing some practice problems to prepare for my statistics exam, and I just want to know if my reasoning is correct on one problem, and if not, I want to know how I should reason through this. The question is as follows:
Let X and Y follow a bivariate normal distribution with means (3, 2), variances(1, 4) and covariance c.
$\begin{bmatrix}X\\Y\end{bmatrix} = N(\begin{bmatrix}3\\2\end{bmatrix}, \begin{bmatrix}1&c\\c&4\end{bmatrix})$
What is the Marginal Distribution of X?
My initial reasoning is that one method I could go about it is the calculation of the joint PDF and integrating it, but that doesn't sound very clean. A cleaner method could possibly be that since the joint is determined by two normal distributions, then $f(X,Y) = f(X)f(Y)$, and $f_x(X)$ (the marginal distribution of X) equals $\int_{-\infty}^{\infty}f(X)f(Y)dY = f(X)\int_{-\infty}^{\infty}f(Y)dY = f(X)$ since $f(Y)$ is a normal distribution, with the final answer being $f(X) = N(\mu_1, \sigma_1^2) = N(3,1)$. This seems to make sense but seems a little bit too simplified. Could anyone give me insight on how to tackle this problem, and if I am wrong anywhere, point it out? Thank you.
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Given a 4-vector $p^\mu$ the Lorentz group acts on it in the vector representation:$$ \tag{1} p^\mu \longrightarrow (J_V[\Lambda])^\mu_{\,\,\nu} p^\nu\equiv \Lambda^\mu_{\,\,\nu} p^\nu. $$However, I can always represent a 4-vector $p^\mu$ using left and right handed spinor indices, writing$$ \tag{2} p_{\alpha \dot{\alpha}} \equiv \sigma^\mu_{\alpha \dot{\alpha}} p_\mu.$$So the question is:
in what representation does the Lorentz group act on $p_{\alpha \dot{\alpha}}$?
There are
a lot of questions about this and related topics around physics.se, with a lot of excellent answers, so let me clear up more specifically what I am asking for.
I already know that the answer to this question is that the transformation law is $$ \tag{3} p_{\alpha \dot{\alpha}} \rightarrow (A p A^\dagger)_{\alpha\dot{\alpha}}$$with $A \in SL(2,\mathbb{C})$ (how is mentioned for example in this answer by Andrew McAddams).I also understand that$$ \tag{4} \mathfrak{so}(1,3) \cong \mathfrak{sl}(2,\mathbb{C}),$$(which is explained for example here by Edward Hughes, here by joshphysics, here by Qmechanic).
So what is missing? Not much really. Two things:
This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user glance
How do I obtain
(3) and what is the specific form of $A$, i.e. its relation with the vector representation $\Lambda^\mu_{\,\,\nu}$?Defining the following$$ (\tilde p) \equiv p^\mu, \qquad \Lambda \equiv \Lambda^\mu_{\,\,\nu},$$$$ \sigma \equiv \sigma_{\alpha \dot \alpha}, \qquad \hat p \equiv p_{\alpha \dot \alpha},$$we can rewrite (1) and (2) in matrix form as$$ \tag{5} \hat p \equiv \sigma \tilde p\rightarrow \sigma \Lambda \tilde p = ( \sigma \Lambda \sigma^{-1}) \hat p,$$however, this disagrees with , so what is wrong with my reasoning? (3) which I know to be right
Why does the transformation law
(3) has a form$$\tag{6} A \rightarrow U^{-1} A U,$$while the usual vector transformation (1) has a form $V \rightarrow \Lambda V$?I suspect this comes from a similar reason to that explained here by Prahar, but I would appreciate a confirmation about this.
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* [[Siril:Tutorial_sequence|Work on a sequence of converted images]]
* [[Siril:Tutorial_sequence|Work on a sequence of converted images]]
* [[Siril:Tutorial_preprocessing|Pre-processing images]]
* [[Siril:Tutorial_preprocessing|Pre-processing images]]
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* [[Siril:Tutorial_manual_registration|Registration (
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* [[Siril:Tutorial_manual_registration|Registration (alignment)]]
* → '''Stacking'''
* → '''Stacking'''
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The final step to do with Siril is to stack the images. Go to the "stacking" tab, indicate if you want to stack all images, only selected images or the best images regarding the value of FWHM previously computed. Siril proposes several algorithms for stacking computation.
The final step to do with Siril is to stack the images. Go to the "stacking" tab, indicate if you want to stack all images, only selected images or the best images regarding the value of FWHM previously computed. Siril proposes several algorithms for stacking computation.
* Sum Stacking
* Sum Stacking
−
This is the simplest algorithm: each pixel in the stack is summed using 32-bit precision, and the result is normalized to 16-bit. The increase in signal-to-noise ratio (SNR) is proportional to <math>\sqrt{N}</math>, where N is the number of images.
+
This is the simplest algorithm: each pixel in the stack is summed using 32-bit precision, and the result is normalized to 16-bit. The increase in signal-to-noise ratio (SNR) is proportional to <math>\sqrt{N}</math>, where Nis the number of images.
* Average Stacking With Rejection
* Average Stacking With Rejection
+
** Sigma Clipping: this is an iterative algorithm which will reject pixels whose distance from median will be farthest than two given values in sigma units (<math>\sigma_{low}</math>, <math>\sigma_{high}</math>).
** Sigma Clipping: this is an iterative algorithm which will reject pixels whose distance from median will be farthest than two given values in sigma units (<math>\sigma_{low}</math>, <math>\sigma_{high}</math>).
** Median Sigma Clipping: this is the same algorithm except than the rejected pixels are replaced by the median value of the stack.
** Median Sigma Clipping: this is the same algorithm except than the rejected pixels are replaced by the median value of the stack.
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This algorithm is mainly used to construct long exposure star-trails images.
This algorithm is mainly used to construct long exposure star-trails images.
Pixels of the image are replaced by pixels at the same coordinates if intensity is greater.
Pixels of the image are replaced by pixels at the same coordinates if intensity is greater.
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In the case of
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In the case of sequence, we used "Winsorized Sigma Clipping" in "Average stacking with rejection" to remove satellite tracks (<math>\sigma_{low}=4</math> and <math>\sigma_{high}=</math>).
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[[File:Siril stacking screen.png]]
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[[File:Siril stacking screen.png]]
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result.
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result .
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The images above picture the result in Siril using the
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The images above picture the result in Siril using the rendering mode. Note the improvement of the signal-to-noise ratio regarding the result given for one frame in the previous [[Siril:Tutorial_preprocessing|step]] (take a look to the sigma value). The increase in SNR is of <math>/.= .\approx \sqrt{} = .</math> and you should try to improve this result adjusting <math>\sigma_{low}</math> and <math>\sigma_{high}</math>.
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[[File:Siril_Comparison_sigma.png|700px]]
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Latest revision as of 10:34, 13 September 2016 Siril processing tutorial Convert your images in the FITS format Siril uses (image import) Work on a sequence of converted images Pre-processing images Registration (Global star alignment) → Stacking Stacking
The final step to do with Siril is to stack the images. Go to the "stacking" tab, indicate if you want to stack all images, only selected images or the best images regarding the value of FWHM previously computed. Siril proposes several algorithms for stacking computation.
Sum Stacking
This is the simplest algorithm: each pixel in the stack is summed using 32-bit precision, and the result is normalized to 16-bit. The increase in signal-to-noise ratio (SNR) is proportional to [math]\sqrt{N}[/math], where [math]N[/math] is the number of images. Because of the lack of normalisation, this method should only be used for planetary processing.
Average Stacking With Rejection Percentile Clipping: this is a one step rejection algorithm ideal for small sets of data (up to 6 images). Sigma Clipping: this is an iterative algorithm which will reject pixels whose distance from median will be farthest than two given values in sigma units ([math]\sigma_{low}[/math], [math]\sigma_{high}[/math]). Median Sigma Clipping: this is the same algorithm except than the rejected pixels are replaced by the median value of the stack. Winsorized Sigma Clipping: this is very similar to Sigma Clipping method but it uses an algorithm based on Huber's work [1] [2]. Linear Fit Clipping: this is an algorithm developed by Juan Conejero, main developer of PixInsight [2]. It fits the best straight line ([math]y=ax+b[/math]) of the pixel stack and rejects outliers. This algorithm performs very well with large stacks and images containing sky gradients with differing spatial distributions and orientations.
These algorithms are very efficient to remove satellite/plane tracks.
Median Stacking
This method is mostly used for dark/flat/offset stacking. The median value of the pixels in the stack is computed for each pixel. As this method should only be used for dark/flat/offset stacking, it does not take into account shifts computed during registration. The increase in SNR is proportional to [math]0.8\sqrt{N}[/math].
Pixel Maximum Stacking
This algorithm is mainly used to construct long exposure star-trails images. Pixels of the image are replaced by pixels at the same coordinates if intensity is greater.
Pixel Minimum Stacking
This algorithm is mainly used for cropping sequence by removing black borders. Pixels of the image are replaced by pixels at the same coordinates if intensity is lower.
In the case of NGC7635 sequence, we first used the "Winsorized Sigma Clipping" algorithm in "Average stacking with rejection" section, in order to remove satellite tracks ([math]\sigma_{low}=4[/math] and [math]\sigma_{high}=3[/math]).
The output console thus gives the following result:
14:33:06: Pixel rejection in channel #0: 0.181% - 1.184% 14:33:06: Pixel rejection in channel #1: 0.151% - 1.176% 14:33:06: Pixel rejection in channel #2: 0.111% - 1.118% 14:33:06: Integration of 12 images: 14:33:06: Pixel combination ......... average 14:33:06: Normalization ............. additive + scaling 14:33:06: Pixel rejection ........... Winsorized sigma clipping 14:33:06: Rejection parameters ...... low=4.000 high=3.000 14:33:07: Saving FITS: file NGC7635.fit, 3 layer(s), 4290x2856 pixels 14:33:07: Execution time: 9.98 s. 14:33:07: Background noise value (channel: #0): 9.538 (1.455e-04) 14:33:07: Background noise value (channel: #1): 5.839 (8.909e-05) 14:33:07: Background noise value (channel: #2): 5.552 (8.471e-05)
After that, the result is saved in the file named below the buttons, and is displayed in the grey and colour windows. You can adjust levels if you want to see it better, or use the different display mode. In our example the file is the stack result of all files, i.e., 12 files.
The images above picture the result in Siril using the Auto-Stretch rendering mode. Note the improvement of the signal-to-noise ratio regarding the result given for one frame in the previous step (take a look to the sigma value). The increase in SNR is of [math]21/5.1 = 4.11 \approx \sqrt{12} = 3.46[/math] and you should try to improve this result adjusting [math]\sigma_{low}[/math] and [math]\sigma_{high}[/math].
Now should start the process of the image with crop, background extraction (to remove gradient), and some other processes to enhance your image. To see processes available in Siril please visit this page. Here an example of what you can get with Siril:
Peter J. Huber and E. Ronchetti (2009), Robust Statistics, 2nd Ed., Wiley Juan Conejero, ImageIntegration, Pixinsight Tutorial
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I will set up a naive static/short run model to examine the case (so this post may be a bit long - I will try to dispense with some algebraic steps). I will use convenient functional forms, which are nevertheless consistent with usual assumptions.
FIRMS There are $i=1,...,n$ identical, price taking firms. In the short run they maximize the objective function
$$A\ln\ell_i - (1+s_f+\xi)w\ell_i\tag{1} $$
where $A$ includes any component of the production function that is fixed in the short run, $\ell_i$ is the amount of labor firm $i$ employs, $s_f$ is the Employer's Social Security Fees (SSF) as a
percentage over the "mixed" wage $w$. $\xi$ is a possible change in this percentage, which I include from the start.The concept of "mixed wage" is central in actual labor markets: in most cases bilateral or union negotiations over the wage are carried in terms of the "mixed wage", not in terms of the "take home" wage.
Profit maximizing behavior will lead to market labor demand
$$L^d= n\cdot\frac {A}{(1+s_f+\xi)w} \tag{2}$$
WORKERS There are $j=1,...,m$ workers, who posses one unit of labor and perform static maximization of the quasilinear utility function
$$U = c + \gamma \ln(1-\ell_j)\;\; s.t\;\; c= (1-s_w+\psi)w\ell_j \tag{3}$$
i.e. there is no consumption-saving decision here. $s_w$ is the "Employee's SSF" and $\psi$ is a possible change of this percentage (a positive $\psi$ implies
lowering of the percentage)Utility maximization leads to
$$L^s = m\cdot \frac{(1-s_w+\psi)w - \gamma}{(1-s_w+\psi)w} \tag{4}$$
Assuming that the labor markets clears, we have
$$L^d = L^s \implies n\cdot\frac {A}{(1+s_f+\xi)w} = m\cdot \frac{(1-s_w+\psi)w - \gamma}{(1-s_w+\psi)w}$$
$$\implies (nA/m)\frac {(1-s_w+\psi)}{(1+s_f+\xi)} = (1-s_w+\psi)w - \gamma$$
$$\implies w^* = \frac {(nA/m)}{(1+s_f+\xi)} + \frac {\gamma}{(1-s_w+\psi)} \tag{5}$$
Equation $(5)$ provides the first major conclusion :
If we increase "Employer's SSF" ($\xi >0$), the equilibrium mixed wage
will fall. But also, if we
decrease "Employee's SSF" ($\psi >0$),
the equilibrium mixed wage will also fall.
This is because the "take-home wage" will
increase for any given level of mixed wage, and so the labor supply curve will shift outwards in the $(w, L)$ space. Of course this result depends critically on labor-market clearing.
What will happen to individual worker's income?
Dividing $(2)$ by $m$ and using the equilibrium wage, equilibrium labor employed
per worker will be
$$\ell_j^* = \frac {nA/m}{(1+s_f+\xi)w^*}$$ and so equilibrium
take-home (disposable) labor income per worker will be
$$DI^*=(1-s_w+\psi)w^*\ell_j^* = (1-s_w+\psi)w^*\frac {nA/m}{(1+s_f+\xi)w^*} \tag{6}$$
$$\implies DI^* = \frac {1-s_w+\psi}{1+s_f+\xi }(nA/m) \tag{7}$$
Let's now start to implement's the adviser's idea. We start by the situation where $\xi=\psi = 0$. We want to determine $\xi$ and $\psi$ so that disposable income increases. This requires
$$DI^* \uparrow \implies \frac {1-s_w+\psi}{1+s_f+\xi } > \frac {1-s_w}{1+s_f}$$
$$\rightarrow DI^* \uparrow \implies \psi > \xi \frac {1-s_w}{1+s_f} \tag {8}$$
Since $(1-sw)/(1+s_f) <1 $ we conclude that
We do not need to decrease the Employee's SSF percentage as much as we
will increase the Employer's SSF percentage, in order to increase the
worker's disposable income. But the decrease should satisfy $(8)$.
But we want also to increase total Social Security Fees collected. Total Social Security Fees are
$$SSF^* = m\cdot\ell_j^*\cdot w^* \cdot (s_f+\xi + s_w - \psi) $$
$$\implies SSF^* = m\cdot \frac {nA/m}{(1+s_f+\xi)w^*} \cdot w^*\cdot (s_f+\xi + s_w - \psi) \tag{9}$$
$$\implies SSF^* = nA\cdot \frac {s_f+\xi + s_w - \psi}{(1+s_f+\xi)} $$
The condition to increase social security fees is
$$SSF^* \uparrow \implies \frac {s_f+\xi + s_w - \psi}{(1+s_f+\xi)} > \frac {s_f+s_w}{(1+s_f)}$$
$$\implies (1+s_f)(\xi-\psi) > \xi(s_f+s_w)$$
$$\implies \xi + s_f\xi - (1+s_f)\psi > s_f\xi + s_w\xi$$
$$\rightarrow SSF^* \uparrow \implies \psi < \xi \frac {1-s_w}{1+s_f} \tag{10}$$
But $(10)$ is the exact opposite condition than $(8)$. So:
There exist no combination of $\xi, \psi$ that will increase worker's
disposable income
and increase total social security collections.
In other words, the adviser's proposal is
infeasible.Of course, I do not claim that this result generalizes to all models -neither do I have at this point a clear view of what are the crucial assumptions on which this infeasibility results rests.
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Nonlinear Eigenvalue Problems for Even Functionals 2007 (English)In: Applicable Analysis: an international journal, ISSN 1563-504X (electronic) 0003-6811 (paper), Vol. 86, no 7, p. 829-849Article in journal (Refereed) Published Abstract [en]
Let $H$ be a Hilbert space and let $g\in C^1(H,\mathbb R)$ be an even Fréchet differentiable functional with completely continuous derivative. We introduce maximin values
$\sigma_k(t)$ which are critical values of $g$ restricted to the sphere
$$ S_t = \left\{ u\in H;\; \frac{1}{2} \|u\|^2 = t \right\}$$
and show that the functions $\sigma_k(t)$ have right and left derivatives and that $\sigma_{k\pm}'(t)$ are eigenvalues of g', i.e. there exist $u_k^{\pm}\in S_t$ such that
$$ g'(u_k^\pm) = \sigma_{k\pm}'(t) u_k^{\pm}.
Applications of the result are given to semilinear elliptic equations.
Place, publisher, year, edition, pages
2007. Vol. 86, no 7, p. 829-849
Keywords [en]
nonlinear eigenvalue problem, Krasnoselskii genus, elliptic semilinear equations.
National Category Mathematical Analysis IdentifiersURN: urn:nbn:se:su:diva-22159DOI: doi:10.1080/00036810701460503ISI: 000250334600004OAI: oai:DiVA.org:su-22159DiVA, id: diva2:188686 2007-12-192007-12-192011-01-11Bibliographically approved
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There's a 59.5125% chance of survival.
Naively, we might have thought there'd be a 55% chance of survival as 55% of the roll results are good. But the 20 is a slightly better result than the 1 is a bad one, so that pushes up the probability a bit. Let's see how.
The approach
The simplest way to tackle this is to look at the probabilities of surviving in exclusive ways, then combine those probabilities. Remember the rules of combining probabilities of multiple events:
If we want to know the probability of (A or B), when A and B don't overlap, we sum the probability of A with the probability of B. (We'll be constructing all of our scenarios below without overlap.) If we want to know the probability of (A and B), we multiply the probability of A with the probability of B. Notation
We'll use a number or a range of numbers in brackets to indicate the probability of that result on a d20. That is, \$[4]=\frac 1 {20}\$, \$[7]=\frac 1 {20}\$, and \$[10-20]=\frac {11} {20}\$. In this manner we'll generally be concerned with four different results: \$[1]\$ (\$\frac 1 {20}\$ chance), \$[2-9]\$ (\$\frac 8 {20}\$), \$[10-19]\$ (\$\frac {10} {20}\$), and \$[20]\$ (\$\frac 1 {20}\$ again).
When we want to indicate a particular sequence of rolls, we'll list them in order: \$[1][1-9]\$ would be read as "the probability of rolling a 1, then some single-digit number."
When we want to indicate a particular selection of rolls, but
don't care about their order * we'll group the unordered results in curly-brackets. Thus \$\{[2-9][10-19][2-9]\}[20]\$ would be read as "the probability of rolling two simple failures and a simple success in any order, followed by a 20." Ways to stabilize after...
The probabilities of stabilizing on the first, second, third, &c. rolls are exclusive: one can stabilize on the first
or on the second, but not both. So we'll determine each of these probabilities and sum them up according to our "or-rule" from above.
1 roll:
$$[20]=\frac 1 {20}$$
2 rolls:
$$[1-19][20]= \frac{19}{20} \times \frac 1 {20}= \frac{19}{400}$$
3 rolls:
$$\begin{align} [10-19][10-19][10-19] = \frac {10} {20} \times \frac{10}{20} \times \frac{10}{20} &= \frac 1 8 \\\text{or} \quad \{[1][10-19]\}[20] = \left\{2 \times \left(\frac 1 {20} \times \frac{10}{20}\right)\right\} \times \frac 1 {20} &= \frac 1 {400} \\\text{or} \quad [2-19][2-19][20] = \frac{18}{20} \times \frac{18}{20} \times \frac 1{20} &= \frac{81}{2000}\end{align}$$
$$\left(\text{total: } \frac 1 8 + \frac 1 {400} + \frac{81}{2000} = \frac{21}{125}\right)$$
4 rolls:
$$\begin{align} \{[10-19][10-19][1-9]\}[10-20]=\left\{3 \times \left( \frac{10}{20} \times \frac{10}{20} \times \frac 9 {20}\right)\right\} \times \frac{11}{20} &= \frac {297}{1600}\\\text{or} \quad \{[2-9][2-9][10-19]\}[20]=\left\{3 \times \left(\frac{8}{20} \times \frac{8}{20} \times \frac {10}{20}\right)\right\} \times \frac {1}{20} &= \frac{3}{250}\end{align}$$
$$\left(\text{total: } \frac {297}{1600} + \frac{3}{250} = \frac{1581}{8000}\right)$$
5 rolls:
$$\{[2-9][2-9][10-19][10-19]\}[10-20]=\left\{6 \times \left(\frac{8}{20} \times \frac{8}{20} \times \frac{10}{20} \times \frac{10}{20}\right)\right\} \times \frac{11}{20}=\frac{33}{250}$$
Summing it all up:
$$\frac 1 {20} + \frac{19}{400} + \frac{21}{125} + \frac{1581}{8000} + \frac{33}{250} = \frac{4761}{8000} = 0.595125$$
Bonus for reading this far: expected time to recovery
With these probabilities in hand it's easy to give an expected time-of-stabilization
for those that do stabilize:
$$t_\text{stable} = \dfrac{\left(\begin{align} 1 & \times \frac 1 {20} \\+2 & \times \frac{19}{400} \\+3 & \times \frac{21}{125} \\+4 & \times \frac{1581}{8000} \\+5 & \times \frac{33}{250}\end{align}\right)}{\frac{4761}{8000}} = 3.5278 \, \text{rounds, or about 21 sec}$$
* - Order is really important. Sometimes. For instance, if we want to know the probability of the first two rolls being a success and a failure, we have to count all the ways that we could get a success then a failure, and add on all the ways we could get a failure then a success.
