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Infinite definite integral
11-08-2014, 11:40 PM
Post: #1
Infinite definite integral
On the WP 34S, if consecutive screen updates increase by what appears to be a constant increment, can I take this to mean that the integral is infinite? If the screen updates are by smaller and smaller increments, does this mean that the integral coverges? Or are there exceptions to both cases?
What prompts this is int(x/(1-sqrt(x)) from 0 to 1. This should be infinite. My WP 34S (at least my WP 34S app on my I-pad) stops at 37.8151, was increasing with increments of ~2.77.
I'm amazed at how close to 1 one can get, and still have a fairly small area, far from infinity....from 0 to 1-1EE12 produces something 50 or 60ish!
11-08-2014, 11:51 PM
Post: #2
RE: Infinite definite integral
The 52 ish answer for that interval from my Prime...
11-09-2014, 12:03 AM
Post: #3
RE: Infinite definite integral
(11-08-2014 11:40 PM)lrdheat Wrote: On the WP 34S, if consecutive screen updates increase by what appears to be a constant increment, can I take this to mean that the integral is infinite? If the screen updates are by smaller and smaller increments, does this mean that the integral coverges? Or are there exceptions to both cases?
No, no and yes
- Pauli
11-09-2014, 05:22 AM
Post: #4
RE: Infinite definite integral
\(\int\frac{x}{1-\sqrt{x}}dx=-\frac{1}{3}\sqrt{x}(2x+3\sqrt{x}+6)-2\log(1-\sqrt{x})+C\)
Quote:I'm amazed at how close to 1 one can get, and still have a fairly small area, far from infinity....
That's due to the last term: \(-2\log(1-\sqrt{x})\)
For \(x=1-\epsilon\) we get:
\(-2\log(1-\sqrt{1-\epsilon})\approx-2\log(1-(1-\frac{1}{2}\epsilon))=-2\log(\frac{1}{2}\epsilon)\)
This grows slowly as \(\epsilon \to 0\).
Quote:from 0 to 1-1EE12 produces something 50 or 60ish!
With \(\epsilon=10^{-12}\) we get approximately:
\(-\frac{11}{3}+2(\log(2)+12\log(10))\approx52.982\)
Kind regards
Thomas
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This is follow-up question to
this one. My aim is to let greek letters behave the same like normal latin letters, when it comes to get them upright. At the moment I'm using
\mathrm to get letters upright in math mode.
So
\mathrm{\delta d} should return an upright
d and an upright delta. A custom command
\upright would be an option too, but I wouldn't know where to start. The important part is, that I could use that command for both, greek and latin letters,
equally - without additional commands like
\deltaup or
\updelta. The reason is that I need to work with different classes, different fonts and even different compilers, so I need to get a most generic solution. Using fontspec/unicode-math (which would solve the problem, is not an option).
From my previous question I learned by
Ulrike Fischer's answer, that there would be no way around using
\deltaup for my case. And she offered the neat solution:
\let\deltait\delta\renewcommand\delta{\ifnum\fam=0 \deltaup\else \deltait\fi}
In case some of my colleagues uses the upgreek package and
\updelta I wrapped an if condition around:
\makeatletter\ifcsname deltaup\endcsname% \let\deltait\delta\renewcommand\delta{\ifnum\fam=0 \deltaup\else \deltait\fi}\else \ifcsname updelta\endcsname \let\deltait\delta\renewcommand\delta{\ifnum\fam=0 \updelta\else \delta\fi} \fi\fi%
Now I wonder whether I could put this into a loop to deal with
all lowercase greek letters. I thought about a list with all letters, but I don't know how to get dynamic input arguments in cases like
\let\deltait\delta.
Of course I could just write the same lines for every letter, but maybe another font than
pxfonts later requires another solution and always redefining the whole alphabet seems cumbersome.
Furthermore jfbu mentioned in his answer/comments that using
\mathrm wouldn't be a good option at all, as I'm changing the font family, which I shouldn't.
MWE
\documentclass{article}\usepackage[utf8x]{luainputenc}\usepackage[T1]{fontenc}\usepackage{pxfonts}\usepackage{siunitx}%\usepackage{isomath}%\usepackage{upgreek}\usepackage[ISO]{pxgreeks}\makeatletter\ifcsname deltaup\endcsname% \let\deltait\delta\renewcommand\delta{\ifnum\fam=0 \deltaup\else \deltait\fi}\else \ifcsname updelta\endcsname \let\deltait\delta\renewcommand\delta{\ifnum\fam=0 \updelta\else \deltait\fi} \fi\fi%\begin{document}\begin{equation} \Phi_{\mathrm{\delta}} = \SI{42}{\micro\Omega} \cdot \delta_{\mathrm{\Phi}} \end{equation} \begin{equation} P_{\mathrm{d}} = \SI{42}{\nano\ampere} \cdot d_{\mathrm{P}}\end{equation} \end{document} |
When you define a base of vectors, in your case $\{ |1,1\rangle, |1,0\rangle, |0,1\rangle, |0,0\rangle \} $ you assign to each vector some phase, e.g. the vector $|1,1\rangle$ has a certain phase that you don't mention explicitly outside the bra-kets. Now, in your calculi you may have superpositions of these vectors of the form
$|\psi\rangle = ae^{i\alpha} |1,1\rangle + be^{i\beta}|1,0\rangle + ce^{i\gamma}|0,1\rangle + de^{i\delta}|0,0\rangle,$
with $a, b, c, d$ positive numbers, i.e. besides the intrinsic phases of the four vectors, they appear in the superposition with complex amplitudes. Thus the amplitude of $|0,0\rangle$ has a difference of phase in comparison with the amplitude of $|1,1\rangle$ and this difference is $\delta - \alpha$.
Returning to your functions, in the first superposition the vectors $|1,1\rangle$ and $|0,0\rangle$ are in phase, while in the 2nd superposition they are in anti-phase, there is a difference $\delta - \alpha = \pi$ between them, i.e. you can write
$ \frac {1}{\sqrt{2}}e^{i \cdot 0} |1,1\rangle + \frac {1}{\sqrt{2}}e^{i\pi}|0,0\rangle$. |
also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling
@heather well, there's a spectrum
so, there's things like New Journal of Physics and Physical Review X
which are the open-access branch of existing academic-society publishers
As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di...
Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago
> A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service”
for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty
> for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing)
@0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals.
@BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work...
@BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions.
Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley.
I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea.
@EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results...
Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town...
@EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit. |
Generalizing Woodin’s extender algebra, cf. e.g. Steel Handbook of set theory, Springer, Berlin, 2010), we isolate the long extender algebra as a general version of Bukowský’s forcing, cf. Bukovský, in the presence of a supercompact cardinal.
We show that either of the following hypotheses imply that there is an inner model with a proper class of strong cardinals and a proper class of Woodin cardinals. 1) There is a countably closed cardinal k ≥ N₃ such that □k and □(k) fail. 2) There is a cardinal k such that k is weakly compact in the generic extension by Col(k, k⁺). Of special interest is 1) with k = N₃ since it follows from PFA by theorems of (...) Todorcevic and Velickovic. Our main new technical result, which is due to the first author, is a weak covering theorem for the model obtained by stacking mice over $K^c ||k.$. (shrink)
We give a direct and elementary proof of the fact that every real-valued probability measure can be approximated—up to an infinitesimal—by a hyperreal-valued one which is regular and defined on the whole powerset of the sample space.
The main result of this paper is the following theorem: Let M be a premouse with a top extender, F. Suppose that (a) M is linearly coarsely iterable via hitting F and its images, and (b) if M * is a linear iterate of M as in (a), then M * is coarsely iterable with respect to iteration trees which do not use the top extender of M * and its images. Then M is coarsely iterable.
We introduce 0• as a sharp for an inner model with a proper class of strong cardinals. We prove the existence of the core model K in the theory “ does not exist”. Combined with work of Woodin, Steel, and earlier work of the author, this provides the last step for determining the exact consistency strength of the assumption in the statement of the 12th Delfino problem pp. 221–224)).
We show that L absoluteness for semi-proper forcings is equiconsistent with the existence of a remarkable cardinal, and hence by [6] with L absoluteness for proper forcings. By [7], L absoluteness for stationary set preserving forcings gives an inner model with a strong cardinal. By [3], the Bounded Semi-Proper Forcing Axiom is equiconsistent with the Bounded Proper Forcing Axiom , which in turn is equiconsistent with a reflecting cardinal. We show that Bounded Martin's Maximum is much stronger than BSPFA in (...) that if BMM holds, then for every X ∈ V , X# exists. (shrink)
Let L[E] be an iterable tame extender model. We analyze to which extent L[E] knows fragments of its own iteration strategy. Specifically, we prove that inside L[E], for every cardinal K which is not a limit of Woodin cardinals there is some cutpoint t K > a>ω1 are cardinals, then ◊$_{K.\lambda }^* $ holds true, and if in addition λ is regular, then ◊$_{K.\lambda }^* $ holds true.
We extend the core model induction technique to a choiceless context, and we exploit it to show that each one of the following two hypotheses individually implies that , the Axiom of Determinacy, holds in the of a generic extension of : every uncountable cardinal is singular, and every infinite successor cardinal is weakly compact and every uncountable limit cardinal is singular.
If the Bounded Proper Forcing Axiom BPFA holds, then Mouse Reflection holds at N₂ with respect to all mouse operators up to the level of Woodin cardinals in the next ZFC-model. This yields that if Woodin's ℙ max axiom (*) holds, then BPFA implies that V is closed under the "Woodin-in-the-next-ZFC-model" operator. We also discuss stronger Mouse Reflection principles which we show to follow from strengthenings of BPFA, and we discuss the theory BPFA plus "NS ω1 is precipitous" and strengthenings (...) thereof. Along the way, we answer a question of Baumgartner and Taylor. [2, Question 6.11]. (shrink)
The current paper proves the results announced in [5]. We isolate a new large cardinal concept, "remarkability." Consistencywise, remarkable cardinals are between ineffable and ω-Erdos cardinals. They are characterized by the existence of "O # -like" embeddings; however, they relativize down to L. It turns out that the existence of a remarkable cardinal is equiconsistent with L(R) absoluteness for proper forcings. In particular, said absoluteness does not imply Π 1 1 determinacy.
We introduce and consider the inner-model reflection principle, which asserts that whenever a statement \\) in the first-order language of set theory is true in the set-theoretic universe V, then it is also true in a proper inner model \. A stronger principle, the ground-model reflection principle, asserts that any such \\) true in V is also true in some non-trivial ground model of the universe with respect to set forcing. These principles each express a form of width reflection in (...) contrast to the usual height reflection of the Lévy–Montague reflection theorem. They are each equiconsistent with ZFC and indeed \-conservative over ZFC, being forceable by class forcing while preserving any desired rank-initial segment of the universe. Furthermore, the inner-model reflection principle is a consequence of the existence of sufficient large cardinals, and lightface formulations of the reflection principles follow from the maximality principle MP and from the inner-model hypothesis IMH. We also consider some questions concerning the expressibility of the principles. (shrink)
Let N be a transitive model of ZFC such that ωN ⊂ N and P(R) ⊂ N. Assume that both V and N satisfy "the core model K exists." Then KN is an iterate of K. i.e., there exists an iteration tree J on K such that J has successor length and $\mathit{M}_{\infty}^{\mathit{J}}=K^{N}$. Moreover, if there exists an elementary embedding π: V → N then the iteration map associated to the main branch of J equals π ↾ K. (This answers (...) a question of W. H. Woodin, M. Gitik, and others.) The hypothesis that P(R) ⊂ N is not needed if there does not exist a transitive model of ZFC with infinitely many Woodin cardinals. (shrink)
We generalize ∇(A), which was introduced in [Sch∞], to larger cardinals. For a regular cardinal κ>ℵ0 we denote by ∇ κ (A) the statement that and for all regular θ>κ, is stationary in It was shown in [Sch∞] that can hold in a set-generic extension of L. We here prove that can hold in a set-generic extension of L as well. In both cases we in fact get equiconsistency theorems. This strengthens results of [Rä00] and [Rä01]. is equivalent with the (...) existence of 0#. (shrink)
We give an optimal lower bound in terms of large cardinal axioms for the logical strength of projective uniformization in conjuction with other regularity properties of projective sets of real numbers, namely Lebesgue measurability and its dual in the sense of category . Our proof uses a projective computation of the real numbers which code inital segments of a core model and answers a question in Hauser.
It is shown in ZF that if $\delta are such that δ and δ + are either both weakly compact or singular cardinals and Ω is large enough for putting the core model apparatus into action then there is an inner model with a Woodin cardinal.
Suppose that there is no transitive model of ZFC + there is a strong cardinal, and let K denote the core model. It is shown that if $\delta$ has the tree property then $\delta^{+K} = \delta^+$ and $\delta$ is weakly compact in K.
We construct, assuming that there is no inner model with a Woodin cardinal but without any large cardinal assumption, a model Kc which is iterable for set length iterations, which is universal with respect to all weasels with which it can be compared, and is universal with respect to set sized premice.
We show that if I is a precipitous ideal on ω₁ and if θ > ω₁ is a regular cardinal, then there is a forcing P = P(I, θ) which preserves the stationarity of all I-positive sets such that in $V^P $ , is a generic iterate of a countable structure . This shows that if the nonstationary ideal on ω₁ is precipitous and $H_\theta ^\# $ exists, then there is a stationary set preserving forcing which increases $\delta _2^1 $ (...) · Moreover, if Bounded Martin's Maximum holds and the nonstationary ideal on ω₁ is precipitous, then $\delta _2^1 $ = u₂ = ω₂. (shrink)
We generalize results of [3] and [1] to hyperprojective sets of reals, viz. to more than finitely many strong cardinals being involved. We show, for example, that if every set of reals in Lω is weakly homogeneously Souslin, then there is an inner model with an inaccessible limit of strong cardinals.
We show that weakly compact cardinals are the smallest large cardinals k where k+ < k+ is impossible provided 0# does not exist. We also show that if k+Kc < k+ for some k being weakly compact , then there is a transitive set M with M ⊨ ZFC + “there is a strong cardinal”.
Continuing [7], we here prove that the Chang Conjecture $(\aleph_3,\aleph_2) \Rightarrow (\aleph_2,\aleph_1)$ together with the Continuum Hypothesis, $2^{\aleph_0} = \aleph_1$ , implies that there is an inner model in which the Mitchell ordering is $\geq \kappa^{+\omega}$ for some ordinal $\kappa$.
We show that weakly compact cardinals are the smallest large cardinals k where k+ < k+ is impossible provided 0# does not exist. We also show that if k+Kc < k+ for some k being weakly compact , then there is a transitive set M with M ⊨ ZFC + “there is a strong cardinal”.
Let 0 < n < ω. If there are n Woodin cardinals and a measurable cardinal above, but $M_{n+1}^{\#}$ doesn't exist, then the core model K exists in a sense made precise. An Iterability Inheritance Hypothesis is isolated which is shown to imply an optimal correctness result for K.
We show that the following conjecture about the universe V of all sets is wrong: for all set-theoretical (i.e., first order) schemata true in V there is a transitive set "reflecting" in such a way that the second order statement corresponding to is true in . More generally, we indicate the ontological commitments of any theory that exploits reflection principles in order to yield large cardinals. The disappointing conclusion will be that our only apparently good arguments for the existence of (...) large cardinals have bad presuppositions. (shrink)
Let n ≥ 3 be an integer. We show that it is consistent (relative to the consistency of n - 2 strong cardinals) that every $\Sigma_n^1-set$ of reals is universally Baire yet there is a (lightface) projective well-ordering of the reals. The proof uses "David's trick" in the presence of inner models with strong cardinals.
We consider certain predicative classes with respect to their bearing on set theory, namely on its semantics, and on its ontological power. On the one hand, our predicative classes will turn out to be perfectly suited for establishing a nice hierarchy of metalanguages starting from the usual set theoretical language. On the other hand, these classes will be seen to be fairly inappropriate for the formulation of strong principles of infinity. The motivation for considering this very type of classes is (...) a reasonable philosophy of set theory. Familiarity is assumed only with basic concepts of both set theory and its philosophy. (shrink)
The current paper proves the results announced in [5]. We isolate a new large cardinal concept, "remarkability." Consistencywise, remarkable cardinals are between ineffable and $\omega$-Erdos cardinals. They are characterized by the existence of "O$^#$-like" embeddings; however, they relativize down to L. It turns out that the existence of a remarkable cardinal is equiconsistent with L absoluteness for proper forcings. In particular, said absoluteness does not imply $\Pi^1_1$ determinacy.
For ordinals α beginning a Σ1 gap in equation image, where equation image is closed under number quantification, we give an inner model-theoretic proof that every thin equation image equivalence relation is equation image in a real parameter from the hypothesis equation image.
It is shown in ZF that if $\delta < \delta^+ < \Omega$ are such that $\delta$ and $\delta^+$ are either both weakly compact or singular cardinals and $\Omega$ is large enough for putting the core model apparatus into action then there is an inner model with a Woodin cardinal.
Let n ≥ 3 be an integer. We show that it is consistent that every σ1n-set of reals is universally Baire yet there is a projective well-ordering of the reals. The proof uses “David’s trick” in the presence of inner models with strong cardinals.
We isolate natural strengthenings of Bounded Martin’s Maximum which we call ${\mathsf{BMM}}^{*}$ and $A-{\mathsf{BMM}}^{*,++}$, and we investigate their consequences. We also show that if $A-{\mathsf{BMM}}^{*,++}$ holds true for every set of reals $A$ in $L$, then Woodin’s axiom $$ holds true. We conjecture that ${\mathsf{MM}}^{++}$ implies $A-{\mathsf{BMM}}^{*,++}$ for every $A$ which is universally Baire. |
This shows you the differences between two versions of the page.
en:fitting [2012/10/15 18:35]
deinega
en:fitting [2012/10/15 18:37] (current)
deinega
Line 26: Line 26: *modified Lorentz term <tex>\frac{\Delta \varepsilon (\omega_p^2 - i\omega\gamma_p')}{\omega_p^2-2i\omega\gamma_p-\omega^2}</tex> *modified Lorentz term <tex>\frac{\Delta \varepsilon (\omega_p^2 - i\omega\gamma_p')}{\omega_p^2-2i\omega\gamma_p-\omega^2}</tex> - Last case does not correspond to any of physical model, but allows to obtain more accurate fittings. + Using modified Lorentz term allows to obtain more accurate fittings. - For example, two (<tex>P=2</tex>) terms of this case are sufficient to fit silicon dielectric function over the wavelength range from 300 to 1000 nm, whereas even a large number of Debye, Drude or Lorentz terms (<tex>a_{p,1}=0</tex>) is inadequate there + For example, two (<tex>P=2</tex>) modified Lorentz terms are sufficient to fit silicon dielectric function over the wavelength range from 300 to 1000 nm, whereas even a large number of Debye, Drude or Lorentz terms (<tex>a_{p,1}=0</tex>) is inadequate there ((A. Deinega and S. John, "Effective optical response of silicon to sunlight in the finite-difference time-domain method," Opt. Lett. 37, 112-114 (2012) ((A. Deinega and S. John, "Effective optical response of silicon to sunlight in the finite-difference time-domain method," Opt. Lett. 37, 112-114 (2012) [[http://www.opticsinfobase.org/ol/abstract.cfm?uri=ol-37-1-112|http]] [[http://www.opticsinfobase.org/ol/abstract.cfm?uri=ol-37-1-112|http]] |
While Ert's and Ben's answers are excellent, they rely on the assumption that you understand how to perform
statistical modelling on time series, which is not a trivial topic. Entire books are written on this and this simple answer is only an quick introduction using linear regression as an example. Time series linear regression Assumptions
As with linear regression of cross-sectional data, time series regression requires a set of assumptions to be met in order to perform statistical inference. The strongest set of assumptions are the Gauss Markov conditions, which are often impractical with real data. A weaker set of assumptions are the asymptotic Least Squares.
Given a model
$$ Y_t = \alpha + \beta_1 X_{1t} + \beta_2 X_{2t} + ... + \epsilon_t$$
We require the following:
Linear model: $Y_t = \alpha + \beta_1X_{1t} + \beta_2X_{2t} + \epsilon_t$ Stationarity in mean, variance and covariance Weak dependence, i.e. correlation tends to zero $ Corr(X_t , X{t + h}) \rightarrow 0 \ \ \text{as} \ \ h \rightarrow \infty$ Weak exogeneity $E[\epsilon_t | X_{it}] = 0$ - note that this is less restrictive than the strict exogeneity assumption, as it does not depend on all times, but only on the particular time $t$ No perfect colinearity - as usual for Gauss Markov conditions $Var(\epsilon_t | X_{it}) = \sigma^2$ - this is less restrictive than Gauss Markov as it is only for the particular $t$ $Cov(\epsilon_t, \epsilon_s | X_t, X_s) = 0$ - less restrictive than Gauss Markov as only depends on time $t$ and $s$.
If assumptions 1 - 5 are satisfied then the parameter $\beta_{LS}$ are consistent. Additional assumptions 6,7 allow to perform inference using the CLT etc...
If all these assumptions are satisfied then you can perform a straightforward linear regression on the model, and use a
t-test to test for the significance of a single variable, or an F-test for the significance of the whole model. Have a look here for a recap of the maths and a Python implementation When assumptions are not met
The reason entire books are written on this topic is that dealing with violations of these assumptions can be difficult, and often requires some subjective decision makings, various statistical tests, pre-processing steps etc...
An example
Consider fitting a linear model on US consumer expenditure % change levels based on % change in income, production, savings and unemployment.
A few checks on the data show that the above assumptions are mostly met. Fitting a linear regression model and performing t-tests on the significance of the variables gives
The fitted model isn't too bad
Source |
Yes, there are many ways to produce a sequence of numbers that are more evenly distributed than random uniforms. In fact, there is a whole field dedicated to this question; it is the backbone of quasi-Monte Carlo (QMC). Below is a brief tour of the absolute basics.
Measuring uniformity
There are many ways to do this, but the most common way has a strong, intuitive, geometric flavor. Suppose we are concerned with generating $n$ points $x_1,x_2,\ldots,x_n$ in $[0,1]^d$ for some positive integer $d$. Define$$\newcommand{\I}{\mathbf 1}D_n := \sup_{R \in \mathcal R}\,\left|\frac{1}{n}\sum_{i=1}^n \I_{(x_i \in R)} - \mathrm{vol}(R)\right| \>,$$where $R$ is a rectangle $[a_1, b_1] \times \cdots \times [a_d, b_d]$ in $[0,1]^d$ such that $0 \leq a_i \leq b_i \leq 1$ and $\mathcal R$ is the set of all such rectangles. The first term inside the modulus is the "observed" proportion of points inside $R$ and the second term is the volume of $R$, $\mathrm{vol}(R) = \prod_i (b_i - a_i)$.
The quantity $D_n$ is often called the
discrepancy or extreme discrepancy of the set of points $(x_i)$. Intuitively, we find the "worst" rectangle $R$ where the proportion of points deviates the most from what we would expect under perfect uniformity.
This is unwieldy in practice and difficult to compute. For the most part, people prefer to work with the
star discrepancy,$$D_n^\star = \sup_{R \in \mathcal A} \,\left|\frac{1}{n}\sum_{i=1}^n \I_{(x_i \in R)} - \mathrm{vol}(R)\right| \>.$$The only difference is the set $\mathcal A$ over which the supremum is taken. It is the set of anchored rectangles (at the origin), i.e., where $a_1 = a_2 = \cdots = a_d = 0$.
Lemma: $D_n^\star \leq D_n \leq 2^d D_n^\star$ for all $n$, $d$. Proof. The left hand bound is obvious since $\mathcal A \subset \mathcal R$. The right-hand bound follows because every $R \in \mathcal R$ can be composed via unions, intersections and complements of no more than $2^d$ anchored rectangles (i.e., in $\mathcal A$).
Thus, we see that $D_n$ and $D_n^\star$ are equivalent in the sense that if one is small as $n$ grows, the other will be too. Here is a (cartoon) picture showing candidate rectangles for each discrepancy.
Examples of "good" sequences
Sequences with verifiably low star discrepancy $D_n^\star$ are often called, unsurprisingly,
low discrepancy sequences.
van der Corput. This is perhaps the simplest example. For $d=1$, the van der Corput sequences are formed by expanding the integer $i$ in binary and then "reflecting the digits" around the decimal point. More formally, this is done with the radical inverse function in base $b$,$$\newcommand{\rinv}{\phi}\rinv_b(i) = \sum_{k=0}^\infty a_k b^{-k-1} \>,$$where $i = \sum_{k=0}^\infty a_k b^k$ and $a_k$ are the digits in the base $b$ expansion of $i$. This function forms the basis for many other sequences as well. For example, $41$ in binary is $101001$ and so $a_0 = 1$, $a_1 = 0$, $a_2 = 0$, $a_3 = 1$, $a_4 = 0$ and $a_5 = 1$. Hence, the 41st point in the van der Corput sequence is $x_{41} = \rinv_2(41) = 0.100101\,\text{(base 2)} = 37/64$.
Note that because the least significant bit of $i$ oscillates between $0$ and $1$, the points $x_i$ for odd $i$ are in $[1/2,1)$, whereas the points $x_i$ for even $i$ are in $(0,1/2)$.
Halton sequences. Among the most popular of classical low-discrepancy sequences, these are extensions of the van der Corput sequence to multiple dimensions. Let $p_j$ be the $j$th smallest prime. Then, the $i$th point $x_i$ of the $d$-dimensional Halton sequence is$$x_i = (\rinv_{p_1}(i), \rinv_{p_2}(i),\ldots,\rinv_{p_d}(i)) \>.$$For low $d$ these work quite well, but have problems in higher dimensions.
Halton sequences satisfy $D_n^\star = O(n^{-1} (\log n)^d)$. They are also nice because they are
extensible in that the construction of the points does not depend on an a priori choice of the length of the sequence $n$.
Hammersley sequences. This is a very simple modification of the Halton sequence. We instead use$$x_i = (i/n, \rinv_{p_1}(i), \rinv_{p_2}(i),\ldots,\rinv_{p_{d-1}}(i)) \>.$$Perhaps surprisingly, the advantage is that they have better star discrepancy $D_n^\star = O(n^{-1}(\log n)^{d-1})$.
Here is an example of the Halton and Hammersley sequences in two dimensions.