But some of our rolls are indistinguishable events, and that's where it gets fun. If we want to count the ways of our first three rolls being two failures and a success we only have three orders we can consider (FFS, FSF, SFF), not the six one might expect (ABC, ACB, BAC, BCA, CAB, CBA).
The count of possible arrangements of n objects with multiplicities \$m_1, m_2, \dots\$ is \$N = \dfrac{n!}{m_1! \times m_2! \times \dots}\$. See if you can spot the three times this pops up!
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My problem is that I am about to write a longer mathematical text, and it will be filled with integrals. Integrals tend to be filled with pesky fractions, square roots and what not.
Personally I feel like LaTeX is spacing things "wrongly" I prefer to have more space in my fractions, and a tad more space after the square roots. Look at the comparison below. The difference is small, but noticable.
How it normally looks
How I prefer it to look
My question is that, I think doing these small fixes manually is bad. So my question is
Should I avoid doing it? I mean is it "wrong"?
And if not, is there a more automatic solution to this?
(Right now I am merely putting in some space after the roots. like
\, )
Here is a smaller MWE, I think the right side looks better than the left.
\documentclass[10pt,a4paper]{minimal}\usepackage{mathtools}\begin{document}\begin{align*}\int \frac{1+x^2}{1+x^4} \mathrm{d}x \qquad & \text{versus} \qquad \int \frac{\,1+x^2\,}{\,1+x^4\,}\, \mathrm{d}x\\\int_1^\infty \frac{\mathrm{d}x}{x\sqrt{-1+\sqrt[n]{x}}} \qquad & \text{versus} \qquad \int_1^\infty \frac{\mathrm{d}x}{\,x\sqrt{-1+\sqrt[n]{x}\,}\,} \end{align*}\end{document}
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Decomposition of infinite-to-one factor codes and uniqueness of relative equilibrium states
Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea
We show that an arbitrary factor map $\pi :X \to Y$ on an irreducible subshift of finite type is a composition of a finite-to-one factor code and a class degree one factor code. Using this structure theorem on infinite-to-one factor codes, we then prove that any equilibrium state $\nu $ on $Y$ for a potential function of sufficient regularity lifts to a unique measure of maximal relative entropy on $X$. This answers a question raised by Boyle and Petersen (for lifts of Markov measures) and generalizes the earlier known special case of finite-to-one factor codes.
Keywords:Class degree, relative equilibrium state, infinite-to-one factor code, relative thermodynamic formalism. Mathematics Subject Classification:Primary: 37B10; Secondary: 37D35, 37A35. Citation:Jisang Yoo. Decomposition of infinite-to-one factor codes and uniqueness of relative equilibrium states. Journal of Modern Dynamics, 2018, 13: 271-284. doi: 10.3934/jmd.2018021
References:
[1] [2] [3]
M. Allahbakhshi, S. Hong and U. Jung,
Structure of transition classes for factor codes on shifts of finite type,
[4] [5]
R. Bowen,
[6]
M. Boyle and K. Petersen, Hidden Markov processes in the context of symbolic dynamics, in
[7] [8] [9] [10] [11]
show all references
References:
[1] [2] [3]
M. Allahbakhshi, S. Hong and U. Jung,
Structure of transition classes for factor codes on shifts of finite type,
[4] [5]
R. Bowen,
[7] [8] [9] [10] [11]
[1] [2] [3] [4] [5]
Florian Rupp, Jürgen Scheurle.
Classification of a class of relative equilibria in three body coulomb systems.
[6] [7]
Alain Chenciner, Jacques Féjoz.
The flow of the equal-mass spatial 3-body problem in the neighborhood of the equilateral relative equilibrium.
[8] [9]
Manfred Denker, Yuri Kifer, Manuel Stadlbauer.
Thermodynamic formalism for random countable Markov shifts.
[10] [11] [12]
Manfred Denker, Yuri Kifer, Manuel Stadlbauer.
Corrigendum to: Thermodynamic formalism for random countable Markov shifts.
[13] [14]
Frederic Laurent-Polz, James Montaldi, Mark Roberts.
Point vortices on the sphere: Stability of symmetric relative equilibria.
[15] [16] [17] [18] [19]
Simon Scott.
Relative zeta determinants and the geometry of the determinant line bundle.
[20]
2018 Impact Factor: 0.295
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Question: How to classify/characterize the phase structure of (quantum) gauge theory?
Gauge Theory (say with a gauge group $G_g$) is a powerful quantum field theoretic(QFT) tool to describe many-body quantum nature (because QFT naturally serves for understanding many-body quantum problem with (quasi-)particle creation/annihilation).
Classification of gauge theory shall be something profound, in a sense that gauge fields (p-form $A_\mu$, $B_{\mu\nu}$, or connections of $G_g$-bundle etc) are just mediators propagating the interactions between matter fields (fermion $\psi$, boson $\phi$). Thus, effectively, we may "integrate out" or "smooth over" the matter fields, to obtain an effective gauge theory described purely by gauge fields ($A_\mu$, $B_{\mu\nu}$, etc).
Characterization of gauge theory should NOT simply rely on its gauge group $G_g$, due to "Gauge symmetry is not a symmetry". We should not classify (its distinct or the same phases) or characterize (its properties) ONLY by the gauge group $G_g$. What I have been taught is that some familiar terms to describe the phase structure of (quantum) gauge theories, are:
(1) confined or deconfined
(2) gapped or gapless
(3) Higgs phase
(4) Coulomb phase
(5) topological or not.
(6) weakly-coupling or strongly-coupling
sub-Question A.:
Is this list above (1)-(6) somehow enough to address the phase structure of gauge theory? What are other important properties people look for to classify/characterize the phase structure of gauge theory? Like entanglement? How?
(for example, in
2+1D gapped deconfined weak-coupling topological gauge theory with finite ground state degeneracy on the $\mathbb{T}^2$ torus describes anyons can be classified/characterized by braiding statistics $S$ matrix (mutual statistics) and $T$ (topological spin) matrix.)
sub-Question B.:
Are these properties (1)-(6) somehow related instead of independent to each other?
It seems to me that
confined of gauge fields implies that the matter fields are gapped? Such as 3+1D Non-Abelian Yang-Mills at IR low energy has confinement, then we have the Millennium prize's Yang–Mills(YM) existence and mass gap induced gapped mass $\Delta>0$ for the least massive particle, both(?) for the matter field or the gauge fields (glueball?). So confinement and gapped mass $\Delta>0$ are related for 3+1D YM theory. Intuitively, I thought confinement $\leftrightarrow$ gapped, deconfinement $\leftrightarrow$ gapless.
However, in 2+1D, condensed matter people study $Z_2$, U(1) spin-liquids, certain kind of 2+1D gauge theory, one may need to ask whether it is (1) confined or deconfined, (2) gapped or gapless, separate issues. So in 2+1D case,
the deconfined can be gapped? the confined can be gapless? Why is that? Should one uses Renormalization group(RG) argument by Polyakov? how 2+1D/3+1D RG flow differently affect this (1) confined or deconfined, (2) gapped or gapless, separate issues?
sub-Question C.: are there known mathematical objects to classify gauge theory?
perhaps, say This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Idear
other than/beyond the recently-more-familiar group cohomology: either topological group cohomology $H^{d+1}(BG_g,U(1))$ using classifying space $BG_g$ of $G_g$, or Borel group cohomology $\mathcal{H}^{d+1}(G_g,U(1))$ recently studied in SPT and topological gauge theory and Dijkgraaf-Witten?
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As far as I know, absorbing of the positive coefficient of $i\epsilon$ in a propagator seems to be a trivial operation without even the need of justification.
In Peskin page 286, he did this:$$k^0\rightarrow k^0(1+i\epsilon)$$$$(k^2-m^2)\rightarrow (k^2-m^2+i\epsilon)$$
In M. Srednicki's Quantum Field Theory, page 51,
The factor in large parentheses is equal to $E^2-\omega^2+i(E^2+\omega^2)\epsilon$, and
we can absorb the positive coefficient in to $\epsilon$ to get $E^2-\omega^2+i\epsilon$.
Why and does this kind of manipulation affect the final result of calculation?
Although $\frac{1}{k^2-m^2+i\epsilon k^2}-\frac{1}{k^2-m^2+i\epsilon}$ is infinitesimal, but the integration of such terms may lead to divergences, and this is my worry.
Also the presence of $k^0$ in the coefficient of $i\epsilon$ could potentially influence the poles of an integrand and consequently influence the validity of Wick Rotation.This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user LYg
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Introduction
In this post, I review the convergence proofs of gradient algorithms. Our main reference is: Leon Bottou, Online learning and stochastic approximations. I rewrite the proofs described in Bottou's paper but with more details about the points which are subtle to me. I tried to write the proofs as clear as possible so as to make them accessible to everyone.
Consider the classical (discrete-time) gradient descent algorithm: \begin{align*} \theta_n = \theta_{n-1} - \gamma_n \nabla_{\theta} f(\theta,\x) |_{\theta = \theta_{n-1}} \end{align*} where $f(\theta,\x)$ is the cost function for a fixed parameter $\theta$. $\x$ represents the full dataset consists of $N$ fixed observations. Therefore $f$ is a function of $\theta$ only - and we define $f: \bR \to \bR$ with argument $\theta$. In the practical setup, the fixed $\theta^*$ is unknown and our goal is to find $\theta^*$. We denote the current value of the parameter estimate with $\theta_{n}$. In this post, we will show that, under certain conditions for $f$, $\theta_n \to \theta^*$ eventually. If we evaluate $x_i$ instead of $\x$ in the gradient-descent update, the convergence result holds. The latter algorithm is called the stochastic gradient descent algorithm and although we won't give its proof, the machinery provided here is sufficient to prove the stochastic version.
Conditions for $f$
We have to assume certain conditions for $f$ to ensure the convergence. These conditions, which are collectively called "general convexity", are as follows.
(i) $f(\theta)$ has a single minimum $\theta^*$(ii) The following holds, \begin{align}\label{condition} \epsilon > 0, \,\,\,\,\,\, \inf_{(\theta - \theta^*)^2 > \epsilon} (\theta - \theta^*) \nabla_\theta f(\theta) > 0 \end{align}
Condition (ii) ensures that, gradient always points the minimum.
Continuous-time gradient descent
Here, we start with the continuous-time gradient descent proof, which is very easy but contains all main steps for the discrete-time proofs.
Consider a differential equation defines the parameter trajectory $\theta(t)$, \begin{align}\label{contGrad} \frac{\text{d}\theta}{\text{d}t} = - \nabla_\theta f(\theta) \end{align} It is more intuitive to think this equation as, \begin{align*} {\text{d}\theta} = - {\text{d}t}\nabla_\theta f(\theta) \end{align*} then we can see $dt$ as the step-size in the discrete-time gradient descent.
Discrete-time gradient descent
Now, we return to the classical discrete-time gradient descent: \begin{align}\label{DiscGradientDesc} \theta_n = \theta_{n-1} - \gamma_n \nabla_{\theta} f(\theta) |_{\theta = \theta_{n-1}} \end{align} Here now we have $\gamma_n$ as the step-size explicitly. The convergence proof relies on same three steps as in the continuous GD proof. However, now ensuring the convergence of a sequence requires more effort.
We note that proof sketch is the same with the Proposition 1. This time we have to define $h_n$ instead of $h(t)$ (emphasizing the time is discrete). In order to ensure that $h_n$ converges, we need a criterion for sequence convergence. To overcome this technical point, Bottou uses a technical lemma. Basically, that lemma proves that, if positive variations of a sequence converges, then it is sufficient to say that the sequence is convergent. Here, we review this lemma first. Then, we analyze the system defined in $\eqref{DiscGradientDesc}$.
Conclusion
The convergence proof stochastic version of the discrete-time gradient descent really mimics the arguments in the discrete-time proof. One has to define a positive stochastic process $h_n(\omega)$ instead of a positive sequence of numbers $h_n$. Then using probabilistic analog of the lemma we presented here, namely the quasimartingale convergence theorem, it is easy to prove the stochastic version by using conditional expectations. By using the machinery and extra details presented here, it is easy to understand the stochastic proof from the Bottou. So, we do not reproduce it here.
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Hi, Can someone provide me some self reading material for Condensed matter theory? I've done QFT previously for which I could happily read Peskin supplemented with David Tong. Can you please suggest some references along those lines? Thanks
@skullpatrol The second one was in my MSc and covered considerably less than my first and (I felt) didn't do it in any particularly great way, so distinctly average. The third was pretty decent - I liked the way he did things and was essentially a more mathematically detailed version of the first :)
2. A weird particle or state that is made of a superposition of a torus region with clockwise momentum and anticlockwise momentum, resulting in one that has no momentum along the major circumference of the torus but still nonzero momentum in directions that are not pointing along the torus
Same thought as you, however I think the major challenge of such simulator is the computational cost. GR calculations with its highly nonlinear nature, might be more costy than a computation of a protein.
However I can see some ways approaching it. Recall how Slereah was building some kind of spaceitme database, that could be the first step. Next, one might be looking for machine learning techniques to help on the simulation by using the classifications of spacetimes as machines are known to perform very well on sign problems as a recent paper has shown
Since GR equations are ultimately a system of 10 nonlinear PDEs, it might be possible the solution strategy has some relation with the class of spacetime that is under consideration, thus that might help heavily reduce the parameters need to consider to simulate them
I just mean this: The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
@ooolb Even if that is really possible (I always can talk about things in a non joking perspective), the issue is that 1) Unlike other people, I cannot incubate my dreams for a certain topic due to Mechanism 1 (consicous desires have reduced probability of appearing in dreams), and 2) For 6 years, my dream still yet to show any sign of revisiting the exact same idea, and there are no known instance of either sequel dreams nor recurrence dreams
@0celo7 I felt this aspect can be helped by machine learning. You can train a neural network with some PDEs of a known class with some known constraints, and let it figure out the best solution for some new PDE after say training it on 1000 different PDEs
Actually that makes me wonder, are the space of all coordinate choices more than all possible moves of Go?
enumaris: From what I understood from the dream, the warp drive showed here may be some variation of the alcuberrie metric with a global topology that has 4 holes in it whereas the original alcuberrie drive, if I recall, don't have holes
orbit stabilizer: h bar is my home chat, because this is the first SE chat I joined. Maths chat is the 2nd one I joined, followed by periodic table, biosphere, factory floor and many others
Btw, since gravity is nonlinear, do we expect if we have a region where spacetime is frame dragged in the clockwise direction being superimposed on a spacetime that is frame dragged in the anticlockwise direction will result in a spacetime with no frame drag? (one possible physical scenario that I can envision such can occur may be when two massive rotating objects with opposite angular velocity are on the course of merging)
Well. I'm a begginer in the study of General Relativity ok? My knowledge about the subject is based on books like Schutz, Hartle,Carroll and introductory papers. About quantum mechanics I have a poor knowledge yet.
So, what I meant about "Gravitational Double slit experiment" is: There's and gravitational analogue of the Double slit experiment, for gravitational waves?
@JackClerk the double slits experiment is just interference of two coherent sources, where we get the two sources from a single light beam using the two slits. But gravitational waves interact so weakly with matter that it's hard to see how we could screen a gravitational wave to get two coherent GW sources.
But if we could figure out a way to do it then yes GWs would interfere just like light wave.
Thank you @Secret and @JohnRennie . But for conclude the discussion, I want to put a "silly picture" here: Imagine a huge double slit plate in space close to a strong source of gravitational waves. Then like water waves, and light, we will see the pattern?
So, if the source (like a Black Hole binary) are sufficent away, then in the regions of destructive interference, space-time would have a flat geometry and then with we put a spherical object in this region the metric will become schwarzschild-like.
if**
Pardon, I just spend some naive-phylosophy time here with these discussions**
The situation was even more dire for Calculus and I managed!
This is a neat strategy I have found-revision becomes more bearable when I have The h Bar open on the side.
In all honesty, I actually prefer exam season! At all other times-as I have observed in this semester, at least-there is nothing exciting to do. This system of tortuous panic, followed by a reward is obviously very satisfying.
My opinion is that I need you kaumudi to decrease the probabilty of h bar having software system infrastructure conversations, which confuse me like hell and is why I take refugee in the maths chat a few weeks ago
(Not that I have questions to ask or anything; like I said, it is a little relieving to be with friends while I am panicked. I think it is possible to gauge how much of a social recluse I am from this, because I spend some of my free time hanging out with you lot, even though I am literally inside a hostel teeming with hundreds of my peers)
that's true. though back in high school ,regardless of code, our teacher taught us to always indent your code to allow easy reading and troubleshooting. We are also taught the 4 spacebar indentation convention
@JohnRennie I wish I can just tab because I am also lazy, but sometimes tab insert 4 spaces while other times it inserts 5-6 spaces, thus screwing up a block of if then conditions in my code, which is why I had no choice
I currently automate almost everything from job submission to data extraction, and later on, with the help of the machine learning group in my uni, we might be able to automate a GUI library search thingy
I can do all tasks related to my work without leaving the text editor (of course, such text editor is emacs). The only inconvenience is that some websites don't render in a optimal way (but most of the work-related ones do)
Hi to all. Does anyone know where I could write matlab code online(for free)? Apparently another one of my institutions great inspirations is to have a matlab-oriented computational physics course without having matlab on the universities pcs. Thanks.
@Kaumudi.H Hacky way: 1st thing is that $\psi\left(x, y, z, t\right) = \psi\left(x, y, t\right)$, so no propagation in $z$-direction. Now, in '$1$ unit' of time, it travels $\frac{\sqrt{3}}{2}$ units in the $y$-direction and $\frac{1}{2}$ units in the $x$-direction. Use this to form a triangle and you'll get the answer with simple trig :)
@Kaumudi.H Ah, it was okayish. It was mostly memory based. Each small question was of 10-15 marks. No idea what they expect me to write for questions like "Describe acoustic and optic phonons" for 15 marks!! I only wrote two small paragraphs...meh. I don't like this subject much :P (physical electronics). Hope to do better in the upcoming tests so that there isn't a huge effect on the gpa.
@Blue Ok, thanks. I found a way by connecting to the servers of the university( the program isn't installed on the pcs on the computer room, but if I connect to the server of the university- which means running remotely another environment, i found an older version of matlab). But thanks again.
@user685252 No; I am saying that it has no bearing on how good you actually are at the subject - it has no bearing on how good you are at applying knowledge; it doesn't test problem solving skills; it doesn't take into account that, if I'm sitting in the office having forgotten the difference between different types of matrix decomposition or something, I can just search the internet (or a textbook), so it doesn't say how good someone is at research in that subject;
it doesn't test how good you are at deriving anything - someone can write down a definition without any understanding, while someone who can derive it, but has forgotten it probably won't have time in an exam situation. In short, testing memory is not the same as testing understanding
If you really want to test someone's understanding, give them a few problems in that area that they've never seen before and give them a reasonable amount of time to do it, with access to textbooks etc.
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Let $X$ be an uncountable set. We define the cocountable topology $\tau$ as the set of all subsets $U\subseteq X$ such that either $U=\emptyset$ or $X\setminus U$ is countable.
I am interested in the conditions under which a set $E\subseteq X$ will be dense in $X$, and I honestly don't know where to start.
I have already shown that $(X,\tau)$ defines a topological space. I know that by definition, a set $E$ is dense in $X$ with respect to $\tau$ if for every $x\in X$ and every open set $U \subseteq X$ with $x\in U$, there exists an element $y\in E$ such that $y\in U$, or equivalently that $E$ is dense in $X$ with respect to $\tau$ if for every nonempty open set $U\subseteq X$, $E\cap U \neq \emptyset$.
I would like to see a complete proof if possible.
EDIT: I first want to show that: If $E\subseteq X$ is dense, then $X\setminus E$ is countable.
I attempted to prove this by the contrapositive (i.e. show that if $X\setminus E$ is uncountable, then $E$ is not dense in $X$). Suppose that $X\setminus E$ is uncountable. Then, $X \setminus E$ is open with respect to $\tau$, and so by definition $X\setminus (X\setminus E)=E$ is closed in $X$ with respect to $\tau$. Then, $E=\overline {E}$, where $\overline {E}$ denotes the closure of $E$. Since $E$ is countable and $X$ is uncountable, $E\neq X$. I am struggling to jump from here to the fact that $E$ is not dense in $X$.
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I'm having trouble with this problem in Ahlfors' Complex Analysis (page 238):
If a vertex of the polygon is allowed to be at $\infty$, what modification does the formula undergo? If in this context $\beta_k = 1$, what is the polygon like?
The formula he is referring to is (probably) $$F(w) = C \int_0^w \prod_{k=1}^n ( w - w_k)^{-\beta_k}dw + C'$$ which maps the unit disk coformally onto a polygon $\Omega$ with outer angles $\{\beta_k \pi\}$. Here $\{w_k\}$ are points on the unit circle, and $C,C'$ are complex constants.
Since $F(w)$ has no dependence on the vertices of $\Omega$ I can't see what modification he is talking about. Shouldn't the exact same formula hold in all cases?
Also, I believe that letting a vertex $z_k$ tend to $\infty$ should always force the corresponding outer angle $\beta_k \pi$ into $ \pi$ (that is $\beta_k \to 1$). Thus, I can't see how a polygon with an infinite vertex $z_k$ could have the corresponding $\beta_k$ other than 1.
Are my conclusions correct? If not, please help me figure this out.
Thanks!
P.S.
I do understand that having some $\beta_k=1$ means that the polygon admits two infinite parallel line segments (?).
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The Tensor Product, Demystified
Previously on the blog, we've discussed a recurring theme throughout mathematics: making new things from old things. Mathematicians do this all the time:
When you have two integers, you can find their greatest common divisor or least common multiple. When you have some sets, you can form their Cartesian product or their union. When you have two groups, you can construct their direct sum or their free product. When you have a topological space, you can look for a subspace or a quotient space. When you have some vector spaces, you can ask for their direct sum or their intersection. The list goes on!
Today, I'd like to focus on a particular way to build a new vector space from old vector spaces:
the tensor product. This construction often come across as scary and mysterious, but I hope to help shine a little light and dispel some of the fear. In particular, we won't talk about axioms, universal properties, or commuting diagrams. Instead, we'll take an elementary, concrete look:
Given two vectors $\mathbf{v}$ and $\mathbf{w}$, we can build a new vector, called the
tensor product $\mathbf{v}\otimes \mathbf{w}$. But what is that vector, really? Likewise, given two vector spaces $V$ and $W$, we can build a new vector space, also called their tensor product $V\otimes W$. But what is that vector space, really? Making new vectors from old
In this discussion, we'll assume $V$ and $W$ are finite dimensional vector spaces. That means we can think of $V$ as $\mathbb{R}^n$ and $W$ as $\mathbb{R}^m$ for some positive integers $n$ and $m$. So a vector $\mathbf{v}$ in $\mathbb{R}^n$ is really just a list of $n$ numbers, while a vector $\mathbf{w}$ in $\mathbb{R}^m$ is just a list of $m$ numbers.
Let's try to make new, third vector out of $\mathbf{v}$ and $\mathbf{w}$. But how? Here are two ideas: We can stack them on top of each other, or we can first multiply the numbers together and
then stack them on top of each other.
The first option gives a new list of $n+m$ numbers, while the second option gives a new list of $nm$ numbers. The first gives a way to build a new space where the dimensions
add; the second gives a way to build a new space where the dimensions multiply. The first is a vector $(\mathbf{v},\mathbf{w})$ in the direct sum $V\oplus W$ (this is the same as their direct product $V\times W$); the second is a vector $\mathbf{v}\otimes \mathbf{w}$ in the tensor product $V\otimes W$.
And that's it!
Forming the tensor product $\mathbf{v}\otimes \mathbf{w}$ of two vectors is
a lot like forming the Cartesian product of two sets $X\times Y$. In fact, that's exactly what we're doing if we think of $X$ as the set whose elements are the entries of $\mathbf{v}$ and similarly for $Y$.
So a tensor product is like a grown-up version of multiplication. It's what happens when you systematically multiply a bunch of numbers together, then organize the results into a list. It's multi-multiplication, if you will.
There's a little more to the story.
Does
every vector in $V\otimes W$ look like $\mathbf{v}\otimes\mathbf{w}$ for some $\mathbf{v}\in V$ and $\mathbf{w}\in W$? Not quite. Remember, a vector in a vector space can be written as a weighted sum of basis vectors, which are like the space's building blocks. This is another instance of making new things from existing ones: we get a new vector by taking a weighted sum of some special vectors!
So a typical vector in $V\otimes W$ is a weighted sum of basis vectors. W
hat are those basis vectors? Well, there must be exactly $nm$ of them, since the dimension of $V\otimes W$ is $nm$. Moreover, we'd expect them to be built up from the basis of $V$ and the basis of $W$. This brings us again to the "How can we construct new things from old things?" question. Asked explicitly: If we have $n$ bases $\mathbf{v}_1,\ldots,\mathbf{v}_n$ for $V$ and if we have $m$ bases $\mathbf{w}_1,\ldots,\mathbf{w}_m$ for $W$ then how can we combine them to get a new set of $nm$ vectors?
This is totally analogous to the construction we saw above: given a list of $n$ things and a list of $m$ things, we can obtain a list of $nm$ things by multiplying them all together. So we'll do the same thing here! We'll simply multiply the $\mathbf{v}_i$ together with the $\mathbf{w}_j$ in all possible combinations,
except "multiply $\mathbf{v}_i$ and $\mathbf{w}_j$ " now means "take the tensor product of $\mathbf{v}_i$ and $\mathbf{w}_j$."
Concretely, a basis for $V\otimes W$ is the set of all vectors of the form $\mathbf{v}_i\otimes\mathbf{w}_j$ where $i$ ranges from $1$ to $n$ and $j$ ranges from $1$ to $m$. As an example, suppose $n=3$ and $m=2$ as before. Then we can find the six basis vectors for $V\otimes W$ by forming a 'multiplication chart.' (The sophisticated way to say this is: "$V\otimes W$ is the free vector space on $A\times B$, where $A$ is a set of generators for $V$ and $B$ is a set of generators for $Y$.")
So $V\otimes W$ is the six-dimensional space with basis
$$\{\mathbf{v}_1\otimes\mathbf{w}_1,\;\mathbf{v}_1\otimes\mathbf{w}_2,\; \mathbf{v}_2\otimes\mathbf{w}_1,\;\mathbf{v}_2\otimes\mathbf{w}_2,\;\mathbf{v}_3\otimes\mathbf{w}_1,\;\mathbf{v}_3\otimes\mathbf{w}_2 \}$$
This might feel a little abstract with all the $\otimes$ symbols littered everywhere. But don't forget—we know exactly what each $\mathbf{v}_i\otimes\mathbf{w}_j$ looks like—it's just a list of numbers!
Which list of numbers? Well,
So what is $V\otimes W$? It's the vector space whose vectors are linear combinations of the $\mathbf{v}_i\otimes\mathbf{w}_j$. For example, here are a couple of vectors in this space:
Well, technically...
Technically, $\mathbf{v}\otimes\mathbf{w}$ is called the
outer product of $\mathbf{v}$ and $\mathbf{w}$ and is defined by $$\mathbf{v}\otimes\mathbf{w}:=\mathbf{v}\mathbf{w}^\top$$ where $\mathbf{w}^\top$ is the same as $\mathbf{w}$ but written as a row vector. (And if the entries of $\mathbf{w}$ are complex numbers, then we also replace each entry by its complex conjugate.) So technically the tensor product of vectors is matrix:
This may seem to be in conflict with what we did above, but it's not! The two go hand-in-hand. Any $m\times n$ matrix can be reshaped into a $nm\times 1$ column vector and vice versa. (So thus far, we've exploiting the fact that $\mathbb{R}^3\otimes\mathbb{R}^2$ is
isomorphic to $\mathbb{R}^6$.) You might refer to this as matrix-vector duality.
It's a little like a
process-state duality. On the one hand, a matrix $\mathbf{v}\otimes\mathbf{w}$ is a process—it's a concrete representation of a (linear) transformation. On the other hand, $\mathbf{v}\otimes\mathbf{w}$ is, abstractly speaking, a vector. And a vector is the mathematical gadget that physicists use to describe the state of a quantum system. So matrices encode processes; vectors encode states. The upshot is that a vector in a tensor product $V\otimes W$ can be viewed in either way simply by reshaping the numbers as a list or as a rectangle.
By the way, this idea of viewing a matrix as a process can easily be generalized to
higher dimensional arrays, too. These arrays are called tensors and whenever you do a bunch of these processes together, the resulting mega-process gives rise to a tensor network. But manipulating high-dimensional arrays of numbers can get very messy very quickly: there are lots of numbers that all have to be multiplied together. This is like multi-multi-multi-multi...plication. Fortunately, tensor networks come with lovely pictures that make these computations very simple. (It goes back to Roger Penrose's graphical calculus.) This is a conversation I'd like to have here, but it'll have to wait for another day! In quantum physics
One application of tensor products is related to the brief statement I made above: "A
vector is the mathematical gadget that physicists use to describe the state of a quantum system." To elaborate: if you have a little quantum particle, perhaps you’d like to know what it’s doing. Or what it’s capable of doing. Or the probability that it’ll be doing something. In essence, you're asking: What’s its status? What’s its state? The answer to this question— provided by a postulate of quantum mechanics—is given by a unit vector in a vector space. (Really, a Hilbert space, say $\mathbb{C}^n$.) That unit vector encodes information about that particle.
The dimension $n$ is, loosely speaking, the number of different things you could observe after making a measurement on the particle. But what if we have two little quantum particles? The state of that two-particle system can be described by something called a
density matrix $\rho$ on the tensor product of their respective spaces $\mathbb{C}^n\otimes\mathbb{C}^n$. A density matrix is a generalization of a unit vector—it accounts for interactions between the two particles.
The same story holds for $N$ particles—the state of an $N$-particle system can be described by a density matrix on an $N$-fold tensor product.
But why the tensor product? Why is it that this construction—out of all things—describes the interactions within a quantum system so well, so naturally? I don’t know the answer, but perhaps the appropriateness of tensor products shouldn't be too surprising. The tensor product itself captures all ways that basic things can "interact" with each other!
Of course, there's lots more to be said about tensor products. I've only shared a snippet of basic arithmetic. For a deeper look into the mathematics, I recommend reading through Jeremy Kun's wonderfully lucid How to Conquer Tensorphobia and Tensorphobia and the Outer Product. Enjoy!
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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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Yes. Let's improve the wording here: that is not just an expression, it's the definition of a dimensionless variable that eases the understanding of the solution of the transient diffusion is a semi-infinite medium. Because there is no clear characteristic length as in a finite diffusion path problem, Buckingham π theorem will tell us the two dimensionless quantities we need are$$\Pi_1 = \frac{z}{\sqrt{D t}}$$ (Or any function of it, like squaring it or dividing it by 4, such as your $\sigma$)
And$$\Pi_2 = \frac{y-y_{A\infty}}{y_{As} - y_{A\infty}}$$
With $s$ being the surface index and $\infty$ indicating a position "at infinity" (often we use $y_{A\infty} = 0$).
Then we conclude $\Pi_2 =\textrm{function}(\Pi_1)$. This particular function can found by solving Fick's second law with the appropriate boundary and initial conditions. In the most common case, it yields a Gauss error function.
Had we be initially interested in a sphere instead of a planar surface, the only major difference would be the use of spherical coordinates instead of a Cartesian frame, which would result in a different functional solution. But the problem of defining the dimensionless variables would be the same, given we assume perfect angular symmetry. The $z$ position would just be substituted by a radial position $r$.
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NOTE: This answer is community wiki: If you feel there is a need for
correction -please update the answer right here.
1. Frequency response: For any system - the (discrete time or continuous time) Fourier Transform of the impulse response is same as the frequency response of it. You can replace $s$ by $j\omega$ to work on this.
2. Stability: System is stable if, the bounded input produces bounded output. The definition of can be found here in the wiki link. For this, the necessary condition to prove is to see if the impulse response be absolutely integrable, i.e., its L^1 norm exist.
For a rational and continuous-time system, the condition for stability is that the region of convergence (ROC) of the Laplace transform includes the imaginary axis.
3. Causality: Any system who's impulse response at sample $s_i$ doesn't require to know any samples presented $s_{i+1}$ or after wards, than the system is causal.
EDIT: Now given the above i am trying to put the calculation as per your system.
$$ H(s)=\frac{s}{(s+1)(s+2)} $$
Now we can replace $s$ with $j\omega$ to get the frequency response in terms of magnitude and phase response.
$$ H(j\omega) = \frac{j\omega}{(j\omega+1)(j\omega+2)} $$$$ H(j\omega) = \frac{j\omega}{2 - \omega^2 + 3j\omega } $$$$ Magnitude = 20 Log_{10} \sqrt { ( H(j\omega) )^2 } $$$$ Magnitude = 20 Log_{10} \sqrt { \frac { (j\omega)^2 }{(2 - \omega^2)^2 + (3j\omega)^2} } $$
NOTE: square is applied separately on real terms than on imaginary terms! I got this understanding from this document here.
Replacing $-j^2$ by $-1$
$$ Magnitude = 20 Log_{10} \sqrt { \frac{-1\cdot\omega^2}{(2 - \omega^2)^2 - (3\omega)^2} } $$
$$ Magnitude = 20 Log_{10} \sqrt { \frac{-1\cdot\omega^2}{(4 - 5\omega^2 +\omega^4)} } $$
$$ Phase = arc tan (real / imaginary) $$
$$ Phase = -1 \cdot \tan^{-1}(\frac {-3\omega}{(2-\omega^2)} ) \ $$
Please correct me if i am wrong.
Another literature that can help you.
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I'm trying to find the roots of $x^3 -2$, I know that one of the roots are $\sqrt[3] 2$ and $\sqrt[3] {2}e^{\frac{2\pi}{3}i}$ but I don't why. The first one is easy to find, but the another two roots?
I need help
Thank you
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I'm trying to find the roots of $x^3 -2$, I know that one of the roots are $\sqrt[3] 2$ and $\sqrt[3] {2}e^{\frac{2\pi}{3}i}$ but I don't why. The first one is easy to find, but the another two roots?
I need help
Thank you
If $\omega^3 = 1$ and $x^3 = 2$ then $(\omega x)^3 = \omega^3 x^3 = 2$.
Possible values of $\omega$ are $e^{\frac{1}{3}2 i \pi}$, $e^{\frac{2}{3}2 i \pi}$ and $e^{\frac{3}{3}2 i \pi}$. This is because $1 = e^{2 i \pi} = (e^{\frac{1}{k} 2 i \pi})^k$.
So the solutions of $x^3 - 2 = 0$ are $e^{\frac{1}{3}2 i \pi} \sqrt[3]{2}$, $e^{\frac{2}{3}2 i \pi} \sqrt[3]{2}$ and $\sqrt[3]{2}$.
Hint: $x^3-2=(x-2^{\frac{1}{3}})(x^2+2^{\frac{1}{3}}x+2^{\frac{2}{3}})$.
Let $x=z\sqrt[3]2$. Then $0=x^3-2=2z^3-2$ and so $z^3=1$. Write $z=e^{i\theta}$ and solve for $\theta$.
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Suppose $x$ and $y$ are integers. Given $2xy+14y=-53-13x$, what does $xy$ equal? The answer is $-15$, but how do I get that? I feel like I should be able to find this.
A "general" way of attacking these types of problems is with a method that could be called "completing the rectangle", analogous to completing the square.
We start with $$2xy + 14y = -53 - 13x$$
Re-order the equation
$$2xy + 14y + 13x = -53$$
Attempt to factor the left hand side into a form that looks like $(ay + b)(cx+y)$
$$2xy + 14y + 13x + 91 - 91 = -53$$
$$(2y +13)(x+7) - 91 = -53$$ $$(2y +13)(x+7)= 38$$
Now we can use the prime factorization of $38 = 2\times19$ to attempt to find solutions to this equation.
One way to accomplish this would be to set $2y+13 = 19$ and $x+7 = 2$. This leaves us with $y = 3$, $x = -5$ and $xy = -15$.
However, another way would be to set $2y + 13 = -19$ and $x+7 = -2$. This leaves us with an equally valid solution of $y = -16$ and $x = -9$.
We can continue with $2y+13 = 1$ and $x+7 = 38$ and even $2y+13 = -1$ and $x+7 = -38$. Note that we cannot use $2y + 13 = \pm2$ or $2y + 13 = \pm38$ because the solution to these equations are non-integer $y$.
This technique works well because there are only finitely many possible solutions and they are each easy enough to check.
We have $\displaystyle 2y(x+7)=-(53+13x)\iff x=2y(x+7)+53+12x$ which is odd
Set $\displaystyle x=2z+1\implies y=-\frac{33+13z}{z+4}\iff y+13=\frac{19}{z+4}$
So, the necessary & sufficient condition for $y$ to be an integer is : $z+4$ divides $19$
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In Minkowski space even-dim (say $d+1$ D) spacetime dimension, we can write fermion-field theory as the Lagrangian:
$$\mathcal{L}=\bar{\psi} (i\not \partial-m)\psi+ \bar{\psi} \phi_1 \psi+\bar{\psi} (i \gamma^5)\phi_2 \psi$$
with Yukawa coupling to scalar term (with $\phi_1$) and pseudoscalar term (with $\phi_2$). Here $\bar{\psi}=\psi^\dagger\gamma^0$. This $\mathcal{L}$ is:
(i) Hermitian
(ii) invariant under Lorentz group SO($d,1$).
In the Euclidean fermion field theory, there are many attempts in the literature to find an analytic continuation from Minkowski to Euclidean field theory, and to satisfy the conditions of:
(i) Hermitian
(ii) invariant under rotational group SO($d+1$)
for example, in Steven Weinberg Vol II,$$\mathcal{L}_E= \psi^\dagger(i\not \partial_E-m)\psi$$
or this PRL paper,
$$\mathcal{L}_E= \psi^\dagger i \gamma^E_5(i\not \partial_E-m)\psi$$
in the case of this PRL paper, $\gamma^E_0=i \gamma_5$, $\gamma^E_5=-i \gamma_0$, and $\gamma^E_j=\gamma^E_j$ for $j=1,2,\dots$
The clifford algebra: ${\gamma_\mu,\gamma_\nu}=2(+,-,-,-,\dots)$, then$${\gamma^E_\mu,\gamma^E_\nu}=2(-,-,-,-,\dots)$$with $x_E=(t_E,x_1,x_2,x_3,\dots)$
I wonder in the Euclidean fermion field theory:
(Q1) Time Reversal and CPT: how does the Time Reversal $T$ work in Euclidean theory, say $T_E$? Is time reversal $T_E$ the same as parity operator $P$ then? Is $T_E$ unitary or anti-unitary? And whether there is CPT theorem still? If partity $P$ and time reversal $T_E$ operators are (almost) the same meaning in Euclidean space, then how CPT theorem in Euclidean field theory version becomes?
(Q2) spin and spin-statistics:
In Minkowski signature, although spin state does not change under parity $P$, but the spin state changes under time reversal $T$ - spin flips direction, (see, for example, Peskin and Schroeder QFT p.68), say in 4D (d+1=4),
$$\xi^s \to (-i \sigma_2)(\xi^s)^* \equiv \xi^{-s}$$
Flip spin twice and measuring the spin eigenvalue by $(\vec{n} \cdot \vec{\sigma})$,
$$(\vec{n} \cdot \vec{\sigma}) \xi^s = s \xi^s \to \\(\vec{n} \cdot \vec{\sigma}) (-i \sigma_2 )\big((-i \sigma_2 )(\xi^s)^*\big)^* = (\vec{n} \cdot \vec{\sigma}) (\mathbf{(-1)}\xi^s) = s (\mathbf{(-1)}\xi^s) $$
will give the
same eigenvalue of $s$, but the statefunction picks up a (-1) sign. This is related to spin-statistics theorem of fermions.
In Euclidean signature, whether there is a defined notion of spin? Does spin state change under parity $P$ and time reversal $T_E$? Is there any kind of spin-statistics theorem in Euclidean signature?
(Q3) pseudoscalar:
In Minkowski, we have a Hermitian pseudoscalar term $\mathcal{L}_{M,pseudo}=\bar{\psi} (i \gamma^5) \psi$,
with:
$$P \mathcal{L}_{M,pseudo} \psi P= -\mathcal{L}_{M,pseudo}$$
$$T \mathcal{L}_{M,pseudo} \psi T= -\mathcal{L}_{M,pseudo}$$
$$C \mathcal{L}_{M,pseudo} C= +\mathcal{L}_{M,pseudo}$$
In Euclidean space, is there a peudoscalar term $\mathcal{L}_{E,pseudo}$ in the version of this Euclidean theory: $\mathcal{L}_E=\psi^\dagger i \gamma^E_5(i\not \partial_E-m)\psi$, such that $\mathcal{L}_{E,pseudo}$ is Hermitian and SO(d+1) rotational invariance, and with
$$P \mathcal{L}_{E,pseudo} \psi P= -\mathcal{L}_{E,pseudo}$$
$$T_E \mathcal{L}_{E,pseudo} \psi T_E= -\mathcal{L}_{E,pseudo}$$
$$C \mathcal{L}_{E,pseudo} C= +\mathcal{L}_{E,pseudo}$$
what is $\mathcal{L}_{E,pseudo}=?$ in the case of
2D (d+1=2) and 4D (d+1=4)?
(Q4) mass gap and chirality:
In Minkowski theory, the mass gap is due to the term:
$$\mathcal{L}_{M,mass}=m \bar{\psi}\psi= m (\psi_L^\dagger \psi_R+\psi_R^\dagger \psi_L)$$
where $\psi_L,\psi_R$ interpreted as massless limit of Left-, Right-handed Weyl spinors.
In Euclidean theory, however, in Steven Weinberg QFT II's theory, the mass term is written as $m \psi^\dagger \psi$, does that mean
$$\mathcal{L}_{E,mass}=m \psi^\dagger \psi=m (\psi_L^\dagger \psi_L+\psi_R^\dagger \psi_R)$$
the mass gap is generated from a forward scattering term( $L\to L, R\to R$)? (This is weird!) Or does the projection $P_L, P_R$ and chirality is defined differently in Euclidean theory? How does this projection $P_L, P_R$ reflect the group structure of
$$SO(4) \simeq SU(2) \times SU(2)/Z_2 \text{ in the case of 4D} (d+1=4)?$$This post imported from StackExchange Physics at 2014-06-04 11:36 (UCT), posted by SE-user Idear
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This is a question that I asked in the mathematics section, but I believe it may get more attention here. I am working on a project dedicated to the quantisation of commuting matrix models. In the appropriate formalism this problem is reduced to a quantisation in a curved space -- the space of commuting matrices. The general prescription for quantisation in curved space involves ambiguity of the Hamiltonian operator proportional to the scalar curvature of the curved space - hence my question.
A set of $p$ commuting $n\times n$ hermitian matrices $X^{\mu}$ for $\mu=1,\dots p$, is parametrised in terms of a set of $p$ diagonal matrices $\Lambda^{\mu}$ and an unitary matrix $U$ via:
$X^{\mu}=U\,\Lambda^{\mu}\,U^{\dagger}~~$ for $~\mu=1\dots p$,
clearly not all degrees of U contribute to this parametrisation, for example a reparametrisation $U'= D\,U$, where $D$ is a diagonal unitary matrix would result in the same set of commuting matrices. In other words only the elements of the quotient space $U(n)\,/\,U(1)^n$, which is the maximal flag manifold $F_n$, contribute to the parametrisation. The metric on the resulting curved manifold can be calculated as a pull-back of the metric on the space of hermitian matrices defined as:
$ds_{X}^2=Tr\,\left( dX^{\mu}dX^{\mu}\right) $ ,
Using that $U^{\dagger}d X^{\mu} U=d\Lambda^{\mu}+[\theta,\Lambda^{\mu}]~~$, where $\theta$ is the Maurer-Cartan form $\theta=U^{\dagger}dU$, one can write the induced metric as:
$ds^2=\sum\limits_{i=1}^n(d\vec\lambda_i)^2+2\sum\limits_{i<j}(\vec\lambda_i-\vec\lambda_j)^2\theta_{ij}\bar{\theta}_{ij}~~$, where $~~\vec \lambda_i =(\Lambda^1_{ii},\dots,\Lambda^p_{ii})$ .