Faure-permuted Halton sequences. A special set of permutations (fixed as a function of $i$) can be applied to the digit expansion $a_k$ for each $i$ when producing the Halton sequence. This helps remedy (to some degree) the problems alluded to in higher dimensions. Each of the permutations has the interesting property of keeping $0$ and $b-1$ as fixed points.
Lattice rules. Let $\beta_1, \ldots, \beta_{d-1}$ be integers. Take$$x_i = (i/n, \{i \beta_1 / n\}, \ldots, \{i \beta_{d-1}/n\}) \>,$$where $\{y\}$ denotes the fractional part of $y$. Judicious choice of the $\beta$ values yields good uniformity properties. Poor choices can lead to bad sequences. They are also not extensible. Here are two examples.
$(t,m,s)$ nets. $(t,m,s)$ nets in base $b$ are sets of points such that every rectangle of volume $b^{t-m}$ in $[0,1]^s$ contains $b^t$ points. This is a strong form of uniformity. Small $t$ is your friend, in this case. Halton, Sobol' and Faure sequences are examples of $(t,m,s)$ nets. These lend themselves nicely to randomization via scrambling. Random scrambling (done right) of a $(t,m,s)$ net yields another $(t,m,s)$ net. The MinT project keeps a collection of such sequences.
Simple randomization: Cranley-Patterson rotations. Let $x_i \in [0,1]^d$ be a sequence of points. Let $U \sim \mathcal U(0,1)$. Then the points $\hat x_i = \{x_i + U\}$ are uniformly distributed in $[0,1]^d$.
Here is an example with the blue dots being the original points and the red dots being the rotated ones with lines connecting them (and shown wrapped around, where appropriate).
Completely uniformly distributed sequences. This is an even stronger notion of uniformity that sometimes comes into play. Let $(u_i)$ be the sequence of points in $[0,1]$ and now form overlapping blocks of size $d$ to get the sequence $(x_i)$. So, if $s = 3$, we take $x_1 = (u_1,u_2,u_3)$ then $x_2 = (u_2,u_3,u_4)$, etc. If, for every $s \geq 1$, $D_n^\star(x_1,\ldots,x_n) \to 0$, then $(u_i)$ is said to be completely uniformly distributed. In other words, the sequence yields a set of points of any dimension that have desirable $D_n^\star$ properties.
As an example, the van der Corput sequence is not completely uniformly distributed since for $s = 2$, the points $x_{2i}$ are in the square $(0,1/2) \times [1/2,1)$ and the points $x_{2i-1}$ are in $[1/2,1) \times (0,1/2)$. Hence there are no points in the square $(0,1/2) \times (0,1/2)$ which implies that for $s=2$, $D_n^\star \geq 1/4$ for all $n$.
Standard references
The Niederreiter (1992) monograph and the Fang and Wang (1994) text are places to go for further exploration. |
Counting Operations in Gaussian Elimination
We have seen from The Gaussian Elimination Algorithm and the Computing the Inverse of a Matrix with Gaussian Elimination pages that solving a system of $n$ linear equations in $n$ unknowns, or finding the inverse (provided that it exists) of a square $n \times n$ matrix requires a lot of arithmetic steps, especially when $n$ is very large. It is thus important to determine the length of the computation in terms of counting the number of operations done in the process.
Consider a system of $n$ linear equations in $n$ unknowns and the corresponding $n \times n$ coefficient matrix $A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}$, the $n \times 1$ solution matrix $x = \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix}$, and the $n \times 1$ constant matrix $b = \begin{bmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{bmatrix}$. Then $Ax = b$. Earlier we saw that at the $(n-1)^{\mathrm{th}}$ step of Gaussian Elimination, we obtain the following system of equations:(1)
The values of the $u$'s and the $g$'s were described on the The Gaussian Elimination Algorithm page. Let $U = \begin{bmatrix} u_{11} & u_{12} & \cdots & u_{nn}\\ 0 & u_{22} & \cdots & u_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & u_{nn}\end{bmatrix}$ be the $n \times n$ coefficient matrix for this equivalent system, and let $g = \begin{bmatrix} g_1\\ g_2\\ \vdots\\ g_n \end{bmatrix}$ be the $n \times 1$ equivalent constant matrix. Then $Ux = g$. Then the following table states the operations count from going from $A$ to $U$ at each step $1, 2, ..., n -1$.
Step Number Number of Additions/Subtractions from $A \to U$ Number of Multiplications from $A \to U$ Number of Divisions from $A \to U$ $1$ $(n - 1)^2$ $(n - 1)^2$ $n - 1$ $2$ $(n - 2)^2$ $(n - 2)^2$ $n - 2$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $1$ $4$ $4$ $2$ $n-1$ $1$ $1$ $1$ Totals: $\mathrm{Total \: Num. \: of \: Add/Sub} = \frac{n(n-1)(2n-1)}{6}$ $\mathrm{Total \: Num. \: of \: Multiplications} = \frac{n(n-1)(2n-1)}{6}$ $\mathrm{Total \: Num. \: of \: Divisions} = \frac{n(n-1)}{2}$
We will count the number of additions/subtractions together and the number of multiplications/divisions together. The total number of additions/subtractions and multiplications/divisions in going from $A \to U$ is:(2)
We will now count the number of additions/ subtractions and the number of multiplications/divisions in going from $b \to g$. We have that:(4)
Lastly we count the number of additions/subtractions and multiplications/divisions for finding the solutions from the back-substitution method.(6)
Therefore the total number of operations to obtain the solution of a system of $n$ linear equations in $n$ variables using Gaussian Elimination is:(8) |
@Secret et al hows this for a video game? OE Cake! fluid dynamics simulator! have been looking for something like this for yrs! just discovered it wanna try it out! anyone heard of it? anyone else wanna do some serious research on it? think it could be used to experiment with solitons=D
OE-Cake, OE-CAKE! or OE Cake is a 2D fluid physics sandbox which was used to demonstrate the Octave Engine fluid physics simulator created by Prometech Software Inc.. It was one of the first engines with the ability to realistically process water and other materials in real-time. In the program, which acts as a physics-based paint program, users can insert objects and see them interact under the laws of physics. It has advanced fluid simulation, and support for gases, rigid objects, elastic reactions, friction, weight, pressure, textured particles, copy-and-paste, transparency, foreground a...
@NeuroFuzzy awesome what have you done with it? how long have you been using it?
it definitely could support solitons easily (because all you really need is to have some time dependence and discretized diffusion, right?) but I don't know if it's possible in either OE-cake or that dust game
As far I recall, being a long term powder gamer myself, powder game does not really have a diffusion like algorithm written into it. The liquids in powder game are sort of dots that move back and forth and subjected to gravity
@Secret I mean more along the lines of the fluid dynamics in that kind of game
@Secret Like how in the dan-ball one air pressure looks continuous (I assume)
@Secret You really just need a timer for particle extinction, and something that effects adjacent cells. Like maybe a rule for a particle that says: particles of type A turn into type B after 10 steps, particles of type B turn into type A if they are adjacent to type A.
I would bet you get lots of cool reaction-diffusion-like patterns with that rule.
(Those that don't understand cricket, please ignore this context, I will get to the physics...)England are playing Pakistan at Lords and a decision has once again been overturned based on evidence from the 'snickometer'. (see over 1.4 ) It's always bothered me slightly that there seems to be a ...
Abstract: Analyzing the data from the last replace-the-homework-policy question was inconclusive. So back to the drawing board, or really back to this question: what do we really mean when we vote to close questions as homework-like?As some/many/most people are aware, we are in the midst of a...
Hi I am trying to understand the concept of dex and how to use it in calculations. The usual definition is that it is the order of magnitude, so $10^{0.1}$ is $0.1$ dex.I want to do a simple exercise of calculating the value of the RHS of Eqn 4 in this paper arxiv paper, the gammas are incompl...
@ACuriousMind Guten Tag! :-) Dark Sun has also a lot of frightening characters. For example, Borys, the 30th level dragon. Or different stages of the defiler/psionicist 20/20 -> dragon 30 transformation. It is only a tip, if you start to think on your next avatar :-)
What is the maximum distance for eavesdropping pure sound waves?And what kind of device i need to use for eavesdropping?Actually a microphone with a parabolic reflector or laser reflected listening devices available on the market but is there any other devices on the planet which should allow ...
and endless whiteboards get doodled with boxes, grids circled red markers and some scribbles
The documentary then showed one of the bird's eye view of the farmlands
(which pardon my sketchy drawing skills...)
Most of the farmland is tiled into grids
Here there are two distinct column and rows of tiled farmlands to the left and top of the main grid. They are the index arrays and they notate the range of inidex of the tensor array
In some tiles, there's a swirl of dirt mount, they represent components with nonzero curl
and in others grass grew
Two blue steel bars were visible laying across the grid, holding up a triangle pool of water
Next in an interview, they mentioned that experimentally the process is uite simple. The tall guy is seen using a large crowbar to pry away a screw that held a road sign under a skyway, i.e.
ocassionally, misshaps can happen, such as too much force applied and the sign snapped in the middle. The boys will then be forced to take the broken sign to the nearest roadworks workshop to mend it
At the end of the documentary, near a university lodge area
I walked towards the boys and expressed interest in joining their project. They then said that you will be spending quite a bit of time on the theoretical side and doddling on whitebaords. They also ask about my recent trip to London and Belgium. Dream ends
Reality check: I have been to London, but not Belgium
Idea extraction: The tensor array mentioned in the dream is a multiindex object where each component can be tensors of different order
Presumably one can formulate it (using an example of a 4th order tensor) as follows:
$$A^{\alpha}_{\beta}_{\gamma,\delta,\epsilon}$$
and then allow the index $\alpha,\beta$ to run from 0 to the size of the matrix representation of the whole array
while for the indices $\gamma,\delta,epsilon$ it can be taken from a subset which the $\alpha,\beta$ indices are. For example to encode a patch of nonzero curl vector field in this object, one might set $\gamma$ to be from the set $\{4,9\}$ and $\delta$ to be $\{2,3\}$
However even if taking indices to have certain values only, it is unsure if it is of any use since most tensor expressions have indices taken from a set of consecutive numbers rather than random integers
@DavidZ in the recent meta post about the homework policy there is the following statement:
> We want to make it sure because people want those questions closed. Evidence: people are closing them. If people are closing questions that have no valid reason for closure, we have bigger problems.
This is an interesting statement.
I wonder to what extent not having a homework close reason would simply force would-be close-voters to either edit the post, down-vote, or think more carefully whether there is another more specific reason for closure, e.g. "unclear what you're asking".
I'm not saying I think simply dropping the homework close reason and doing nothing else is a good idea.
I did suggest that previously in chat, and as I recall there were good objections (which are echoed in @ACuriousMind's meta answer's comments).
@DanielSank Mostly in a (probably vain) attempt to get @peterh to recognize that it's not a particularly helpful topic.
@peterh That said, he used to be fairly active on physicsoverflow, so if you really pine for the opportunity to communicate with him, you can go on ahead there. But seriously, bringing it up, particularly in that way, is not all that constructive.
@DanielSank No, the site mods could have caged him only in the PSE, and only for a year. That he got. After that his cage was extended to a 10 year long network-wide one, it couldn't be the result of the site mods. Only the CMs can do this, typically for network-wide bad deeds.
@EmilioPisanty Yes, but I had liked to talk to him here.
@DanielSank I am only curious, what he did. Maybe he attacked the whole network? Or he toke a site-level conflict to the IRL world? As I know, network-wide bans happen for such things.
@peterh That is pure fear-mongering. Unless you plan on going on extended campaigns to get yourself suspended, in which case I wish you speedy luck.
4
Seriously, suspensions are never handed out without warning, and you will not be ten-year-banned out of the blue. Ron had very clear choices and a very clear picture of the consequences of his choices, and he made his decision. There is nothing more to see here, and bringing it up again (and particularly in such a dewy-eyed manner) is far from helpful.
@EmilioPisanty Although it is already not about Ron Maimon, but I can't see here the meaning of "campaign" enough well-defined. And yes, it is a little bit of source of fear for me, that maybe my behavior can be also measured as if "I would campaign for my caging". |
Talk:Gamma-function $\Gamma$-function
$ \newcommand{\Re}{\mathop{\mathrm{Re}}} $
A transcendental function $\Gamma(z)$ that extends the values of the factorial $z!$ to any complex number $z$. It was introduced in 1729 by L. Euler in a letter to Ch. Goldbach, using the infinite product $$ \Gamma(z) = \lim_{n\rightarrow\infty}\frac{n!n^z}{z(z+1)\ldots(z+n)} = \lim_{n\rightarrow\infty}\frac{n^z}{z(1+z/2)\ldots(1+z/n)}, $$ which was used by L. Euler to obtain the integral representation (Euler integral of the second kind, cf. Euler integrals) $$ \Gamma(z) = \int_0^\infty x^{z-1}e^{-x} \rd x, $$ which is valid for $\Re z > 0$. The multi-valuedness of the function $x^{z-1}$ is eliminated by the formula $x^{z-1}=e^{(z-1)\ln x}$ with a real $\ln x$. The symbol $\Gamma(z)$ and the name gamma-function were proposed in 1814 by A.M. Legendre.
If $ $ and $ $, $ $ the gamma-function may be represented by the Cauchy–Saalschütz integral:
$ $
In the entire plane punctured at the points $ $ the gamma-function satisfies a Hankel integral representation:
$ $
where $ $ and $ $ is the branch of the logarithm for which $ $; the contour $ $ is represented in Fig. a. It is seen from the Hankel representation that $ $ is a meromorphic function. At the points $ $, $ $ it has simple poles with residues $ $.
Figure: g043310a
Contents Fundamental relations and properties of the gamma-function.
1) Euler's functional equation:
$ $
or
$ $
$ $, $ $ if $ $ is an integer; it is assumed that $ $.
2) Euler's completion formula:
$ $
In particular, $ $;
$ $
if $ $ is an integer;
$ $
3) Gauss' multiplication formula:
$ $
If $ $, this is the Legendre duplication formula.
4) If $ $ or $ $, then $ $ can be asymptotically expanded into the Stirling series:
$ $
$ $
where $ $ are the Bernoulli numbers. It implies the equality
$ $
$ $
In particular,
$ $
More accurate is Sonin's formula [6]:
$ $
5) In the real domain, $ $ for $ $ and it assumes the sign $ $ on the segments $ $, $ $ (Fig. b).
Figure: g043310b
The graph of the function $ $.
For all real $ $ the inequality
$ $
is valid, i.e. all branches of both $ $ and $ $ are convex functions. The property of logarithmic convexity defines the gamma-function among all solutions of the functional equation
$ $
up to a constant factor.
For positive values of $ $ the gamma-function has a unique minimum at $ $ equal to $ $. The local minima of the function $ $ form a sequence tending to zero as $ $.
Figure: g043310c
The graph of the function $ $.
6) In the complex domain, if $ $, the gamma-function rapidly decreases as $ $,
$ $
7) The function $ $ (Fig. c) is an entire function of order one and of maximal type; asymptotically, as $ $,
$ $
where
$ $
It can be represented by the infinite Weierstrass product:
$ $
which converges absolutely and uniformly on any compact set in the complex plane ($ $ is the Euler constant). A Hankel integral representation is valid:
$ $
where the contour $ $ is shown in Fig. d.
Figure: g043310d
$ $
G.F. Voronoi [7] obtained integral representations for powers of the gamma-function.
In applications, the so-called poly gamma-functions — $ $-th derivatives of $ $ — are of importance. The function (Gauss' $ $-function)
$ $
$ $
is meromorphic, has simple poles at the points $ $ and satisfies the functional equation
$ $
The representation of $ $ for $ $ yields the formula
$ $
where
$ $
This formula may be used to compute $ $ in a neighbourhood of the point $ $.
$ $
The functions $ $ and $ $ are transcendental functions which do not satisfy any linear differential equation with rational coefficients (Hölder's theorem).
The exceptional importance of the gamma-function in mathematical analysis is due to the fact that it can be used to express a large number of definite integrals, infinite products and sums of series (see, for example, Beta-function). In addition, it is widely used in the theory of special functions (thehypergeometric function, of which the gamma-function is a limit case, cylinder functions, etc.), in analytic number theory, etc.
References
[1] E.T. Whittaker, G.N. Watson, "A course of modern analysis" , Cambridge Univ. Press (1952) [2] H. Bateman (ed.) A. Erdélyi (ed.) , Higher transcendental functions , 1. The gamma function. The hypergeometric functions. Legendre functions , McGraw-Hill (1953) [3] N. Bourbaki, "Elements of mathematics. Functions of a real variable" , Addison-Wesley (1976) (Translated from French) [4] , Math. anal., functions, limits, series, continued fractions , Handbook Math. Libraries , Moscow (1961) (In Russian) [5] N. Nielsen, "Handbuch der Theorie der Gammafunktion" , Chelsea, reprint (1965) [6] N.Ya. Sonin, "Studies on cylinder functions and special polynomials" , Moscow (1954) (In Russian) [7] G.F. Voronoi, "Studies of primitive parallelotopes" , Collected works , 2 , Kiev (1952) pp. 239–368 (In Russian) [8] E. Jahnke, F. Emde, "Tables of functions with formulae and curves" , Dover, reprint (1945) (Translated from German) [9] A. Angot, "Compléments de mathématiques. A l'usage des ingénieurs de l'electrotechnique et des télécommunications" , C.N.E.T. (1957) Comments
The $ $-analogue of the gamma-function is given by
$ $
$ $ References
[a1] E. Artin, "The gamma function" , Holt, Rinehart & Winston (1964) [a2] R. Askey, "The $ $-Gamma and $ $-Beta functions" Appl. Anal. , 8 (1978) pp. 125–141 How to Cite This Entry:
Gamma-function.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Gamma-function&oldid=25554 |
Exercise: Find the derivative of the function. f(z)=\frac{1}{z^2+1}.
By the quotient rule, f'(z)=\frac{-2z}{(z^2+1)^2}
Exercise: Find the first and second derivatives of the function. y=\cos(x^2).
By the chain rule, y'=-2x\sin(x^2). Again, by the product rule and the chain rule, y''=-2\sin(x^2)-4x^2\cos(x^2).
Exercise: Find all points on the graph of the function f(x)=2\sin x+\sin^2x at which the tangent line is horizontal.
Calculate the derivative, f'(x)=2\cos x+2\sin x\cos x=2\cos x(1+\sin x). The points at which the tangent line is horizontal are those whose derivatives are zero. Solve f'(x)=0, we get x=(k+\frac{1}{2})\pi, where k\in \mathbb{Z}.
Exercise: Find dy/dx by implicit differentiation. y\cos x=x^2+y^2.
Take derivative on both sides, we have y'\cos x-y\sin x=2x+2yy'. Hence y'=\frac{2x+y\sin x}{\cos x-2y}
Exercise: If f(x)+x^2[f(x)]^3=10 and f(1)=2, find f'(1).
Take derivative on both sides, we have f'(x)+2x[f(x)]^3+3x^2[f(x)]^2f'(x)=0. Let x=1. Get f'(1)=-\frac{16}{25}. |
Proof attempt:
Let there be another clopen set $S$ in which is a proper subset of $\mathbb{R}$. Hence, $ S^c \neq \emptyset $.
We can assert the following statements:
No point of $S$ lies in $S^c$.
No point of $\overline{S}$ lies in $S^c$ [Since the set $S$ is closed, no point of the derived set of $S$ is a member of $S^c$].
No point of $S^c$ lies in $S$.
No point of $\overline{S^c}$ lies in $S$. [$S$ is both open and closed. Hence its complement, i.e. $S^c$ is closed and open. (complement of a closed set is an open set and vice versa).]
Therefore, $S \cap \overline{S^c} =\emptyset$ and $\overline{S} \cap S^c= \emptyset $. Therefore, $S$ and $S^c$ are separated sets. Again, $S\cup S^c =\mathbb{R}$. Therefore, $\mathbb{R}$ is disconnected, a contradiction.
[ $\mathbb{R}$ is connected, since for any $x, y \in \mathbb{R}$ $\implies$ $z \in \mathbb{R}$, where $z$ is any point such that $x<z<y$.]
Is it correct?
EDIT:
I don't know how to react (every answer being downvoted by someone with a better understanding of the subject than mine). I now try to write up a proof ( although very much unoriginal and basically a copy-paste from Rudin).
We can all agree on the fact (regarding $\mathbb{R}$) that for any two numbers $x, y \in \mathbb{R}$ with $x<y$, every number between $x$ and $y$ belongs to $\mathbb{R}$ (can we?).
Suppose, $\mathbb{R}$ can be written as the union of two non-empty separated sets $A$ and $B$ (i.e. by the very definition of separated sets, $A \cap \overline{B}= \emptyset$ and $ \overline{A} \cap B= \emptyset )$
We pick $x \in A$ and $y \in B$ with $x<y$. Define $z= \sup(A \cap [x,y])$. $z$ is going to be a limit point of $A$, $z \in \overline{A}$, therefore $z \notin B$. $\ $ $x\leq z<y$. If $z\notin A$, clearly $z\notin \mathbb{R}$. Again, if $z\in A$, we can find some $t$ between $z$ and $y$ such that $t\notin B$ [since $z$ is not a limit point of $B$]. Consequently, $t \notin \mathbb{R}$.
Being not a union of two separated subsets, $\mathbb{R}$ is connected. |
Definition. Let be a set equipped with a -algebra . A measure on (or on , or simply on if such that
1). .
2). (Countable additivity) if is a sequence of disjoint sets in , then .
2)’. if are disjoint sets in , then ,
because one can take for . A function that satisfies 1) and 2)’ but not necessarily 2) is called a finitely additive measure.
If is a set and is a -algebra, is called a measurable space and the sets in are called measurable sets. If is a measure on , then is called a measure space.
Let be a measure space. If (which implies that for all since , is called finite. If where and for all , the set is said to be -finite for . More generally, if where and for all , the set is said to be -finite for . If for each with there exists with and , is called semifinite.
Remark. Every -finite measure is semifinite, but not conversely.
Examples. 1) Let be any nonempty set, , and any function from to . Then determines a measure on by the formula . is semifinite iff for every , and is -finite iff is semifinite and is countable.
a. If for all , is called counting measure;
b. If for some , is defined by and for , is called the point mass or Dirac measure at .
2) Let be an uncountable set, and let be the -algebra of countable or co-countable sets. The function on defined by if is countable and if is co-countable is easily seen to be a measure.
3) Let be an infinite set and . Define if is finite, if is infinite. Then is a finitely additive measure but not a measure.
Theorem. Let be a measure space.
a). (Monotonicity) If and , then $\mu(E)\le\mu(F)$.
b). (Subadditivity) If , then .
c). (Continuity from below) If and , then .
d). (Continuity from above) If , , and , then .
Proof. a) If , then .
b) Let and for . Then the are disjoint and for all . Therefore, by a)
.
c) Setting , we have
.
d) Let ; then , , and . By c) then
.
Since , we may subtract it from both sides to yield the desired result.
Remark. The condition in d) could be replaced by for some , as the first ‘s can be discarded from the sequence without affecting the intersection. However, some finiteness assumption is necessary, as it can happen that for all but .
Definition. If is a measure space, a set such that is called a null set. By subadditivity, any countable union of null sets is a null set, a fact which we shall use frequently. If a statement about points is true except for in some every . (If more precision is needed, we shall speak of a -null set, or -almost everywhere).
If and , then by monotonicity provided that , but in general it need not be true that . A measure whose domain includes all subsets of null sets is called complete. Completeness can sometimes obviate annoying technical points, and it can always be achieved by enlarging the domain of , as follows.
Theorem. Suppose that is a measure space. Let and and for some . Then is a -algebra, and there is a unique extension of to a complete measure on .
Proof. Since and are closed under countable unions, so is . If where and , we can assume that (otherwise, replace and by and ). Then , so . But and , so that . Thus is a -algebra.
If as above, we set . This is well defined, since if where $F_j\subset N_j\in\mathcal N$, then $E_1\subset E_2\cup N_2$ and so , and likewise . It is easily verified that is a complete measure on , and that is the only measure on that extends .
Remark. The measure is called the completion of , and is called the completion of with respect to .
“Whenever you meet a ghost, don’t run away, because the ghost will capture the substance of your fear and materialize itself out of your own substance… it will take over all your own vitality… So then, whenever confronted with a ghost, walk straight into it and it will disappear.” ~ Allan Watts
References:
[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 24-27.
[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf. |
In the book "Quantum Field theory and the Standard Model" by Matthew Schwartz, the author states:
In quantum field theory, we generally work in the Heisenberg picture, where all time dependence is in operators such as $\phi$ and $a_p$. For free fields, the creation and annihilation operators for each momentum $\bf{p}$ in the quantum field are just those of a simple harmonic oscillator. These operators should satisfy $a_p(t) = e^{-i\omega_p t}a_p$ and $a_p^\dagger(t) = e^{i\omega_p t}a_p^\dagger$, where $a_p$ and $a_p^\dagger$ are time independent. Then we can
definea quantum scalar field as
$$\phi_0({\bf{x}}, t)=\int \dfrac{d^3 p}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_p}}(a_p e^{-ipx}+a_p^\dagger e^{ipx})$$
with $p^\mu = (\omega_p, {\bf{p}})$ and $\omega_p = |{\bf{p}}|$.
Now I first thought the author was presenting the KG field as just one example. I also thought that this form of writing $\phi$ was only for the KG field, after all it was developed in order to make the field satisfy the massless KG equation.
By this statement, it seems that the author implies this is valid for
all quantum scalar fields. Is this true? I mean, this exact same expansion, with creation and annihilation operators defined in a Fock Space, is valid for all quantum scalar fields?
If so, what distinguishes one field from the other?
And why it would be reasonable for this to be so general as that, if it was derived from a very simple case? |
For reference, here is figure 5.21, which is originally from NACA report 116 (1921):
Each curve represents a wing with an aspect ratio as labeled. In modern terms, $C_a=100C_\mathrm{L}$ and $C_w=100C_\mathrm{D}$, where $C_\mathrm{L}$ is lift coefficient and $C_\mathrm{D}$ is drag coefficient.
Anderson derives eq. (5.66), which scales the drag coefficient of a given wing (subscript 2) to that of a wing with aspect ratio of 5 (subscript 1):\begin{equation}C_{D,1}=C_{D,2}+\frac{C_{L}^2}{\pi e}\Bigg(\frac{1}{5}-\frac{1}{\mathrm{AR}_2}\Bigg)\end{equation}
The original NACA report states that this formula has "been found to hold for distributions of lift [that] do not deviate too much from elliptical ones, although strictly speaking [it applies] only to the latter." I'm not sure why Anderson doesn't include this important point, but the implication is that we can safely take $e=1$.