Now I need the Riemann curvature of the above metric. It seems that it is convenient to work in tetrad formalism, using tetrads $E_{ij}=|\vec\lambda_i-\vec\lambda_j|\,\theta_{ij}$, for $i<j$. The problem is that $d E_{ij}$ will now contain a term proportional to $(\theta_{ii}-\theta_{jj})\wedge\theta_{ij}$ and since $\theta_{ii}$ are not part of the basis the spin curvature cannot be written easily without using the explicit parametrisation of $U(n)$. Intuitively, I know that the scalar curvature should depend only on the lambdas ($\vec\lambda_i$), and I have verified that explicitly for $SU(2)$ and $SU(3)$, however a general result seems to require some invariant way to express the pullback of the term $(\theta_{ii}-\theta_{jj})\wedge\theta_{ij}$ on the submanifold spanned by the off diagonal $\theta$'s.
I was wondering if mathematicians have explored the manifold of commuting hermitian matrices. In fact even a reference to a convenient parametrisation of the maximal flag manifold $F_n$ would greatly help me in deriving a general expression for the scalar curvature. Any comments/suggestions are welcomed. This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user vesofilev
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Treasury futures are actually really complicated... There are complete books dedicated to this topic (e.g., The Treasury Bond Basis) and really good sell-side research papers ("Understanding Treasury Bond Futures" by Salomon Brothers) that I highly recommend.You're actually very much on the right track, but I'll try to paint a somewhat complete picture....
Not saying this trade won't work, but there's certainly no guarantee that it will...Given that QE will stop in October is well teleported at this point and has been expected since last year, you'd think this should be fully priced in.Last year, when the "tapering" talk started, Treasuries did sell off quite a bit, but has since rallied all the way back. ...
This is a basic fact about futures trading and the storage of commodities.The phrase that was used by futures traders in the old days (and probably still today) was "the contango is limited by the carrying cost, there is no limit to the backwardation". This means that for example if spot gold is at 1200, gold dated one year from now cannot possibly sell ...
There is actually a lot of art involved. The most simplistic framework is as follows:The first step is to obtain a list of FOMC meeting dates. These are available currently for 2015 and 2016 here. If you're interested for rate expectations beyond 2016, you'd need to "guess" the meeting dates in the future based on past patterns.The next step is to ...
You are missing the futures basis and roll cost. Futures expire, and need to be rolled into the new expiry. The basis is not static and can vary considerably, depending on the specific underlying and contract. Quants may have a hard time to appreciate this but the basis is not at all fully quantifiable at all times: It can hugely vary entirely due to shifts ...
I will talk about equity futures. Commodity futures can be slightly different, as I briefly point out.Equity futures are standardised exchange-traded instruments.Futures on stock indices are especially liquid. The reason for that, is that one cannot simply buy/sell an index as he would buy/sell a single stock because an index is merely a reference value, ...
Using the following data from 12/18/16:Jan 2017 Fed funds futures =9936,Jan 2018 Fed Funds futures =9877implies that 99.36-98.77 = 59bp of hikes are built in for 2017.IF you assume the only two possibilities are 2 hikes or 3 hikes (meaning, 50bp or 75bp of hikes, assuming each hike would be 25bp), then by simple linear interpolation the probability of 3 ...
The best theoretical model for pricing vix futures and options is a variance gamma model.However in practice that class of models is difficult to get robust results...In practice, most floor traders in vix products base their hedging off of the SPX option chain. Vix is calculated from those options, in the first place, so this approach makes intuitive ...
Think of it like a forward trade on the settlement price. If you are buying with a TAS you are agreeing to go long the futures contract at the settlement price (+/- the offset), and whoever you trade with is agreeing to go short at the same price. It is guaranteed because the exchange becomes the counterparty for both traders and there is a margin deposit....
Forward delta is 1 (defined as change in the value of the forward with respect to an instantaneous change in the price of the underlying, holding everything else constant).However for a meaningful discussion of the differences in forward and futures pricing, the forward price delta of forwards should be considered and it is exp(r(T-t)).Though the delta ...
Just take something like$$\frac{\log{\frac{F_j}{F_i}}}{t_j - t_i} \times 365$$where $t_i$ denotes the expiry (or alternatively delivery) date of future $i$. The annualization is so you can compare different futures.
Not really a quant question, but a quick search led to this from the CME: http://www.cmegroup.com/market-data/files/CME_Group_Settlement_Procedures.pdf. Unfortunately it depends on the contract, for example:Equity Futures: For S&P and NASDAQ, the settlement price of the lead* month contract is the midpoint of the closing range determined based on pit ...
VIX is calculated from a basket of SPX options, and VIX futures expire into following expiration, e.g. September VIX futures that will expire next Wednesday will use SPX October options chain to calculate settlement value. If $B$ is the value of the basket then VIX value at expiration is $\sqrt{ B }$. Then VIX futures price is the expectation of the basket $...
Put simply, VIX is a spot index (fair value to a variance swap on SPX of constant maturity) that you cannot own as a security. Market participants create futures for you to trade. Futures trade higher than the VIX -- if you long VIX futures, you lose when the futures contract converges to VIX. You therefore have a negative roll-down. VIX ETF doesn't avoid ...
For reference, note that execution strategies for some types of futures contracts can be very different from equities. An example are Interest Rate Futures, e.g. here.The main reason lies in microstructure differences. For some more details see the white paper "US Treasury Futures Roll Microstructure Basics" by Quantitative Brokers (I have no affiliation ...
Yes. Although sometimes people mean the Euro/Dollar currency pair which can cause confusion.Besides the daily mark-to-market, the counter-party risk is also removed through the clearing house for the futures.No. Eurodollar and FRA are not the same as swaps. A Eurodollar fixes an interest rate for a three month period in the future whereas a swap represents ...
Your question is an important one, but I am not aware of any particularly satisfying answer. There are several papers on this issue -- see Luo and Zhang 2009 and Zhang et al 2010, just for example.One thing to note is that VIX futures are not always in contango -- after large jumps in the VIX, they can even be in rather steep backwardation. I have heard ...
Your questions about contango in VIX futures have close analogies inoptions too. The Black & Scholes model suggests that all timeframes and all strikes should have the same implied volatility, butthey don't. I think one of the reasons is that the B&S modelassumes that stock returns are distributed in a normal (gaussian)distribution, but ...
VIX is a measure of volatility -- something that changes explicitly with uncertainty. The chances of uncertainty arising tomorrow, is lower than the chances of uncertainty increasing in the longer term. A long-dated option should therefore have more "potential uncertainty" baked into the price.When pricing normal futures, the price is a martingale, the ...
This part of your postIn addition, on expiry day the holder (...)is wrong.[Short Story]Due to the daily variation margins calculated by the clearing house on each market close, you have already received/coughed up what you should upon expiry. If the contract is cash-settled, the story thus ends here. In case of physical delivery however, although ...
Treasury bond futures are surprisingly complicated - this is an attempt at a short explanation, it will obviously gloss over some details, but hopefully gives you a flavour of how they are priced.The most important fact is that the underlying is not a single bond, but a basket of bonds. For example, the US Treasury Bond Futures contract spec says that you ...
Based on the your comments, I believe the issue lies with what you consider to be "carry." The reality is that there's no consensus. So let's take mini steps.We'll start with what rates guys consider as "pure carry." In this most classical and fairly strict definition, carry is the deterministic component of expected returns – you know exactly what it is ...
This is a surprisingly complicated question that encompasses many moving parts. Without knowing exactly what your objectives are, it's a bit difficult to offer concrete advice, so I'll provide some general comments below.Mechanically, you earn the total return when you buy and hold a real bond or a bond ETF. By contrast, bond futures are financed ...
Futures are in "zero net supply", or "for every long there is a short", which means that at any time there are investors who are long a certain number of contracts and other investors who are short an (exactly matching!) number of contracts. This number is called the Open Interest. It starts at zero when the exchange introduces a new contract (like Sep 2019 ...
No I believe there is no directional predictive value derived from looking at divergences between futures and their underlying price value. The reason for divergences are of the no-arbitrage argument type. Futures could be arbitraged (and are immediately if such arbitrage opportunities surface, even those opportunities may only fill the stomach of a single ...
You could compute index dividend yield from ATM options using linearized put-call parity (assuming index options are European.)The present value of the dividend payment is:$PV(div) = P - C + (S - K) + K(e^{rT} - 1)$where $r$ is interest rate to the option expiration and $T$ is time to maturity in years. Then the implied dividend is:$d = \frac{PV(div)}{...
As you pointed out there are many ways to adjust for the roll overs. Hence, I guess you would agree that there is no one-size-fits-all answer to this. It really depends on the usage of the data:First think about how the trades in your back test are structured. If they are longer-term trades and you hold over roll overs then think what you would do if you ...
Short Answer : Futures don't have GreeksLong Answer : Assuming a non strictly mathematical (i.e. false) point of view.Well, having Greeks on VIX Futures is not relevant, VIX value is itself a "Greek" (and Futures don't have Greeks).Sensitivity toPrice of the Underlying : Insensitive (ν = 0)Volatility of the Underlying : Delta Δ = 1 (to Volatility of ...
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Difference between revisions of "ORD is Mahlo"
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If there is a pseudo [[uplifting]] (proof in that article) cardinal, or indeed, merely a pseudo $0$-uplifting cardinal, then there is a transitive set model of ZFC with a [[reflecting]] cardinal and consequently also a transitive model of ZFC plus $\text{Ord}$ is Mahlo.<cite>HamkinsJohnstone:ResurrectionAxioms</cite>
If there is a pseudo [[uplifting]] (proof in that article) cardinal, or indeed, merely a pseudo $0$-uplifting cardinal, then there is a transitive set model of ZFC with a [[reflecting]] cardinal and consequently also a transitive model of ZFC plus $\text{Ord}$ is Mahlo.<cite>HamkinsJohnstone:ResurrectionAxioms</cite>
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{{references}}
{{references}}
Revision as of 09:32, 10 October 2019
The assertion
$\text{Ord}$ is Mahlo is the scheme expressing that the proper class REG consisting of all regular cardinals is a stationary proper class, meaning that it has elements from every definable (with parameters) closed unbounded proper class of ordinals. In other words, the scheme asserts for every formula $\varphi$, that if for some parameter $z$ the class $\{\alpha\mid \varphi(\alpha,z)\}$ is a closed unbounded class of ordinals, then it contains a regular cardinal. If $\kappa$ is Mahlo, then $V_\kappa\models\text{Ord is Mahlo}$. Consequently, the existence of a Mahlo cardinal implies the consistency of $\text{Ord}$ is Mahlo, and the two notions are not equivalent. Moreoever, since the ORD is Mahlo scheme is expressible as a first-order theory, it follows that whenever $V_\gamma\prec V_\kappa$, then also $V_\gamma$ satisfies the Levy scheme. Consequently, if there is a Mahlo cardinal, then there is a club of cardinals $\gamma\lt\kappa$ for which $V_\gamma\models\text{Ord is Mahlo}$.
A simple compactness argument establishes that $\text{Ord}$ is Mahlo is equiconsistent over $\text{ZFC}$ with the existence of an inaccessible reflecting cardinal. On the one hand, if $\kappa$ is an inaccessible reflecting cardinal, then since $V_\kappa\prec V$ it follows that any class club definable in $V$ with parameters below $\kappa$ will be unbounded in $\kappa$ and hence contain $\kappa$ as an element and consequently contain an inaccessible cardinal. On the other hand, if $\text{Ord}$ is Mahlo is consistent, then every finite fragment of the theory asserting that $\kappa$ is an inaccessible reflecting cardinal (which is after all asserted as a scheme) is consistent, and hence by compactness the whole theory is consistent.
If there is a pseudo uplifting (proof in that article) cardinal, or indeed, merely a pseudo $0$-uplifting cardinal, then there is a transitive set model of ZFC with a reflecting cardinal and consequently also a transitive model of ZFC plus $\text{Ord}$ is Mahlo.[1]
The Vopěnka principle implies that $Ord$ is Mahlo: every club class contains a regular cardinal and indeed, an extendible cardinal and more. It is relatively consistent that GBC and the generic Vopěnka principle holds, yet $Ord$ is not Mahlo. It is relatively consistent that ZFC and the generic Vopěnka scheme holds, yet $Ord$ is not definably Mahlo and not even $∆_2$-Mahlo. In such a model, there can be no $Σ_2$-reflecting cardinals and therefore also no remarkable cardinals.
References Hamkins, Joel David and Johnstone, Thomas A. Resurrection axioms and uplifting cardinals., 2014. www arχiv bibtex Gitman, Victoria and Hamkins, Joel David. A model of the generic Vopěnka principle in which the ordinals are not Mahlo., 2018. arχiv bibtex
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Difference between revisions of "De Bruijn-Newman constant"
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where
where
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:<math>\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{s/2} \Gamma(s/2) \zeta(s)</math>
+
:<math>\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{s/2} \Gamma(s/2) \zeta(s)</math>
is the Riemann xi function. In particular, <math>z</math> is a zero of <math>H_0</math> if and only if <math>\frac{1}{2} + \frac{iz}{2}</math> is a non-trivial zero of the Riemann zeta function. Thus, for instance, the Riemann hypothesis is equivalent to all the zeroes of <math>H_0</math> being real, and [https://en.wikipedia.org/wiki/Riemann%E2%80%93von_Mangoldt_formula Riemann-von Mangoldt formula] (in the explicit form given by Backlund) gives
is the Riemann xi function. In particular, <math>z</math> is a zero of <math>H_0</math> if and only if <math>\frac{1}{2} + \frac{iz}{2}</math> is a non-trivial zero of the Riemann zeta function. Thus, for instance, the Riemann hypothesis is equivalent to all the zeroes of <math>H_0</math> being real, and [https://en.wikipedia.org/wiki/Riemann%E2%80%93von_Mangoldt_formula Riemann-von Mangoldt formula] (in the explicit form given by Backlund) gives
−
:<math>\displaystyle N_0(T) - (\frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} - \frac{7}{8})| < 0.137 \log (T/2) + 0.443 \log\log(T/2) + 4.350 </math>
+
:<math>\displaystyle N_0(T) - (\frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} - \frac{7}{8})| < 0.137 \log (T/2) + 0.443 \log\log(T/2) + 4.350 </math>
for any <math>T > 4</math>, where <math>N_0(T)</math> denotes the number of zeroes of <math>H_0</math> with real part between 0 and T.
for any <math>T > 4</math>, where <math>N_0(T)</math> denotes the number of zeroes of <math>H_0</math> with real part between 0 and T.
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It is known that <math>\xi</math> is an entire function of order one ([T1986, Theorem 2.12]). Hence by the fundamental solution for the heat equation, the <math>H_t</math> are also entire functions of order one for any <math>t</math>.
It is known that <math>\xi</math> is an entire function of order one ([T1986, Theorem 2.12]). Hence by the fundamental solution for the heat equation, the <math>H_t</math> are also entire functions of order one for any <math>t</math>.
+ +
Let <math>\sigma_{max}(t)</math> denote the largest imaginary part of a zero of <math>H_t</math>, thus <math>\sigma_{max}(t)=0</math> if and only if <math>t \geq \Lambda</math>. It is known that the quantity <math>\frac{1}{2} \sigma_{max}(t)^2 + t</math> is non-increasing in time whenever <math>\sigma_{max}(t)>0</math> (see [KKL2009, Proposition A]. In particular we have
Let <math>\sigma_{max}(t)</math> denote the largest imaginary part of a zero of <math>H_t</math>, thus <math>\sigma_{max}(t)=0</math> if and only if <math>t \geq \Lambda</math>. It is known that the quantity <math>\frac{1}{2} \sigma_{max}(t)^2 + t</math> is non-increasing in time whenever <math>\sigma_{max}(t)>0</math> (see [KKL2009, Proposition A]. In particular we have
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* [https://terrytao.wordpress.com/2018/03/02/polymath15-fifth-thread-finishing-off-the-test-problem/ Polymath15, fifth thread: finishing off the test problem?], Terence Tao, Mar 2, 2018.
* [https://terrytao.wordpress.com/2018/03/02/polymath15-fifth-thread-finishing-off-the-test-problem/ Polymath15, fifth thread: finishing off the test problem?], Terence Tao, Mar 2, 2018.
* [https://terrytao.wordpress.com/2018/03/18/polymath15-sixth-thread-the-test-problem-and-beyond/ Polymath15, sixth thread: the test problem and beyond], Terence Tao, Mar 18, 2018.
* [https://terrytao.wordpress.com/2018/03/18/polymath15-sixth-thread-the-test-problem-and-beyond/ Polymath15, sixth thread: the test problem and beyond], Terence Tao, Mar 18, 2018.
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== Other blog posts and online discussion ==
== Other blog posts and online discussion ==
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* [https://github.com/km-git-acc/dbn_upper_bound Github repository]
* [https://github.com/km-git-acc/dbn_upper_bound Github repository]
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== Test problem ==
== Test problem ==
See [[Polymath15 test problem]].
See [[Polymath15 test problem]].
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== Wikipedia and other references ==
== Wikipedia and other references ==
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* [G2004] Gourdon, Xavier (2004), [http://numbers.computation.free.fr/Constants/Miscellaneous/zetazeros1e13-1e24.pdf The <math>10^{13}</math> first zeros of the Riemann Zeta function, and zeros computation at very large height]
* [G2004] Gourdon, Xavier (2004), [http://numbers.computation.free.fr/Constants/Miscellaneous/zetazeros1e13-1e24.pdf The <math>10^{13}</math> first zeros of the Riemann Zeta function, and zeros computation at very large height]
* [KKL2009] H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathematics, 22 (2009), 281–306. [http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.164.5595&rep=rep1&type=pdf Citeseer]
* [KKL2009] H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathematics, 22 (2009), 281–306. [http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.164.5595&rep=rep1&type=pdf Citeseer]
−
* [N1976] C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61 (1976), 246–251.
+
* [N1976] C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61 (1976), 246–251.
* [P2017] D. J. Platt, [http://www.ams.org/journals/mcom/2017-86-307/S0025-5718-2017-03198-7/ Isolating some non-trivial zeros of zeta], Math. Comp. 86 (2017), 2449-2467.
* [P2017] D. J. Platt, [http://www.ams.org/journals/mcom/2017-86-307/S0025-5718-2017-03198-7/ Isolating some non-trivial zeros of zeta], Math. Comp. 86 (2017), 2449-2467.
* [P1992] G. Pugh, [https://web.viu.ca/pughg/thesis.d/masters.thesis.pdf The Riemann-Siegel formula and large scale computations of the Riemann zeta function], M.Sc. Thesis, U. British Columbia, 1992.
* [P1992] G. Pugh, [https://web.viu.ca/pughg/thesis.d/masters.thesis.pdf The Riemann-Siegel formula and large scale computations of the Riemann zeta function], M.Sc. Thesis, U. British Columbia, 1992.
* [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. [https://arxiv.org/abs/1801.05914 arXiv:1801.05914]
* [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. [https://arxiv.org/abs/1801.05914 arXiv:1801.05914]
* [T1986] E. C. Titchmarsh, The theory of the Riemann zeta-function. Second edition. Edited and with a preface by D. R. Heath-Brown. The Clarendon Press, Oxford University Press, New York, 1986. [http://plouffe.fr/simon/math/The%20Theory%20Of%20The%20Riemann%20Zeta-Function%20-Titshmarch.pdf pdf]
* [T1986] E. C. Titchmarsh, The theory of the Riemann zeta-function. Second edition. Edited and with a preface by D. R. Heath-Brown. The Clarendon Press, Oxford University Press, New York, 1986. [http://plouffe.fr/simon/math/The%20Theory%20Of%20The%20Riemann%20Zeta-Function%20-Titshmarch.pdf pdf]
+ + Latest revision as of 17:37, 30 April 2019
For each real number [math]t[/math], define the entire function [math]H_t: {\mathbf C} \to {\mathbf C}[/math] by the formula
[math]\displaystyle H_t(z) := \int_0^\infty e^{tu^2} \Phi(u) \cos(zu)\ du[/math]
where [math]\Phi[/math] is the super-exponentially decaying function
[math]\displaystyle \Phi(u) := \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3 \pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u}).[/math]
It is known that [math]\Phi[/math] is even, and that [math]H_t[/math] is even, real on the real axis, and obeys the functional equation [math]H_t(\overline{z}) = \overline{H_t(z)}[/math]. In particular, the zeroes of [math]H_t[/math] are symmetric about both the real and imaginary axes. One can also express [math]H_t[/math] in a number of different forms, such as
[math]\displaystyle H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du[/math]
or
[math]\displaystyle H_t(z) = \frac{1}{2} \int_0^\infty e^{t\log^2 x} \Phi(\log x) e^{iz \log x}\ \frac{dx}{x}.[/math]
In the notation of [KKL2009], one has
[math]\displaystyle H_t(z) = \frac{1}{8} \Xi_{t/4}(z/2).[/math]
De Bruijn [B1950] and Newman [N1976] showed that there existed a constant, the
de Bruijn-Newman constant [math]\Lambda[/math], such that [math]H_t[/math] has all zeroes real precisely when [math]t \geq \Lambda[/math]. The Riemann hypothesis is equivalent to the claim that [math]\Lambda \leq 0[/math]. Currently it is known that [math]0 \leq \Lambda \lt 1/2[/math] (lower bound in [RT2018], upper bound in [KKL2009]).