Now proceed as follows: Select a curve from figure 5.21 and plug the corresponding aspect ratio into eq. (5.66). Then pick any $(C_w,C_a)$ pair off the curve and plug those into the equation. Plot the resultant $C_{w,1}$ on a figure against the same $C_{a}$. Continue with as many coordinate pairs as you want from the same curve; plot them all with the same symbol. Then select another curve and repeat but using a different symbol. Include a legend. The result should look something like figure 5.22: |
Topological Quotients
Recall from the Final Topologies page that if $X$ is a set and $\{ Y_i : i \in I \}$ is a collection of topological spaces then the final topology induced by $\{ f_i : i \in I \}$ on $X$ is the finest topology $\tau$ which makes the maps $f_i : Y_i \to X$ continuous for all $i \in I$.
We saw that if $\tau$ is the final topology induced by the collection of maps $\{ f_i : i \in I \}$ can be given by:(1)
That is, the final topology induced by $\{ f_i : i \in I \}$ is equal to the collection of subsets of $X$ whose inverse images are contained in every topology $\tau_i$ for $i \in I$.
We will now look at a special type of topology on a set known as a quotient topology but we will first need to define a special map known as a quotient map.
Definition: Let $(X, \tau)$ be a topological space and let $\sim$ be an equivalence relation on $X$. For each $x \in X$, denote the equivalence class of $x$ by $[x] = \{ y \in X : x \sim y \}$ and let $X \: / \sim$ denote the set of all of these equivalence classes. The Quotient Map with respect to the set $X$ and the equivalence relation $\sim$ is defined as the map $q : X \to X \: / \sim$ defined for all $x \in X$ by $q(x) = [x]$.
We are now ready to define a quotient topology on the set of equivalence classes $X \: / \sim$ where $\sim$ is an equivalence relation on $X$.
Definition: Let $(X, \tau)$ be a topological space and let $\sim$ be an equivalence relation on $X$. The Quotient Topology on $X \: / \sim$ is the final topology induced by the quotient map $q : X \to X \: / \sim$. Then the resulting Topological Quotient is the topological space of the set $X \: / \sim$ with the quotient topology. Equivalently, the quotient topology on $X \: / \sim$ is the finest topology which maps the quotient map $q : X \to X \: / \sim$ continuous.
Proposition 1: Let $(X, \tau)$ be a topological space and $\sim$ be an equivalence relation on $X$. Furthermore, let $X \: / \sim$ have the quotient topology with respect to $\sim$. Then $U \subseteq X \: / \sim$ is open in $X \: / \sim$ if and only if $\displaystyle{\bigcup_{[x] \in U} [x]}$ is open in $X$. Proof:By definition, the quotient topology $\tau^*$ on $X \: / \sim$ is the final topology on the quotient map $q : X \to X \: / \sim$ and can be described as the following set: $\Rightarrow$ Suppose that $U \subseteq X \: / \sim$ is open in $X \: / \sim$. Then $q^{-1}(U)$ is open in $X$. But: So $\bigcup_{[x] \in U} [x]$ is open in $X$. $\Leftarrow$ Let $U \subseteq X \: / \sim$ and suppose that $\bigcup_{[x] \in U} [x]$ is open in $X$. Then $q \left ( \bigcup_{[x] \in U} [x] \right ) = U$. So $q^{-1}(U)$ is open and since the quotient topology is the finest topology making $q : X \to X \: / \sim$ continuous we have that $U$ is open in $X \: / \sim$. $\blacksquare$ |
First, let's take a look at the first law, which says that during any thermodynamic process,
$$dU = \delta Q - \delta W$$where $dU$ is the change in the total internal energy of the system, $\delta Q$ is the amount of heat added to the system, and $\delta W$ is the mechanical work done by the system.
The reason we typically use $\delta$ instead of $d$ on the right hand side is because "heat" and "work" are not so-called
state functions. What that means is that even if you have complete knowledge of the current state of the system, there is no way to know how much mechanical work was performed (or how much heat was added) in order to bring it to its current condition. Writing the right hand side as $dQ-dW$ would misleadingly imply that there are functions $Q$ and $W$ which have corresponding differentials $dQ$ and $dW$; this is emphatically not the case. Such functions don't exist.
This can be demonstrated by observing that if you take your system through a cyclic process, so the state of the system at the end is precisely the same as it was in the beginning, then the total amount of heat added (and the total amount of work done) is generically nonzero - therefore "total heat" (and "total work") cannot possibly be functions which are determined by the state of the system at any given time.
Mathematically, this can be expressed by the statement$$\sum_{\text{cyclic process}} \delta Q \neq 0$$
It was historically noticed by Clausius that although the above is true, in certain (so-called reversible) processes, we find that
$$\sum_{\text{rev. cyclic process}} \frac{\delta Q}{T} = 0 $$
This implies that for reversible processes, while there is no state function whose differential is $\delta Q$, there
is a state function whose differential is $\frac{\delta Q}{T}$. We call this function $S$, and say that for reversible processes,
$$dS = \frac{\delta Q}{T} \implies \delta Q = T dS$$
Similarly,
$$\sum_{\text{rev. cyclic process}} \frac{\delta W}{p} = 0$$
where $p$ is the pressure. Therefore, there exists a state function (which we'll call $V$) such that during reversible processes,$$\frac{\delta W}{p} = dV \implies \delta W = p dV$$
It follows that for reversible processes, we can rewrite the first law as follows:
$$dU = TdS - pdV$$
The second term is easy enough to interpret - $V$ is simply the volume of the system at a particular time, and $pdV$ can be identified as the mechanical work being done by the pressure being exerted on the walls of the box, in complete agreement with Newtonian physics. The first term is less familiar; $S$ is called the
entropy of the system, but at this point it doesn't have an immediately obvious physical interpretation besides being a valid state function.
The development of statistical thermodynamics gives us a much clearer interpretation for $S$. From a macroscopic point of view, we describe a system by macroscopic quantities like total energy and volume. Such quantities define a so-called
macrostate for our system.
If you zoom in to the level of atoms and molecules, however, the system becomes much more complicated. A
microstate is specified by providing microscopic quantities like the position and momentum of each atom or molecule.
Each macrostate of the system corresponds to many, many microstates. For instance, it's easy to imagine rearranging the particles of a non-interacting gas in such a way as to create a totally different microstate with precisely the same pressure, temperature, total energy, etc. Let $\Omega_i$ be the number of microstates which correspond to some macrostate $i$; then the
entropy $S_i$ of that macrostate is simply $k_B \ln(\Omega_i)$. In other words, the entropy is the (log of the) number of ways you could microscopically rearrange your system without changing anything on the macroscopic level.
From there, we argue that in equilibrium, the system will occupy the macrostate which corresponds to the largest number of microstates - the macrostate of maximal entropy.
Why is the change of entropy related to the change of heat?
From a thermodynamic perspective, the answer is "because that's how it is defined." It is the state function which can be related to the change in heat by dividing by the temperature of the system, and its existence was first noticed empirically.
From a microscopic/statistical perspective, the answer is that the number of accessible microstates of a system depends on how much total energy is divided among the system's constituents. Loosely speaking, if I give you and four of your friends a handful of pennies, the number of ways you could divide up those pennies is related to how many pennies I give you. In the same way, the entropy of a system is related to its total internal energy.
From there, there are two ways I could increase the total internal energy of the system - I could allow it to do work, or I could provide it with heat. Performing work equates to a change in volume; adding heat equates to a change in entropy.
Does an increase of entropy produce an increase of heat? It does. But why?
No, not always. You could increase the entropy of a system without adding heat. For instance, you could increase the number of particles in the system, or increase the size of the box. In the coin analogy, if I give you the same number of coins but you add another five friends, then the number of possible arrangements goes up.
Does it mean that a system which undergoes a decrease of entropy (at the expense of its surroundings) requires less energy to reproduce?
When a system loses entropy, then that means that it occupies a macrostate corresponding to fewer possible microscopic rearrangements. It could be because the system loses energy (fewer coins), or it could be because there are fewer particles (some of your friends go home), or it could be because the volume of the system decreases (my analogy doesn't really apply here).
Put it in another way: does the second law mean that, as the entropy increases in an isolated system, every system is increasingly rising its internal energy?
No, it definitely doesn't mean that. Imagine a hot system $H$ in thermal contact with a cold system $C$. Energy will flow from $H$ to $C$, and it will be conserved because whatever energy is lost by $H$ is gained by $C$.
However, $H$ will also lose
entropy. Since we have $dU_H + dU_C = 0$ and that $dU=TdS$ (we assume the volumes don't change), $$dS_H + dS_C = \frac{dU_H}{T_H} + \frac{dU_C}{T_C} = dU_H\left(\frac{1}{T_H}-\frac{1}{T_C}\right)>0$$
because $dU_C=-dU_H$, $T_H > T_C$, and $dU_H<0$ (because $H$ is losing energy). Therefore, the total energy has been conserved ($dU_H+dU_C=0$) but the total entropy has increased ($dS_H+dS_C > 0$). |
The Jordan Curve Theorem
Definition: Let $f : S^1 \to \mathbb{R}^2$ be an embedding. A Jordan Curve or Plane Simple Closed Curve is the image $f(S^1)$.
In other words, if $f : S^1 \to \mathbb{R}^2$ is an embedding of the circle in $\mathbb{R}^2$, then $f(S^1)$ will be a closed loop that does not intersect itself nontrivially. Some examples of Jordan curves are illustrated below:
Before we state the Jordan curve theorem, we will first state an important lemma that can be used to prove the result.
Lemma 1: Let $\alpha, \beta : [0, 1] \to [0, 1] \times [0, 1]$ be paths and suppose that: a) $\alpha(0) \in \{ 0 \} \times [0, 1]$. b) $\alpha(1) \in \{ 1 \} \times [0, 1]$. c) $\beta(0) \in [0, 1] \times \{ 0 \}$. d) $\beta(1) \in [0, 1] \times \{ 1 \}$.
Then $\alpha(I) \cap \beta (I) \neq \emptyset$, that is, the graphs of $\alpha$ and $\beta$ intersect in at least one point on $[0, 1] \times [0, 1]$.
We now state the Jordan curve theorem.
Theorem (The Jordan Curve Theorem): Let $f : S^1 \to \mathbb{R}^2$ be an embedding so that $f(S^1)$ is a Jordan curve. Then $\mathbb{R}^2 \setminus f(S^1)$ has two components - one of which is bounded and the other of which is unbounded.
Intuitively, the Jordan curve theorem is clear but it actually takes quite a bit of work to prove. We will omit the proof for the time being. |
also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling
@heather well, there's a spectrum
so, there's things like New Journal of Physics and Physical Review X
which are the open-access branch of existing academic-society publishers
As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di...
Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago
> A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service”
for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty
> for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing)
@0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals.
@BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work...
@BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions.
Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley.
I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea.
@EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results...
Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town...
@EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit. |
I will explain these two spaces in the context of fluid dynamics. In fluid dynamics, flow velocity components can be expressed as the derivative of scalar stream function. Interestingly, the structure of the stream function is analogous to Hamilton's equation. This similarity was realized in the 80's which gave a new direction to the field.Configuration ...
In case someone else bumps into the same question, I think I found the answer:The Euler's theorem states that if $f$ is a homogeneous function ofdegree $n$ in the variables $x_i$, then$$\sum_i x_i\frac{\partial f}{\partial x_i}=nf.$$So for example, if $f$ is a function of two variables $x_1, x_2$ and it is homogeneous, say to 3rd degree, in ...
When a photon interacts with an atom, three things can happen:elastic scattering, the photon keeps its energy and changes angle (mirror reflection)https://en.wikipedia.org/wiki/Elastic_scatteringinelastic scattering, the photon keeps part of its energy and changes angle (photon transferring vibrational and rotational energies to the molecules, heat up ...
Angular velocity is a vector quantity, and angular speed is defined as its magnitude (and therefore a scalar quantity).If you are talking about uniform circular motion, the angular velocity after one complete rotation is not zero, but rather a constant (non-zero) vector.
Angular velocity is a Vector quantity and angular speed is a scalar quantity.angular speed may be can constant but angular velocity not necessarily remaining constant as $\mathbf{\omega}=\mathbf{r}\text{x}\mathbf{v}$.But $\mathbf{v}$ changing with time due to changing direction.
Homogeneity is a property of composition. If a system is made of the same parts everywhere, then it is considered homogenous. Homogenity has to do with the smallest units that have identical composition or character. The central question here is one of identity. This allows a multi-component system to be described by homogeneity as well. Even a multi-...
I think there is something wrong with the problem. By convention the direction of the electric field is the direction of the force that a positive charge would experience if placed in the field. In this case no external force is needed to move the charge from A to B as it would naturally be accelerated by the field. Only the field does work.If the charge ...
What your book is saying, for moving the block from B to A you need to do some work, and this work will be same and opposite of the work done by electrostatic force. Here external force, which provide by you and internal mean electrostatic.You can think about this situation in one more way, if charge to be at rest, at every point on line Ab=force by ...
When you compare capacitors with discharge tubes, you are comparing those vacuum capacitors sealed in glass tubes with gas discharge tubes. There are many other types of capacitors that do not look like discharge tubes thus are not comparable.Of course the gas pressure and type in the tubes play a big role. In a vacuum capacitor, the residual gas pressure ...
Exponentials differentiate and integrate better than trig functions, and in general are “easier to combine” than working with trig functions, v.g. complex impedance in a circuit. Taking the real part at the end “brings you back” to the physical fields, voltages, currents etc.
By "physical values" I assume you mean observables, i.e. quantities that one can measure in real life. Observables are always real number ($\mathbb{R}$) -- at least so far. If you manage to measure a $3\mathrm{i}$ long slab of wood, let me know.Complex numbers enter physical problems in two ways:1:They are integral part of a theory (e.g. quantum ...
The shortest answer is that a quantum system is any system that obeys the laws of quantum mechanics. This means that:The system's state at a given time is described by a vector in acomplex vector space. This vector is called the system's wave function.The system's wave function evolves over time following the Schrodinger wave equation.Each attribute of ...
It came up because I was thinking about a state $|\psi\rangle$ which is defined to be the (unique, say) eigenstate of some observable $\hat{o}$ with eigenvalue $\lambda$. Then we see that the time-evolved state $|\psi(t)\rangle = U(t,0) |\psi\rangle$ can be characterized as the eigenstate of the operator $\hat{o}_R(t)$This property is necessary for the ...
Yes, they are (for the most part) the same, typically being used interchangeably.And really, both are kind of a misnomer, because things typically not considered "analytic" or "exact" forms are, in fact, very often "exact" in the literal sense. There is nothing "inexact" about, say, the expression$$e = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\...
Free parameters are not predefined but must be estimated by theory or experimentally. Or it can be a parameter used in fitting a dataset with an expression. The free parameters are varied to get a good fit to the data.G in Newton's gravitational equation is a free parameter and has been measured but not with high accuracy.
I pronounce it “nine point eight zero meters per second per second”, but there isn’t a standard pronunciation for physical quantities. Another pronunciation would be “nine point eight oh meters per second squared”.Only the seconds are being squared. Units are written using the usual mathematical precedence rules. You are probably aware that $ab^2$ means $a(...
More generally, if the wave equation $\Box y=0$ is satisfied for a scalar field $$\mathbb{R}^{n+1}~\ni~ (\vec{x},t)\quad \stackrel{y}{\mapsto} \quad y(\vec{x},t) ~\in~ \mathbb{R},$$ with spacetime $\mathbb{R}^{n+1}$ as domain, and with a 1-dimensional target space $\mathbb{R}$, we speak of the wave equation in $n+1$-dimensional spacetime, or equivalently, ...
The 1D wave equation is called like that because it has only one independent space variable, $x$. That's it. The 2D equation has two variables, etc.You are correct that an oscillating rope sweeps out a 2-dimensional plane; in fact, by moving one end in a circle instead of up and down, you can make it occupy a 3D region. But the equation doesn't care about ...
Threads refer to the helical ridge that wraps around a screw—tornillo in Spanish. Each complete rotation of a screw with 10 threads per inch advances it one tenth of an inch.As mentioned in the comments, 126 inches is the circumference of the circle swept out by the handle—the distance you move your hand to put the screw through one full rotation.
A field configuration that solves the equations of motion of a theory is topologically stable if it cannot be continuously transformed to the vacuum keeping the energy finite.The reason for the name is the fact that topologically stable solutions can be classified by a topological quantum number.
By no first-order changes Feynman means that the first-order functional derivative vanishes, or equivalently, the path is stationary.By the way, no first-order changes is a common talking point of Feynman. Listen e.g. to 46:48-48:48 in the talk The Character of Physical Law, part 4, where he makes similar remarks about the principle of least action.
Any continuous and differentiable function $f(x)$ can be expressed as a Taylor series:$$ f(x_0+\delta x) = f(x_0) + \frac{\mathrm{d}f}{\mathrm{d}x}\bigg|_{x_0}\delta x + \frac{1}{2}\frac{\mathrm{d} x^2f}{\mathrm{d}^2x}\bigg|_{x_0}\delta x^2 + \dots + \frac{1}{n!}\frac{\mathrm{d}^nf}{\mathrm{d}^nx}\bigg|_{x_0}\delta x^n .$$Each of these terms are called of ... |
Is there any possibility to have a robot with
less than six degrees of freedom & still be able to achieve any position & orientation around end effector?
Short answer: No.
Long answer: In how many dimensions you are defining 'any'?
The best way to understand it is actually through math.
Let's say you have a desired position $$p=\left[\begin{matrix}x\\y\\z\end{matrix}\right]$$ and an orientation $$\omega=\left[\begin{matrix}\chi\\\psi\\\zeta\end{matrix}\right]$$ in three dimensions. Now let's say we have a robot with six degrees of freedom $$Q=\left[\begin{matrix}q1\\q2\\q3\\q4\\q5\\q6\end{matrix}\right]$$.
Assuming you know your position and orientation you are trying to find your joint angles. This can be written in matrix form.
$$\left[\begin{matrix}q1\\q2\\q3\\q4\\q5\\q6\end{matrix}\right]= F\left[\begin{matrix}x\\y\\z\\\chi\\\psi\\\zeta\end{matrix}\right]$$.
Where F a 6x6 matrix with your inverse kinematics functions.
Notice how you have 6 knowns (your position and orientation) and six unknown (your joint angles).
Basically you have a 6-by-6 system to solve. If you have fewer degrees of freedom then you have fewer unknowns which means that you can't solve independently. Meaning your position and orientation are linearly dependent.
For example changing your position $x$, will change your $\chi$ or your $\psi$ or even your $y$. Imagine trying to move a glass of water in a straight line and suddenly the glass turns over and then back again.
So think a robot with less than six degrees of freedom as a robot that you can't control its behavior at any time rather than a robot that can't reach a certain position at any time.
Now if you have a task that does not require to move along and rotate around an axis (e.g. $z$ & $\zeta$) then you can ignore them and now you have a 4-by-4 system to worry about. Therefore, for this 2D task, a robot with four degrees of freedom can achieve 'any' position and orientation. Similarly, if you put a constraint on the force along the $x$ direction, you now have an $F_x$ variable that entered your system making it a 7-by-7 system that requires seven degrees of freedom to achieve 'any' position and orientation.
Generally, 6 degrees of freedom is the standard robot because it can control its end effector along each dimension independently. Any fewer degrees of freedom can still be useful in some more limited tasks. Higher degrees of freedom can tackle more complicated tasks, though in those case there is a selection whether all parameters will be satisfied independently or no.
The bottom line is, for general cases in a 3D environment you need 6 degrees of freedom.
I tried to keep my answer short without using too much math but let me know if you are still unsure about the effect the number of degrees of freedom have on the position and orientation of the end effector.
In abstract theory, yes. You can have a body constrained to a path (1 Dof, forward and back along the path) and the path can loop through all possible orientations. Likewise you can constrain the body to a higher order manifold (a surface allows 2 or 3Dof depending on the physical mechanism of constraint).
Mechanisms that we can build have other requirements (like supporting loads and forces) that limit what we can do.
Here's a simple example of a practical mechanism. Say you have a 2 DoF mechanism that controls orientation and distance along a line. The accuracy of the distance, and orientation, is limited to the precision of your encoders. You can replace that 2Dof mechanism with a 1DoF screw who's pitch is equal to the linear encoder resolution. As the output spins on the screw, it can reproduce every position and orientation that the 2Dof mechanism could. The cost is that you have to spin around a full revolution to move one step in the linear direction.
Another good example is a car. A 2DoF mechanism on a 3DoF plane. You can achieve any position and orientation with a car, but you may need to do some loops or back-and-forth driving to get there.
So, keeping in mind that a 6DoF robot has limited positions and orientations that it can reach, and limited resolution, Yes, there is the possibility that a 5 or fewer Dof machine could achieve the same range of position and orientations.
Physically Speaking
No.
Even most 6-axis (6DoF) industrial robots are unable to achieve full range of motion within their work envelope (or the work envelope is artificially restricted as more commonly seen). 7-axis (7DoF) robotic arms (commonly know as collaborative robots) have a much greater flexibility; They are generally only restricted by physical interference with their own casting (body).
In the below link, the author discusses the relative restrictions on modeling an Nth degree robotic arm. Specifically, as to how difficult it is to model how the casting restricts End-of-arm-Tooling in both position and orientation, as the mathematics can produce a valid result that results in an unattainable position (robot physically intersecting itself). |
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Isomorphisms are very important in mathematics, and we can no longer put off talking about them. Intuitively, two objects are 'isomorphic' if they look the same. Category theory makes this precise and shifts the emphasis to the 'isomorphism' - the
way in which we match up these two objects, to see that they look the same.
For example, any two of these squares look the same after you rotate and/or reflect them:
An isomorphism between two of these squares is a
process of rotating and/or reflecting the first so it looks just like the second.
As the name suggests, an isomorphism is a kind of morphism. Briefly, it's a morphism that you can 'undo'. It's a morphism that has an inverse:
Definition. Given a morphism \(f : x \to y\) in a category \(\mathcal{C}\), an inverse of \(f\) is a morphism \(g: y \to x\) such that
and
I'm saying that \(g\) is 'an' inverse of \(f\) because in principle there could be more than one! But in fact, any morphism has at most one inverse, so we can talk about 'the' inverse of \(f\) if it exists, and we call it \(f^{-1}\).
Puzzle 140. Prove that any morphism has at most one inverse. Puzzle 141. Give an example of a morphism in some category that has more than one left inverse. Puzzle 142. Give an example of a morphism in some category that has more than one right inverse.
Now we're ready for isomorphisms!
Definition. A morphism \(f : x \to y\) is an isomorphism if it has an inverse. Definition. Two objects \(x,y\) in a category \(\mathcal{C}\) are isomorphic if there exists an isomorphism \(f : x \to y\).
Let's see some examples! The most important example for us now is a 'natural isomorphism', since we need those for our databases. But let's start off with something easier. Take your favorite categories and see what the isomorphisms in them are like!
What's an isomorphism in the category \(\mathbf{3}\)? Remember, this is a free category on a graph:
The morphisms in \(\mathbf{3}\) are paths in this graph. We've got one path of length 2:
$$ f_2 \circ f_1 : v_1 \to v_3 $$ two paths of length 1:
$$ f_1 : v_1 \to v_2, \quad f_2 : v_2 \to v_3 $$ and - don't forget - three paths of length 0. These are the identity morphisms:
$$ 1_{v_1} : v_1 \to v_1, \quad 1_{v_2} : v_2 \to v_2, \quad 1_{v_3} : v_3 \to v_3.$$ If you think about how composition works in this category you'll see that the only isomorphisms are the identity morphisms. Why? Because there's no way to compose two morphisms and get an identity morphism unless they're both that identity morphism!
In intuitive terms, we can only move from left to right in this category, not backwards, so we can only 'undo' a morphism if it doesn't do anything at all - i.e., it's an identity morphism.
We can generalize this observation. The key is that \(\mathbf{3}\) is a poset. Remember, in our new way of thinking a
preorder is a category where for any two objects \(x\) and \(y\) there is at most one morphism \(f : x \to y\), in which case we can write \(x \le y\). A poset is a preorder where if there's a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(x = y\). In other words, if \(x \le y\) and \(y \le x\) then \(x = y\). Puzzle 143. Show that if a category \(\mathcal{C}\) is a preorder, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(g\) is the inverse of \(f\), so \(x\) and \(y\) are isomorphic. Puzzle 144. Show that if a category \(\mathcal{C}\) is a poset, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then both \(f\) and \(g\) are identity morphisms, so \(x = y\).
Puzzle 144 says that in a poset, the only isomorphisms are identities.
Isomorphisms are a lot more interesting in the category \(\mathbf{Set}\). Remember, this is the category where objects are sets and morphisms are functions.
Puzzle 145. Show that every isomorphism in \(\mathbf{Set}\) is a bijection, that is, a function that is one-to-one and onto. Puzzle 146. Show that every bijection is an isomorphism in \(\mathbf{Set}\).
So, in \(\mathbf{Set}\) the isomorphisms are the bijections! So, there are lots of them.
One more example:
Definition. If \(\mathcal{C}\) and \(\mathcal{D}\) are categories, then an isomorphism in \(\mathcal{D}^\mathcal{C}\) is called a natural isomorphism.
This name makes sense! The objects in the so-called 'functor category' \(\mathcal{D}^\mathcal{C}\) are functors from \(\mathcal{C}\) to \(\mathcal{D}\), and the morphisms between these are natural transformations. So, the
isomorphisms deserve to be called 'natural isomorphisms'.
But what are they like?
Given functors \(F, G: \mathcal{C} \to \mathcal{D}\), a natural transformation \(\alpha : F \to G\) is a choice of morphism
$$ \alpha_x : F(x) \to G(x) $$ for each object \(x\) in \(\mathcal{C}\), such that for each morphism \(f : x \to y\) this naturality square commutes:
Suppose \(\alpha\) is an isomorphism. This says that it has an inverse \(\beta: G \to F\). This \(beta\) will be a choice of morphism
$$ \beta_x : G(x) \to F(x) $$ for each \(x\), making a bunch of naturality squares commute. But saying that \(\beta\) is the inverse of \(\alpha\) means that
$$ \beta \circ \alpha = 1_F \quad \textrm{ and } \alpha \circ \beta = 1_G .$$ If you remember how we compose natural transformations, you'll see this means
$$ \beta_x \circ \alpha_x = 1_{F(x)} \quad \textrm{ and } \alpha_x \circ \beta_x = 1_{G(x)} $$ for all \(x\). So, for each \(x\), \(\beta_x\) is the inverse of \(\alpha_x\).