The
Polymath15 project seeks to improve the upper bound on [math]\Lambda[/math]. The current strategy is to combine the following three ingredients: Numerical zero-free regions for [math]H_t(x+iy)[/math] of the form [math]\{ x+iy: 0 \leq x \leq T; y \geq \varepsilon \}[/math] for explicit [math]T, \varepsilon, t \gt 0[/math]. Rigorous asymptotics that show that [math]H_t(x+iy)[/math] whenever [math]y \geq \varepsilon[/math] and [math]x \geq T[/math] for a sufficiently large [math]T[/math]. Dynamics of zeroes results that control [math]\Lambda[/math] in terms of the maximum imaginary part of a zero of [math]H_t[/math]. Contents [math]t=0[/math]
When [math]t=0[/math], one has
[math]\displaystyle H_0(z) = \frac{1}{8} \xi( \frac{1}{2} + \frac{iz}{2} ) [/math]
where
[math]\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \zeta(s)[/math]
is the Riemann xi function. In particular, [math]z[/math] is a zero of [math]H_0[/math] if and only if [math]\frac{1}{2} + \frac{iz}{2}[/math] is a non-trivial zero of the Riemann zeta function. Thus, for instance, the Riemann hypothesis is equivalent to all the zeroes of [math]H_0[/math] being real, and Riemann-von Mangoldt formula (in the explicit form given by Backlund) gives
[math]\displaystyle \left|N_0(T) - (\frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} - \frac{7}{8})\right| \lt 0.137 \log (T/2) + 0.443 \log\log(T/2) + 4.350 [/math]
for any [math]T \gt 4[/math], where [math]N_0(T)[/math] denotes the number of zeroes of [math]H_0[/math] with real part between 0 and T.
The first [math]10^{13}[/math] zeroes of [math]H_0[/math] (to the right of the origin) are real [G2004]. This numerical computation uses the Odlyzko-Schonhage algorithm. In [P2017] it was independently verified that all zeroes of [math]H_0[/math] between 0 and 61,220,092,000 were real.
[math]t\gt0[/math]
For any [math]t\gt0[/math], it is known that all but finitely many of the zeroes of [math]H_t[/math] are real and simple [KKL2009, Theorem 1.3]. In fact, assuming the Riemann hypothesis,
all of the zeroes of [math]H_t[/math] are real and simple [CSV1994, Corollary 2].
It is known that [math]\xi[/math] is an entire function of order one ([T1986, Theorem 2.12]). Hence by the fundamental solution for the heat equation, the [math]H_t[/math] are also entire functions of order one for any [math]t[/math].
Because [math]\Phi[/math] is positive, [math]H_t(iy)[/math] is positive for any [math]y[/math], and hence there are no zeroes on the imaginary axis.
Let [math]\sigma_{max}(t)[/math] denote the largest imaginary part of a zero of [math]H_t[/math], thus [math]\sigma_{max}(t)=0[/math] if and only if [math]t \geq \Lambda[/math]. It is known that the quantity [math]\frac{1}{2} \sigma_{max}(t)^2 + t[/math] is non-increasing in time whenever [math]\sigma_{max}(t)\gt0[/math] (see [KKL2009, Proposition A]. In particular we have
[math]\displaystyle \Lambda \leq t + \frac{1}{2} \sigma_{max}(t)^2[/math]
for any [math]t[/math].
The zeroes [math]z_j(t)[/math] of [math]H_t[/math] obey the system of ODE
[math]\partial_t z_j(t) = - \sum_{k \neq j} \frac{2}{z_k(t) - z_j(t)}[/math]
where the sum is interpreted in a principal value sense, and excluding those times in which [math]z_j(t)[/math] is a repeated zero. See dynamics of zeros for more details. Writing [math]z_j(t) = x_j(t) + i y_j(t)[/math], we can write the dynamics as
[math] \partial_t x_j = - \sum_{k \neq j} \frac{2 (x_k - x_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math] [math] \partial_t y_j = \sum_{k \neq j} \frac{2 (y_k - y_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math]
where the dependence on [math]t[/math] has been omitted for brevity.
In [KKL2009, Theorem 1.4], it is shown that for any fixed [math]t\gt0[/math], the number [math]N_t(T)[/math] of zeroes of [math]H_t[/math] with real part between 0 and T obeys the asymptotic
[math]N_t(T) = \frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} + \frac{t}{16} \log T + O(1) [/math]
as [math]T \to \infty[/math] (caution: the error term here is not uniform in t). Also, the zeroes behave like an arithmetic progression in the sense that
[math] z_{k+1}(t) - z_k(t) = (1+o(1)) \frac{4\pi}{\log |z_k|(t)} = (1+o(1)) \frac{4\pi}{\log k} [/math]
as [math]k \to +\infty[/math].
Threads Polymath proposal: upper bounding the de Bruijn-Newman constant, Terence Tao, Jan 24, 2018. Polymath15, first thread: computing H_t, asymptotics, and dynamics of zeroes, Terence Tao, Jan 27, 2018. Polymath15, second thread: generalising the Riemann-Siegel approximate functional equation, Terence Tao and Sujit Nair, Feb 2, 2018. Polymath15, third thread: computing and approximating H_t, Terence Tao and Sujit Nair, Feb 12, 2018. Polymath 15, fourth thread: closing in on the test problem, Terence Tao, Feb 24, 2018. Polymath15, fifth thread: finishing off the test problem?, Terence Tao, Mar 2, 2018. Polymath15, sixth thread: the test problem and beyond, Terence Tao, Mar 18, 2018. Polymath15, seventh thread: going below 0.48, Terence Tao, Mar 28, 2018. Polymath15, eighth thread: going below 0.28, Terence Tao, Apr 17, 2018. Polymath15, ninth thread: going below 0.22?, Terence Tao, May 4, 2018. Polymath15, tenth thread: numerics update, Rudolph Dwars and Kalpesh Muchhal, Sep 6, 2018. Polymath15, eleventh thread: Writing up the results, and exploring negative t, Terence Tao, Dec 28, 2018. Effective approximation of heat flow evolution of the Riemann xi function, and a new upper bound for the de Bruijn-Newman constant, Terence Tao, Apr 30, 2019. Other blog posts and online discussion Heat flow and zeroes of polynomials, Terence Tao, Oct 17, 2017. The de Bruijn-Newman constant is non-negative, Terence Tao, Jan 19, 2018. Lehmer pairs and GUE, Terence Tao, Jan 20, 2018. A new polymath proposal (related to the Riemann hypothesis) over Tao's blog, Gil Kalai, Jan 26, 2018. Code and data Writeup
Here are the Polymath15 grant acknowledgments.
Test problem Zero-free regions
See Zero-free regions.
Wikipedia and other references Bibliography [A2011] J. Arias de Reyna, High-precision computation of Riemann's zeta function by the Riemann-Siegel asymptotic formula, I, Mathematics of Computation, Volume 80, Number 274, April 2011, Pages 995–1009. [B1994] W. G. C. Boyd, Gamma Function Asymptotics by an Extension of the Method of Steepest Descents, Proceedings: Mathematical and Physical Sciences, Vol. 447, No. 1931 (Dec. 8, 1994),pp. 609-630. [B1950] N. C. de Bruijn, The roots of trigonometric integrals, Duke J. Math. 17 (1950), 197–226. [CSV1994] G. Csordas, W. Smith, R. S. Varga, Lehmer pairs of zeros, the de Bruijn-Newman constant Λ, and the Riemann hypothesis, Constr. Approx. 10 (1994), no. 1, 107–129. [G2004] Gourdon, Xavier (2004), The [math]10^{13}[/math] first zeros of the Riemann Zeta function, and zeros computation at very large height [KKL2009] H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathematics, 22 (2009), 281–306. Citeseer [N1976] C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61 (1976), 246–251. [P2017] D. J. Platt, Isolating some non-trivial zeros of zeta, Math. Comp. 86 (2017), 2449-2467. [P1992] G. Pugh, The Riemann-Siegel formula and large scale computations of the Riemann zeta function, M.Sc. Thesis, U. British Columbia, 1992. [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. arXiv:1801.05914 [T1986] E. C. Titchmarsh, The theory of the Riemann zeta-function. Second edition. Edited and with a preface by D. R. Heath-Brown. The Clarendon Press, Oxford University Press, New York, 1986. pdf
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The conclusion merely counts dimensions of vector spaces. However, it is not generally true.
The most basic properties of matrix multiplication show that the linear transformation represented by the matrix $\mathbb{H}=X(X^\prime X)^{-}X^\prime$ satisfies
$$\mathbb{H}^2 = \left(X(X^\prime X)^{-}X^\prime\right)^2=X(X^\prime X)^{-}(X^\prime X)(X^\prime X)^{-}X^\prime=\mathbb{H},$$
exhibiting it as a projection operator. Therefore its complement
$$\mathbb{Q} = 1 - \mathbb{H}$$
(as given in the question) also is a projection operator. The trace of $\mathbb{H}$ is its rank $h$ (see below), whence the trace of $\mathbb{Q}$ equals $n-h$.
From its very formula it is apparent that $\mathbb{H}$ is the matrix associated with the composition of two linear transformations $$\mathbb{J}=(X^\prime X)^{-}X^\prime$$ and $X$ itself. The first ($\mathbb{J}$) transforms the $n$-vector $y$ into the $p$-vector $\hat\beta$. The second ($X$) is a transformation from $\mathbb{R}^p$ to $\mathbb{R}^n$ given by $\hat y = X\hat \beta$. Its rank cannot exceed the smaller of those two dimensions, which in a least squares setting is always $p$ (but could be less than $p$, whenever $\mathbb{J}$ is not of full rank). Consequently the rank of the composition $\mathbb{H}=X\mathbb{J}$ cannot exceed the rank of $X$.
The correct conclusion, then, is
$\text{tr} (\mathbb{Q}) = n-p$ if and only if $\mathbb{J}$ is of full rank; and in general $n \ge \text{tr} (\mathbb{Q}) \ge n-p$. In the former case the model is said to be "identifiable" (for the coefficients of $\beta$).
$\mathbb{J}$ will be of full rank if and only if $X^\prime X$ is invertible.
Geometric interpretation
$\mathbb{H}$ represents the orthogonal projection from $n$-vectors $y$ (representing the "response" or "dependent variable") onto the space spanned by the columns of $X$ (representing the "independent variables" or "covariates"). The difference $\mathbb{Q}=1-\mathbb{H}$ shows how to decompose any $n$-vector $y$ into a sum of vectors $$y = \mathbb{H}(y) + \mathbb{Q}(y),$$ where the first can be "predicted" from $X$ and the second is perpendicular to it. When the $p$ columns of $X$ generate a $p$-dimensional space (that is, are not collinear), the rank of $\mathbb{H}$ is $p$ and the rank of $\mathbb{Q}$ is $n-p$, reflecting the $n-p$ additional dimensions of variation in the response that are not represented within the independent variables. The trace gives an algebraic formula for these dimensions.
Linear Algebra Background
A projection operator on a vector space $V$ (such as $\mathbb{R}^n$) is a linear transformation $\mathbb{P}:V\to V$ (that is, an
endomorphism of $V$) such that $\mathbb{P}^2=\mathbb{P}$. This makes its complement $\mathbb{Q}=1-\mathbb{P}$ a projection operator, too, because
$$\mathbb{Q}^2 = \left(1 - \mathbb{P}\right)^2 = 1 - 2\mathbb{P} + \mathbb{P}^2 = 1-2\mathbb{P}+\mathbb{P} = \mathbb{Q}.$$
All projections fix every element of their images, for whenever $v\in \text{Im}(\mathbb{P})$ we may write $v = \mathbb{P}(w)$ for some $w\in V$, whence $$w = \mathbb{P}(v) = \mathbb{P}^2(v) = \mathbb{P}(\mathbb{P}(v)) = \mathbb{P}(w).$$
Associated with any endomorphism $\mathbb{P}$ of $V$ are two subspaces: its
kernel $$\text{ker}(\mathbb{P}) = \{v\in v\,|\, \mathbb{P}(v)=0\}$$ and its image $$\text{Im}(\mathbb{P}) = \{v\in v\,|\, \exists_{w\in V} \mathbb{P}(w)=v\}.$$ Every vector $v\in V$ can be written in the form $$v = w+u$$ where $w\in \text{Im}(\mathbb{P})$ and $u\in \text{Ker}(\mathbb{P})$. We may therefore construct a basis $E \cup F$ for $V$ for which $E \subset \text{Ker}(\mathbb{P})$ and $F \subset \text{Im}(\mathbb{P})$. When $V$ is finite-dimensional, the matrix of $\mathbb{P}$ in this basis will therefore be in block-diagonal form, with one block (corresponding to the action of $\mathbb{P}$ on $E$) all zeros and the other (corresponding to the action of $\mathbb{P}$ on $F$) equal to the $f$ by $f$ identity matrix, where the dimension of $F$ is $f$. The trace of $\mathbb{P}$ is the sum of the values on the diagonal and therefore must equal $f\times 1 = f$. This number is the rank of $\mathbb{P}$: the dimension of its image.
The trace of $1-\mathbb{P}$ equals the trace of $1$ (equal to $n$, the dimension of $V$) minus the trace of $\mathbb{P}$.
These results may be summarized with the assertion that
the trace of a projection equals its rank.
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The Annals of Mathematical Statistics Ann. Math. Statist. Volume 43, Number 5 (1972), 1412-1427. Likelihood Ratio Tests for Sequential $k$-Decision Problems Abstract
Sequential tests of separated hypotheses concerning the parameter $\theta$ of a Koopman-Darmois family are studied from the point of view of minimizing expected sample sizes pointwise in $\theta$ subject to error probability bounds. Sequential versions of the (generalized) likelihood ratio test are shown to exceed the minimum expected sample sizes by at most $M \log \log \underline{\alpha}^{-1}$ uniformly in $\theta$, where $\underline{\alpha}$ is the smallest error probability bound. The proof considers the likelihood ratio tests as ensembles of sequential probability ratio tests and compares them with alternative procedures by constructing alternative ensembles, applying a simple inequality of Wald and a new inequality of similar type. A heuristic approximation is given for the error probabilities of likelihood ratio tests, which provides an upper bound in the case of a normal mean.
Article information Source Ann. Math. Statist., Volume 43, Number 5 (1972), 1412-1427. Dates First available in Project Euclid: 27 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aoms/1177692374 Digital Object Identifier doi:10.1214/aoms/1177692374 Mathematical Reviews number (MathSciNet) MR343501 Zentralblatt MATH identifier 0262.62045 JSTOR links.jstor.org Citation
Lorden, Gary. Likelihood Ratio Tests for Sequential $k$-Decision Problems. Ann. Math. Statist. 43 (1972), no. 5, 1412--1427. doi:10.1214/aoms/1177692374. https://projecteuclid.org/euclid.aoms/1177692374
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Queries ping a certain computer server at random times, on average $\lambda$ arriving per second. The server can respond to one per second and those that can’t be serviced immediately are queued up. What is the average wait time per query? Clearly if $\lambda \ll 1$, the average wait time is zero. But if $\lambda > 1$, the queue grows indefinitely and the answer is infinity! Here, we give a simple derivation of the general result — (9) below.
Follow @efavdb Follow us on twitter for new submission alerts! Introduction
The mathematics of queue waiting times — first worked out by Agner Krarup Erlang — is interesting for two reasons. First, as noted above, queues can exhibit phase-transition like behaviors: If the average arrival time is shorter than the average time it takes to serve a customer, the line will grow indefinitely, causing the average wait time to diverge. Second, when the average arrival time is less than the service time, waiting times are governed entirely by fluctuations — and so can’t be estimated well using mean-field arguments. For example, in the very low arrival rate limit, the only situation where anyone would ever have to wait at all is that where someone else happens to arrive just before them — an unlucky, rare event.
Besides being interesting from a theoretical perspective, an understanding of queue formation phenomena is also critical for many practical applications — both in computer science and in wider industry settings. Optimal staffing of a queue requires a careful estimate of the expected customer arrival rate. If too many workers are staffed, the average wait time will be nicely low, but workers will largely be idle. Staff too few, and the business could enter into the divergent queue length regime — certainly resulting in unhappy customers and lost business (or dropped queries). Staffing just the right amount requires a sensitive touch — and in complex cases, a good understanding of the theory.
In order to derive the average wait time for queues of different sorts, one often works within the framework of Markov processes. This approach is very general and elementary, but requires a bit of effort to develop the machinery needed get to the end results. Here, we demonstrate an alternative, sometimes faster approach that is based on writing down an integral equation for the wait time distribution. We consider only a simple case — that where the queue is serviced by only one staff member, the customers arrive at random times via a Poisson process, and each customer requires the same time to service, one second.
Integral equation formulation
Suppose the $N$-th customer arrives at time $0$, and let $P(t)$ be the probability that this customer has to wait a time $t\geq 0$ before being served. This wait time can be written in terms of the arrival and wait times of the previous customer: If this previous customer arrived at time $t^{\prime}$ and has to wait a time $w$ before being served, his service will conclude at time $t = t^{\prime} + w + 1$. If this is greater than $0$, the $N$-th customer will have to wait before being served. In particular, he will wait $t$ if the previous customer waited $w = t – t^{\prime} – 1$.
The above considerations allow us to write down an equation satisfied by the wait time distribution. If we let the probability that the previous customer arrived at $t^{\prime}$ be $A(t^{\prime})$, we have (for $t > 0$)
$$ \tag{1} \label{int_eqn} P(t) = \int_{-\infty}^{0^-} A(t^{\prime}) P(t – t^{\prime} – 1) d t^{\prime} = \int_{-\infty}^{0^-} \lambda e^{\lambda t^{\prime}} P(t – t^{\prime} – 1) d t^{\prime} $$ Here, in the first equality we’re simply averaging over the possible arrival times of the previous customer (which had to occur before the $N$-th, at $0$), multiplying by the probability $P(t – t^{\prime} – 1)$ that this customer had to wait the amount of time $w$ needed so that the $N$-th customer will wait $t$. We also use the symmetry that each customer has the same wait time distribution at steady state. In the second equality, we have plugged in the arrival time distribution appropriate for our Poisson model.
To proceed, we differentiate both sides of (\ref{int_eqn}) with respect to $t$,
$$ \tag{2} \label{int2} P^{\prime}(t) = \int_{-\infty}^{0^-} \lambda e^{\lambda t^{\prime}} \frac{d}{dt}P(t – t^{\prime} – 1) d t^{\prime} = – \int_{-\infty}^{0^-} \lambda e^{\lambda t^{\prime}} \frac{d}{dt^{\prime}}P(t – t^{\prime} – 1) d t^{\prime}. $$ The second equality follows after noticing that we can switch the parameter being differentiated in the first. Integrating by parts, we obtain $$ P^{\prime}(t) = \lambda \left [P(t) – P(t-1) \right], \tag{3} \label{sol} $$ a delay differential equation for the wait time distribution. This could be integrated numerically to get the full solution. However, our interest here is primarily the mean waiting time — as we show next, it’s easy to extract this part of the solution analytically. Probability of no wait and the mean wait time
We can obtain a series of useful relations by multiplying (\ref{sol}) by powers of $t$ and integrating. The first such expression is obtained by multiplying by $t^1$. Doing this and integrating its left side, we obtain
$$\tag{4} \label{int3} \int_{0^{+}}^{\infty} P^{\prime}(t) t dt = \left . P(t) t \right |_{0^{+}}^{\infty} – \int_{0^+}^{\infty} P(t) dt = 1 – P(0). $$ Similarly integrating its right side, we obtain
$$\tag{5} \label{int4}
\lambda \int_{0^{+}}^{\infty} t \left [P(t) – P(t-1) \right] = \lambda [ \overline{t} – \overline{(t + 1)} ] = – \lambda. $$ Equating the last two lines, we obtain the probability of no wait, $$ \tag{6} \label{int5} P(0) = 1 – \lambda. $$ This shows that when the arrival rate is low, the probability of no wait goes to one — an intuitively reasonable result. On the other hand, as $\lambda \to 1$, the probability of no wait approaches zero. In between, the idle time fraction of our staffer (which is equal to the probability of no wait, given a random arrival time) grows linearly, connecting these two limits.
To obtain an expression for the average wait time, we carry out a similar analysis to that above, but multiply (\ref{sol}) by $t^2$ instead. The integral on left is then
$$ \tag{7} \label{int6} \int_{0^{+}}^{\infty} P^{\prime}(t) t^2 dt = \left . P(t) t^2 \right |_{0^{+}}^{\infty} – 2\int_{0^+}^{\infty} P(t) t dt = – 2 \overline{t}. $$ Similarly, the integral at right is $$ \tag{8} \label{fin_int} \lambda \int_{0^{+}}^{\infty} t^2 \left [P(t) – P(t-1) \right] = \lambda \overline{ t^2} – \overline{ (t + 1)^2} = – \lambda (2 \overline{t} +1). $$ Equating the last two lines and rearranging gives our solution for the average wait, $$ \tag{9} \label{fin} \overline{t} = \frac{\lambda}{2 (1 – \lambda)}. $$ As advertised, this diverges as $\lambda \to 1$, see illustration in the plot below. It’s very interesting that even as $\lambda$ approaches this extreme limit, the line is still empty a finite fraction of the time — see (\ref{int5}). Evidently a finite idle time fraction can’t be avoided, even as one approaches the divergent $\lambda = 1$ limit. Conclusions and extensions
To carry this approach further, one could consider the case where the queue feeds $k$ staff, rather than just one. I’ve made progress on this effort in certain cases, but have been stumped on the general problem. One interesting thing you can intuit about this $k$-staff version is that one approaches the mean-field analysis as $k\to \infty$ (adding more staff tends to smooth things over, resulting in a diminishing of the importance of the randomness of the arrival times). This means that as $k$ grows, we’ll have very little average wait time for any $\lambda<1$, but again divergent wait times for any $\lambda \geq 1$ -- like an infinite step function. Another direction one could pursue is to allow the service times to follow a distribution. Both cases can also be worked out using the Markov approach -- references to such work can be found in the link provided in the introduction.
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Trig Basics
Despite its reputation as being "high math", trigonometry can be very simple. At least its use on the SAT or ACT is simple—and that's all we're worried about here.
To use trig, you need two things:
a triangle with a right angle (AKA, a right triangle) one of the other parts labeled (either an angle or a leg-length)
If you have these two things, you can figure out the rest of the parts fairly easily.