In short: if \(\alpha\) is a natural isomorphism then \(\alpha\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\).
But the converse is true, too! It takes a
little more work to prove, but not much. So, I'll leave it as a puzzle. Puzzle 147. Show that if \(\alpha : F \Rightarrow G\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\), then \(\alpha\) is a natural isomorphism.
Doing this will help you understand natural isomorphisms. But you also need examples!
Puzzle 148. Create a category \(\mathcal{C}\) as the free category on a graph. Give an example of two functors \(F, G : \mathcal{C} \to \mathbf{Set}\) and a natural isomorphism \(\alpha: F \Rightarrow G\). Think of \(\mathcal{C}\) as a database schema, and \(F,G\) as two databases built using this schema. In what way does the natural isomorphism between \(F\) and \(G\) make these databases 'the same'. They're not necessarily equal!
We should talk about this. |
M 3: a new muon missing momentum experiment to probe (g – 2) μ and dark matter at Fermilab Abstract
Here, new light, weakly-coupled particles are commonly invoked to address the persistent $$\sim 4\sigma$$ anomaly in $$(g-2)_\mu$$ and serve as mediators between dark and visible matter. If such particles couple predominantly to heavier generations and decay invisibly, much of their best-motivated parameter space is inaccessible with existing experimental techniques. In this paper, we present a new fixed-target, missing-momentum search strategy to probe invisibly decaying particles that couple preferentially to muons. In our setup, a relativistic muon beam impinges on a thick active target. The signal consists of events in which a muon loses a large fraction of its incident momentum inside the target without initiating any detectable electromagnetic or hadronic activity in downstream veto systems. We propose a two-phase experiment, M$^3$ (Muon Missing Momentum), based at Fermilab. Phase 1 with $$\sim 10^{10}$$ muons on target can test the remaining parameter space for which light invisibly-decaying particles can resolve the $$(g-2)_\mu$$ anomaly, while Phase 2 with $$\sim 10^{13}$$ muons on target can test much of the predictive parameter space over which sub-GeV dark matter achieves freeze-out via muon-philic forces, including gauged $$U(1)_{L_\mu - L_\tau}$$.
Authors: Princeton Univ., Princeton, NJ (United States) Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Publication Date: Research Org.: Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Sponsoring Org.: USDOE Office of Science (SC), High Energy Physics (HEP) (SC-25) OSTI Identifier: 1439466 Report Number(s): arXiv:1804.03144; FERMILAB-PUB-18-087-A Journal ID: ISSN 1029-8479; 1667037; TRN: US1900618 Grant/Contract Number: AC02-07CH11359 Resource Type: Journal Article: Accepted Manuscript Journal Name: Journal of High Energy Physics (Online) Additional Journal Information: Journal Volume: 2018; Journal Issue: 9; Journal ID: ISSN 1029-8479 Publisher: Springer Berlin Country of Publication: United States Language: English Subject: 79 ASTRONOMY AND ASTROPHYSICS; 46 INSTRUMENTATION RELATED TO NUCLEAR SCIENCE AND TECHNOLOGY; 72 PHYSICS OF ELEMENTARY PARTICLES AND FIELDS; Fixed target experiments Citation Formats
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, and Whitbeck, Andrew.
M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab. United States: N. p., 2018. Web. doi:10.1007/JHEP09(2018)153.
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, & Whitbeck, Andrew.
M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab. United States. doi:10.1007/JHEP09(2018)153.
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, and Whitbeck, Andrew. Wed . "M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab". United States. doi:10.1007/JHEP09(2018)153. https://www.osti.gov/servlets/purl/1439466.
@article{osti_1439466,
title = {M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab}, author = {Kahn, Yonatan and Krnjaic, Gordan and Tran, Nhan and Whitbeck, Andrew}, abstractNote = {Here, new light, weakly-coupled particles are commonly invoked to address the persistent $\sim 4\sigma$ anomaly in $(g-2)_\mu$ and serve as mediators between dark and visible matter. If such particles couple predominantly to heavier generations and decay invisibly, much of their best-motivated parameter space is inaccessible with existing experimental techniques. In this paper, we present a new fixed-target, missing-momentum search strategy to probe invisibly decaying particles that couple preferentially to muons. In our setup, a relativistic muon beam impinges on a thick active target. The signal consists of events in which a muon loses a large fraction of its incident momentum inside the target without initiating any detectable electromagnetic or hadronic activity in downstream veto systems. We propose a two-phase experiment, M$^3$ (Muon Missing Momentum), based at Fermilab. Phase 1 with $\sim 10^{10}$ muons on target can test the remaining parameter space for which light invisibly-decaying particles can resolve the $(g-2)_\mu$ anomaly, while Phase 2 with $\sim 10^{13}$ muons on target can test much of the predictive parameter space over which sub-GeV dark matter achieves freeze-out via muon-philic forces, including gauged $U(1)_{L_\mu - L_\tau}$.}, doi = {10.1007/JHEP09(2018)153}, journal = {Journal of High Energy Physics (Online)}, issn = {1029-8479}, number = 9, volume = 2018, place = {United States}, year = {2018}, month = {9} } Citation information provided by Web of Science
Web of Science
Figures / Tables: left) and vector ( right) forces that couple predominantly to muons. In both cases, a relativistic muon beam is incident on a fixed target and scatters coherently off a nucleus to produce the new particle as initial- ormore » |
This question already has an answer here:
I just found this site today and I asked a few questions but I didn't know how to use the latex syntax, so my questions were difficult to understand. Can anyone tell me how to use this Latex program? thanks
This question already has an answer here:
I just found this site today and I asked a few questions but I didn't know how to use the latex syntax, so my questions were difficult to understand. Can anyone tell me how to use this Latex program? thanks
Expanding on Dilaton's comment:
The site (and most other stackexchange sites where such is appropriate) use the exact same implementation of MathJax for Latex markup. Basically, just enclose tex within
$ (inline) or
$$ (set apart equations). Most basic commands work, like
\frac and
\sin.
You can also begin some Latex environments without dollar signs. For example
\begin{align}...\end{align} is a valid construct that will have its contents parsed by the tex parser.
As for how to use Latex itself, independent of this site, well that is a large enough topic to have its own stackexchange site. For basic stuff to get started, try this other FAQ topic. For site-specific abilities, I can think of no better guide than what they have over at math.SE, found here.
Finally, you can always click the
Edit button on any post to see how it was written. You are not committed to making any edits. |
Properties of the Complex Exponential Function
Recall from The Complex Exponential Function page that if $z = x + yi \in \mathbb{C}$ then we defined the complex exponential function by:(1)
We will now look at some basic properties regarding the complex exponential function - many of which are analogous to the real exponential function but we must indeed prove these for the complex case!
Proposition 1: If $z, w \in \mathbb{C}$ then: a) $e^{z} \neq 0$. b) $e^{\pi i} = -1$ (Euler's Formula). c) $e^z = 1$ if and only if $z = 2k\pi i$ for some $k \in \mathbb{Z}$. d) $e^{z + w} = e^z \cdot e^w$. e) $\mid e^z \mid = e^x$ (where $z = x + yi$. f) $e^0 = 1$, $e^{\frac{\pi}{2}i} = i$, $e^{\pi i} = -1$, and $e^{\frac{3\pi}{2}i} = -i$. Proof of a)Let $z = x + yi \in \mathbb{C}$. Then $x, y \in \mathbb{R}$, and $e^z = e^x (\cos y + i \sin y )$. Notice that $e^x > 0$ for all $x \in \mathbb{R}$, so we will look at the other factor, $\cos y + i \sin y$. This factor equals zero if and only if $\cos y = 0$ and $\sin y = 0$. But no such $y \in \mathbb{R}$ exists. So $\cos x + i \sin y \neq 0$ for all $y \in \mathbb{R}$. Thus $e^z \neq 0$. $\blacksquare$ Proof of b)Substituting $x = \pi i$ into the complex exponential function yields: Therefore $e^{\pi i} = -1$. $\blacksquare$ Proof of c)$\Rightarrow$ Let $z = x + yi$ and suppose that $e^z = 1$. Then $e^x (\cos y + i \sin y) = 1$. So $\sin y = 0$, i.e., $y = k\pi$ for some $k \in \mathbb{Z}$. Since $e^x \cos y = 1$ and $e^x > 0$ we must have that $\cos y > 0$ and so $y = 2k\pi $ for some [[$ k \in \mathbb{Z}$ in which case we have that $e^x = 0$, i.e., $x = 0$. So $z = 2k\pi i$ for some $k \in \mathbb{Z}$. $\Leftarrow$ Suppose that $z = 2k \pi i$ for some $k \in \mathbb{Z}$. Then: Proof of d)Let $z = x_1 + y_1i, w = x_2 + y_2i \in \mathbb{C}$. Then: Proof of e)Let $z = x + yi \in \mathbb{C}$. Then: We know that $x \in \mathbb{R}$ and $e^x > 0$ so $\mid e^z \mid = e^x$. $\blacksquare$ Proof of f)We have that: |
Hint Applet (Trigonometric Substitution) Sample Problem Flash Applets embedded in hint portion of WeBWorK questions Example Sample Problem with trigSubWW.swf embedded
A standard WeBWorK PG file with an embedded applet has six sections:
A tagging and description section, that describes the problem for future users and authors, An initialization section, that loads required macros for the problem, A problem set-up sectionthat sets variables specific to the problem, An Applet link sectionthat inserts the applet and configures it, (this section is not present in WeBWorK problems without an embedded applet) A text section, that gives the text that is shown to the student, and An answer, hint and solution section, that specifies how the answer(s) to the problem is(are) marked for correctness, gives hints after a given number of tries and gives a solution that may be shown to the student after the problem set is complete.
The sample file attached to this page shows this; below the file is shown to the left, with a second column on its right that explains the different parts of the problem that are indicated above. Immediately is a screenshot of the WeBWorK problem with the swf file showing the step by step hints.
<flash>file=trigSubWWpic.swf|height=400px|width=550px</flash> Other applet sample problems: GraphLimit Flash Applet Sample Problem GraphLimit Flash Applet Sample Problem 2 Derivative Graph Matching Flash Applet Sample Problem
PG problem file Explanation ##DESCRIPTION ##KEYWORDS('integrals', 'trigonometric','substitution') ## DBsubject('Calculus') ## DBchapter('Techniques of Integration') ## DBsection('Trigonometric Substitution') ## Date('8/20/11') ## Author('Barbara Margolius') ## Institution('Cleveland State University') ## TitleText1('') ## EditionText1('2010') ## AuthorText1('') ## Section1('') ## Problem1('20') ##ENDDESCRIPTION ######################################## # This work is supported in part by the # National Science Foundation # under the grant DUE-0941388. ########################################
This is the
The description is provided to give a quick summary of the problem so that someone reading it later knows what it does without having to read through all of the problem code.
All of the tagging information exists to allow the problem to be easily indexed. Because this is a sample problem there isn't a textbook per se, and we've used some default tagging values. There is an on-line list of current chapter and section names and a similar list of keywords. The list of keywords should be comma separated and quoted (e.g., KEYWORDS('calculus','derivatives')).
DOCUMENT(); loadMacros( "PGstandard.pl", "AppletObjects.pl", "MathObjects.pl", "parserFormulaUpToConstant.pl", );
This is the
The
# Set up problem TEXT(beginproblem()); $showPartialCorrectAnswers = 1; $a = random(2,9,1); $a2 = $a*$a; $a3 = $a2*$a; $a4 = $a2*$a2; $a4_3 = 3*$a4; $a2_5 = 5*$a2; $funct = FormulaUpToConstant("-sqrt{$a2-x^2}/{x}-asin({x}/{$a})");
This is the
The
################################### # Create link to applet ################################### $appletName = "trigSubWW"; $applet = FlashApplet( codebase => findAppletCodebase("$appletName.swf"), appletName => $appletName, appletId => $appletName, setStateAlias => 'setXML', getStateAlias => 'getXML', setConfigAlias => 'setConfig', maxInitializationAttempts => 10, height => '550', width => '595', bgcolor => '#e8e8e8', debugMode => 0, onInit => 'ggbOnInit', ); ################################### # Configure applet ################################### $applet->configuration(qq {<xml><trigString>sin</trigString></xml>}); $applet->initialState(qq {<xml><trigString>sin</trigString></xml>}); TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); <script> if (navigator.appVersion.indexOf("MSIE") > 0) { document.write("<div width='3in' align='center' style='background:yellow'> You seem to be using Internet Explorer.<br/> It is recommended that another browser be used to view this page.</div>"); } </script> END_TEXT HEADER_TEXT(qq! <script language="javascript"> function ggbOnInit(param) { if (param == "$appletName") { applet_loaded(param,1); // report that applet is ready. ww_applet_list[param].safe_applet_initialize(2); } } </script> ! );
This is the
Those portions of the code that begin the line with
You must include the section that follows
The lines
TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); <script> if (navigator.appVersion.indexOf("MSIE") > 0) { document.write("<div width='3in' align='center' style='background:yellow'> You seem to be using Internet Explorer. <br/>It is recommended that another browser be used to view this page.</div>"); } </script> END_TEXT
The text between the
BEGIN_TEXT Evaluate the indefinite integral. $BR \[ \int\frac{\sqrt{$a2 - x^2}}{x^2}dx \] $BR \{ans_rule( 60) \} END_TEXT ################################## Context()->texStrings;
This is the
################################### # # Answers # ## answer evaluators ANS( $funct->cmp() ); TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); $showHint=5; Context()->normalStrings; TEXT(hint( $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( debug =>0, reinitialize_button => 0, includeAnswerBox=>0, )) )); ################################## Context()->texStrings; SOLUTION(EV3(<<'END_SOLUTION')); $BBOLD Solution: $EBOLD $PAR To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] $BR$BR Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). $BR$BR With the substitution \[x = {$a}\sin\theta\] \[dx = {$a}\cos\theta \; d\theta\] $BR$BR Therefore: \[\int\frac{\sqrt{$a2 - x^2}}{x^2}dx= \int \frac{{$a}\cos\theta\sqrt{$a2 - {$a2}\sin^2\theta}} {{$a2}\sin^2\theta} \; d\theta\] \[=\int \frac{\cos^2\theta}{\sin^2\theta} \; d\theta\] \[=\int \cot^2\theta \; d\theta\] \[=\int \csc^2\theta-1 \; d\theta\] \[=-\cot\theta-\theta+C\] $BR$BR Substituting back in terms of \(x\) yields: \[-\cot\theta-\theta+C =-\frac{\sqrt{$a2-x^2}}{x}-\sin^{-1}\left(\frac{x}{$a}\right)+C \] so \[ \int\frac{\sqrt{$a2 - x^2}}{x^2}dx =-\frac{\sqrt{$a2-x^2}}{x}-\sin^{-1}\left(\frac{x}{$a}\right)+C\] END_SOLUTION Context()->normalStrings; ################################## ENDDOCUMENT();
This is the
The
License
The Flash applets developed under DUE-0941388 are protected under the following license: Creative Commons Attribution-NonCommercial 3.0 Unported License. |
I'm reading "The Indian Buffet Process: An Introduction and Review" by Griffiths and Ghahramani and wanted to confirm my understanding of one of the terms they use. On page 1188, they say that the "Discrete($\theta$) is the multiple-outcome analogue of a Bernoulli event. . ." It sounds like this is a Categorical distribution, which is a "generalized Bernoulli . . . or less precisely, a 'discrete distribution. . .'" The Wiki shows two parameters for Cat($K$,
p). Additionally, the key words on the Wiki are "less precisely," and I'm not sure how to relate the single-parameter of the Discrete() to the two-parameters of the Categorical(). A layman's explanation would be helpful.
I'm reading "The Indian Buffet Process: An Introduction and Review" by Griffiths and Ghahramani and wanted to confirm my understanding of one of the terms they use. On page 1188, they say that the "Discrete($\theta$) is the multiple-outcome analogue of a Bernoulli event. . ." It sounds like this is a Categorical distribution, which is a "generalized Bernoulli . . . or less precisely, a 'discrete distribution. . .'" The Wiki shows two parameters for Cat($K$,
This is a multinomial distribution with parameter $\mathbf{\theta}$. You are right that this is a categorical distribution per the Wiki page.
In page 1187, the paper mentions that $\mathbf{\theta}$ is a multinomial distribution over those $K$ classes. This means that the dimension of $\mathbf{\theta}$ is $1 \times K$. You can think of this as a line of $K$ seats, each of the seats have the probability to be 1 or 0. For example, the probability of seat 1 being 1 is $\theta_1$, the probability of seat 2 being 1 is $\theta_2$,...,the probability of seat $K$ being 1 is $\theta_K$.
When you understand how the distribution of $\mathbf{\theta}$ works. This section of the paper is discussing the finite mixture model. The "mixture model" means that the distribution is a combination of multiple components. In this paper, there are possible $K$ components. Each of the components have probability of $\theta_k$ to be included into this mixture model, $1 \leq k \leq K$. Every component is a distribution, such as a normal distribution with different mean and standard deviation. Normal distribution is the most frequent used distribution per my experiences. You can use other type of distribution if needed. Like discusses above,
$c_i|\theta \sim Discrete(\theta)$
is a method of choosing which component can be included. $\theta$ can be adjusted to increase or decrease the probability.
$\theta|\alpha \sim Dirichlet(\frac{\alpha}{K}, \frac{\alpha}{K}, \ldots, \frac{\alpha}{K})$
is the prior of $\theta$. The Dirichlet distribution is conjugate to the multinomial distribution. There are many resources online discussing the pattern of Dirichlet distribution corresponding to difference $\theta$. For example, page 4 on http://mayagupta.org/publications/FrigyikKapilaGuptaIntroToDirichlet.pdf This is a three dimensional Dirichlet distribution, that is $K =3$.
For example, the set up can be:
\begin{align*} x_i|c_i,\Theta &\sim N \bigg( \begin{bmatrix} \mu_1 \\ \mu_2 \\ \mu_3 \end{bmatrix}_{c_i}, \begin{bmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \end{bmatrix}_{c_i}^{-1} \bigg) \\ \begin{bmatrix} \mu_1 \\ \mu_2 \\ \mu_3 \end{bmatrix}_{c_i=1} &\sim N(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}) \\ \begin{bmatrix} \mu_1 \\ \mu_2 \\ \mu_3 \end{bmatrix}_{c_i=1} &\sim N(\begin{bmatrix} 20 \\ 20 \\ 20 \end{bmatrix}, \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}) \\ \begin{bmatrix} \mu_1 \\ \mu_2 \\ \mu_3 \end{bmatrix}_{c_i=1} &\sim N(\begin{bmatrix} 50 \\ 50 \\ 50 \end{bmatrix}, \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}) \\ \begin{bmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \end{bmatrix}_{c_i=1} &\sim Wish_{p=2}(W_0=\begin{bmatrix} 10 & 1 & 1 \\ 1 & 10 & 1 \\ 1 & 1 & 10 \end{bmatrix}) \\ \begin{bmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \end{bmatrix}_{c_i=2} &\sim Wish_{p=2}(W_0=\begin{bmatrix} 10 & 1 & 1 \\ 1 & 10 & 1 \\ 1 & 1 & 10 \end{bmatrix}) \\ \begin{bmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \end{bmatrix}_{c_i=3} &\sim Wish_{p=2}(W_0=\begin{bmatrix} 10 & 1 & 1 \\ 1 & 10 & 1 \\ 1 & 1 & 10 \end{bmatrix}) \\ \boldsymbol{\pi}|\alpha &\sim Dir(\alpha_1, \alpha_2, \alpha_3) \\ c_i|\boldsymbol{\pi} &\sim Multi(\boldsymbol{\pi}) \end{align*}
Here is R code for this mixture of three normal distribution with Dirichlet prior:
#library(rgl)threeDim = function(concen1, concen2, concen3) { stor = NULL S=1000 for (i in 1:S){ pi = rdirichlet(n=1, alpha=c(concen1, concen2, concen3)) tag = rmultinom(1, 1, prob=pi) # parameter for x1 sigma1 = matrix(c(1,1,1,1,1,1,1,1,1), ncol=3) wishSigma1 = matrix(c(10, 1, 1, 1, 10, 1, 1, 1, 10), ncol=3) mu1 = rmvnorm(n=1, mean=c(1, 1, 1), sigma=sigma1) phi1 = matrix(rWishart(1, 3, wishSigma1), ncol = 3) # parameter for x2 sigma2 = matrix(c(1,1,1,1,1,1,1,1,1), ncol=3) wishSigma2 = matrix(c(10, 1, 1, 1, 10, 1, 1, 1, 10), ncol=3) mu2 = rmvnorm(n=1, mean=c(20, 20, 20), sigma=sigma2) phi2 = matrix(rWishart(1, 3, wishSigma2), ncol = 3) # parameter for x3 sigma3 = matrix(c(1,1,1,1,1,1,1,1,1), ncol=3) wishSigma3 = matrix(c(10, 1, 1, 1, 10, 1, 1, 1, 10), ncol=3) mu3 = rmvnorm(n=1, mean=c(50, 50, 50), sigma=sigma3) phi3 = matrix(rWishart(1, 3, wishSigma3), ncol = 3) x1 = rmvnorm(n=1, mean=mu1, sigma=solve(phi1)) # sigma is covariance matrix x2 = rmvnorm(n=1, mean=mu2, sigma=solve(phi2)) # sigma is covariance matrix x3 = rmvnorm(n=1, mean=mu3, sigma=solve(phi3)) # sigma is covariance matrix x = tag[1]*x1 + tag[2]*x2 + tag[3]*x3 stor = rbind(stor, x) } return(stor)}
Here are some of the plots regarding changing of the Dirichlet Prior. You will see how this change of prior can affect the Discrete($\theta$) and the component patterns. |
Cross Product
The cross product is another form of vector multiplication. Unlike the dot product, the cross product results in a vector instead of a scalar. Furthermore, the cross product is defined only in $\mathbb{R}^3$.
Definition: Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, the Cross Product denoted $\vec{u} \times \vec{v}$ results in a new vector that is perpendicular to both $\vec{u}$ and $\vec{v}$. The formula to calculate the cross product is $\vec{u} \times \vec{v} = (u_{2}v_{3} - u_{3}v_{2}, -u_{1}v_{3} + u_{3}v_{1}, u_{1}v_{2} - u_{2}v_{1})$.
For example, consider the vectors $\vec{u} = (1, 2, 3)$ and $\vec{v} = (2, 3, 4)$. To calculate the cross product $\vec{u} \times \vec{v}$, all we need to do is apply the formula to get $\vec{u} \times \vec{v} = (-1, 2, -1)$.
Of course, this formula is rather difficult to remember, so we will use somewhat of a cheat as a memory device. First set up a $2 \times 3$ matrix $A$ where the first row represents components from $\vec{u}$ and the second row represents components from $\vec{v}$, that is:(1)
The first component of the cross product will be the determinant of the $2 \times 2$ matrix that results from deleting the first column of $A$, that is $\begin{vmatrix} u_2 & u_3\\ v_2 & v_3\end{vmatrix} = u_2v_3 - u_3v_2$. The second component of the cross product will be the negativeof the determinant of the $2 \times 2$ matrix that results from deleting the second column of $A$, that is $-\begin{vmatrix} u_1 & u_3\\ v_1 & v_3\end{vmatrix} = -u_1v_3 + u_3v_1$. The third component of the cross product will be the determinant of the $2 \times 2$ matrix that results from deleting the third column of $A$, that is $\begin{vmatrix} u_1 & u_2\\ v_1 & v_2\end{vmatrix} = u_1v_2 - u_2v_1$.
Thus we derive the cross product of two vectors to be $\vec{u} \times \vec{v} = \left ( \begin{vmatrix} u_2 & u_3\\ v_2 & v_3\end{vmatrix}, -\begin{vmatrix} u_1 & u_3\\ v_1 & v_3\end{vmatrix} ,\begin{vmatrix} u_1 & u_2\\ v_1 & = v_2\end{vmatrix} \right )= (u_{2}v_{3} - u_{3}v_{2}, -u_{1}v_{3} + u_{3}v_{1}, u_{1}v_{2} - u_{2}v_{1})$.
Example 1 Find the cross product of $\vec{u} = (2, 3)$ and $\vec{v} = (3, 4)$.
Since the cross product is only defined in 3-space, there is no solution since $\vec{u}, \vec{v} \in \mathbb{R}^2$.
Example 2 Find the cross product given the vectors $\vec{u} = (2, 3, 1)$ and $\vec{v} = (3, -2, 1)$.
All we need to do is apply out formula as follows:(2)
Properties of the Cross Product
Theorem 1: Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, the dot products $\vec{u} \cdot (\vec{u} \times \vec{v}) = 0$ and $\vec{v} \cdot (\vec{u} \times \vec{v}) = 0$. Proof:Let $\vec{u}, \vec{v} \in \mathbb{R}^3$ and expand the formulas for the dot and cross product to get that:
Note that the same proof can be applied for showing that $\vec{v} \cdot (\vec{u} \times \vec{v}) = 0$.
Theorem 2: For three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, the cross product $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$. Proof:Consider the following $2 \times 3$ matrix: $\begin{bmatrix}u_{1} & u_{2} & u_{3} \\ u_{2}v_{3} - u_{3}v_{2}& -v_{1}w_{3} + v_{3}w_{1} & v_{1}w_{2} - v_{2}w_{1} \end{bmatrix}$. If we take the cross product with this matrix we obtain that:
Note that we did not expand this out in its fully entirety, however, feel free to verify.
One important aspect of the cross product that we should touch upon is that the associative property does NOT hold. If we have three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, often times $\vec{u} \times (\vec{v} \times \vec{w}) ≠ (\vec{u} \times \vec{v}) \times \vec{w}$.
We can easily see this with unit vectors. We note that:(5)
Furthermore, if we rearrange the vectors:(6)
Therefore there is clearly no associativity.
Similarly, the commutative property for the cross product of vectors does not always hold, that is $\vec{u} \times \vec{v} \neq \vec{v} \times \vec{u}$. This can easily be proven with the determinant method.