Look at the graphic above There are 3 angles (of course): $\angle A$ , $\angle B$ , and $\angle C$ . Note that $\angle C$ is the right angle. There are 3 legs or sides (of course): $a$ , $b$ , $c$ The leg-labels are "Opposite", "Adjacent", and "Hypotenuse" The "Hypotenuse" will always be the same, it's the hypotenuse! "Opposite" and "Adjacent" will change depending on which angle we're talking about. The labels shown are in relation to $\angle A$. We know this because starting with $\angle A$, we can see that the "Adjacent" leg is indeed adjacent (next to it) to the angle. And the "Opposite" leg is directly opposite from the angle. Basic Trig Functions The chart above should be read such as $sinx={opp \over adj }$ meaning the $sin$ of $\angle x$ is equal to the opposite side divided by the adjacent side. An Example
Suppose we knew that side b was 32 feet and we're asked to find out the length of side C (the hypotenuse).
First, take a look at what we know. We know… $\angle C = 90^\circ$ Side b is 32 feet. We have two parts, that's all we need. Next, decide if we want to use $sin$, $cos$, or $tan$. Focus on $\angle C$ because that's the one we know. We know the adjacent side is 32 and wantto know the hypotenuse. So, we're dealing with Adjacent and Hypotenuse. The Trig Functions chart tells us Adjacent and Hypotenuse go with Cosine. Our formula would then look like the one below…
$cos90={32 \over hyp}$
From here, it's just a matter of punching in some numbers on a calculator to get the cosine of 90, then doing some simple algebra to solve for the hypotenuse. The steps would be…
Punch in 32 and hit "cosine" on your scientific calculator. It'll show you 0.848. The formula would then read…
$0.848={32 \over hyp}$
To solve for $hyp$, cross multiply like this: $0.848 \times hyp = 32 \times 1$ and you'll get…
$0.848 hyp = 32$
To solve, simply isolate the variable ($hyp$) by dividing both sides by 0.848
$32 \div 0.848 = 37.735$ and that's your answer. The hypotenuse is $37.735$ feet long.
A note
If we needed to go farther, we could. We could now use the Pythagorean Theorem ($a^+b^2=c^2$) to find the other side's length. Combing Trig and Pythag, we can find all of the angles and leg-lengths of a right triangle from only two initial measurements.
Mnemonic Devices for Trig
Below are some tricks to help remember the basic trig functions.
SohCahToa — the ancient Indian princess of geometry. This puts the parts in order (S=sin, C=cosine, T=tangent, o=opposite, h=hypotenuse, a=adjacent) Sally can tell Oscar has a hat on always. This sentence puts the parts in a different order.
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Regularity of the extremal solution for a fourth-order elliptic problem with singular nonlinearity
1.
Institute of Contemporary Mathematics, Henan University, School of Mathematics and Information Science, Henan University, Kaifeng 475004, China
2.
School of Mathematics and Information Science, Henan University, Kaifeng 475004, China
$\beta \Delta^{2}u-\tau \Delta u=\frac{\lambda}{(1-u)^{p}} \mbox{in} B$,
$0 < u \leq 1 \mbox{in} B$, $u=\Delta u=0 \mbox{on} \partial B$,
where $B$ is the unit ball in $R^{n}$, $\lambda>0$ is a parameter, $\tau>0, \beta>0,p>1$ are fixed constants. By Hardy-Rellich inequality, we find that when $p$ is large enough, the critical dimension is 13.
Mathematics Subject Classification:Primary 35B45; Secondary 35J4. Citation:Baishun Lai, Qing Luo. Regularity of the extremal solution for a fourth-order elliptic problem with singular nonlinearity. Discrete & Continuous Dynamical Systems - A, 2011, 30 (1) : 227-241. doi: 10.3934/dcds.2011.30.227
References:
[1]
S. Agmon, A. Douglis and L. Nirenberg,
[2]
C. Cowan, P. Esposito, N. Ghoussoub and A. Moradifam,
[3]
M. G. Crandall and P. H. Rabinawitz,
[4]
P. Esposito, N. Ghoussoub and Y. Guo,
[5] [6]
Z. M. Guo and J. C. Wei,
[7] [8] [9] [10] [11] [12]
F. Rellich,
[13]
show all references
References:
[1]
S. Agmon, A. Douglis and L. Nirenberg,
[2]
C. Cowan, P. Esposito, N. Ghoussoub and A. Moradifam,
[3]
M. G. Crandall and P. H. Rabinawitz,
[4]
P. Esposito, N. Ghoussoub and Y. Guo,
[5] [6]
Z. M. Guo and J. C. Wei,
[7] [8] [9] [10] [11] [12]
F. Rellich,
[13]
[1]
Craig Cowan, Pierpaolo Esposito, Nassif Ghoussoub.
Regularity of extremal solutions in fourth order nonlinear eigenvalue problems on general domains.
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Hiroyuki Hirayama, Mamoru Okamoto.
Well-posedness and scattering for fourth order nonlinear Schrödinger type equations at the scaling critical regularity.
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A. Aghajani, S. F. Mottaghi.
Regularity of extremal solutions of semilinaer fourth-order elliptic problems with general nonlinearities.
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Jaime Angulo Pava, Carlos Banquet, Márcia Scialom.
Stability for the modified and fourth-order Benjamin-Bona-Mahony equations.
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Feliz Minhós, João Fialho.
Existence and multiplicity of solutions in fourth order BVPs with unbounded nonlinearities.
[7] [8] [9]
Tokushi Sato, Tatsuya Watanabe.
Singular positive solutions for a fourth order elliptic problem in $R$.
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John R. Graef, Lingju Kong, Min Wang.
Existence of multiple solutions to a discrete fourth order periodic boundary value problem.
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John B. Greer, Andrea L. Bertozzi.
$H^1$ Solutions of a class of fourth order nonlinear equations for image processing.
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Chunhua Jin, Jingxue Yin, Zejia Wang.
Positive periodic solutions to a nonlinear fourth-order differential equation.
[13]
Takahiro Hashimoto.
Existence and nonexistence of nontrivial solutions of some nonlinear fourth order elliptic equations.
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M. Ben Ayed, K. El Mehdi, M. Hammami.
Nonexistence of bounded energy solutions for a fourth order equation on thin annuli.
[15]
John R. Graef, Johnny Henderson, Bo Yang.
Positive solutions to a fourth order three point boundary value problem.
[16]
Craig Cowan.
Uniqueness of solutions for elliptic systems and fourth order equations involving a parameter.
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Rui-Qi Liu, Chun-Lei Tang, Jia-Feng Liao, Xing-Ping Wu.
Positive solutions of Kirchhoff type problem with singular and critical nonlinearities in dimension four.
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Qilin Xie, Jianshe Yu.
Bounded state solutions of Kirchhoff type problems with a critical exponent in high dimension.
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Nodal minimal partitions in dimension $3$.
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I am trying to represent myself quotient groups and I'm having trouble seeing what the kernel of a homomorphism : $\Phi: G \rightarrow G/H$ is (be it a ring homomorphism or a group homomorphism).
I understand that $\mathbb{Z}/n\mathbb{Z}$ is a quotient group that is cyclic and that the kernel of the application: $\Pi : \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$ are all the elements of $n\mathbb{Z}$.
But in the more general case I have trouble truly visualizing the kernel of a quotient group:
Let us take the 2 following examples:
In the case of a group homomorphism, Let G be a group and H be a subgroup of G: $\Xi : G \rightarrow G/H$, $\Xi(g) \mapsto gH$ I know that its kernel is H but I don't understand why.
In the case of a ring homomorphism: If $(A,+,*)$ is a ring and $I$ is a two-sided ideal of $A$. Let the ring homomorphism $\chi: A \rightarrow A/I$, $\chi(a) \mapsto a+I$ I know that its kernel is I but I don't understand why.
Thank you for any clarification
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This is false for $n\geq 4$. Consider the Grassmannian $\mathrm{Gr}(2,n)$ of all two-dimensional subspaces of $\mathbb{R}^n$, and recall that $\mathrm{Gr}(2,n)$ is a compact manifold of dimension $2n-4$. For each $\varphi\in\mathrm{GL}_n(\mathbb{Q})$, let$$S_\varphi = \{A\in \mathrm{Gr}(2,n) \mid \varphi u=v \text{ for some linearly independent }u,v\in A\}.$$By the Baire category theorem, $\mathrm{Gr}(2,n)$ cannot be expressed as a union of countably many closed, nowhere dense sets. Therefore, it suffices to prove that each $S_\varphi$ is closed and nowhere dense in $\mathrm{Gr}(2,n)$.
To that end, we decompose $S_\varphi$ as a disjoint union $T_\varphi \uplus U_\varphi$, where
$T_\varphi$ is the set of all $A\in S_\varphi$ for which $\varphi(A) \ne A$, and
$U_\varphi$ is the set of all $A\in S_\varphi$ for which $\varphi(A) = A$.
It suffices to prove that $T_\varphi$ and $U_\varphi$ are closed and nowhere dense in $\mathrm{Gr}(2,n)$.
Claim. $T_\varphi$ is either empty or is a submanifold of $\mathrm{Gr}(2,n)$ of dimension $n-1$.
Proof: Suppose $T_\varphi$ is nonempty. If $A\in T_\varphi$, then $A\cap\varphi^{-1}(A)$ is a one-dimensional subspace of $A$, and this contains exactly one pair $\{u,-u\}$ of unit vectors. Such a $u$ has the property that $u,\varphi u\in A$ and $\{u,\varphi u,\varphi^2 u\}$ are linearly independent. Let$$\widetilde{T_\varphi} = \{u\in \mathbb{R}^n : \|u\|=1\text{ and }u,\varphi u,\varphi^2 u\text{ are linearly independent}\}.$$Then $\widetilde{T_\varphi}$ is an open subset of the unit $(n-1)$-sphere in $\mathbb{R}^n$ and the map $p\colon \widetilde{T_\varphi}\to T_\varphi$ defined by $p(u) = \mathrm{Span}\{u,\varphi u\}$ is a degree two covering map, which proves the claim. $\square$
Since $n-1 < 2n-4$ for $n\geq 4$, this gives us the following.
Corollary. $T_\varphi$ is closed and nowhere dense in $\mathrm{Gr}(2,n)$ as long as $n\geq 4$.
Claim. $U_\varphi$ is a union of finitely many submanifolds of $\mathrm{Gr}(2,n)$, all of dimension at most $n-2$.
Proof: We separate the possible $A\in U_\varphi$ into three types, based on the eigenvalues of the restriction of $\varphi$ to $A$: The restriction of $\varphi$ to $A$ has two distinct real eigenvalues $\lambda,\mu$. The restriction of $\varphi$ to $A$ has one real eigenvalue $\lambda$ and is not diagonalizable. The restriction of $\varphi$ to $A$ has two complex eigenvalues $\lambda,\overline{\lambda}$.
In each case the eigenvalues of the restriction must also be eigenvalues of $\varphi$, of which there are only finitely many. Our strategy is to analyze the set of all $A$ of a given type corresponding to a given eigenvalue or pair of eigenvalues.
For type (1), let $\lambda$ and $\mu$ be distinct real eigenvalues of $\varphi$, and let $E_\lambda$ and $E_\mu$ be the corresponding eigenspaces. Then any $A$ corresponding to $\lambda$ and $\mu$ can be written uniquely as the sum of a one-dimensional subspace of $E_\lambda$ and a one-dimensional subspace of $E_\mu$. If $\dim(E_\lambda) = d_\lambda$ and $\dim(E_\mu) = d_\mu$, then the set of all such $A$ is homeomorphic to $\mathrm{Gr}(1,d_\lambda) \times \mathrm{Gr}(1,d_\mu)$, which is a manifold of dimension $d_\lambda+d_\mu - 2$. In particular, since $d_\lambda+d_\mu \leq n$, the set of all such $A$ for a given pair $\lambda,\mu$ is a submanifold of $\mathrm{Gr}(2,n)$ of dimension at most $n-2$.
For type (2), let $\lambda$ be a real eigenvalue of $\varphi$ with higher algebraic multiplicity than geometric multiplicity. Let $E_\lambda$ be the eigenspace for $\lambda$ and let $E_\lambda'$ be the nullspace of $(\varphi-\lambda I)^2$. Then any $A$ of type (2) corresponding to $\lambda$ has one-dimensional image in $E_\lambda'/E_\lambda$ and is entirely determined by this image. If $\dim(E_\lambda) = d_\lambda$ and $\dim(E_\lambda') = d_\lambda'$, then the set of all such $A$ is homeomorphic to $\mathrm{Gr}(1,d_\lambda'-d_\lambda)$, which is a manifold of dimension $d_\lambda'-d_\lambda - 1$. In particular, since $d_\lambda'-d_\lambda \leq n-1$, the set of all such $A$ for a given $\lambda$ is a submanifold of $\mathrm{Gr}(2,n)$ of dimension at most $n-2$.
For type (3), let $\lambda$ be a complex eigenvalue of $\varphi$, and let $E_\lambda$ be the eigenspace for $\lambda$ in $\mathbb{C}^n$. Then any $A$ of type (3) corresponding to $\lambda$ is obtained by taking a subspace of $E_\lambda$ of complex dimension one and taking the real part of each vector. If $\dim_{\mathbb{C}}(E_\lambda) = d_\lambda$, then the set of all such $A$ is homeomorphic to the complex Grassmannian $\mathrm{Gr}_{\mathbb{C}}(1,d_\lambda)$, which is a manifold of real dimension $2d_\lambda-2$. In particular, since $2d_\lambda \leq n$, the set of all such $A$ for a given $\lambda$ is a submanifold of $\mathrm{Gr}(2,n)$ of dimension at most $n-2$. $\square$
Corollary. $U_\varphi$ is closed and nowhere dense in $\mathrm{Gr}(2,n)$ for all $n\geq 3$.
Incidentally, what's going on here from an algebraic perspective should be roughly that each $S_\varphi$ is an algebraic subvariety of $\mathrm{Gr}(2,n)$ of dimension $n-1$, with $T_\varphi$ being the set of regular points of $S_\varphi$ and $U_\varphi$ being its set of singular points, but we don't need to know any of that to provide a topological proof that it's nowhere dense in $\mathrm{Gr}(2,n)$.
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In Emery's book "Stochastic calculus in manifolds", he shows how to make sense of integrals of the form $$ \int \langle\Theta_t, \mathbf{d} X_t\rangle,$$ where $X$ is a semimartingale on a manifold $M$ and $\Theta$ is a predictable process above $X$ taking values in the second order tangent bundle $\tau M$ (the bundle of second order differential operators without constant term). See also "An invitation to second order stochastic differential geometry".
In the above, $\mathbf{d}X_t$ is the second order vector $$\mathbf{d}X_t = dX^i_t D_i +\frac{1}{2}d [X^i_t, X^j_t] D_{ij},$$ the existence of which is "only metaphysical" (to speak with Emery's words) and the actual definition of the integral above does not use this "identity"; instead it is defined axiomatically.
Now I wonder: What is the relation of this to the theory of rough paths? In particular, can one make rigorous sense of $\mathbf{d}X_t$ as a covariant object using the theory of rough paths on a manifold, i.e. for a rough path, is its (second order?) infinitesimal increment somehow an element of the second order tangent bundle at a point? Can one then make sense
pathwise of the integral above?
And what is actually a good covariant definition of a rough path on a manifold in this setting? I must admit that I have a hard time understanding rough paths in a covariant setting.
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I've recently read Yan LeCuns comment on 1x1 convolutions:
In Convolutional Nets, there is no such thing as "fully-connected layers". There are only convolution layers with 1x1 convolution kernels and a full connection table.
It's a too-rarely-understood fact that ConvNets don't need to have a fixed-size input. You can train them on inputs that happen to produce a single output vector (with no spatial extent), and then apply them to larger images. Instead of a single output vector, you then get a spatial map of output vectors. Each vector sees input windows at different locations on the input.
In that scenario, the "fully connected layers" really act as 1x1 convolutions.
I would like to see a simple example for this.
Example
Assume you have a fully connected network. It has only an input layer and an output layer. The input layer has 3 nodes, the output layer has 2 nodes. This network has $3 \cdot 2 = 6$ parameters. To make it even more concrete, lets say you have a ReLU activation function in the output layer and the weight matrix
$$ \begin{align} W &= \begin{pmatrix} 0 & 1 & 1\\ 2 & 3 & 5\\ \end{pmatrix} \in \mathbb{R}^{2 \times 3}\\ b &= \begin{pmatrix}8\\ 13\end{pmatrix} \in \mathbb{R}^2 \end{align} $$
So the network is $f(x) = ReLU(W \cdot x + b)$ with $x \in \mathbb{R}^3$.
How would the convolutional layer have to look like to be the same? What does LeCun mean with "full connection table"?
I guess to get an equivalent CNN it would have to have exactly the same number of parameters. The MLP from above has $2 \cdot 3 + 2 = 8$ parameters.
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I am assuming you are working over a number field $K$. Then, $f(x,y)$ describes a conic section. If you homogenize it by adding suitable powers of $z$ to every monomial, you get a ternary quadratic form, which can always be diagonalized by suitable change of variables to obtain something of the form $Q(x,y,z) = ax^2 + by^2 + cz^2$. You are then asking when the quadratic form is isotropic (ie, there is a triple $(x,y,z)\ne (0,0,0)$ such that $ax^2 + by^2 + cz^2 = 0$).
This is the theory of ternary quadratic forms, which is contained in the theory of quaternion algebras. There are many sources on this online (I'd suggest looking up quadratic forms)
Since the
Hasse Principle holds for quadratic forms, $Q = 0$ has a solution over a number field $K$ if and only it has solutions over every $K_\nu$, where $\nu$ is a place of $K$ and $K_\nu$ is the completion of $K$ at $\nu$. If $K = \mathbb{Q}$, then $\{K_\nu\}_\nu = \{\mathbb{Q}_p\}_{p\text{ prime}}\cup\{\mathbb{R}\}$. (Here $\mathbb{Q}_p$ is the field of $p$-adic numbers).
Over $\mathbb{R}$, $Q = 0$ has solutions if and only if $a,b,c$ are
not all positive and not all negative.
At any finite place $\nu$, you can scale $Q$ by some number so that $a,b,c$ are all $\nu$-adically integral, and one of them is $-1$. WLOG suppose $c = -1$. In this case, $Q = 0$ has a solution iff the Hilbert symbol $(a,b)_p = 1$. If $K = \mathbb{Q}$ you can find formulas for it here: https://en.wikipedia.org/wiki/Hilbert_symbol
This is pretty much as good as good as it gets.
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I have a linear system of equations of size mxm, where m is large. However, the variables that I'm interested in are just the first n variables (n is small compared to m). Is there a way I can approximate the solution for the first m values without having to solve the entire system? If so, would this approximation be faster than solving the full linear system?
As others have pointed out, this is difficult to do with a direct solver. That said, it isn't that hard to do with iterative solvers. To this end, note that most iterative solvers in one way or another minimize the error with respect to some norm. Oftentimes, this norm is either induced by the matrix itself, but sometimes it is also just the l2 vector norm. But that doesn't have to be the case: you can choose which norm you want to minimize the error (or residual) in, and you could, for example, choose a norm in which you weigh the components you care about with 1 and all others with 1e-12, i.e. for example something like $|| x ||^2 = \sum_{i=1}^5 x_i^2 + $(1e-24)$\sum_{i=6}^N x_i^2$ and corresponding scalar product. Then write all steps of the iterative solver with respect to this norm and scalar product, and you get an iterative solver that pays significantly more attention to the vector elements you care about than to the others.
The question of course is whether you need fewer iterations than with the norm/scalar product that weighs all components equally. But that should indeed be the case: let's say you only care about the five first vector elements. Then you should need at most five iterations to reduce the error by a factor of 1e12 since five iterations is what is needed for the 5x5 system that describes them. That's not a proof but I'm pretty certain that you should indeed get away with a far smaller number of iterations if the weight in the norm (1e-12 above) is smaller than the tolerance with which you want to solve the linear system iteratively.
Forming the Schur complement
Suppose that you have permuted and partitioned your matrix into the form
$$A=\left(\begin{array}{cc}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right),$$
such that $A_{22}$ contains your degrees of freedom of interest and is much smaller than $A_{11}$, then one can form the Schur complement
$$ S_{22} := A_{22} - A_{21} A_{11}^{-1} A_{12},$$
either through a partial right-looking LU factorization or the explicit formula, and then $S_{22}$ can be understood in the following sense:
$$S_{22} x = y \;\;\rightarrow\;\; \left(\begin{array}{cc}A_{11} & A_{12}\\ A_{21} & A_{22}\end{array}\right) \left(\begin{array}{c}\star\\ x\end{array}\right)=\left(\begin{array}{c}0\\ y\end{array}\right),$$
where $\star$ represents the 'uninteresting' portion of the solution. Thus, provided a right-hand side which is only nonzero in the degrees of freedom of the Schur complement $S_{22}$, we need only solve against $S_{22}$ in order to get the portion of the solution corresponding to those degrees of freedom.
Computational complexity in unstructured dense case
Setting $N$ to the height of $A$ and $n$ to the height of $A_{22}$, then the standard method for computing $S_{22}$ is to first factor $L_{11} U_{11} := A_{11}$ (let's ignore pivoting for now) in roughly $2/3 (N-n)^3$ work, then to form
$$ S_{22} := A_{22} - (A_{21} U_{11}^{-1})(L_{11}^{-1} A_{12}) = A_{22} - A_{21} A_{11}^{-1} A_{12}$$
using two triangle solves requiring $n(N-n)^2$ work each, and then performing the update to $A_{22}$ in $2n^2 (N-n)$ work.
Thus, the total work is roughly $2/3 (N-n)^3 + 2n(N-n)^2 + 2n^2 (N-n)$. When $n$ is very small, $N-n \approx N$, so the cost can be seen to be roughly $2/3 N^3$, which is the cost of a full factorization.
The benefit is that, if there is a very large number of right-hand sides to be solved with the same system of equations, then $S_{22}$ could potentially be reused a large number of times, where each solve would only require $2n^2$ work (rather than $2N^2$ work) if $S_{22}$ is factored.