Theorem 3 (Anticommutativity of the Cross Product): Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, the following cross products are equal $\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$.
We will not prove Theorem 3, but it can be done using determinants.
Theorem 4 (Distributivity of the Cross Product): Given three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, the distributive property holds such that $\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}$ and $(\vec{u} + \vec{v}) \times \vec{w} = \vec{u} \times \vec{w} + \vec{v} \times \vec{w}$.
Once again we will omit the proof of this theorem.
Theorem 5: Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, and some scalar $k$, it follows that $k(\vec{u} \times \vec{v}) = (k\vec{u}) \times \vec{v} = \vec{u} \times (k\vec{v})$.
Like the last two theorems, we will omit the proof of this theorem as it is rather straight-forward using determinants.
Theorem 6: Given the two vectors $\vec{u}, \vec{0} \in \mathbb{R}^3$, it follows that $\vec{u} \times \vec{0} = \vec{0} \times \vec{u} = \vec{0}$. Proof:The cross product creates a vector that is perpendicular to both the vectors cross product multiplied together. However, the zero vector has no length or direction. Hence, there is no vector that is perpendicular to both some vector $\vec{u}$ and the zero vector $\vec{0}$. For convention, we say the result is the zero vector, as it can be assigned any direction because it has no magnitude. $\blacksquare$
Like the other proofs, theorem 6 can also be shown using determinants.
Theorem 7: Given a vector $\vec{u} \in \mathbb{R}^3$, it follows that $\vec{u} \times \vec{u} = \vec{0}$. Proof:This proof is also intuitive. The zero vector can be assigned any direction despite having no magnitude. Hence, it is the only vector that is perpendicular to itself from the cross product. $\blacksquare$ |
Defining Curvilinear Coordinates for Anisotropic Materials
A lot of materials have anisotropic properties and, in many cases, the anisotropy follows the shape of the material. The COMSOL Multiphysics® software offers different methods for defining curvilinear coordinate systems. Here, we discuss the concepts of each and when to use which method.
Anisotropic Properties
Anisotropic properties are found in a wide variety of areas, such as rock formations with anisotropic seismic properties; liquid crystals used in LCD displays; materials for the aerospace industry, which must be lightweight and still withstand high loads; or soft tissues that biomedical replacements should mimic for optimal performance.
The Basics of Curvilinear Systems
In a previous blog post, we saw how to use the Curvilinear Coordinates interface and how to apply it to account for anisotropic thermal conductivity.
Let’s revisit this application and consider a carbon-fiber-reinforced polymer. The woven fibers embedded in an epoxy matrix have high thermal conductivity along the fiber axis and low conductivity in the cross section. It is almost impossible to express the anisotropy referring to the well-known Cartesian coordinate system. If we had a coordinate system that follows the fibers, it would be straightforward to specify the anisotropic properties.
Woven fibers in an epoxy matrix.
How can such a coordinate system be determined? Physically, there are numerous effects that result in a vector field following the shape of the geometry; for instance, flow through the fibers or heat conduction from one end to the other or even a bundle of current-carrying wires that produce a magnetic field. These are precisely the methods that are used in the COMSOL® software to compute the curvilinear system. All methods compute a vector field, \mathbf{v}, which forms the
first basis vector. Since most applications require a normalized vector field, COMSOL Multiphysics automatically normalizes by dividing with |\mathbf{v}|. A second vector field is specified manually, and one of the Cartesian coordinates is often a good choice. Starting from this, the second basis vector \mathbf{e}_2 is reconstructed to ensure that it is perpendicular to \mathbf{e}_1 and normalized. The cross product of these two gives the third base vector, \mathbf{e}_3.
Internally, the software uses the Cartesian coordinate system (\mathbf{e}_x,\ \mathbf{e}_y,\ \mathbf{e}_z) for computation and converts all quantities referring to a different coordinate system (\mathbf{e}_1,\ \mathbf{e}_2,\ \mathbf{e}_3). A direction given by a vector, \mathbf{F}=(F_1,\ F_2,\ F_3), in an arbitrary coordinate system can always be transformed into Cartesian coordinates, as follows:
e_{11} & e_{12} & e_{13} \\
e_{21} & e_{22} & e_{23} \\
e_{31} & e_{32} & e_{33}
\end{matrix}\right)\cdot\left(\begin{matrix} F_1\\F_2\\F_3\end{matrix}\right)=\mathbf{M}\left(\begin{matrix} F_1\\F_2\\F_3\end{matrix}\right)
Here, \mathbf{M} is the transformation matrix. For the inverse transformation, simply use the inverse, \mathbf{M}^{-1}, and if (\mathbf{e}_1,\ \mathbf{e}_2,\ \mathbf{e}_3) is orthonormal then \mathbf{M}^{-1}=\mathbf{M}^{T}.
Next, I will illustrate the different methods available in COMSOL Multiphysics that can be used to calculate a curvilinear coordinate system, which include the:
Diffusion method Adaptive method Flow method Elasticity method
Let’s pick a single fiber and have a closer look.
Diffusion Method
The diffusion method solves Laplace’s equation: -\Delta U=0. The solution, U, is a scalar potential, and its gradient forms the first base vector. Because you solve for a single scalar potential only, this method is computationally inexpensive. The direction of the vector field is specified with the inlet and outlet boundary conditions. If the geometry is a closed loop, you can set the
Jump boundary condition on an interior boundary to specify the direction.
The diffusion method is equivalent to solving the stationary heat conduction equation with constant temperatures at the inlet and outlet boundaries. The temperature gradient then forms the first base vector, as illustrated below.
Curvilinear coordinate system (arrows), temperature gradient (streamlines), and temperature (surface). Adaptive Method
The adaptive method, which is similar to the diffusion method, is based on Laplace’s equation as well. In addition, the resulting vector field is adapted to the shape of the geometry such that the streamline density is constant across the cross section of the geometry. This formulation is used to model multiturn coils (bundles of wires) in 3D magnetic applications with the AC/DC Module, an add-on to COMSOL Multiphysics. For multiturn coils, the current density should be roughly constant in the cross section, since it is assumed that each wire carries the same current and the wires are evenly spaced.
Flow Method
Here, you solve the incompressible Stokes equation for a vector field and a scalar. Thus, this method is the most computationally expensive. The boundary conditions are the same as for the diffusion method. A physical analogy would be incompressible creeping flow with a constant normal velocity at the inlet and a fixed pressure at the outlet. The resulting velocity field gives you the first base vector.
Curvilinear coordinate system (arrows), velocity field (streamlines), and pressure (surface). Elasticity Method
The elasticity method solves the following eigenvalue equation:
where \mathbf{e} is the vector field, \mathbf{I} the identity matrix, and \lambdathe eigenvalue.
This method is slightly cheaper in terms of computational costs compared to the flow method because you solve for a vector field only. This difference in performance is more apparent in 2D models. The inlet and outlet boundary conditions are identical, \mathbf{e}\times \mathbf{n}=0. Prior to the implementation of the adaptive method, this method was used for modeling multiturn coils because it provided the best results in terms of constant streamline density in cross section.
Curvilinear coordinate system (arrows), coil direction (gray streamlines) and magnetic flux density (red streamlines).
Apart from these predefined methods, the COMSOL® software also provides a user-defined input, as usual. You may encounter other scenarios where you want to implement curvilinear coordinates manually, such as anisotropic hyperelastic material for modeling collagenous soft tissue in arterial walls.
Which Method Should I Choose?
At first glance, all of the methods lead to the same coordinate system. However, some shapes require special attention, and the choice of the method can lead to significantly different results when the coordinate system is used for a physical application. Pay attention to geometries that have at least one of the following characteristics.
Curvature
Let’s take a closer look at the streamlines for the various methods. Remember, the streamlines follow the vector field, which defines the first base vector. They start from equidistant points but follow different paths, as can be seen below.
Diffusion method: Streamlines follow the “shortest” path. Adaptive method: Evenly distributed streamlines. Elasticity method: Streamlines tend to accumulate at the convex bend. Flow method: Streamlines tend to accumulate at the convex bend.
If bends are very sharp, the characteristics of each method are more distinctive, and also the adaptive method can develop nonuniform streamline density.
Variable Cross Section
In this case, the elasticity method can fail, and you obtain an eigenvalue with eigenvectors that do not produce the required coordinate system. Alternatively, you may have to manually search for the correct eigenvalue. We can see that the streamlines also do not follow the shape perfectly at the upper part of the geometry. Similar, but not as distinctive, behavior is observed with the diffusion and adaptive methods. The flow method provides the best results here but is also the most computationally expensive.
Diffusion method. Adaptive method. Elasticity method. Flow method. Streamlines along the center plane of the geometry. Application to Heat Transfer
Let’s go back to our model with the fibers that have an anisotropic thermal conductivity of 60\ W/mK in the fiber direction and 4\ W/mK perpendicular to this. If these directions coincide with the axes of the coordinate system, the thermal conductivity — a second-order tensor — has 0 off-diagonal elements.
k_{xx} & k_{xy} & k_{xz} \\
k_{yx} & k_{yy} & k_{yz} \\
k_{zx} & k_{zy} & k_{zz}
\end{matrix}\right)=\left(\begin{matrix}
60 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{matrix}\right)
To be able to use this diagonal form, the curvilinear coordinate system for the fibers must be calculated prior to solving for the heat transfer. Because the geometry doesn’t have sharp bends or a changing cross section, the diffusion method gives a fast solution for the curvilinear coordinate system.
After that, you can refer to this coordinate system in the
Heat Transfer in Fibers node. The anisotropy of the thermal conductivity can be defined in the Materials node, with the syntax k=\{60, 4, 4\}. Alternatively, you can select the user-defined input in the associated heat transfer node. Definition of anisotropy in the associated heat transfer node.
In the model, a boundary heat source in the form of a Gaussian pulse is applied to the center of the geometry and the temperature spreads along the fibers.
Streamlines indicate the vector field obtained with the Curvilinear Coordinates interface.
If you want to visualize, for example, the
xx-component of the thermal conductivity (k_{xx}), keep in mind that you plot the xx-component in Cartesian coordinates (\mathbf{e}_x,\ \mathbf{e}_y,\ \mathbf{e}_z). Then, the thermal conductivity tensor, k, for the fibers is of nondiagonal form, according to the transformation described above. The local base vector system (\mathbf{e}_1,\ \mathbf{e}_2,\ \mathbf{e}_3), which was used to define k, is now varying through space and so does k_{xx}. In this model, you can plot the components of the thermal conductivity vector in a Slice plot, for example. You can select them from the Expressions menu in the corresponding Settings window or simply type
ht.kxx (where
ht is the tag for the
Heat Transfer in Solids interface that was used for this model). Concluding Remarks
This blog post has described the different methods for defining the curvilinear coordinate systems that are available in COMSOL Multiphysics as well as when to choose one over another.
In summary, the adaptive method provides the best solution for most cases at comparatively low computational costs. The diffusion method has even lower computational costs and is suitable for simple geometries without bends or varying cross sections. The other methods can be advantageous in some cases and are interesting for certain applications.
Diffusion method Pros: low computational cost Cons: computed vector field tends to take the shortest path in bends Adaptive method Pros: low computational cost, provides the best solution for most cases Cons: variable cross sections not always handled perfectly Elasticity method Pros: computational cost lower than the flow method, better representation of moderate bends than diffusion method Cons: often requires manual selection of the eigenvalue and is not robust in all cases Flow method Pros: robust method, supports cross-section changes and sharp bends Cons: computational cost is often greater
To try the carbon fiber model presented here, click the button below. Note that to download the MPH-files for this example, you need a COMSOL Access account and valid software license.
Editor’s note: This blog post was updated on December 5, 2018, to include information on the adaptive method. Comments (11) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
The Baire Category Theorem
Math Online's 1500th Page - 7:56 PM CST on September 28th, 2015 The Baire Category Theorem
Lemma 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. If $A$ is a nowhere dense set then for every $U \in \tau$ there exists a $B \subseteq U$ such that $A \cap \bar{B} = \emptyset$.
Theorem 1 (The Baire Category Theorem): Every complete metric space is of the second category. Proof:Let $X$ be a complete metric space. Then every Cauchy sequence $(x_n)_{n=1}^{\infty}$ of elements from $X$ converges in $X$. Suppose that $X$ is of the first category. Then there exists a countable collection of nowhere dense sets $A_1, A_2, ... \subset X$ such that:
\begin{align} \quad X = \bigcup_{i=1}^{\infty} A_i \end{align}
Let $U \subset X$. For each nowhere dense set $A_i$, $i \in \{ 1, 2, ... \}$ there exists a set $B_i \subset U$ such that $A_i \cap \bar{B_i} = \emptyset$. Let $B_1(x_1, r) \subset U$ be a ball contained in $U$ such that $A_1 \cap \bar{B_1} = \emptyset$. Let $B_2 \left ( x_2, \frac{r}{2} \right ) \subseteq B_1$ be a ball contained in $B_1$ whose radius is $\frac{r}{2}$ and such that $A_2 \cap \bar{B_2} = \emptyset$. Repeat this process. For each $n \in \{ 2, 3, ... \}$ let $B_n \left (x_n, \frac{r}{n} \right )$ be a ball contained in $B_{n-1}$ whose radius is $\frac{r}{n}$ and such that $A_n \cap \bar{B_n} = \emptyset$ and such that $A_n \cap \bar{B_n} = \emptyset$. The sequence $(x_n)_{n=1}^{\infty}$ is Cauchy since as $n \in \mathbb{N}$ gets large, the elements $x_n$ are very close. Since $X$ is a complete metric space, we must have that this Cauchy sequence therefore converges to some $p \in X$, i.e., $\lim_{n \to \infty} x_n = p$. Now notice that $x \in \bar{B_n}$ for all $n \in \mathbb{N}$ because if not, then there exists an $m \in \mathbb{N}$ such that $x \not \in \bar{B_n}$ for all $n \geq m$. Hence $(\bar{B_n})^c$ is open and so there exists an open ball $X$ such that $x \in B \subset (\bar{B_n})^c$ but then $\lim_{x \to \infty} x_n \neq x$ because $x_m \not \in B$ for all $m \geq n$. Since $x \in \bar{B_n}$ for all $n \in \mathbb{N}$ then since $A_n \cap \bar{B_n} = \emptyset$ we must have that then [[$ x \in |
What is the meaning of $V(D,G)$? How do we get these expectation parts?I was trying to understand it following this article: Understanding Generative Adversarial Networks
(D.Seita), but, after many tries, I still can't understand how he got from $\sum_{n=1}^{N} \log D(x)$ to $\mathbb{E}(\log(D(x))$.
What is the meaning of $V(D,G)$? How do we get these expectation parts?I was trying to understand it following this article: Understanding Generative Adversarial Networks
To understand this equation first you need to understand the context in which it is first introduced. We have two neural networks (i.e. $D$ and $G$) that are playing a minimax game. This means that they have competing goals. Let's look at each one separately:
Generator
Before we start, you should note that throughout the whole paper the notion of the data-generating distribution is used; in short the authors will refer to the samples through their underlying distributions, i.e. if a sample $a$ is drawn from a distribution $p_a$, we'll denote this as $a \sim p_a$. Another way to look at this is that $a$
follows distribution $p_a$.
The generator ($G$) is a neural network that produces samples from a distribution $p_g$. It is trained so that it can bring $p_g$ as close to $p_{data}$ as possible so that samples from $p_g$ become indistinguishable to samples from $p_{data}$. The catch is that it never gets to actually see $p_{data}$. Its inputs are samples $z$ from a noise distribution $p_z$.
Discriminator
The discriminator ($D$) is a simple binary classifier that tries to identify which class a sample $x$ belongs to. There are two possible classes, which we'll refer to as the
fake and the real. Their respective distributions are $p_{data}$ for the real samples and $p_g$ for the fake ones (note that $p_g$ is actually the distribution of the outputs of the generator, but we'll get back to this later).
Since it is a simple binary classification task, the discriminator is trained on a binary cross-entropy error:
$$ J^{(D)} = H(y, \hat y) = H(y, D(x)) $$
where $H$ is the cross-entropy $x$ is sampled either from $p_{data}$ or from $p_g$ with a probability of $50\%$. More formally:
$$ x \sim \begin{cases} p_{data} \rightarrow & y = 1, & \text{with prob 0.5}\\ p_g \;\;\;\,\rightarrow & y = 0, & \text{otherwise} \end{cases} $$
We consider $y$ to be $1$ if $x$ is sampled from the
real distribution and $0$ if it is sampled from the fake one. Finally, $D(x)$ represents the probability with which $D$ thinks that $x$ belongs to $p_{data}$. By writing the cross-entropy formula we get:
$$ H(y, D(x)) = \mathbb{E}_y[-log \; D(x)] = \frac{1}{N} \sum_{i=1}^{N}{ \; y_i \; log(D(x_i))} $$
where $N$ is the size of the dataset. Since each class has $N/2$ samples we can split this sum into two parts: $$ = - \left[ \frac{1}{N} \sum_{i=1}^{N/2}{ \; y_i \; log(D(x_i))} + \frac{1}{N} \sum_{i=N/2}^{N} \; (1 - y_i) \; log((1 - D(x_i))) \right] $$
The first of the two terms represents the the samples from the $p_{data}$ distribution, while the second one the samples from the $p_g$ distribution. Since all $y_i$ are equally likely to occur, we can convert the sums into expectations:
$$ = - \left[ \frac{1}{2} \; \mathbb{E}_{x \sim p_{data}}[log \; D(x)] + \frac{1}{2} \; \mathbb{E}_{x \sim p_{g}}[log \; (1 - D(x))] \right] $$
At this point, we'll ignore $2$ from the equations since it's constant and thus irrelevant when optimizing this equation. Now, remember that samples that were drawn from $p_g$ were actually outputs from the generator (obviously this affects only the second term). If we substitute $D(x), x \sim p_g$ with $D(G(z)), z \sim p_z$ we'll get:
$$ L_D = - \left[\; \mathbb{E}_{x \sim p_{data}}[log \; D(x)] + \; \mathbb{E}_{z \sim p_{z}}[log \; (1 - D(G(z)))] \right] $$
This is the final form of the discriminator loss.
Zero-sum game setting
The discriminator's goal, through training, is to minimize its loss $L_D$. Equivalently, we can think of it as trying to maximize the opposite of the loss:
$$ \max_D{[-J^{(D)}]} = \max_D \left[\; \mathbb{E}_{x \sim p_{data}}[log \; D(x)] + \; \mathbb{E}_{z \sim p_{z}}[log \; (1 - D(G(z)))] \right] $$
The generator however, wants to maximize the discriminator's uncertainty (i.e. $J^{(D)}$), or equivalently minimize $-J^{(D)}$.
$$ J^{(G)} = - J^{(D)} $$
Because the two are tied, we can summarize the whole game through a
value function $V(D, G) = -J^{(D)}$. At this point I like to think of it like we are seeing the whole game through the eyes of the generator. Knowing that $D$ tries to maximize the aforementioned quantity, the goal of $G$ is:
$$ \min_G\max_D{V(D, G)} = \min_G\max_D \left[\; \mathbb{E}_{x \sim p_{data}}[log \; D(x)] + \; \mathbb{E}_{z \sim p_{z}}[log \; (1 - D(G(z)))] \right] $$
Disclaimer:
This whole endeavor (on both my part and the authors' part) was to provide a mathematical formulation to training GANs. In practice there are many tricks that are invoked to effectively train a GAN, that are not depicted in the above equations. |
This is a continuation of Quantum algorithm for linear systems of equations (HHL09): Step 2 - What is $|\Psi_0\rangle$?
In the paper:
Quantum algorithm for linear systems of equations (Harrow, Hassidim & Lloyd, 2009), the details of the actual implementation of the algorithm is not given. How exactly the states $|\Psi_0\rangle$ and $|b\rangle$ are created, is sort of a " black-box" (see pages 2-3).
$$|\Psi_0\rangle = \sqrt{\frac{2}{T}}\sum_{\tau = 0}^{T-1}\sin \frac{\pi (\tau+\frac{1}{2})}{T}|\tau\rangle$$
and $$|b\rangle = \sum_{1}^{N}b_i|i\rangle$$
where $|\Psi_0\rangle$ is the initial state of the clock register and $|b\rangle$ is the initial state of the Input register.
(Say)
I want to carry out their algorithm on the IBM $16$-qubit quantum computer. And I want to solve a certain equation $\mathbf{Ax=b}$ where $\mathbf{A}$ is a $4\times 4$ Hermitian matrix with real entries and $\mathbf{b}$ is a $4\times 1$ column vector with real entries.
Let's take an example:
$$\mathbf{A} = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 5 & 6 \\ 3 & 5 & 1 & 7 \\ 4 & 6 & 7 & 1 \end{bmatrix}$$
and
$$\mathbf{b}=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$$
Given the dimensions of $\mathbf{A}$ and $\mathbf{b}$, we should need $\lceil{\log_2 4\rceil}=2$ qubits for the input register and another $6$ qubits for the clock register assuming we want the eigenvalues to be represented with $90\%$ accuracy and up to $3$-bit precision for the eigenvalues (this has been discussed here previously). So total $2+6+1=9$ qubits will be needed for this purpose (the extra $1$ qubit is the ancilla).
Questions:
Using this information, is it
possibleto create the initial states $|\Psi_0\rangle$ and $|b\rangle$ on the IBM $16$ qubit version?
If you think $4\times 4$ is too large to be implemented on the IBM quantum computers you could even show an
exampleof initial state preparation for a $2\times 2$ Hermitian matrix $\mathbf{A}$ (or just give a reference to such an example).
I simply want to get a
general idea about whether this can be done (i.e. whether it is possible) on the IBM 16-qubit quantum computer, and for that which gates will be necessary. If not the IBM 16-qubit quantum computer, can the QISKit simulator used for recreating the initial state preparation of $|\Psi_0\rangle$ and $|b\rangle$ in the HHL algorithm? Is there any other better alternative to go about this? |
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J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV
(Elsevier, 2013-04-10)
The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (|y| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Multiplicity dependence of the average transverse momentum in pp, p-Pb, and Pb-Pb collisions at the LHC
(Elsevier, 2013-12)
The average transverse momentum <$p_T$> versus the charged-particle multiplicity $N_{ch}$ was measured in p-Pb collisions at a collision energy per nucleon-nucleon pair $\sqrt{s_{NN}}$ = 5.02 TeV and in pp collisions at ...
Production of $K*(892)^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$ =7 TeV
(Springer, 2012-10)
The production of K*(892)$^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$=7 TeV was measured by the ALICE experiment at the LHC. The yields and the transverse momentum spectra $d^2 N/dydp_T$ at midrapidity |y|<0.5 in ...
Directed flow of charged particles at mid-rapidity relative to the spectator plane in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(American Physical Society, 2013-12)
The directed flow of charged particles at midrapidity is measured in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV relative to the collision plane defined by the spectator nucleons. Both, the rapidity odd ($v_1^{odd}$) and ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
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 ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ... |
Table of Contents
The Closed Set Definition of Continuous Maps on Topological Spaces
Recall from the Continuous Maps on Topological Spaces page that we say $f : X \to Y$ is continuous at $a \in X$ if there exists local bases $\mathcal B_a$ of $a$ and $\mathcal B_{f(a)}$ of $f(a)$ such that for every $B \in \mathcal B_{f(a)}$ there exists a $B' \in \mathcal B_{a}$ such that:(1)
On the The Open Neighbourhood Definition of Continuous Maps on Topological Spaces page we saw that $f : X \to Y$ is continuous at $a \in X$ if for every open neighbourhood $V$ of $f(a)$ there exists an open neighbourhood $U$ of $a$ such that:(2)
On the Equivalent Statements Regarding Continuous Maps on Topological Spaces page that we saw that that the following statements are equivalent:
(1)$f : X \to Y$ is continuous on all of $X$. (2)For every open set $V$ in $Y$ we have that $f^{-1}(V)$ is open in $X$. (3)For every basis $\mathcal B_Y$ of $Y$ we have that for all $B \in \mathcal B_Y$ that $f^{-1}(B)$ is open in $X$.
We will now look at yet another equivalent definition of a map being continuous on all of $X$ with regards to closed sets. We must first prove the following lemma though.
Lemma 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. Then for all $S \subseteq Y$ we have that $f^{-1}(S^c) = (f^{-1}(S))^c$. Proof:Let $x \in f^{-1}(S^c)$. Then $f(x) \in S^c$ so $f(x) \not \in S$ and $x \not \in f^{-1}(S)$ so $x \in (f^{-1}(S))^c$. Therefore $f^{-1}(S^c) \subseteq (f^{-1}(S))^c$. Now let $x \in (f^{-1}(S))^c$. Then $x \not \in f^{-1}(S)$ so $f(x) \not \in S$ so $f(x) \in S^c$ and $x \in f^{-1}(S^c)$. Therefore $f^{-1}(S^c) \supseteq (f^{-1}(S))^c$. Since $f^{-1}(S^c) \subseteq (f^{-1}(S))^c$ and $f^{-1}(S^c) \supseteq (f^{-1}(S))^c$ we have that then:
Theorem 2: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. Then $f$ is continuous on all of $X$ if and only if for all closed sets $V$ in $Y$ we have that $f^{-1}(V)$ is closed in $X$. Proof:$\Rightarrow$ Suppose that $f$ is continuous on all of $X$. Then for all open sets $V$ in $Y$ we have that $f^{-1} (V)$ is an open set in $X$. Let $V$ now be a closed set in $Y$. Then $V^c$ is an open set in $Y$ and so $f^{-1}(V^c)$ is open in $X$. But by Lemma 1: So $(f^{-1}(V))^c$ is open in $X$ which implies that $f^{-1}(V)$ is closed in $X$. Since $V$ is an arbitrary closed set in $Y$ we see that for every closed set $V$ in $Y$ we have that $f^{-1}(V)$ is closed in $X$. $\Leftarrow$ Suppose that for every closed set $V$ in $Y$ we have that $f^{-1}(V)$ is closed in $X$. Let $V$ instead be an open set in $Y$. Then $V^c$ is closed in $Y$ so $f^{-1}(V^c)$ is closed in $X$ and by Lemma 1: So $(f^{-1}(V))^c$ is closed in $X$ which implies that $f^{-1}(V)$ is open in $X$. Since $V$ is an arbitrary open set in $Y$ we see that for every open set $V$ in $Y$ we have that $f^{-1}(V)$ is open in $X$. Therefore $f$ is continuous on all of $X$. $\blacksquare$ |
Let $G= (V,E)$ be a given directed weighted graph, and $s,t$ two specified nodes, so that there is no negative cycle reachable from $s$, and $t$ is reachable from $s$.
We're looking for the shortest s-t-path.
By regarding this problem as a special flow network, we can express this using linear programming as follows :
Minimize the function: $$ \sum_{e\in E} c_e\cdot x_e $$
Under the constraints: $$ \forall v\in V-\{s,t\}: \sum_{e\in in(v)}x_e -\sum_{e\in out(v)} x_e = 0 \\ \sum_{e\in out(s)} x_e = 1 \\ \sum_{e\in in(t)} x_e = 1 \\ \sum_{e\in in(s)} x_e = 0 \\ \sum_{e\in out(t)} x_e = 0 \\ \forall e\in E: x_e \ge 0 $$
Here $c_e$ are the weights of the edges, $in(v)$ are all edges going into $v$, and $out(v)$ are all edges starting in $v$.
About the correctness:
Let $S$ be a solution to above linear program. As the constraints are the same as for network flows (if we see each edge as having infinite capacity), $S$ is a flow on $G$.
Given that every $s-t$-path fulfills the constraints, we can conclude that if $S$ is an $s-t$-path, then it is an $s-t$-path with minimal costs.
Further, $S$ can't contain any cycle with positive weight, as we could simply remove this cycle from $S$ and end up with a lower-cost solution that fulfills the constraints.
Finally, $S$ has to be an $s-t$-path. Let's assume $S$ wasn't. Given that $S$ is a flow, we know that it has to contain some $s-t$-path. So let $M$ be the set of all $s-t$-paths that $S$ contains.
Then there is a path with minimal costs in $M$ (as $S$ is cycle-free, and therefore there can only be finitely many elements in $M$). Let this $s-t$-path be $p_{min}$.
If we let $c: M\to \mathbb R $ now be the map that assigns each $s-t$-path in $M$ its cost, we obtian the following inequality: $$1\cdot p_{min} \le \sum_{p\in M} \lambda_p \cdot p \qquad\text{ given that } \forall p\in M: \lambda_p \ge 0 , \sum_{p\in M} \lambda_p = 1$$
Therefore, if $S$ would contain multiple $s-t$-paths, it wouldn't be minimal.
Thus we can conclude, that $S$ is a minimal $s-t$-path.
My question now is: Did I miss something in the proof?
Addendum:
It turns out that the above proof needs that each $s-t$-path has a unique cost. Otherwise, there might be no single $s-t$-path with minimal costs. In this case, it can't be shown that the solution $S$ only contains a single of those paths with minimal costs.
However, in this case, by the reasoning above, it still is true that every path in $S$ is optimal. So in this case, we can just pick any path in $S$ as solution (which we can find using e.g. DFS)
Some final (off-topic) remarks:
This whole procedure seems like a lot for a less efficient method to obtain a shortest path. What caught my eye were the following two properties of this algorithm: $(i)$ Besides the non-negativity constraint, all constraints are actually equalities. $(ii)$ The algorithm should be easily adaptable to also allow for negative cycles (by putting some linear restriction on the solution, e.g. that the path mustn't be too long, etc.) |
Model Problem
We consider the numerical approximation of the inverse problem for the linear advection-diffusion equation,
with $d$ being the diffusivity of the material and $v$ the direction of the advection.
Given a final time $T>0$ and a target function $u^\ast$ the aim is to identify the initial condition $u_0$ such that the solution, at time $t=T$, reaches the target $u^*$ or gets as close as possible to it. We assume that the initial condition $u_0$ is characterized as a combined set of sparse sources. This means that $u_0$ is a linear combination of unitary deltas with certain and possibly different weights, i.e:
We formulate the inverse problem using optimal control techniques. In particular, we consider the minimization of the following functional:
Space and time discretization
Letting $\textbf{u} : [0, T] \rightarrow \mathbb{R}^s$ where $s$ is the number of grid points on $\Omega$, we can write a general finite element (FE) discretization of the diffusion–advection equation in \eqref{modeleq} in a compact form as:
In order to get a time discretized version of the previous equation, we apply implicit Euler method with stepsize $\Delta t := T/N$ where $N$ is the total number of time steps. The numerical approximations to the solution are given by the vectors $\textbf{u}^n \approx \textbf{u}(t_n) \in \mathbb{R}^s$ with respect to the index $n=i\Delta t$ for $i=1,2,..,N$. Therefore, the fully discrete version of model equation is as follows,
Adjoint Algorithm for sparse source identification
The algorithm to be presented in this work for the sparse source identification of the linear diffusion-advection equation based on the adjoint methodology consists of two steps. Firstly, we use the adjoint methodology to identify the locations of the sources. Secondly, a least squares fitting is applied to find the corresponding intensities of the sources.
We have considered here a two-dimensional example with several sources to be identified in a multi-model environment. This means that the left half ($\Omega_1 = [0,1] \times [0,1]$) and the right half ($\Omega_2 = [1,2] \times [0,1]$) of the domain are modelled with different equations. In particular, the heat equation is used on $\Omega_1$ and the diffusion–advection equation is used on $\Omega_2$.
The initialization parameters look as follows:
N=30; %% space discretization points in y-directiondx=1/(N+1); %% mesh sizet0=0; %% initial timetf=0.1; %% final timen=5; %% time discretization pointsdt=(tf-t0)/n; %% stepsizeTOL=1e-5; %% stopping toleranced1=0.05; %% diffusivity of the material on the left sudomaind2=0.05; %% diffusivity of the material on the left sudomain%% advection componentsvx=0;vy=-3;tau=dx^4; %% regularization parameterepsilon=0.1; %% stepsize of the gradient descent method
We now compute the FE discretization matrices $M$, $A$ and $V$ that are respectively the mass matrix, the stiffness matrix and the advection matrix. For the FE discretization we assume equidistant structured meshes. In particular we use triangular elements and the classical pyramidal test functions are employed.
[M,A,V] = computeFEmatrices(N,d1,d2,vx,vy); %% Compute FE discretization matrices
A reference initial condition is chosen and we compute using the FE discretization specified above and implicit Euler in time its corresponding final state at time $T$. This final state will be considered the initial data of the inverse problem to be solved and we name it the target function $u^*$ as mentioned previously.
U0_ref = initial_deltas(N); %% computes reference initial condition[U_target,u_target] = compute_target(U0_ref,N,n,dt,M,A,V); %% Compute target distribution
We now call the algorithm that estimates the initial condition $u_0$ using as a initial data the target function $u^*$. As mentioned before, this algorithm consists of two steps.
Firstly, the classical adjoint methodology that minimizes the functional $J(u_0)$ subject to the diffusion-advection equation is used. The iterative optimization algorithm employed is the classical gradient descent method. However, although this iterative procedure finds quite accurately the locations of the sources, it does not recover the sparse character of the initial condition. This is not suprising because the recovered initial data comes from solving the adjoint problem which is basically a diffusive process that smoothes out its state. Consequently, a second procedure is needed to project the obtained non sparse initial condition into the set of admissible sparse solutions.
As the initial condition $u_0$ is assumed to be a linear combination between the locations and the intensities, once we have fixed the locations using the adjoint methodology we can solve a least squares problem to get the remaining intensities. We assemble a matrix $\textbf{L} \in \mathbb{R}^{s \times l}$ where at each column we have the forward solution for a single unitary delta placed at each of the locations already identified. We then solve the following linear system of equations for the vector of unknowns $\alpha = (\alpha_1, \alpha_2, …, \alpha_l)^T$:
to find the intensities vector $\alpha$.
Adjoint algorithm for sparse source identification (algorithm 4)
U0 = SparseIdentification(u_target,TOL,dt,n,N,M,A,V,epsilon,tau);
Finally, the final state at $T$ is computed using as a initial condition the estimated sparse sources identified with our algorithm.
Compute final state with the recovered initial condition
[UF,u_final] = compute_target(U0,N,n,dt,M,A,V);
We can see the evolution recovered
We now visualize the numerical results. Plots on the left side show the reference initial solution and the given target. Similarly, plots on the right side show the recovered initial condition and the distribution at the final time $T$ produced by the recovered initial sources.
One can observe the difference between the two models (the heat equation on $\Omega_1$ and the diffusion-advection on $\Omega_2$) in the two figures at the bottom where the initial sources on $\Omega_2$ move downwards at the same time as they dissipate while the initial sources on $\Omega_1$ only dissipate without displacement.
xplot = linspace(0,2,2*N+3); %% space grid w.r.t component xyplot = linspace(0,1,N+2); %% space grid w.r.t component y%%figure('unit','norm','pos',[0.25 0.1 0.5 0.8])subplot(3,2,1)surf(xplot,yplot,U0_ref)shading interp;colorbar;colormap jettitle('Reference initial state (front view)')subplot(3,2,2)surf(xplot,yplot,U0)shading interp;colorbar;colormap jettitle('Recovered initial state (front view)')subplot(3,2,3)pcolor(xplot,yplot,U0_ref)shading interp;colorbar;colormap jettitle('Reference initial state (above view)')subplot(3,2,4)pcolor(xplot,yplot,U0)shading interp;colorbar;colormap jettitle('Recovered initial state (above view)')subplot(3,2,5)pcolor(xplot,yplot,U_target)shading interp;colorbar;colormap jettitle('Given target u^*')subplot(3,2,6)pcolor(xplot,yplot,UF)shading interp;colorbar;colormap jettitle('Recovered final state') |
This is a continuation of Quantum algorithm for linear systems of equations (HHL09): Step 2 - What is $|\Psi_0\rangle$?
In the paper:
Quantum algorithm for linear systems of equations (Harrow, Hassidim & Lloyd, 2009), the details of the actual implementation of the algorithm is not given. How exactly the states $|\Psi_0\rangle$ and $|b\rangle$ are created, is sort of a " black-box" (see pages 2-3).
$$|\Psi_0\rangle = \sqrt{\frac{2}{T}}\sum_{\tau = 0}^{T-1}\sin \frac{\pi (\tau+\frac{1}{2})}{T}|\tau\rangle$$
and $$|b\rangle = \sum_{1}^{N}b_i|i\rangle$$
where $|\Psi_0\rangle$ is the initial state of the clock register and $|b\rangle$ is the initial state of the Input register.
(Say)
I want to carry out their algorithm on the IBM $16$-qubit quantum computer. And I want to solve a certain equation $\mathbf{Ax=b}$ where $\mathbf{A}$ is a $4\times 4$ Hermitian matrix with real entries and $\mathbf{b}$ is a $4\times 1$ column vector with real entries.
Let's take an example:
$$\mathbf{A} = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 5 & 6 \\ 3 & 5 & 1 & 7 \\ 4 & 6 & 7 & 1 \end{bmatrix}$$
and
$$\mathbf{b}=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$$
Given the dimensions of $\mathbf{A}$ and $\mathbf{b}$, we should need $\lceil{\log_2 4\rceil}=2$ qubits for the input register and another $6$ qubits for the clock register assuming we want the eigenvalues to be represented with $90\%$ accuracy and up to $3$-bit precision for the eigenvalues (this has been discussed here previously). So total $2+6+1=9$ qubits will be needed for this purpose (the extra $1$ qubit is the ancilla).
Questions:
Using this information, is it
possibleto create the initial states $|\Psi_0\rangle$ and $|b\rangle$ on the IBM $16$ qubit version?
If you think $4\times 4$ is too large to be implemented on the IBM quantum computers you could even show an
exampleof initial state preparation for a $2\times 2$ Hermitian matrix $\mathbf{A}$ (or just give a reference to such an example).
I simply want to get a
general idea about whether this can be done (i.e. whether it is possible) on the IBM 16-qubit quantum computer, and for that which gates will be necessary. If not the IBM 16-qubit quantum computer, can the QISKit simulator used for recreating the initial state preparation of $|\Psi_0\rangle$ and $|b\rangle$ in the HHL algorithm? Is there any other better alternative to go about this? |
These exercises are not tied to a specific programming language. Example implementations are provided under the Code tab, but the Exercises can be implemented in whatever platform you wish to use (e.g., Excel, Python, MATLAB, etc.).
###Exercise 1 - Time Evolution of Particle in 1D Box
The ultimate goal of this exercise is to create an animation depicting how the wavefunction of a particle in a one-dimensional, infinite square well (aka particle in a box) evolves in time. The results will, of course, depend on the choice of initial wavefunction, so choosing an interesting, nontrivial wavefunction is the first step. The exercise is broken into several parts to walk one through the procedure.
#### Part 1 - Initial Wavefunction
Come up with a nontrivial initial wavefunction. Your choice here will determine what your final animation looks like. Assume the particle begins localized in the center of the box. Once you have an appropriate initial wavefunction, create a plot of its square modulus. Here are some important points you should consider when crafting an initial wavefunction:
* Make sure your initial wavefunction satisfies the boundary conditions. For this exercise, that means the wavefunction must vanish at the edges of the box.
* Be sure to normalize your initial wavefunction.
#### Part 2 - Expanding in Stationary States
The next step is to express your chosen initial wavefunction as a linear combination of the stationary states. This means you will be writing your wavefunction as a series expansion in the stationary states. First calculate the expansion coefficients using Equation (4) from the theory section. Once this is done, your initial wavefunction can be written in terms of the stationary states $\psi_{n}(x)$ as $$\Psi(x,0) = \sum_{n=0}^{\infty}c_{n}\psi_{n}(x).$$
To use this result we will need to truncate this expansion, i.e. only keep a finite number of terms. Determine how many terms are needed to obtain an accurate reproduction of the initial wavefunction and plot the square modulus of the result. It should look identical to your plot from the previous part.
#### Part 3 - Time Evolution
In the previous part, you approximated your initial wavefunction as a partial sum of stationary states. Now you are ready to determine how your wavefunction depends on time. To do so, simply add the appropriate time dependence factor to each term in the sum in accordance with Equation (3) from the theory section. Define your time-dependent wavefunction and plot its square modulus with $t = 0$. Again, this plot should be identical to those of the previous two sections.
#### Part 4 - Creating an Animation
Now you can see what your wavefunction looks like at _any_ time, but it is first important to get a sense of how much counts as "a long time." We also want our final animation to be periodic and loop continuously. Both of these issues are addressed by noting that for a particle in a box, there is a special amount of time $T$ called the _revival time_ after which the wavefunction will have returned to its initial state. Roughly speaking, this is the quantum analogue to the time it takes a classical ball, bouncing around in a box, to bounce back and forth and return to its original position with its original velocity. For a particle in a box, the revival time is given by $$T = \frac{4ma^{2}}{\pi \hbar},$$ where $a$ is the width of the box.
Try plotting the square modulus of your time-dependent partial sum at various fractions of the revival time. If you superimpose plots at different times, you should be able to see a graduate evolution. Some wavefunctions will evolve more rapidly and in a more complex manner than others, so depending on your choice of initial wavefunction you may need to superimpose plots at closer times.
The final step is to divide the revival time into a large number of time steps and create a plot of the square modulus of your time-dependent partial sum at each time step. These are the individual frames of your animation. How many time steps/frames you need depends on the complexity of your initial wavefunction. The more frames you add, the smoother your animation will be, but the longer it will take to generate.
Lastly, string your frames together to produce the finalized animation!
###Exercise 2
Repeat the procedure outlined in Exercise 1 for a particle trapped in a quantum harmonic oscillator potential. Note that the basic steps remain the same, but now the stationary states have changed. Can you think about what else has changed? |
The Cauchy-Riemann Theorem
Recall from the Analytic/Holomorphic Complex Functions page that if $A \subseteq \mathbb{C}$ is open,
Theorem 1 (The Cauchy-Riemann Theorem): Let $A \subseteq \mathbb{C}$ be an open set, $z_0 \in A$, and let $f : A \to \mathbb{C}$ where $f = u + iv$ ($u, v : \mathbb{R}^2 \to \mathbb{R}$). Then $f$ is analytic at $z_0$ if and only there is a neighbourhood $\mathcal N \subseteq A$ of $z_0$ for wihch: 1) $\displaystyle{\frac{\partial u}{\partial x}}$, $\displaystyle{\frac{\partial u}{\partial y}}$, $\displaystyle{\frac{\partial v}{\partial x}}$, and $\displaystyle{\frac{\partial v}{\partial y}}$ all exist and are continuous on $\mathcal N$ 2) The equations $\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$ and $\displaystyle{\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}}$, called the Cauchy-Riemann Equations, are satisfied on all of $\mathcal N$.
Moreover we have that $\displaystyle{f'(x + iy) = \frac{\partial u}{\partial x} (x, y) + i \frac{\partial v}{\partial x} (x, y) = \frac{\partial v}{\partial y} (x, y) - i \frac{\partial u}{\partial y}(x, y)}$.
Partial Proof:$\Rightarrow$ Suppose that $f$ is analytic at $z_0$. Then there exists a neighbourhood $\mathcal N$ of $z_0$ such that $f$ is complex differentiable on $\mathcal N$. In particular, the limit $\displaystyle{f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}}$ exists. Since this limit exists, the limit also exists along any path as $z \to z_0$ and so the existence of these partial derivatives on $\mathcal N$ is given. If $z_0 = x_0 + iy_0$ and $z = x + iy_0$ then consider this limit as $x \to x_0$ (so that $z \to z_0$): And more generally, $\displaystyle{f'(z) = \frac{\partial u}{\partial x} (z) + i \frac{\partial v}{\partial x} (z)}$ $(*)$ for all $z \in \mathcal N$ since $f$ is analytic at $z_0$. Similarly we consider the limit as $z = x_0 + iy$ where $y \to y_0$: And more generally, $\displaystyle{f'(z) = -i \frac{\partial u}{\partial y} (z) + \frac{\partial v}{\partial y} (z)}$ $(**)$ for all $z \in \mathcal N$ since $f$ is analytic at $z_0$. Comparing $(*)$ and $(**)$ gives us the following equality which holds for all $z \in \mathcal N$: Therefore we have that $\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$ and $\displaystyle{\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}}$, (the Cauchy-Riemann equations) are satisfied on $\mathcal N$ All that remains to show in this direction is that the partial derivatives of $u$ and $v$ are continuous on $\mathcal N$ which we currently do not have the tools to prove, so we will come back to this later on. $\square$ The Complex Inverse Function Theorem
The extremely important inverse function theorem that is often taught in advanced calculus courses appears in many different forms. One of such forms arises for complex functions. We will state (but not prove) this theorem as it is significant nonetheless.
Theorem 2 (The Complex Inverse Function Theorem): Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ and $z_0 \in A$. If $f$ is analytic at $z_0$ and $f'(z_0) \neq 0$ then there exists open neighbourhoods $U$ of $z_0$ and $V$ of $f(z_0)$ for which $f : U \to V$ is a bijection and the inverse function (which exists since $f$ restricted to $U$ is bijective), $f^{-1} : V \to U$ is analytic at $f(z_0)$ with $\displaystyle{\frac{d}{dw} f^{-1} (w) = \frac{1}{f'(z)}}$ where $w = f(z)$.
It is important to note that the complex inverse function tells us that provided that $f'(z_0) \neq 0$ then there exists a LOCAL analytic inverse function of $f$. It may be that no global inverse function exists even if $f'(z_0) \neq 0$ for every $z_0 \in \mathbb{C}$.
Example 1 Determine whether or not the function $f(z) = \overline{z}$ is analytic on a subset $A$ of $\mathbb{C}$.
Let $z = x + yi$. Then if $f = u + iv$ then $f(z) = f(x + yi) = x - yi$ and so $u(x, y) = x$ and $v(x, y) = -y$. Notice that for all $z \in \mathbb{C}$ that:(4)
So $\displaystyle{\frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}}$, i.e., the Cauchy-Riemann equations are not satisfied for all $z \in \mathbb{C}$ and so $f$ is analytic nowhere. |
Bob has $2$ cranes and $M$ available containers ($1 \leq C_i \leq M$) and he has to do $N$ transports from one container to another given an input list (the order of the transports must be respected).
To execute a transport from container $C_i$ to $C_j,$ Bob must choose a crane and move it to $C_i$, pick up the supplies, and then move it to $C_j$, where the crane remains until executing another transport.
The distance between two containers $C_i$ and $C_j$, is always $|i - j|$. We consider that each crane starts at the position of the first container it interacts with.
Since the cranes run on fuel, Bob would like to know, given the transport list, which crane to use for which transport, in order to conserve as much fuel as possible, by minimizing the total distance traveled by the two cranes. Print the minimum distance.
The input file, two_cranes.in, contains, on its first line, two numbers, $N$ and $M$, representing the number of transports Bob has to make, and the total number of available containers. On the next $N$ lines, it will contain one pair $[C_i, C_j]$ per line, representing the start and end positions for each container for the respective transport.
Example test:
$\text{3 10}$
$\text{2 4}$
$\text{5 4}$
$\text{9 8}$
Answer : $5$.
Explanation:
Arbitrarily start with
crane 1 at position $2$ and travel to $4$. ($D_0 = | 2- 4| = 2$).
Proceed using
crane 1 for the second transport as well. Note that the current position of crane 1 is $4$ (after completing the first transport). So now it travels to positon $5$ then to position $4$, for a total distance of $D_1 = |4 - 5| + |5-4| = 2.$
Finally, use
crane 2 for the last transport, so the last distance is $D_2 = |9 - 8| =1$.
Hence, the final answer is $5 = D_0 +D_1 + D_2$.
My attemptS:
First of, I've mistaken this problem to be much easier than it is, hitting it with a straight greedy approach that seemed to fail a lot of test cases. I'd simply go along and choose the closest crane to the $C_i$ of the current transport and increment the distance with the appropriate cost.
Since this didn't work, I figured out that this should actually be solved using dynamic programming, but that's a particularly weak spot of mine. Initially, I found a very simple recurrence relation, but that only seemed to work on two test cases, so I guess it was dumb luck.
I would store a $dp[]$ int array, initialize it with $\infty$, then overwrite $dp[0]$ with the $C_i$ of the first transport. After that, I'd start from $i = 2$, process each transport and let $dp[i] = \min(dp[i - 1], dp[i - 2]) + cost(transport) + 1$ and finally output $dp[N]$.
Alas, this failed as well.
My last approach was trying to find some recurrence relation that I would later optimize using dynamic programming, but I couldn't actually put the pieces together for quite some time.
Data limits
$1 \leq N, M \leq 10^3$
$1 \leq C_i, C_j \leq M \text{ for every such pair}$
$C_i \neq C_j$ for every transport!
Time limits: C++ - $0.3s$, Java - $0.6s$
This problem is just practice, but the course I have it from doesn't give a solution. I've spent too much time on this already, I just want to know the solution, no hints please. |
Provable by Gentzen Proof System iff Negation has Closed Tableau/Set of Formulas Theorem
Let $\mathscr G$ be instance 1 of a Gentzen proof system.
$\bar U$ has a closed semantic tableau Proof Necessary Condition Suppose first that $U$ is an axiom of $\mathscr G$.
Hence, so does $\bar U$, by definition of logical complement.
Next, suppose that the last step in proving $U$ was an instance of the $\alpha$-rule.
That is, for some $\alpha$-formula $\mathbf A$ and corresponding $\mathbf A_1, \mathbf A_2$:
$U = U_1 \cup U_2 \cup \left\{{\mathbf A}\right\}$
where $U_1 \cup \left\{{\mathbf A_1}\right\}$ and $U_2 \cup \left\{{\mathbf A_2}\right\}$ are $\mathscr G$-theorems.
By induction hypothesis, the logical complements:
$\bar U_1 \cup \left\{{\bar{\mathbf A}_1}\right\}$ $\bar U_2 \cup \left\{{\bar{\mathbf A}[email protected]}\right\}$
Now by Superset of Unsatisfiable Set is Unsatisfiable, so are:
$\bar U' := \bar U_1 \cup \bar U_2 \cup \left\{{\bar{\mathbf A}_1}\right\}$ $\bar U'' := \bar U_1 \cup \bar U_2 \cup \left\{{\bar{\mathbf A}_1}\right\}$
By De Morgan's Laws, it follows that:
$\bar{\mathbf A}$ is equivalent to $\bar{\mathbf A}_1 \lor \bar{\mathbf A}_2$.
Because $\bar U'$ and $\bar U''$ have closed tableaus, so does $\bar U$.
Finally, suppose that the last step in proving $U$ was an instance of the $\beta$-rule.
That is, for some $\beta$-formula $\mathbf B$ and corresponding $\mathbf B_1, \mathbf B_2$:
$U = U_1 \cup \left\{{\mathbf B}\right\}$
where $U' := U_1 \cup \left\{{\mathbf B_1, \mathbf B_2}\right\}$ is a $\mathscr G$-theorem.
By induction hypothesis, the logical complement:
$\bar U' = \bar U_1 \cup \left\{{\bar{\mathbf B}_1, \bar{\mathbf B}_2}\right\}$
By De Morgan's Laws, it follows that:
$\bar{\mathbf B}$ is equivalent to $\bar{\mathbf B}_1 \land \bar{\mathbf B}_2$.
Because $\bar U'$ has a closed tableau, so does $\bar U$.
The result follows by the Second Principle of Mathematical Induction.
$\Box$
Sufficient Condition
Let $T$ be a closed tableau for $\bar U$.
Suppose $T$ has only one node.
But then $U$ also contains a complementary pair.
Hence, it is an axiom of $\mathscr G$.
First the case that an $\alpha$-formula $\mathbf A$ was selected.
Define, for convenience, $\bar U' = \bar U \setminus \left\{{\mathbf A}\right\}$.
Then the rest of $T$ is a semantic tableau for $\bar U' \cup \left\{{\mathbf A_1, \mathbf A_2}\right\}$.
By induction hypothesis, this means that $U' \cup \left\{{\bar{\mathbf A}_1, \bar{\mathbf A}_2}\right\}$ is a $\mathscr G$-theorem.
Hence, $U = U' \cup \left\{{\bar{\mathbf A}}\right\}$ is a $\mathscr G$-theorem, as desired.
Now the case that a $\beta$-formula $\mathbf B$ was selected.
Define, for convenience, $\bar U' = \bar U \setminus \left\{{\mathbf B}\right\}$.
Each of these has fewer nodes than $T$.
By induction hypothesis, this means that $U' \cup \left\{{\bar{\mathbf B}_1}\right\}$ and $U' \cup \left\{{\bar{\mathbf B}_2}\right\}$ are $\mathscr G$-theorems.
Hence, $U = U' \cup \left\{{\bar{\mathbf B}}\right\}$ is a $\mathscr G$-theorem, as desired.