Computational complexity in the (typical) sparse case
If your sparse system arose from some type of finite-difference or finite-element approximation, then sparse-direct solvers will almost certainly be able to exploit some of the structure; 2d systems can be solved with $O(N^{3/2})$ work and $O(N \log N)$ storage, while 3d systems can be solved with $O(N^2)$ work and $O(N^{4/3})$ storage. The factored systems can then be solved with the same amount of work as the storage requirements.
The point of bringing up the computational complexities is that, if $n \approx \sqrt{N}$ and you have a 2d system, then since the Schur complement will likely be dense, the solve complexity given the factored Schur complement will be $O(n^2) = O(N)$, which is only missing a logarithmic factor versus solving the full system! In 3d, it requires $O(N)$ work instead of $O(N^{4/3})$.
It is thus important to keep in mind that, in your case where $n=\sqrt{N}$, there will only be significant savings if you're working in several dimensions and have many right-hand sides to solve. The model reduction approach
Since Paul asked, I'll talk about what happens if you use projection-based model reduction methods on this problem. Suppose that you could come up with a projector $\mathbf{P}$ such that the range of $\mathbf{P}$, denoted $\mathcal{R}(\mathbf{P})$, contains the solution to your linear system $\mathbf{Ax} = \mathbf{b}$, and has dimension $k$, where $k$ is the number of unknowns for which you wish to solve in a linear system.
A singular value decomposition of $\mathbf{P}$ will yield the following partitioned matrix:
$$\mathbf{P} = \left[ \begin{array}{cc}\mathbf{V} & * \end{array} \right]\left[\begin{array}{cc}\mathrm{diag}(\mathbf{1}_{k}) & \mathbf{0} \\ \mathbf{0} & \mathbf{0}\end{array}\right]\left[\begin{array}{c} \mathbf{W}^{T} \\ *\end{array}\right].$$
The matrices obscured by stars matter for other things (like estimating error, etc.), but for now, we'll avoid dealing with extraneous details. It follows that
$$\mathbf{P} = \mathbf{VW}^{T}$$
is a full rank decomposition of $\mathbf{P}$.
Essentially, you'll solve the system
$$\mathbf{PAx} = \mathbf{Pb}$$
in a clever way, because $\mathbf{V}$ and $\mathbf{W}$ also have the property that $\mathbf{W}^{T}\mathbf{V} = \mathbf{I}$. Multiplying both sides of $\mathbf{PAx} = \mathbf{Pb}$ by $\mathbf{W}^{T}$ and letting $\mathbf{y} = \mathbf{V}\widehat{\mathbf{x}}$ be an approximation for $\mathbf{x}$ yields
$$\mathbf{W}^{T}\mathbf{A}\widehat{\mathbf{x}} = \mathbf{W}^{T}\mathbf{b}.$$
Solve for $\widehat{\mathbf{x}}$, premultiply it by $\mathbf{V}$, and you have $\mathbf{y}$, your approximation for $\mathbf{x}$.
Why the Schur complement approach is probably better
For starters, you have to pick $\mathbf{P}$ somehow. If the solution to $\mathbf{Ax} = \mathbf{b}$ is in $\mathcal{R}(\mathbf{P})$, then $\mathbf{y} = \mathbf{x}$, and $\mathbf{y}$ isn't an approximation. Otherwise, $\mathbf{y} \neq \mathbf{x}$, and you introduce some approximation error. This approach doesn't really leverage at all the structure you mentioned wanting to exploit. If we pick $\mathbf{P}$ such that its range is the standard unit basis in the coordinates of $\mathbf{x}$ you want to calculate, the corresponding coordinates of $\mathbf{y}$ will have errors in them. It's not clear how you'd want to pick $\mathbf{P}$. You could use an SVD of $\mathbf{A}$, for instance, and select $\mathbf{P}$ to be the product of the first $k$ left singular vectors of $\mathbf{A}$ and the adjoint of the first $k$ right singular vectors of $\mathbf{A}$, assuming that singular vectors are arranged in decreasing order of singular value. This choice of projector would be equivalent to performing proper orthogonal decomposition on $\mathbf{A}$, and it would minimize the L$_{2}$-error in the approximate solution.
In addition to introducing approximation errors, this approach also introduces three extra matrix multiplies on top of the linear solve of the smaller system and the work needed to calculate $\mathbf{V}$, and $\mathbf{W}$.
Unless you're solving the same linear system a lot, only changing the right hand side, and $\mathbf{P}$ is still a "good" projection matrix for all of those systems, those extra costs will probably make solving the reduced system more expensive than solving your original system.
The drawbacks are much like JackPoulson's approach, except that you're not quite leveraging the structure that you mentioned.
The long answer is...sort of.
You can re-arrange your system of equations such that the farthest right $k$ columns are the variables which you wish to solve for.
Step 1: Perform Gaussian Elimination so that the matrix is upper triangular. Step 2: solve by back substitution for only the first (last) $k$ variables which you are interested in
This will save you the computational complexity of having to solve for the last $n-k$ variables via back-substitution, which could be worth it if $n$ is as large as you say. Keep in mind that a fair amount of work will still have to be done for step 1.
Also, keep in mind that restricting the order in which you are going to perform back-substituion may restrict the form of the matrix (it takes away the ability to exchange columns) which could
possibly lead to an ill conditioned system, but I am not sure about that - just something to keep in mind.
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Feynman diagrams provide a very compact and intuitive way of representing interactions between particles. These diagrams can be included into LaTeX documents thanks to a few packages. One of the older packages is
feynmf which uses MetaPost in order to generate the diagrams. More recently, a new package called Ti
kZ-Feynman has been published which uses Ti kZ in order to generate Feynman diagrams.
Contents
Ti
kZ-Feynman is a LaTeX package allowing Feynman diagrams to be easily generated within LaTeX with minimal user instructions and without the need of external programs. It builds upon the Ti kZ package and its graph drawing algorithms in order to automate the placement of many vertices. Ti kZ-Feynman still allows fine-tuned placement of vertices so that even complex diagrams can be generated with ease.
Currently, Ti
kZ-Feynman is too new to have made it into ShareLaTeX's installation, but we are working to get it included soon. In the meantime, it is possible to include the package files manually in a ShareLaTeX project as shown in this template.
After installing the package, the Ti
kZ-Feynman package can be loaded with
\usepackage{tikz-feynman} in the preamble. It is recommend that you also specify the version of Ti
kZ-Feynman to use with the
compat package option:
\usepackage[compat=1.0.0]{tikz-feynman}. This ensures that any new versions of Ti
kZ-Feynman do not produce any undesirable changes without warning.
Feynman diagrams can be declared with the
\feynmandiagram command. It is analogous to the
\tikz command from Ti
kZ and requires a final semi-colon (
;) to finish the environment. For example, a simple
s-channel diagram is:
\feynmandiagram [horizontal=a to b] { i1 -- [fermion] a -- [fermion] i2, a -- [photon] b, f1 -- [fermion] b -- [fermion] f2, };
Let's go through this example line by line:
\feynmandiagram introduces the Feynman diagram and allows for optional arguments to be given in the brackets
[<options>]. In this instance,
horizontal=a to b orients the algorithm outputs such that the line through vertices
a and
b is horizontal.
i1,
a and
i2) and connecting them with edges
--. Just like the
\feynmandiagram command above, each edge also take optional arguments specified in brackets
[<options>]. In this instance, we want these edges to have arrows to indicate that they are fermion lines, so we add the
fermion style to them. As you will see later on, optional arguments can also be given to the vertices in exactly the same way.
a and
b with an edge styled as a photon. Since there is already a vertex labelled
a, the algorithm will connect it to a new vertex labeled
b.
f1 and
f2. It re-uses the previously labelled
b vertex.
;) is important.
The name given to each vertex in the graph does not matter. So in this example,
i1,
i2 denote the initial particles;
f1,
f2 denotes the final particles; and
a,
b are the end points of the propagator. The only important aspect is that what we called
a in line 2 is also
a in line 3 so that the underlying algorithm treats them as the same vertex.
The order in which vertices are declared does not matter as the default algorithm re-arranges everything. For example, one might prefer to draw the fermion lines all at once, as with the following example (note also that the way we named vertices is completely different):
\feynmandiagram [horizontal=f2 to f3] { f1 -- [fermion] f2 -- [fermion] f3 -- [fermion] f4, f2 -- [photon] p1, f3 -- [photon] p2, };
As a final remark, the calculation of where vertices should be placed is usually done through an algorithm written in Lua. As a result, LuaTeX is required in order to make use of these algorithms. If LuaTeX is not used, Ti
kZ-Feynman will default to a more rudimentary algorithm and will warn the user instead.
So far, the examples have only used the
photon and
fermion styles. The Ti
kZ-Feynman package comes with quite a few extra styles for edges and vertices which are all documented over in the package documentation. For example, it is possible to add momentum arrows with
momentum=<text>, and in the case of end vertices, the particle can be labelled with
particle=<text>. To demonstrate how they are used, we take the generic
s-channel diagram from earlier and make it a electron-positron pairs annihilating into muons:
\feynmandiagram [horizontal=a to b] { i1 [particle=\(e^{-}\)] -- [fermion] a -- [fermion] i2 [particle=\(e^{+}\)], a -- [photon, edge label=\(\gamma\), momentum'=\(k\)] b, f1 [particle=\(\mu^{+}\)] -- [fermion] b -- [fermion] f2 [particle=\(\mu^{-}\)], };
In addition to the style keys documented below, style keys from Ti
kZ can be used as well:
\feynmandiagram [horizontal=a to b] { i1 [particle=\(e^{-}\)] -- [fermion, very thick] a -- [fermion, opacity=0.2] i2 [particle=\(e^{+}\)], a -- [red, photon, edge label=\(\gamma\), momentum'={[arrow style=red]\(k\)}] b, f1 [particle=\(\mu^{+}\)] -- [fermion, opacity=0.2] b -- [fermion, very thick] f2 [particle=\(\mu^{-}\)], };
For a list of all the various styles that Ti
kZ provides, have a look at the Ti kZ manual; it is extremely thorough and provides many usage examples.
By default, the
\feynmandiagram and
\diagram commands use the
spring layout algorithm to place all the edges. The
spring layout algorithm attempts to `spread out' the diagram as much as possible which—for most simpler diagrams—gives a satisfactory result; however in some cases, this does not produce the best diagram and this section will look at alternatives. There are three main alternatives:
draw=none. The algorithm will treat these extra edges in the same way, but they are simply not drawn at the end;
The underlying algorithm treats all edges in exactly the same way when calculating where to place all the vertices, and the actual drawing of the diagram (after the placements have been calculated) is done separately. Consequently, it is possible to add edges to the algorithm, but prevent them from being drawn by adding
draw=none to the edge style.
This is particularly useful if you want to ensure that the initial or final states remain closer together than they would have otherwise as illustrated in the following example (note that
opacity=0.2 is used instead of
draw=none to illustrate where exactly the edge is located).
% No invisible to keep the two photons together \feynmandiagram [small, horizontal=a to t1] { a [particle=\(\pi^{0}\)] -- [scalar] t1 -- t2 -- t3 -- t1, t2 -- [photon] p1 [particle=\(\gamma\)], t3 -- [photon] p2 [particle=\(\gamma\)], };
% Invisible edge ensures photons are parallel \feynmandiagram [small, horizontal=a to t1] { a [particle=\(\pi^{0}\)] -- [scalar] t1 -- t2 -- t3 -- t1, t2 -- [photon] p1 [particle=\(\gamma\)], t3 -- [photon] p2 [particle=\(\gamma\)], p1 -- [opacity=0.2] p2, };
The graph drawing library from Ti
kZ has several different algorithms to position the vertices. By default,
\diagram and
\feynmandiagram use the
spring layout algorithm to place the vertices. The
spring layout attempts to spread everything out as much as possible which, in most cases, gives a nice diagram; however, there are certain cases where this does not work. A good example where the
spring layout doesn't work are decays where we have the decaying particle on the left and all the daughter particles on the right.
% Using the default spring layout \feynmandiagram [horizontal=a to b] { a [particle=\(\mu^{-}\)] -- [fermion] b -- [fermion] f1 [particle=\(\nu_{\mu}\)], b -- [boson, edge label=\(W^{-}\)] c, f2 [particle=\(\overline \nu_{e}\)] -- [fermion] c -- [fermion] f3 [particle=\(e^{-}\)], };
% Using the layered layout \feynmandiagram [layered layout, horizontal=a to b] { a [particle=\(\mu^{-}\)] -- [fermion] b -- [fermion] f1 [particle=\(\nu_{\mu}\)], b -- [boson, edge label'=\(W^{-}\)] c, c -- [anti fermion] f2 [particle=\(\overline \nu_{e}\)], c -- [fermion] f3 [particle=\(e^{-}\)], };
You may notice that in addition to adding the
layered layout style to
\feynmandiagram, we also changed the order in which we specify the vertices. This is because the
layered layout algorithm does pay attention to the order in which vertices are declared (unlike the default
spring layout); as a result,
c--f2, c--f3 has a different meaning to
f2--c--f3. In the former case,
f2 and
f3 are both on the layer below
c as desired; whilst the latter case places
f2 on the layer above
c (that, the same layer as where the W-boson originates).
In more complicated diagrams, it is quite likely that none of the algorithms work, no matter how many invisible edges are added. In such cases, the vertices have to be placed manually. Ti
kZ-Feynman allows for vertices to be manually placed by using the
\vertex command.
The
\vertex command is available only within the
feynman environment (which itself is only available inside a
tikzpicture). The
feynman environment loads all the relevant styles from Ti
kZ-Feynman and declares additional Ti kZ-Feynman-specific commands such as
\vertex and
\diagram. This is inspired from PGFPlots and its use of the
axis environment.
The
\vertex command is very much analogous to the
\node command from Ti
kZ, with the notable exception that the vertex contents are optional; that is, you need not have
{<text>} at the end. In the case where
{} is specified, the vertex automatically is given the
particle style, and otherwise it is a usual (zero-sized) vertex.
To specify where the vertices go, it is possible to give explicit coordinates though it is probably easiest to use the
positioning library from Ti
kZ which allows vertices to be placed relative to existing vertices. By using relative placements, it is possible to easily tweak one part of the graph and everything will adjust accordingly—the alternative being to manually adjust the coordinates of every affected vertex.
Finally, once all the vertices have been specified, the
\diagram* command is used to specify all the edges. This works in much the same way as
\diagram (and also
\feynmandiagram), except that it uses an very basic algorithm to place new nodes and allows existing (named) nodes to be included. In order to refer to an existing node, the node must be given in parentheses.
This whole process of specifying the nodes and then drawing the edges between them is shown below for the muon decay:
\begin{tikzpicture} \begin{feynman} \vertex (a) {\(\mu^{-}\)}; \vertex [right=of a] (b); \vertex [above right=of b] (f1) {\(\nu_{\mu}\)}; \vertex [below right=of b] (c); \vertex [above right=of c] (f2) {\(\overline \nu_{e}\)}; \vertex [below right=of c] (f3) {\(e^{-}\)}; \diagram* { (a) -- [fermion] (b) -- [fermion] (f1), (b) -- [boson, edge label'=\(W^{-}\)] (c), (c) -- [anti fermion] (f2), (c) -- [fermion] (f3), }; \end{feynman} \end{tikzpicture}
The
feynmf package lets you easily draw Feynman diagrams in your LaTeX documents. All you need to do is specify the vertices, the particles and the labels, and it will automatically layout and draw your diagram for you.
Let's start with a quick example:
\begin{fmffile*}{diagram} \begin{fmfgraph}(40,25) \fmfleft{i1,i2} \fmfright{o1,o2} \fmf{fermion}{i1,v1,o1} \fmf{fermion}{i2,v2,o2} \fmf{photon}{v1,v2} \end{fmfgraph} \end{fmffile*}
The
fmffile* environment must be put around all of your Feynman diagrams. You can use fmffile environment for multiple diagrams, so you can put one around your whole document and forget about it. The second argument to the fmffile environment tells LaTeX where to write the files that it uses to store the diagram. You can name this whatever you want, but you need to run metafont on your diagram between LaTeX runs in order for your diagram to show up (ShareLaTeX does this automatically):
The 'fmfgraph' environment starts a Feynman diagram, and the figures in brackets afterwards specify the width and height of the diagram.
The first thing you need to do is specify your external vertices, and where they should be positioned. You can name your vertices anything you like, and say where they should be positioned with the commands
\fmfleft, \fmfright, \fmftop, \fmfbottom. For example
% Creates two vertices on the left called i1 and i2 \fmfleft{i1,i2} % Creates two vertices on the right called o1 and o2 \fmfright{o1,o2}
You can connect up vertices with the
\fmf, which will create new vertices if you pass in names that haven't been created yet. For example
% Will create a fermion line between i1 and % the newly created v1, and between v1 and o1. \fmf{fermion}{i1,v1,o1} % Will create a photon line between v1 and the newly created v2 \fmf{photon}{v2,v2}
A vertex can be labelled using the
\fmflabel command, which takes two arguments: the label to apply to the vertex, and the name of the vertex to apply it to. For example, in the above diagram, if we add in the following labels, we get the updated diagram below:
Note that math mode can used inside the vertex labels, as we have done above.
We've seen the 'photon' and 'fermion' line styles above, but the
feynmf package support many more.
Appearance Name(s) gluon, curly dbl_curly dashes scalar, dashes_arrow dbl_dashes dbl_dashes_arrow dots ghost, dots_arrow dbl_dots dbl_dots_arrow phantom phantom_arrow vanilla, plain fermion, electron, quark, plain_arrow double, dbl_plain double_arrow, heavy, dbl_plain_arrow boson, photon, wiggly dbl_wiggly zigzag dbl_zigzag
For more information see:
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ProblemFind the Fourier transformation of $u(x) = \frac{1}{1+x^2}$
I want $\int_\mathbb R e^{-itx} \frac{1}{1+x^2} dx$. Let $f(z) = e^{-itz} \frac{1}{1+z^2}$, $z \in \mathbb C$, let's integrate this over the semi-circunference of radius $R$ and the line $[-R, R]$.
First, the integral on the circumference vanishes (with the change $z = Re^{i\theta}$)
$$\left|\int_{\gamma_R} \frac{e^{-izt}}{z^2+1}dz\right| = \left|\int_0^\pi \frac{e^{-itRe^{i\theta}}}{R^2e^{2i\theta}+1}Rie^{i\theta}d\theta\right| \le \int_0^\pi \frac{R}{|R^2e^{2i\theta}+1|}d\theta \le \int_0^\pi\frac1R d\theta= \frac\pi R$$ That goes to $0$ as $R \to \infty$.
The integral on the whole path is $2\pi i \text{ Res }(f, i)$, and since the integral on the circumference vanishes we have
$$\int_\mathbb R \frac{e^{-ixt}}{1+x^2} = 2 \pi i \text{ Res }(f, i)$$
I know that
If $z_0 \neq \infty$ is a pole of order $m$ for $f$, then set $g(z) = (z-z_0)^mf(z)$ and $\displaystyle \text{ Res }(f, z_0) = \frac{g^{(m-1)}(z_0)}{(m-1)!}$
Since $z_0 = i$ is a pole of order $1$, I set $\displaystyle g(z) = \frac{e^{-izt}}{z+i}$ and the residue should be $\displaystyle g(i) = \frac{e^t}{2i}$
And the integral then is $$\int_\mathbb R \frac{e^{-ixt}}{1+x^2} = \pi e^t$$
I know this is wrong (the fourier transform isn't even bounded) but I don't know where the error is. I am pretty sure the residue is wrong, but why?
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Let $G_1$, $G_2$ be arbitrary groups, and $H$ be any group with homomorphisms $\theta_1:G_1\rightarrow H$, $\theta_2:G_1\rightarrow H$. Show that there exist a group $G$ and homomorphisms $\beta_1:G_1\rightarrow G$, $\beta_2:G_2\rightarrow G$, and a unique homomorphism $\theta:G\rightarrow H$ such that $\beta_1\theta = \theta_1$ and $\beta_2\theta = \theta_2$.
Here is my work so far:
Basically I have shown the existence of two homomorphisms taking $G \rightarrow H$. But I'm struggling to prove uniqueness.
Let $X$ be the generating set of both $G_1$ and $G_2$. Let $F$ be the free group on $X$. Define $G:= F/(ker\pi_1 \cup ker\pi_2)^{F} = F/ker\pi_3$. Here $\pi_1$ and $\pi_2$ are induced homomorphisms from $F$ to $G_1$ and $G_2$, respectively.
By Von Dyck's theorem, there exist surjective homomorphisms $\beta_i:G_i \rightarrow G$ where $i\in\{1,2\}$, defined as $f\pi_i\beta_i = f\pi_3$.
Define $\theta'_i:G\rightarrow H$ as follows: $f\pi_3\theta'_i := f\pi_i\theta_i$. Then both $\theta'_1$ and $\theta'_2$ are homomorphisms.
Now in order to prove uniqueness of $\theta$, I must show $\theta'_1=\theta'_2$, and this is where I'm stuck.
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Let us consider monic polynomials of $n^{\text{th}}$ degree $p(t) = t^n + a_{n-1} t^{n-1} + \dots + a_1 t + a_0$ with real coefficients. We know the roots are continuous functions of the coefficients. Let us define a map $F: \mathbb R^n \to \mathbb R_+^n$ by \begin{align*} F: (a_{n-1}, \dots, a_0) \mapsto (r_1, \dots, r_n) \mapsto (|r_1|, \dots, |r_n|), \end{align*} where $r_j = r_j(a_{n-1}, \dots, a_0)$ are roots of the polynomial and $|\cdot|$ denotes the modulus of a complex number. Clearly $F$ is a continuous function. Let us take the Euclidean norm on $\mathbb R^n$. Suppose for some fixed coefficients $b = (b_{n-1}, \dots, b_0)$, the roots are all nonzero and let $(c_1, \dots, c_n)$ be the roots. if we take an open ball $B_b(\varepsilon)$ around $b$, will the image $F[ B_b(\varepsilon) ]$ contain a cube, i.e., $(|c_1|-\varepsilon', |c_1|+\varepsilon') \times (|c_2|-\varepsilon', |c_2|+\varepsilon') \times \dots \times (|c_n|-\varepsilon', |c_n|+\varepsilon')$ for some possible smaller $\varepsilon'$?