The result now follows from the Second Principle of Mathematical Induction.
$\blacksquare$ |
Answer
$4 \times 10^6$
Work Step by Step
$\dfrac{a^m}{a^n}=a^{m-n}, a\ne0$ Divide the corresponding parts while using the rule above to obtain: $=\dfrac{5.2}{1.3} \times \dfrac{10^{13}}{10^7} \\=4 \times 10^{13-7} \\=4 \times 10^6$
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Continuity of Addition, Scalar Mult., and Mult. on Normed Algebras
Continuity of Addition, Scalar Multiplication, and Multiplication on Normed Algebras
Proposition 1: Let $\mathfrak{A}$ be a normed algebra. Then addition, scalar multiplication, and vector multiplication are all continuous. Proof:Every normed linear space is a topological vector space and so addition and scalar multiplication on $\mathfrak{A}$ are continuous. All that we need to show is that multiplication is continuous. Let $f : \mathfrak{A} \times \mathfrak{A} \to \mathfrak{A}$ be defined for all $(a, b) \in \mathfrak{A} \times \mathfrak{A}$ by $f(a, b) = a \cdot b$. Let $(a_0, b_0) \in \mathfrak{A} \times \mathfrak{A}$. We aim to show that $f$ is continuous at $(a_0, b_0)$ and hence continuous on all of $\mathfrak{A} \times \mathfrak{A}$. Let $\epsilon > 0$ be given. If $\| a - a_0 \| < \epsilon$ and $\| b - b_0 \| < \epsilon$ we have that:
\begin{align} \quad \| f(a, b) - f(a_0, b_0) \| &= \| a \cdot b - a_0 \cdot b_0 \| \\ &= \| a \cdot b - a_0 \cdot b + a_0 \cdot b - a_0 \cdot b_0 \| \\ &= \| (a - a_0) \cdot b + a_0(b - b_0) \| \\ &\leq \| a - a_0 \| \| b \| + \| a_0 \| \| b - b_0 \| \\ &< \epsilon (\| b \| + \| a_0 \|) \\ &< \epsilon (\| b - b_0 + b_0 \| + \| a_0 \|) \\ &< \epsilon (\| b - b_0 \| + \| b_0 \| + \| a_0 \|) \\ &< \epsilon ( \epsilon + \| a_0 \| + \| b_0 \|) \\ &< \epsilon^2 + \epsilon (\| a_0 \| + \| b_0 \|) \end{align}
Since the above equality holds true for all $\epsilon > 0$ we see that $f$ is continuous at $(a_0, b_0)$, and is thus continuous on all of $\mathfrak{A} \times \mathfrak{A}$, i.e., multiplication on $\mathfrak{A}$ is continuous. $\blacksquare$ |
The Uni. Bound. Con. Theorem for Point. Con. Seqs. of Functs.
The Uniform Bounded Convergence Theorem for Pointwise Convergent Sequences of Functions
Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Suppose that: 1) $(f_n(x))_{n=1}^{\infty}$ is uniformly bounded. 2) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$.
Then $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E f}$.
We say that a sequence of functions $(f_n(x))_{n=1}^{\infty}$ is uniformly bounded on $E$ if there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $n \in \mathbb{N}$ and for all $x \in E$ we have that $|f_n(x)| \leq M$. Proof:Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of uniformly bounded Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. We have already noted that the limit function $f(x)$ is bounded and Lebesgue measurable. So we break this proof up into two cases. Case 1:If $m(E) = 0$ then once again for each $n \in \mathbb{N}$ we have that $\displaystyle{\int_E f_n = 0}$ and $\displaystyle{\int_E f = 0}$ and the conclusion to the theorem holds. Case 2:Let $m(E) > 0$. Let $\epsilon > 0$ be given. Since $(f_n(x))_{n=1}^{\infty}$ is uniformly bounded on $E$ we have that there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $n \in \mathbb{N}$ and for all $x \in E$:
\begin{align} \quad | f_n(x) | \leq M \end{align}
Furthermore, since $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$ we also have that $|f(x)| \leq M$ as well. So for all $n \in \mathbb{N}$ and for all $x \in E$:
\begin{align} \quad |f_n(x) - f(x)| \leq |f_n(x)| + |f(x)| \leq M + M = 2M \end{align}
Now, by Egoroff's Theorem, since $E$ is Lebesegue measurable, $m(E) < \infty$, and $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$, for $\displaystyle{\epsilon_0 = \frac{\epsilon}{4M} > 0}$ there exists a closed set $F \subseteq E$ such that:
\begin{align} \quad m(E \setminus F) < \epsilon_0 = \frac{\epsilon}{4M} \quad \mathrm{and} \quad f_n \to f \: \mathrm{uniformly \: on} \: F \end{align}
Since $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on $F$, for $\displaystyle{\epsilon_1 = \frac{\epsilon}{2 m(E)} > 0}$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that then for all $x \in F$:
\begin{align} \quad | f_n(x) - f(x) | < \epsilon_1 = \frac{\epsilon}{2m(E)} \quad (**) \end{align}
So from the additivity and monotonicity properties of the Lebesgue integral for bounded Lebesgue measurable functions we have that for all $n \geq N$:
\begin{align} \quad \biggr \lvert \int_E f_n - \int_E f \biggr \rvert = \biggr \lvert \int_E (f_n - f) \biggr \rvert \leq \int_E |f_n - f| & \leq \int_F |f_n - f| + \int_{E \setminus F} |f_n - f| \\ & \leq \int_F \epsilon_1 + \int_{E \setminus F} 2M \\ & \leq \int_F \frac{\epsilon}{2m(E)} + \int_{E \setminus F} 2M \\ & \leq \frac{\epsilon}{2m(E)} m(E) + 2M \cdot m(E \setminus F) \\ & < \frac{\epsilon}{2} + 2M \frac{\epsilon}{4M} \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & < \epsilon \end{align}
Therefore $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E f}$. $\blacksquare$ |
This puzzle is all about making 4 with 4 ones, but with certain constraints.
Allowed operations:
$-$ Subtraction $-a$ Negation $\times$ Multiplication $\div$ Division $\sqrt{a}$ Square Root $\sqrt[b]{a}$ Arbitrary Roots $!$ Factorial $.\!a$ Decimal $.\!\overline{a}$ Recurring decimal Note: Arbitrary roots must use 1s. Easy: Give me four examples of making 4 with 4 ones without addition. Medium: Give me two examples of making 4 with 4 ones without addition, negation, or factorial. Hard: Give me two examples of making 4 with 4 ones without addition and factorial, using only one subtraction and one negation.
(In each of the above cases, do not just circumvent the ban on addition by doing $a -- b$.) |
Open and Closed Sets in Topological Subspaces
Recall from the Topological Subspaces page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then we can define a topology on $\tau_A$ on $A$ called the subspace topology on $A$ (from $X$) explicitly defined as:(1)
Together, $(A, \tau_A)$ form what is called a topological subspace.
Note that by definition, a set $V \subseteq A$ is open in $A$ with the subspace topology if and only if there exists an open set $U$ of $X$ such that:(2)
So, what can be said about the closed sets in $A$? The following theorem gives us exactly what we'd expect.
Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. A set $C \subseteq A$ is closed in $A$ if and only if there exists a closed set $D$ of $X$ such that $C = A \cap D$. Proof:$\Rightarrow$ Let $C \subseteq A$ be closed in $A$. Then $A \setminus C$ is open in $A$. So, there exists an open set $U$ in $X$ such that: Taking the complement of both sides with respect to $A$ shows us that: So let $D = U^c$. Then $D$ is closed in $X$ and $C = A \cap D$. $\Leftarrow$ Suppose that there exists a closed set $D$ of $X$ such that $C = A \cap D$. Then: $D^c$ is an open set in $X$, so $A \cap D^c$ is open in $A$. In other words, $A \setminus C$ is open in $A$, so $C$ is closed in $A$. $\blacksquare$ |
Research Open Access Published: Regularity of weak solutions of the Cauchy problem to a fractional porous medium equation Boundary Value Problems volume 2015, Article number: 28 (2015) Article metrics
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Abstract
This paper concerns the regularity of the weak solutions of the Cauchy problem to a fractional porous medium equation with a forcing term. In the recent work (Fan
et al. in Comput. Math. Appl. 67:145-150, 2014), the authors established the existence of the weak solution and the uniqueness of the weak energy solution. In this paper, we show that the every nonnegative bounded weak energy solution is indeed a strong solution. Introduction
This is a sequel of the previous work [1]. We continue to investigate the Cauchy problem to a fractional porous medium equation with a forcing term
where the integer \(m>0\), the forcing term \(f(x,t)\in C (0,\infty; L^{1}({\mathbb{R}}^{N}) )\). Furthermore, we assume that \(u_{0}(x)\) is a bounded and integrable function.
Recently, the fractional porous medium equation \(u_{t}+(-\Delta)^{\frac{\alpha}{2}}u^{m}= 0\) (\(m>0\), \(\alpha>0\)) has been investigated by Pablo
et al. in [2] for \(\alpha =1\) and in [3] for general case \(0 < \alpha< 2\). Systematic and satisfactory results on the weak solutions to the Cauchy problem to the fractional porous medium problem have been obtained, including the existence, uniqueness, comparison principle, and regularity to the suitable weak solutions.
The interest in studying the fractional diffusion in modeling diffusive processes has a wide literature, especially studying the long-range diffusive interaction in porous medium type propagation and infinitesimal generators of stable Lévy processes [4, 5]. We would like to refer to the survey papers [6] and [7] for the fractional operators and to [8–10] for fractional partial differential equations. For instance, a method based on the Jacobi-tau approximation for solving multi-term time-space-fractional partial differential equations. A spectral-tau algorithm is based on the Jacobi operational matrix for a numerical solution of time fractional diffusion-wave equations. (See [11] for details.) On the other side, there is much literature on the porous medium equations (see the classical book [12] and references [1, 10, 13–18] and references therein).
Before we state the main results in this paper, we present some definitions as regards the fractional operator and the weak solutions to the problem (1.1). The nonlocal operator square root Laplacian operator can be illustrated in the following three items.
The first one is that
using the Fourier transform for any function
g in the Schwartz class.
where \(C(N)= \pi^{-\frac {N+1}{2}}\Gamma (\frac {N+1}{2} )\) is the normalization constant, \(\Gamma(s)\) is the gamma function.
Thirdly, we can define the nonlocal operator \((-\Delta)^{\frac{1}{2}}\) using the so-called Dirichlet to Neumann operator, which is introduced in [21] by Caffarelli and Silvestre. For any given smooth bounded function \(g(x)\), which is defined in \(\mathbb{R}^{N}\), we consider the harmonic extension \(v=v(x,y)\) in \(\mathbb{R}_{+}^{N+1}\), denoting \(v=E(g)\), which is the unique smooth bounded solution to the following problem:
Then we claim that \((-\Delta_{x})^{\frac{1}{2}}\) as the operator
where \(\Delta_{x}\) is the Laplacian acting only on the
x variables, and \(\Delta_{x,y}\) acts on all the variables \((x,y)\). In fact,
Furthermore, it is easy to check that
T is indeed a positive operator by integration by parts argument.
The third way to define the square root of Laplacian developed in [21] was proved to be powerful in dealing with the nonlocal operator. For any given function \(|u|^{m-1}u\), we consider its harmonic extension and denote \(w=E(|u|^{m-1}u)\); then the extension function satisfies the following problem:
where Ω is the upper half-space \(\mathbb{R}^{N+1}_{+}\), \(\bar{x}=(x_{1}, x_{2},\ldots, x_{n}, y)\), \(y>0\), and we denote the boundary of Ω by \(\Gamma=\mathbb{R}^{N}\times\{0\}\).
Fan
et al. [1] established the existence of the weak solution and the uniqueness of the weak energy solution to the problem (1.1). The weak solution and the weak energy solution of the Cauchy problem are defined as follows.
Multiplying the equation \(\Delta w=0\) in (1.5) by a test function \(\varphi\in C_{0}^{1} (\bar{\Omega} \times[0,T) )\) and integrating by parts with respect to \(\bar{x}\), we get
with \(u=\operatorname{Tr} (|w|^{\frac{1}{m}-1}w )\).
Then a pair \((u,w)\) of functions is called a weak solution to the problem (1.5) if \(w\in L^{1} ((0,T);W^{1,1}_{\mathrm{loc}}(\Omega) )\), \(u=\operatorname{Tr} (|w|^{\frac{1}{m}-1}w )\in L^{1}(\Gamma\times(0,T))\), and \(u_{0}(x)\in L^{1}(\mathbb{R}^{N})\cap L^{\infty}(\mathbb{R}^{N})\), the equality (1.6) holds for any \(\varphi\in C_{0}^{1}(\bar{\Omega}\times[0,T))\).
Furthermore, if a pair weak solution pair satisfies \(w\in L^{2} ([0,T];H^{1}(\Omega ) )\), we call the weak solution a weak energy solution.
In a previous work [1], Fan
et al. obtained the following existence and uniqueness results. Theorem 1.1 Assume that \(m>0\), \(f\in C (0,\infty; L^{1}({\mathbb{R}}^{N}) )\) and the data \(u_{0}\in L^{1}({\mathbb{R}}^{N})\cap L^{\infty}({\mathbb{R}}^{N})\), then there exists a unique weak energy solution \((w,u)\) to the problem (1.4), and \(u\in C ([0,\infty);L^{1}({\mathbb{R}}^{N}) )\cap L^{\infty}({\mathbb{R}}^{N}\times[0,\infty))\).
We are ready to announce the main result, proved in the next section.
Theorem 1.2 Every nonnegative bounded weak energy solution to the problem (1.1) is a strong solution. Moreover, for every \(t\in(0,T]\), we have for all \(m\geq 1\), where \(C=C (T,m,\|u_{0}\|_{L^{1}({\mathbb{R}}^{N})} )\). Proof of the main result
In this section, we prove Theorem 1.2. Namely, it suffices to show that the time partial derivative of
u is an \(L^{1}\) function. As the first step, we will show that the time-increment quotients are bounded in \(L^{1}(\Gamma)\), where \(\Gamma={\mathbb{R}}^{N}\times\{0\}\), and thus the limit is still in \(L^{1}({\mathbb{R}}^{N})\). Time-increment quotients of the solution are bounded in \(L^{1}({\mathbb{R}}^{N})\) Proposition 2.1 for every \(t> 0\) and \(h>0\). Proof
In fact, if follows from [22] that we have
The same type of estimate can be obtained for the time-increment quotients of \(u^{m}\) under the condition of \(m\geq1\). Since \(u\in L^{\infty}\), we have
where \(C=m\cdot\max \{u^{m-1}(\cdot,t+h),u^{m-1}(\cdot,t) \}\).
Define \(\delta^{h}u^{m}=\frac{1}{h} (u^{m}(\cdot,t+h)-u^{m}(\cdot ,t) )\), we get
for \(m\geq1\). □
Regularity of time derivative of solution Proposition 2.2 If u is a weak solution to the problem (1.1), then Proof
Due to the lack of regularity of
u in time, we also use the Steklov average of u instead of u itself. Step 1. Assume \(\partial_{t}u\in L^{1}({\mathbb{R}}^{N})\), then the following identity can be obtained by multiplying the equation (1.1) by a test function φ and integrating in \({\mathbb{R}}^{N}\times(0,\infty)\),
Thus, we can rewrite the weak formulation (1.6) into the following form via the Steklov average \(u^{h}\):
Step 2. Let \(\varphi=\zeta\partial_{t}(u^{m})^{h}\), where \(\zeta\in C^{\infty}_{0}((0,\infty))\), \(0\leq\zeta\leq1\), \(\zeta(t)=1\), for \(t\in[t_{1},t_{2}]\) is a cutoff function. Inserting φ as a test function in (2.4) yields
It follows from the inequality
in [2] and the facts \(\delta^{h}u^{m}\in L^{1}({{\mathbb{R}}^{N}})\), \(u^{m}\in L^{2}_{\mathrm{loc}}((0,\infty);H^{1/2}_{0})\) that
Then the proof is completed by passing to the limit \(h\rightarrow0\). □
Proof of Theorem 1.2
The estimate (2.1) implies
for all \(0<\tau\leq t\leq T<\infty\). Furthermore, we can deduce that \(u\in \mathit{BV}((\tau,T);L^{1}({\mathbb{R}}^{N}))\).
It follows from the fact in [23] that \(u\in W^{1,1} ((\tau,T);L^{1}({\mathbb{R}}^{N}) )\), provided that
for some \(v\in W^{1,1} ((\tau,T);L^{1}({\mathbb{R}}^{N}) )\), \(p\in L^{1}_{\mathrm{loc}}({\mathbb{R}})\).
Let \(v=u^{\frac{m+1}{2}}\), we claim that \(u^{\frac{m+1}{2}}\in W^{1,1}((\tau,T);L^{1}({\mathbb{R}}^{N}))\).
In fact, since \(\partial_{t}u^{(m+1)/2}\in L^{2}_{\mathrm{loc}} ((0,\infty);L^{2}({\mathbb{R}}^{N}) )\), we can get \(\partial_{t}u^{(m+1)/2}\in L^{1} (K\times(\tau,T) )\) for any compact set \(K\subset\subset{\mathbb{R}}^{N}\).
In addition, we have the estimate
for \(m\geq1\). Hence, we have \(\partial_{t}u^{(m+1)/2}\in L^{1} ((\tau,T);L^{1}({\mathbb{R}}^{N}) )\), and therefore \(\partial_{t}u\in L^{1} ((\tau,T); L^{1}({\mathbb{R}}^{N}) )\) for all \(0<\tau<T<\infty\). Hence, (1.7) can be obtained by passing \(h\rightarrow0\) in (2.1). We complete the proof of Theorem 1.2.
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119(2), 473-502 (1995) Acknowledgements
The authors would like to thank the referees for helpful suggestions to update our paper. Shan Li is supported in part by NSFC grant (No. 71372189), Project funded by China Postdoctoral Science Foundation (No. 2013M542285) and Social sciences planning project in Sichuan province (No. SC14TJ06).
Additional information Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The article is a joint work of two authors who contributed equally to the final version of the paper. All authors read and approved the final manuscript. |
Let $$f(x)=\left\{ \begin{array}{l l} x+2 & \quad ,x\in\mathbb{Q}\\ 6-x & \quad ,x\notin\mathbb{Q} \end{array} \right.$$ then $$\lim_{x \to 0}f(x)$$ does not exist.
By limit definition.
I see that I should choose $\varepsilon_0=1$ but I don't see how do I continue..
Thanks |
This question already has an answer here:
The question is this.
Let $(s_n)$ be a sequence such that $$\left|s_{n+1}-s_n\right| < 2^{-n}, \forall n \in\mathbb N$$ Prove that $(s_n)$ is a Cauchy sequence and hence convergent.
My proof is below.
Proof. Let $\epsilon > 0 $ and $N = \dfrac{\ln\left(\frac{1}{\epsilon}\right)}{\ln(2)}$. Then, $\forall n > N = \dfrac{\ln(\frac{1}{\epsilon})}{\ln(2)}$ implies $$n\ln(2) > \ln\left(\frac{1}{\epsilon}\right)$$ $$2^n > \frac{1}{\epsilon}$$ $$\epsilon > \frac{1}{2^n} > \left|s_{n+1}-s_n\right|$$Q.E.D.
Is this ok?? |
I would like to understand the probabilistic interpretation of the following elliptic problem with mixed Dirichlet-Neumann boundary conditions:
Let $B := \{ x \in \mathbb{R}^n, \quad \| x \|_2 \leq 1 \}, n \geq 1$. We assume that the boundary of $B$ is composed of two disjoint subsets : $\partial B = \Gamma_N \cup \Gamma_D$. Consider $\lambda>0,\ f \in \mathcal{C}(\bar{B}), g \in \mathcal{C}(\partial B)$ and $u$ satisfying \begin{align} \lambda u - \frac{1}{2} \Delta u & = f \quad \mbox{in} \quad B,\\ u & = g \quad \mbox{in} \quad \Gamma_D,\\ \frac{\partial u}{\partial n} & = 0 \quad \mbox{in} \quad \Gamma_N. \end{align}
I believe that it is a stopping problem for a reflected Wiener process, in the sense that \begin{equation} u(x) = \mathbb{E} \left ( \int_0^\tau \exp(-\lambda t) f(\beta(t)) dt \right ) + \mathbb{E} \exp(-\lambda \tau) g(\beta(\tau)) \end{equation} where \begin{align} & d \beta(t) \in \partial \mathbf{1}_\bar{B} (\beta(t)) + d w(t)\\ & \tau = \inf \{ t>0, \quad \beta(t) \in \Gamma_D \}. \end{align} here $\mathbf{1}_\bar{B}(x) = 0$ if $x \in \bar{B}$ and $\infty$ otherwise. Also, $w(t)$ is Wiener process. If it is correct, how to establish it rigorously?
thanks for your help!
\begin{equation} \star \star \star \end{equation} Let me comment the purely Dirichlet case / stopped case. We keep working with the domain $B$ and the functions $f$ and $g$. Consider $u$ satisfying \begin{align} \lambda u - \frac{1}{2} \Delta u & = f \quad \mbox{in} \quad B,\\ u & = g \quad \mbox{in} \quad \partial B. \end{align} So, if we define \begin{equation} \tau_B = \inf \{ t > 0, \quad w(t) \in \partial B \} \end{equation} which is known to satisfy $\mathbb{E} \tau_B = \frac{1-|x|^2}{n} < \infty$ then Dynkin formula (rigorously speaking we need $u \in \mathcal{C}^2$) tells us that \begin{equation} u(x) = \mathbb{E} \exp(-\lambda \tau_B) g(x+w(\tau_B)) + \mathbb{E} \int_0^{\tau_B} \exp(-\lambda s) f(x+w(s)) ds. \end{equation} \begin{equation} \star \star \star \end{equation} |
Scientific posters present technical information and are intended for congress or presentations with colleagues. Since LaTeX is the most natural choice to typeset scientific documents, one should be able to create posters with it. This article explains how to create posters with latex
Contents
The two main options when it comes to writing scientific posters are
tikzposter and beamerposter. Both offer simple commandsto customize the poster and support large paper formats. Below, you can see a side-to-side comparison of the output generated by both packages (tikzposter on the left and beamerposter on the right).
Tikzposter is a document class that merges the projects
fancytikzposter and tikzposter and it's used to generate scientific posters in PDF format. It accomplishes this by means the TikZ package that allows a very flexible layout.
The preamble in a tikzposter class has the standard syntax.
\documentclass[24pt, a0paper, portrait]{tikzposter} \usepackage[utf8]{inputenc} \title{Tikz Poster Example} \author{ShareLaTeX Team} \date{\today} \institute{ShareLaTeX Institute} \usetheme{Board} \begin{document} \maketitle \end{document}
The first command,
\documentclass[...]{tikzposter} declares that this document is a
tikzposter. The additional parameters inside the brackets set the font size, the paper size and the orientation; respectively. The available font sizes are:
12pt, 14pt, 17pt, 20pt and
24pt. The possible paper sizes are:
a0paper, a1paper and
a2paper. There are some additional options, see the further reading section for a link to the documentation.
The commands
title,
author,
date and
institute are used to set the author information, they are self-descriptive.
The command
\usetheme{Board} sets the current theme, i.e. changes the colours and the decoration around the text boxes. See the reference guide for screenshots of the available themes.
The command
\maketitle prints the title on top of the poster.
The body of the poster is created by means of text blocks. Multi-column placement can be enabled and the width can be explicitly controlled for each column, this provides a lot of flexibility to customize the look of the final output.
\documentclass[25pt, a0paper, portrait]{tikzposter} \usepackage[utf8]{inputenc} \title{Tikz Poster Example} \author{ShareLaTeX Team} \date{\today} \institute{ShareLaTeX Institute} \usepackage{blindtext} \usepackage{comment} \usetheme{Board} \begin{document} \maketitle \block{~} { \blindtext } \begin{columns} \column{0.4} \block{More text}{Text and more text} \column{0.6} \block{Something else}{Here, \blindtext \vspace{4cm}} \note[ targetoffsetx=-9cm, targetoffsety=-6.5cm, width=0.5\linewidth ] {e-mail \texttt{sharelatex@sharelatex.com}} \end{columns} \begin{columns} \column{0.5} \block{A figure} { \begin{tikzfigure} \includegraphics[width=0.4\textwidth]{images/lion-logo.png} \end{tikzfigure} } \column{0.5} \block{Description of the figure}{\blindtext} \end{columns} \end{document}
In
tikzposter the text is organized in blocks, each block is created by the command
\block{}{} which takes two parameters, each one inside a pair of braces. The first one is the title of the block and the second one is the actual text to be printed inside the block.
The environment
columns enables multi-column text, the command
\column{} starts a new column and takes as parameter the relative width of the column, 1 means the whole text area, 0.5 means half the text area and so on.
The command
\note[]{} is used to add additional notes that are rendered overlapping the text block. Inside the brackets you can set some additional parameters to control the placement of the note, inside the braces the text of the note must be typed.
The standard LaTeX commands to insert figures don't work in
tikzposter, the environment
tikzfigure must be used instead.
The package
beamerposter enhances the capabilities of the standard beamer class, making it possible to create scientific posters with the same syntax of a beamer presentation.
By now there are not may themes for this package, and it is slightly less flexible than tikzpopster, but if you are familiar with beamer, using beamerposter don't require learning new commands.
Note: In this article a special
beamerposter theme will be used. The theme "Sharelatex" is based on the theme "Dreuw" created by Philippe Dreuw and Thomas Deselaers, but it was modified to make easier to insert the logo and print the e-mail address at the bottom of the poster. Those are hard-coded in the original themes.
Even though this article explains how to typeset a poster in LaTeX, the easiest way is to use a template as start point. We provide several in the ShareLaTeX templates page
The preamble of a
beamerposter is basically that of a beamer presentation, except for an additional command.
\documentclass{beamer} \usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage{times} \usepackage{amsmath,amsthm, amssymb, latexsym} \boldmath \usetheme{Sharelatex} \usepackage[orientation=portrait,size=a0,scale=1.4,debug]{beamerposter} \title[Beamer Poster]{ShareLaTeX example of the beamerposter class} \author[sharelatexteam@sharelate.com]{ShareLaTeX Team} \institute[Sharelatex University]{The ShareLaTeX institute, Learn faculty} \date{\today} \logo{\includegraphics[height=7.5cm]{SharelatexLogo}}
The first command in this file is
\documentclass{beamer}, which declares that this is a beamer presentation. The theme "Sharelatex" is set by
\usetheme{Sharelatex}. There are some beamer themes on the web, most of them can be found in the web page of the beamerposter authors.