As I commented, your function $F$ is not well-defined. However, your question still makes sense without it. You can ask, given a multiset of nonzero complex numbers $c_1,\dots,c_n$ invariant under conjugation and $\epsilon>0$, does there exist $\epsilon'>0$ such that for any $r_1,\dots,r_n\in\mathbb{R}$ with $|r_k-|c_k||<\epsilon'$ for each $k$, there exist $d_1,\dots,d_n\in\mathbb{C}$ also invariant under conjugation such that $|d_k|=r_k$ for each $i$ and coefficients of the monic polynomial of with the $d_k$ as roots are within $\epsilon$ of the coefficients of the monic polynomial with the $c_k$ as roots. (The invariance under conjugation condition here corresponds to the requirement that the polynomial has real coefficients.)
The answer to this question is no. For instance, let $n=2$, and $(c_1,c_2)=(i,-i)$ and consider $r_1=1+\epsilon'/2$ and $r_2=1$. Notice then that if $|d_k|=r_k$ for $k=1,2$, then $d_1$ and $d_2$ cannot be conjugates of each other, and so they must both be real in order to be invariant under conjugation. It follows that there are no such $d_1$ and $d_2$ that give rise to a polynomial close to the polynomial $t^2+1$ with roots $c_1$ and $c_2$.
Or, in more concrete terms, there is no polynomial with real coefficients close to $t^2+1$ whose roots have distinct absolute values, since any nearby polynomial has no real roots and hence the roots must be a complex conjugate pair.
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Let $a_1, a_2, \ldots$ be an infinite set of non-negative samples taken from a distribution $P_0(a)$, and write
$$\tag{1} \label{problem} S = 1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \ldots. $$ Notice that if the $a_i$ were all the same, $S$ would be a regular geometric series, with value $S = \frac{1}{1-a}$. How will the introduction of $a_i$ randomness change this sum? Will $S$ necessarily converge? How is $S$ distributed? In this post, we discuss some simple techniques to answer these questions.
Note: This post covers work done in collaboration with my aged p, S. Landy.
Follow @efavdb Follow us on twitter for new submission alerts! Introduction — a stock dividend problem
To motivate the sum (\ref{problem}), consider the problem of evaluating the total output of a stock that pays dividends each year in proportion to its present value — say $x \%$. The price dynamics of a typical stock can be reasonably modeled as a geometric random walk$^1$:
$$\label{prod} \tag{2} price(t) = price(t-1) * a_t, $$ where $a_t$ is a random variable, having distribution $P_0(a_t)$. Assuming this form for our hypothetical stock, its total lifetime dividends output will be $$\tag{3} x \times \sum_{t = 0}^{\infty} price(t) = x \times price(0) \left ( 1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \ldots \right) $$ The inner term in parentheses here is precisely (\ref{problem}). More generally, a series of this form will be of interest pretty much whenever geometric series are: Population growth problems, the length of a cylindrical bacterium at a series of time steps$^2$, etc. Will the nature of these sums change dramatically through the introduction of growth variance?
To characterize these types of stochastic geometric series, we will start below by considering their moments: This will allow us to determine the average value of (\ref{problem}), it’s variance etc. This approach will also allow us to determine a condition that is both necessary and sufficient for the sum’s convergence. Following this, we will introduce an integral equation satisfied by the $P(S)$ distribution. We demonstrate its application by solving the equation for a simple example.
The moments of $S$
To solve for the moments of $S$, we use a trick similar to that used to sum the regular geometric series: We write
$$\tag{4} \label{trick} S = 1 + a_1 + a_1 a_2 + \ldots \equiv 1 + a_1 T, $$ where $T = 1 + a_2 + a_2 a_3 + \ldots.$ Now, because we assume that the $a_i$ are independent, it follows that $a_1$ and $T$ are independent. Further, $S$ and $T$ are clearly distributed identically, since they take the same form. Subtracting $1$ from both sides of the above equation, these observations imply $$\tag{5} \label{moments} \overline{(S-1)^k} = \sum_j {k \choose j} (-1)^j \overline{S^{k-j}} = \overline{ a^k S^k} = \overline{a^k} \ \overline{S^k}. $$ This expression can be used to relate the moments of $S$ to those of $a$ — a useful result, whenever the distribution of $a$ is known, allowing for the direct evaluation of its moments.
To illustrate, let us get the first couple of moments of $S$, using (\ref{moments}). Setting $k=1$ above, we obtain
$$\tag{6} \label{mean} \overline{S -1} = \overline{a} \overline{S} \ \ \to \ \ \overline{S} = \frac{1}{1 – \overline{a}} $$ The right side here looks just like the usual geometric sum result, with $a$ replaced by its average value. Similarly, setting $k =2$ in (\ref{moments}), we can solve for the second moment of $S$. Subtracting the square of the first gives the following expression for the sum’s variance, $$\tag{7} \label{var} var(S) = \frac{var(a)}{(1 – \overline{a})^2(1 – \overline{a^2})}. $$ As one might intuit, the variance of $S$ is proportional to the variance of $a$.
Expressions (\ref{mean}) and (\ref{var}) are the most practical results of this post: They provide formal general expressions for the mean and variance for a sum of form (\ref{problem}). They can be used to provide a statistical estimate and error bar for a sum of form $S$ in any practical context. It is interesting/nice that the mean takes such a natural looking form — one that many people likely make use of already, without putting much thought into.
The expressions above are also of some theoretical interest: Note, for example, that as $\overline{a} \to 1$ from below, the average value of $S$ diverges, and then becomes negative as $a$ goes above this value. This is clearly impossible, as $S$ is a sum of positive terms. This indicates that $S$ has no first moment whenever $\overline{a} \geq 1$, while (\ref{mean}) holds whenever $\overline{a} < 1$. Similarly, (\ref{var}) indicates that the second moment of $S$ exists and is finite whenever $\overline{a^2} < 1$. In fact, this pattern continues for all $k$: $\overline{S^k}$ exists and is finite if and only if $\overline{a^k} < 1$ -- a result that can be obtained from (\ref{moments}). A rigorous and elementary proof of these statements can be found in an earlier work by Szabados and Szekeley$^3$. The simple moment equation (\ref{moments}) can also be found there.
Condition for the convergence of $S$
A simple condition for the convergence of $S$ can also be obtained using (\ref{moments}). The trick is to consider the limit as $k$ goes to zero of the $k$-th moments. This gives, for example, the average of $1$ with respect to $P(S)$. If this is finite, then the distribution of $P$ is normalizable. Otherwise, $S$ must diverge: Setting $k = \epsilon$ in (\ref{moments}), expanding to first order in $\epsilon$ gives
$$\tag{8} \label{approximate_log} \overline{ \exp [\epsilon \log (S -1) ]} \sim \overline{ 1 + \epsilon \log (S -1) } \sim \overline{ 1 + \epsilon \log S } \ \overline{ 1 + \epsilon \log a}. $$ Solving for $\overline{1}_S$, the average of $1$ with respect to $P(S)$, gives $$\tag{9} \overline{1}_S = \frac{\overline{\log( 1 – \frac{1}{S})}}{\log a} + O(\epsilon). $$ Like the integer moment expressions above, the right side here is finite up to the point where its denominator diverges. That is, the series will converge, if and only if $\overline{\log a} < 0$, a very simple condition$^4$. Integral equation for the distribution $P(S)$
We have also found that one can sometimes go beyond solving for the moments of $S$, and instead solve directly for its full distribution: Integrating (\ref{trick}) over $a$ gives
$$\tag{10} \label{int} P(S_0) = \int da P_0(a) \int dS P(S) \delta(1+ a S – S_0) = \int \frac{da}{a} P_0(a) P \left (\frac{S_0 -1}{a} \right). $$ This is a general, linear integral equation for $P(S)$. At least in some cases, it can solved in closed-form. An example follows. Uniformly distributed $a_i$
To demonstrate how one might solve the equation (\ref{int}), we consider here the case where the $a_i$ are uniform on $[0,1]$. In this case, writing $a = \frac{S_0 -1}{v}$, (\ref{int}) goes to
$$\tag{11} \label{int2} P(S_0) = \int_{S_0-1}^{\infty} P\left (v\right) \frac{1}{v}dv. $$ To progress, we differentiate with respect to $S_0$, which gives $$\tag{12} \label{delay} P^{\prime} (S_0)\equiv – \frac{1 }{S_0 -1}\times P\left (S_0 -1\right). $$ Equation (\ref{delay}) is a delay differential equation. It can be solved through iterated integrations: To initiate the process, we note that $P(S_0)$ is equal to zero for all $S_0< 1$. Plugging this observation into (\ref{delay}) implies that $P(S_0) \equiv J$ -- some constant -- for $S \in (1,2)$. Continuing in this fashion, repeated integrations of (\ref{delay}) gives $$\tag{13} P (S_0) = \begin{cases} J, \ \ \ S_0 \in (1,2) \\ J[1 - \log (S_0 -1)], \ \ \ S_0 \in (2,3) \\ J \left [ 1 - \log(S_0 - 1) + Li_2(2-S_0) + \frac{ \log(S_0 - 2)}{\log(S_0 - 1)} - Li_2(-1) \right ], \ \ S_0 \in (3,4) \\ \ldots, \end{cases} $$ where $Li_2$ is the polylogarithm function. In practice, to find $J$ one can solve (\ref{delay}) numerically, requiring $P(S)$ to be normalized. The figure below compares the result to a simulation estimate, obtained via binning the results of 250,000 random sums of form (\ref{problem}). The two agree nicely. Discussion
Consideration of this problem was motivated by a geometric series of type (\ref{problem}) that arose in my work at Square. In this case, I was interested in understanding the bias and variance in the natural estimate (\ref{mean}) to this problem. After some weeks of tinkering with S Landy, I was delighted to find that rigorous, simple results could be obtained to characterize these sums, the simplest being the moment and convergence results above. We now realize that these particular issues have already been well- (and better-)studied, by others$^3$.
As for the integral equation approach, we have not found any other works aimed at solving this problem in general. The method discussed in the example above can be used for any $P_0(a)$ that is uniform over a finite segment. We have also found solutions for a few other cases. Unfortunately, we have so far been unable to obtain a formal, general solution in closed form. However, we note that standard iterative approaches can always be used to estimate the solution to (\ref{int}). Finally, in cases where all moments exist, these can also be used to determine $P$.
References and comments
[1] For a discussion on the geometric random walk model for stocks, see here.
[2] Elongated bacteria — eg., e. coli — grow longer at an exponential rate — see my paper on how cell shape affects growth rates. Due to randomness inherent in the growth rates, bacteria populations will have a length distribution, similar in form to $P(S)$.
[3] “An exponential functional of random walks” by Szabados and Szekeley, Journal of Applied Probability 2003.
[4] Although we have given only a hand-waving argument for this result, the authors of [3] state — and give a reference for — the fact that it can be proven using the law of large numbers: By independence of the $a_i$, the $k$-th term in the series approaches $(\overline{\log a})^k$ with probability one, at large $k$. Simple convergence criteria then give the result.
[5] The moment equation (\ref{moments}) can also be obtained from the integral equation (\ref{int}), where it arrises from the application of the convolution theorem.
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DOI:10.1128/AEM.01277-08 ABSTRACT
Network models offer computationally efficient tools for estimating the variability of single-cell lag phases. Currently, optical methods for estimating the variability of single-cell lag phases use single-cell inocula and are technically challenging. A Bayesian network model incorporating small uncertain inocula addresses these limitations.
It is possible to measure the variability of growth from single cells by microscopy (4, 5), but these methods are laborious. Optical density measurements provide a rapid method for generating growth data from small inocula. However, the cell density for which a significant optical measurement is obtained is quite high. The detection threshold for this optical method is typically ∼10
7 cells ml −1. Conclusions concerning growth from a single cell to a population of ∼10 7 cells must be extrapolated from growth observed after the detection threshold (3). This extrapolation adds to measurement uncertainty when trying to estimate the variability of growth from single cells.
There is a consensus among microbiologists that bacterial cells undergo a period of adjustment (lag phase) when placed in a new environment. After adjustment, an exponentially growing population is established which, ultimately, reaches an upper limit (stationary phase). This population limit is caused by the depletion of nutrients or by the accumulation of waste products of metabolism. Growth is not observed during the lag phase. The lag phase and subsequent growth before the stationary phase can be modeled using a biphasic linear function,$$mathtex$$\[\mathrm{ln}(N_{t}){=}\mathrm{ln}(n){+}{\mu}{\times}\mathrm{max}(t{-}L)\]$$mathtex$$(1) where
N t is the population size at time t, nis the inoculum size, μ is the specific growth rate, and Lis the population lag time.
It has been shown that the population lag phase, under the assumptions of the biphasic linear model, is related to the lag phase of the individual cells which make up the inoculum (1). A bacterial population at time
t, grown from an inoculum consisting of n cells, can be represented by$$mathtex$$\[N_{t}{=}\ \begin{array}{l}n\\{\sum}\\i{=}1\end{array}e^{{\mu}{\times}\mathrm{max}(t{-}L_{i},0)}\]$$mathtex$$(2) where μ is the specific growth rate for cells (we assume that this is constant within the cell inoculum). Note that the lag phases of individual cells in the inoculum, L i, are identically and independently distributed random variables. The natural logarithm of the cell population for a sufficiently long time (when tis greater than the maximum of L) is $$mathtex$$\[\mathrm{ln}(N_{t}){=}\mathrm{ln}(n){+}{\mu}\ \left[t{-}\left({-}\frac{1}{{\mu}}\mathrm{ln}\frac{\begin{array}{l}n\\{\sum}\\i{=}1\end{array}e^{{-}{\mu}{\times}L_{i}}}{n}\right)\right]\]$$mathtex$$(3) Then, from the biphasic growth model, the population lag time i K, arising from an initial inoculum of size n n, is $$mathtex$$\[K_{n}{=}{-}\frac{1}{{\mu}}\mathrm{ln}\frac{\begin{array}{l}n\\{\sum}\\i{=}1\end{array}e^{{-}{\mu}{\times}L_{i}}}{n}\]$$mathtex$$(4) and the time it takes to establish a population with size Nin the exponential phase is $$mathtex$$\[t_{b}{=}\frac{\mathrm{ln}(N_{b})}{{\mu}}{-}\frac{\mathrm{ln}(n)}{{\mu}}{+}K_{n}\]$$mathtex$$(5) b
The Bioscreen (Labsystems, Finland) system is an automated optical density reader. Bacterial suspensions are dosed into wells arranged in a honeycomb pattern in a plate. Each well can hold 400 μl of cell suspension, and the turbidity of the suspension increases as the density of bacterial cells in the suspension increases. The increase of turbidity can be correlated with cell numbers. There are 200 wells in a Bioscreen plate, and the measurement of population growth using this system can generate sufficient data for estimating the variability of single-cell lag phases. If each well initially holds one cell, then for each well$$mathtex$$\[t_{b}{=}\frac{\mathrm{ln}(N_{b})}{{\mu}}{+}L\]$$mathtex$$(6) This can be interpreted as the distribution of times,
t b, for a population of wells and is equivalent to a shifted form of the distribution, L, for individual cell lag times (2).
It is a challenge experimentally to place exactly one cell in all of the wells in a Bioscreen honeycomb plate. If it is possible to introduce one cell into 50% of the wells and two cells into the rest of the wells, then
t b has a mixture distribution$$mathtex$$\[p(t_{b}){=}0.5\ \left[\frac{\mathrm{ln}(N_{b})}{{\mu}}{+}L\right]{+}0.5\left[\frac{\mathrm{ln}(N_{b})}{{\mu}}{-}\frac{\mathrm{ln}(2)}{{\mu}}{+}K_{2}\right]\]$$mathtex$$(7) where the terms in brackets are symbolic representations for distributions that correspond to monodisperse inocula. Generalizing to an uncertain number of cells in each well gives$$mathtex$$\[p(t_{b}){=}\ \begin{array}{l}{\sum}\\n{=}1\\{\infty}\end{array}p(n)\ \left[\frac{\mathrm{ln}(N_{b})}{{\mu}}{-}\frac{\mathrm{ln}(n)}{{\mu}}{+}K_{n}\right]\]$$mathtex$$(8) where p( n) is the distribution of inoculum sizes. With this construction, efficient statistical methods combined with Bayesian methodology can be used to establish the distribution of lag times of single cells, p( L), for small i n. The scheme computes the Bayesian posterior for p( L) based on prior beliefs and on observed information for i tfrom the Bioscreen system. Prior beliefs concerning b Nand μ can be obtained from independent calibration experiments, so these variables include only slight uncertainty. b
Practical schemes introduce an uncertain number of cells into each well of a Bioscreen plate by serial dilution of a volume containing a known density of cells. This inoculation leads to a Poisson distribution for the number of cells in a well, and a significant number of wells will be empty. In this case, a well that does not show any signs of growth during an experimental period could still contain a cell (or cells) if this cell has a long lag time. Equation 8 is not sufficient for estimating the lag-phase distribution of cells unless the wells in a Bioscreen plate are certainly not empty (i.e.,
n is >0).
To address this problem, we can consider the Bioscreen experiment a combination of two events. The first event is the initial inoculation of a well with an uncertain number of cells which is determined by a Poisson distribution and parameterized by the expectation of the cell number θ. The second event is uncertain growth from cells in the well which is quantified by the specific growth rate μ and the distribution of single-cell lag times (
L); a finite distribution parameterized by α, within the range (0, d), is a convenient representation for prior belief concerning L. Note that α can be a set of parameters; for example, if L has a one-sided truncated exponential distribution (0 < L ≤ d), then α is the set containing both d and the exponential rate. We reexpress equation 1 as$$mathtex$$\[p(t_{b}{\vert}{\mu},{\theta},{\alpha}){=}p(0)(L_{0}){+}\ \begin{array}{l}{\sum}\\n{=}1\\{\infty}\end{array}p(n)\ \left[\frac{\mathrm{ln}(N_{b})}{{\mu}}{-}\frac{\mathrm{ln}(n)}{{\mu}}{+}K_{n}\right]\]$$mathtex$$(9) where L 0 is a delta function at a t b of d{ p[ L 0( t= b d)] = 1}. We introduce tas the duration of the Bioscreen experiment, and if a population of size c Nhad been established before or at b t, in each well, the observed time taken to establish this population is equal to c t. If a population with size b Nhas not been established at time b t, then c tis uniformly distributed between b tand c d.
This provides a complete prescription for the estimation of the Bayesian posterior distribution for
L based on a Bioscreen observation (i.e., values for t b). The posterior distribution is suitable for inclusion in quantitative risk assessments. Note that equations 2 to 9 are derived from equation 1. ACKNOWLEDGMENTS
We acknowledge support from BBSRC United Kingdom.
We thank J. Baranyi and A. Metris for their helpful discussions during the preparation of the manuscript.
FOOTNOTES Received 9 June 2008. Accepted 9 September 2008.
↵▿ Published ahead of print on 19 September 2008.
American Society for Microbiology
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Let us define a predicate
$$T(i,k_1,...,k_n)\in\{True, False\}$$Where $T(i,k_1,...,k_n)$ means "using the $i$ first values of $S$, we can find $n$ disjoint subsets with sums $k_1, ... k_n$". The answer you are looking for is $T(|S|,K,..,K)$ where $K$ appears $n$ times. We have the following recurrence formula : $$T(i,k_1,...,k_n) = \left\{\begin{matrix} T(i-1,k_1,...,k_n) & \text{// we don't use value $x_i$} \\ \vee T(i-1,k_1-x_i,...,k_n) \text{ if $x_i \geq k_1$} &\text{// we put value $x_i$ in set 1}\\ \vdots \\ \vee T(i-1,k_1,...,k_n-x_i) \text{ if $x_i \geq k_n$} &\text{// we put value $x_i$ in set n}\\ \end{matrix}\right.$$This formula is a big boolean "or", I don't know how I could make it look better. We also the following initialisation : $\forall i, T(i,0,...,0) = True$. You can see it as a big $n+1$-dimensional array, where the first dimension has length $|S|$, and the others have length $K$, which gives $O(K^n|S|)$ values. In that array, all the values can be computed from the other values in time $O(n)$ (the big "or" is over $n+1$ values). Therefore, the worst case complexity is $O(nK^n|S|)$.
About implementation, it might be easier to do with recursively with memorization. However, you're gonna blow up the stack really fast, so you might want to think of something iterative.
To get the actual values in each set, run a backtracking algorithm once you have that truth array.
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The Wiener–Khinchin-Einstein theorem states that the auto-correlation $(r_{xx}(\tau))$ and spectral density $(S(f))$ are Fourier duals, i.e.
$$r_{xx}(\tau) = \int^{+\infty}_{-\infty} S(f) \exp\left( 2\pi i \tau f \right)\:df$$
This relationship has several assumptions:
-The process must be stationary, i.e the spatial correlation depend only upon the the distance between two points ($\tau$) and not the orientation ($\vec{\tau}$). For Gaussian processes this requires the first and second order moments to be stationary.
My question is:
Does this relationship hold for non-Gaussian processes? And if so are there any assumptions or restrictions in place regarding its use? For if it holds for non-Gaussian processes, is it suffient for the process Weak or wide-sense stationarity, i.e. only the first and second order moment need to be stationary and higher order moments can be non-stationary, or are there more strict conditions? My gut feeling is the former since the equation requires no-higher order information, however I'm not sure if that is the case.
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Let $(X,d)$ a metric space, $\alpha >0$ (fixed ) and $T: X \rightarrow X$ a map such that exist $n \in N$, where :
$$ d(T^n x , T^n y) \leq \alpha^n d(x,y), \forall x,y \in X$$
Define $h(x,y) = [d^2 (x,y) + \frac{1}{\alpha^2} d(Tx,ty)+...+ \frac{1}{\alpha^{2(n-1)}} d(T^{n-1}x,T^{n-1}y)]^{1/2}$. I want to prove that $h $ is a metric on $X$. My problem is the triangle inequality ... I am getting anywhere .. Someone could give me a help?
Thanks in advance
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