The command
\usepackage[orientation=portrait,size=a0,scale=1.4,debug]{beamerposter}
Imports the
beamerposter package with some special parameters: the orientation is set to
portrait, the poster size is set to
a0 and the fonts are scaled to
1.4. The poster sizes available are a0, a1, a2, a3 and a4, but the dimensions can be arbitrarily set with the options
width=x,height=y.
The rest of the commands set the standard information for the poster: title, author, institute, date and logo. The command
\logo{} won't work in most of the themes, and has to be set by hand in the theme's .sty file. Hopefully this will change in the future.
Since the document class is
beamer, to create the poster all the contents must be typed inside a
frame environment.
\documentclass{beamer} \usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage{times} \usepackage{amsmath,amsthm, amssymb, latexsym} \boldmath \usetheme{Sharelatex} \usepackage[orientation=portrait,size=a0,scale=1.4]{beamerposter} \title[Beamer Poster]{ShareLaTeX example of the beamerposter class} \author[sharelatexteam@sharelate.com]{ShareLaTeX Team} \institute[Sharelatex University] {The ShareLaTeX institute, Learn faculty} \date{\today} \logo{\includegraphics[height=7.5cm]{SharelatexLogo}} \begin{document} \begin{frame}{} \vfill \begin{block}{\large Fontsizes} \centering {\tiny tiny}\par {\scriptsize scriptsize}\par {\footnotesize footnotesize}\par {\normalsize normalsize}\par ... \end{block} \end{block} \vfill \begin{columns}[t] \begin{column}{.30\linewidth} \begin{block}{Introduction} \begin{itemize} \item some items \item some items ... \end{itemize} \end{block} \end{column} \begin{column}{.48\linewidth} \begin{block}{Introduction} \begin{itemize} \item some items and $\alpha=\gamma, \sum_{i}$ ... \end{itemize} $$\alpha=\gamma, \sum_{i}$$ \end{block} ... \end{column} \end{columns} \end{frame} \end{document}
Most of the content in the poster is created inside a
block environment, this environment takes as parameter the title of the block.
The environment
columns enables multi-column text, the environment
\column starts a new columns and takes as parameter the width of said column. All LaTeX units can be used here, in the example the column width is set relative to the text width.
Tikzposter themes
Default Rays Basic Simple Envelope Wave Board Autumn Desert
For more information see |
The Fundamental Groups of Discrete and Indiscrete Topological Spaces
The Fundamental Groups of Discrete and Indiscrete Topological Spaces
Theorem 1: Let $X$ be a topological space equipped with the discrete topology. Then for any $x \in X$, $\pi_1(X, x) = \{ [c_x] \}$. Proof:If $X$ has the discrete topology then every subset of $X$ is an open set. So the path components of $X$ are the singleton sets $\{ x \} \subseteq X$. We know that the fundamental group of a single point is the trivial group, and so:
\begin{align} \quad \pi_1(X, x) = \{ [c_x] \} \quad \blacksquare \end{align}
Theorem 2: Let $X $] be a topological space equipped with the indiscrete topology. Then for any [[$ x \in X$, $\pi_1(X, x) = \{ [c_x] \}$. Proof:If $X$ has the indiscrete topology then the only open subsets of $X$ are $\emptyset$ and the whole set $X$. Let $x \in X$ and let $\alpha, \beta : I \to X$ be loops such that $\alpha(0) = \alpha(1) = x$ and $\beta(0) = \beta(1) = x$. Define a function $H : X \times I \to X$ by:
\begin{align} \quad H(s, t) = \left\{\begin{matrix} \alpha(x) & \mathrm{if} \: 0 \leq s \leq 1, t = 0\\ \beta(x) & \mathrm{if} \: 0 \leq s \leq 1, t = 1 \\ x & \mathrm{else} \end{matrix}\right. \end{align}
Then $H$ is a trivially a continuous map from $X$ having the indiscrete topology. Furthermore:
\begin{align} \quad H_0(s) = H(s, 0) = \alpha (x) \end{align}(4)
\begin{align} \quad H_1(s) = H(s, 1) = \beta (x) \end{align}
And:
\begin{align} \quad H_t(0) = H(0, t) = x = H(1, t) = H_t(1) \end{align}
So $\alpha$ and $\beta$ are homotopic relative to $\{ 0, 1 \}$, so indeed, the fundamental group is:
\begin{align} \quad \pi_1(X, x) = \{ [c_x] \} \quad \blacksquare \end{align} |
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In this chapter we learned about left and right adjoints, and about joins and meets. At first they seemed like two rather different pairs of concepts. But then we learned some deep relationships between them. Briefly:
Left adjoints preserve joins, and monotone functions that preserve enough joins are left adjoints.
Right adjoints preserve meets, and monotone functions that preserve enough meets are right adjoints.
Today we'll conclude our discussion of Chapter 1 with two more bombshells:
Joins
are left adjoints, and meets are right adjoints.
Left adjoints are right adjoints seen upside-down, and joins are meets seen upside-down.
This is a good example of how category theory works. You learn a bunch of concepts, but then you learn more and more facts relating them, which unify your understanding... until finally all these concepts collapse down like the core of a giant star, releasing a supernova of insight that transforms how you see the world!
Let me start by reviewing what we've already seen. To keep things simple let me state these facts just for posets, not the more general preorders. Everything can be generalized to preorders.
In Lecture 6 we saw that given a left adjoint \( f : A \to B\), we can compute its right adjoint using joins:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$ Similarly, given a right adjoint \( g : B \to A \) between posets, we can compute its left adjoint using meets:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ In Lecture 16 we saw that left adjoints preserve all joins, while right adjoints preserve all meets.
Then came the big surprise: if \( A \) has all joins and a monotone function \( f : A \to B \) preserves all joins, then \( f \) is a left adjoint! But if you examine the proof, you'l see we don't really need \( A \) to have
all joins: it's enough that all the joins in this formula exist:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$Similarly, if \(B\) has all meets and a monotone function \(g : B \to A \) preserves all meets, then \( f \) is a right adjoint! But again, we don't need \( B \) to have
all meets: it's enough that all the meets in this formula exist:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ Now for the first of today's bombshells: joins are left adjoints and meets are right adjoints. I'll state this for binary joins and meets, but it generalizes.
Suppose \(A\) is a poset with all binary joins. Then we get a function
$$ \vee : A \times A \to A $$ sending any pair \( (a,a') \in A\) to the join \(a \vee a'\). But we can make \(A \times A\) into a poset as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Then \( \vee : A \times A \to A\) becomes a monotone map, since you can check that
$$ a \le a' \textrm{ and } b \le b' \textrm{ implies } a \vee b \le a' \vee b'. $$And you can show that \( \vee : A \times A \to A \) is the left adjoint of another monotone function, the
diagonal
$$ \Delta : A \to A \times A $$sending any \(a \in A\) to the pair \( (a,a) \). This diagonal function is also called
duplication, since it duplicates any element of \(A\).
Why is \( \vee \) the left adjoint of \( \Delta \)? If you unravel what this means using all the definitions, it amounts to this fact:
$$ a \vee a' \le b \textrm{ if and only if } a \le b \textrm{ and } a' \le b . $$ Note that we're applying \( \vee \) to \( (a,a') \) in the expression at left here, and applying \( \Delta \) to \( b \) in the expression at the right. So, this fact says that \( \vee \) the left adjoint of \( \Delta \).
Puzzle 45. Prove that \( a \le a' \) and \( b \le b' \) imply \( a \vee b \le a' \vee b' \). Also prove that \( a \vee a' \le b \) if and only if \( a \le b \) and \( a' \le b \).
A similar argument shows that meets are really right adjoints! If \( A \) is a poset with all binary meets, we get a monotone function
$$ \wedge : A \times A \to A $$that's the
right adjoint of \( \Delta \). This is just a clever way of saying
$$ a \le b \textrm{ and } a \le b' \textrm{ if and only if } a \le b \wedge b' $$ which is also easy to check.
Puzzle 46. State and prove similar facts for joins and meets of any number of elements in a poset - possibly an infinite number.
All this is very beautiful, but you'll notice that all facts come in pairs: one for left adjoints and one for right adjoints. We can squeeze out this redundancy by noticing that every preorder has an "opposite", where "greater than" and "less than" trade places! It's like a mirror world where up is down, big is small, true is false, and so on.
Definition. Given a preorder \( (A , \le) \) there is a preorder called its opposite, \( (A, \ge) \). Here we define \( \ge \) by
$$ a \ge a' \textrm{ if and only if } a' \le a $$ for all \( a, a' \in A \). We call the opposite preorder\( A^{\textrm{op}} \) for short.
I can't believe I've gone this far without ever mentioning \( \ge \). Now we finally have really good reason.
Puzzle 47. Show that the opposite of a preorder really is a preorder, and the opposite of a poset is a poset. Puzzle 48. Show that the opposite of the opposite of \( A \) is \( A \) again. Puzzle 49. Show that the join of any subset of \( A \), if it exists, is the meet of that subset in \( A^{\textrm{op}} \). Puzzle 50. Show that any monotone function \(f : A \to B \) gives a monotone function \( f : A^{\textrm{op}} \to B^{\textrm{op}} \): the same function, but preserving \( \ge \) rather than \( \le \). Puzzle 51. Show that \(f : A \to B \) is the left adjoint of \(g : B \to A \) if and only if \(f : A^{\textrm{op}} \to B^{\textrm{op}} \) is the right adjoint of \( g: B^{\textrm{op}} \to A^{\textrm{ op }}\).
So, we've taken our whole course so far and "folded it in half", reducing every fact about meets to a fact about joins, and every fact about right adjoints to a fact about left adjoints... or vice versa! This idea, so important in category theory, is called
duality. In its simplest form, it says that things come on opposite pairs, and there's a symmetry that switches these opposite pairs. Taken to its extreme, it says that everything is built out of the interplay between opposite pairs.
Once you start looking you can find duality everywhere, from ancient Chinese philosophy:
to modern computers:
But duality has been studied very deeply in category theory: I'm just skimming the surface here. In particular, we haven't gotten into the connection between adjoints and duality!
This is the end of my lectures on Chapter 1. There's more in this chapter that we didn't cover, so now it's time for us to go through all the exercises. |
Dear asmaier, you shouldn't view $\vec L = \vec x \times \vec p$ as a primary "definition" of the quantity but rather as a nontrivial result of a calculation.
The angular momentum is
defined as the quantity that is conserved because of the rotational symmetry - and this definition is completely general, whether the physical laws are quantum, relativistic, both, or nothing, and whether or not they're mechanics or field theory.
To derive a conserved charge, one may follow the Noether's procedure that holds for any pairs of a symmetry and a conservation law:
http://en.wikipedia.org/wiki/Noether_charge
In particular, the angular momentum has no problem to be evaluated in relativity - when the background is rotationally symmetric. The fact that you write $\vec L$ as a vector is just a bookkeeping device to remember the three components. More naturally, even outside relativity, you should imagine$$ L_{ij} = x_i p_j - x_j p_i $$i.e. $L_{ij}$ is an antisymmetric tensor with two indices. Such a tensor, or 2-form, may be mapped to a 3-vector via $L_{ij} = \epsilon_{ijk} L_k$ but it doesn't have to be. And in relativity, it shouldn't. So in relativity, one may derive the angular momentum $L_{\mu\nu}$ which contains the 3 usual components $yz,zx,xy$ (known as $x,y,z$ components of $\vec L$) as well as 3 extra components $tx,ty,tz$ associated with the Lorentz boosts that know something about the conservation of the velocity of the center-of-mass.
Incidentally, the general $x\times p$ Ansatz doesn't get any additional "gamma" or other corrections at high velocities. It's because you may imagine that it's the generator of rotations, and rotations are translations (generated by $\vec p$) that linearly depend on the position $x$. So the formula remains essentially unchanged. In typical curved backgrounds which still preserve the angular momentum, the other non-spatial components of the relativistic angular momentum tensor are usually not preserved because the background can't be Lorentz-boost-symmetric at the same moment. |
The Annals of Statistics Ann. Statist. Volume 9, Number 4 (1981), 822-833. On Distributions Determined by Random Variables Distributed Over the $n$-Cube Abstract
The distribution function of a random variable of the form $\sum^n_{i = 1} a_i Y_1 Y_2 \cdots Y_i$ where $a_i > 0$ and $0 \leq Y_i \leq 1$ is considered. A geometric argument is used to obtain the distribution function as a repeated integral. The result is used first to obtain the distribution function of a linear combination of variables defined over the simplex $X_i \geq 0, \sum^n_{i = 1} X_i \leq 1$. As a second application the distribution of certain quadratic forms over the simplex is obtained. This result yields as a special case the distribution of the internally studentized extreme deviate; the cases of normal and exponential samples are considered in detail and the required distributions obtained.
Article information Source Ann. Statist., Volume 9, Number 4 (1981), 822-833. Dates First available in Project Euclid: 12 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aos/1176345522 Digital Object Identifier doi:10.1214/aos/1176345522 Mathematical Reviews number (MathSciNet) MR619285 Zentralblatt MATH identifier 0475.60012 JSTOR links.jstor.org Citation
Currie, Iain D. On Distributions Determined by Random Variables Distributed Over the $n$-Cube. Ann. Statist. 9 (1981), no. 4, 822--833. doi:10.1214/aos/1176345522. https://projecteuclid.org/euclid.aos/1176345522 |
Criterion for the Adjoint of a Bounded Linear Operator to be Injective
Recall from The Adjoint of a Bounded Linear Operator Between Banach Spaces page that if $X$ and $Y$ are Banach spaces and if $T : X \to Y$ is a bounded linear operator then the adjoint of $T$ is the linear operator $T^* : Y^* \to X^*$ defined for all $f \in Y^*$ by:(1)
We noted that $T^*$ is also a bounded linear operator from $Y^*$ to $X^*$ and that $\| T^* \| = \| T \|$.
The following proposition tells us exactly then $T^*$ is injective.
Proposition 1: Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. Then the adjoint $T^* : Y^* \to X^*$ is injective if and only if $T(X)$ is a dense subspace of $Y$. Proof:$\Rightarrow$ Let $T^* : Y^* \to X^*$ be injective. Note that $T(X)$ is always a subspace of $Y$. So $\overline{T(X)}$ is a closed subspace of $Y$. Suppose that $T(X)$ is not dense in $Y$. Then $\overline{T(X)} \neq Y$. Let $y_0 \in Y \setminus \overline{T(X)}$. Then $\delta = d(y_0, \overline{T(X)}) > 0$. By one of the corollaries on the Corollaries to the Hahn-Banach Theorem page there exists an $F \in Y^*$ such that $F(y_0) = \delta$ and $F(y) = 0$ for all $y \in \overline{T(X)}$. So for every $x \in X$ we have that $T(x) \in T(X)$ and so $F(T(x)) = 0$. But observe that $0 \in Y^*$ is such that for all $x \in X$, $0(T(x)) = 0$. Therefore $T^*(F) = T^*(0)$. Since $T^*$ is injective, $F = 0$. But this is not true since $F(y_0) = \delta \neq 0$. So the assumption that $T(X)$ is not dense is false. Hence $T(X)$ is a dense subspace of $Y$. $\Leftarrow$ Let $T(X)$ be a dense subspace of $Y$. Let $f \in \ker (T^*)$. Then $f \in Y^*$ and $T^*(f) = 0$, i.e., $f \circ T = 0$. So for every $x \in X$ we have that $f(T(x)) = 0$. So $T(x) \in \ker f$ for every $x \in X$. Now since $T(X)$ is dense in $Y$ we have that for each $y \in Y$ there exists a sequence $(x_n)$ in $X$ for which $(T(x_n))$ converges to $y$. Since $f$ is continuous, $(f(T(x_n))$ converges to $f(y)$. But for $(f(T(x_n)) = (0)$. So $f(y) = 0$. Since this is true for every $y \in Y$ we have that $f = 0$. Therefore: Hence $T^*$ is injective. $\blacksquare$ |
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Equation
A mathematical
equation is an expression containing at least one variable (=unknown value) and an "equals sign" ( = ) with a mathematical expression on each side of it. [1] [2] [3] The equals sign says that both sides are exactly the same value. An equation can be as simple as x=0, or as hard as 4(3y^99)+76=42÷3x or harder.
There are two kinds of mathematical equations:
The kind of equation that is either true or false; these are also called identities. For example: [math]2 \cdot (x+4)=2x+8 \rightarrow true[/math] The kind of equation that lets you calculate the value of one or several variables. The equation is only true if the variable(s) have that value. For example: [math]2 \cdot x=8 \rightarrow x=4[/math]
The second kind is often used to solve problems in which you have to know the value of some variables. For example,
[math] \text{if}\, 2x = 8, \, x = \frac{8}{2} = 4. [/math] Types of equations An algebraic equation is an equation in which both sides are polynomials. These are further classified by degree: A Diophantine equation is an equation where the unknowns are required to be integers A differential equation is a functional equation involving derivatives of the unknown functions Related pages An equation is an equality that is true only for certain values of the variable. Trigonometric identities. Topics in trigonometry. "A statement of equality between two expressions. Equations are of two types, identitiesand conditional equations(or usually simply "equations")". « Equation», in Mathematics Dictionary, Glenn James et Robert C. James (éd.), Van Nostrand, 1968, 3 ed. 1st ed. 1948, p. 131. Une équation est une égalité entre deux expressions mathématiques, donc une formule de la forme A = B, où les deux membres A et B de l'équation sont des expressions où figurent une ou plusieurs variables, représentées par des lettres. ÉQUATION, mathématique - Encyclopædia Universalis |
The hash function, by definition of rate, processes messages in $r$ bit long blocks. Generate about $2^{c/2}$ ($r$ bit) message blocks until you a pair of messages, say $m$ and $m'$ such that you get a collision in the $c$ bits associated with capacity. That is $T_c(\pi(iv \oplus (0^c \| m)) = T_c(\pi(iv \oplus (0^c \| m'))$ where $T_c$ is a function that truncates* the $n$ bit output of $\pi$ to $c$ bits, $iv$ is the internal state of the hash function prior to absorbing any message, $0^c$ is a string of c zero bits, and $\|$ is the concatenation operator.
The internal state of two instances of the hash function after one absorbs $m$ and the other $m'$ now have identical capacity bits, but $\pi(iv \oplus (0^c \| m)) \neq \pi(iv \oplus (0^c \| m'))$ because $m \neq m'$ and $\pi$ is a permutation. Select any value $m_2$ for the second message block for the first instance of the hash function. Then select $m_2'$ for the second block of the second hash such that $m_2 \oplus m_2' = \pi(iv \oplus (0^c \| m)) \oplus \pi(iv \oplus (0^c \| m'))$.
Prior to absorbing the second message block both instances have just their capacity bits identical. You select the second message block to cancel the remaining $r$ unequal bits before the second time $\pi$ gets applied. Since the inputs to $\pi$ are the same, so are the outputs. Now you have a full internal collision. For this reason, just requesting the output of the hash function at this point will yield an output collision. But a collision in a hash function's internal state is more powerful than just a collision between outputs. You can keep generating collisions by choosing a different $m_2$ or by extending the input to both hash functions with new identical message blocks.
The reason you (may) need $2r$ bits of input is so you can cancel the difference in the non-capacity part of the sponge's state. A sponge based hash can be defined such that the last step isn't necessary. Note that once you have a collision in the capacity bits it's trivial to choose the next message block values such that you have a collision.
The algorithm can be defined such that instead of remembering all $n$ bits of output from $\pi$ it only stores the capacity bits. This changes the algorithm. The outputs of two hash functions that differ only by this detail will be different, but the collision resistance of the two is the same.
It may be defined such that performs state updates as in $h_{i} = \pi(h_{i-1} \oplus (0^c \| m_i))$. Where $h$ is the $n$ bit state of the hash function, $m_i$ is the $i$th r-bit long message block, and $0^c \| m_i$ is n bits because $|0^c| + |m_i| = c + r = n$.
Or it could also be defined to do updates as in $h_i = T_c(\pi((h_{i-1} \| m_i))$. Where $h_i$ is instead $c$ bits long.
Intuitively the difference is that the latter definition overwrites the new message block bits into the sponge state and the former XORs bits into the sponge state. Whether a sponge based hash function is defined one way or the other has no affect on its collision resistance.
TLDR: Just brute force the first message block such that the capacity bits match. Then just pick the second message block to cancel out the remaining r-bit difference.
* In this case discarding the last $r$ bits. I chose to assume in this answer that the hash function was defined such that the rate bits of the sponger were the last $r$ bits and the capacity bits were the first $c$ bits) |
STATISTICS
Category : 7th Class
Learning Objectives:
DATA:
A collection of numerical figures giving some particular type of information is called data.
Example: The marks obtained by 10 students in a class test (out of 50) are:
24, 43, 49, 38, 36, 36, 31, 40, 42, 15.
Raw data: Data obtained in original form is known as raw data. Data given in above example is raw data. ARRAY:
Arranging the data in ascending or descending order is known as array. Like in above example on arranging the above data in ascending order will be as:
15, 23, 31, 36, 36, 38, 40, 42, 43, 49.
It is known as array.
TABULATION OF DATA:
Arranging the data in form of table is known as tabulation of data.
OBSERVATION:
Each numerical figure in a data is called observation.
FREQUENCY OF AN OBSERVATION:
The number of times a particular observation occurs is called its frequency. In above example 36 occurs two times so its frequency is 2.
STATISTICS:
Statistics deals with the collection, presentation, analysis and interpretation of numerical data.
MEAN OF UNGROUPED DATA:
\[\text{Mean:}\,\,\frac{\text{Sum of the given observations}}{\text{No}\text{. of given observations}}\]
Example: Find the mean of the numbers
7, 6, 8, 9, 5, 4, 3, 7, 8, 2
Solution: Sum of the given numbers = 7 + 6 + 8 + 9 + 5 + 4 + 3 + 7 + 8 + 2 = 59
No. of given observation = 10
\[\text{Mean}=\frac{59}{10}=5.9\]
MEAN OF TABULATED DATA:
If the frequency of \[n\] observation \[{{x}_{1}},{{x}_{2}},{{x}_{3}}........{{x}_{n}}\] are \[{{f}_{1}},{{f}_{2}},{{f}_{3}}.........{{f}_{n}}\] respectively then
\[\text{Mean}=\frac{({{f}_{1}}{{x}_{1}}+{{f}_{2}}{{x}_{2}}+{{f}_{3}}{{x}_{3}}.....{{f}_{x}}{{x}_{n}})}{({{f}_{1}}+{{f}_{2}}+{{f}_{3}}.....{{f}_{n}})}=\frac{\Sigma ({{f}_{i}}\times {{x}_{i}})}{\Sigma {{f}_{i}}}\]
\[\Sigma =\]Greek letter showing summation
Example: The following table shows the weight of 15 workers in a factory.
60
63
66
72
75
4
5
3
1
2
Calculate the mean weight.
Solution: For calculating mean weight, we prepare following table.
60
4
240
63
5
315
66
3
198
72
1
72
75
2
150
\[\Sigma {{f}_{1}}=15\]
\[\Sigma {{f}_{1}}{{x}_{1}}=975\]
Example: Calculate the mean for the following data:
24
25
19
15
20
22
3
5
4
5
2
3
Solution: We may prepare the table given below:
15
5
5
75
19
4
9
76
20
2
11
40
22
3
14
66
24
3
17
72
25
5
22
125
\[N=\Sigma {{f}_{i}}=22\]
\[\Sigma ({{f}_{i}}\times {{x}_{i}})\]\[=454\]
Here, \[N=\Sigma {{F}_{i}}=22\]Which is even.
\[\text{Mean}=\frac{\Sigma ({{f}_{i}}\times {{x}_{i}})}{\Sigma {{f}_{i}}}=\frac{454}{22}=20.63\]
MODE
Putting the same observations together and counting them is not easy if the number of observations is large. In such cases we tabulate the data. Tabulation can begin by putting tally marks and finding the frequency, as we did in grouped data.
Look at the following example:
Following are the margins of victory in the football matches of a league.
1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 1, 1, 2, 3, 2, 6, 4, 3, 2, 1, 1, 4, 2, 1, 5, 3, 3, 2, 3, 2, 4, 2, 1, 2
Find the mode of this data.
Let us put the data in a tabular form:
1
\[\bcancel{|\,\,|\,\,|\,\,|} \,\,|\,\,|\,\,|\,\,|\]
9
2
\[\bcancel{|\,\,|\,\,|\,\,|} \bcancel{|\,\,|\,\,|\,\,|}|\,\,|\,\,|\,\,|\]
14
3
\[\bcancel{|\,\,|\,\,|\,\,|}\,\,\,|\,\,|\]
7
4
\[\bcancel{|\,\,|\,\,|\,\,|}\]
5
5
\[|\,\,|\,\,|\]
3
6
\[|\,\,|\]
2
Looking at the table, we can quickly say that 2 is the 'mode' since 2 has occurred the highest number of times.
Thus, the most frequent victory margin is 2 goals in the league.
MEDIAN:
After arranging the given data in ascending or descending order, the value of middle-most observation is called median of the data.
MEDIAN OF UNGROUPED DATA:
Arrange the data in ascending or descending order. If the total no of observation are n.
Case 1: When \[n\] is odd:
Median \[=\left( \frac{n+1}{2} \right)\text{th}\]observation value
Case 2: When \[n\] is even
Media\[=\frac{1}{2}\left\{ \frac{n}{2}th+\left( \frac{n}{2}+1 \right)\text{th}\,\text{observation}\,\text{value} \right\}\]
Example: The runs scored by 1 members of cricket teams
38, 29, 54, 36, 78, 82, 59, 38, 16, 40, 60
Find the median score.
Solution: On arranging the numbers in ascending order
16, 29, 36, 38, 38, 40, 54, 59, 60, 78, 82
Here n = 11 which is odd
\[\therefore \] Median score
\[=\frac{1}{2}(n+1)th\]term value
\[=\frac{1}{2}(11+1)=\frac{1}{2}\times 12=\]6th term value.
So median = 40.
Example: The weight of 8 students (in kg) are:
32, 38, 40, 30, 41, 35, 43, 30.
Find the median weight.
Solution: Arranging the weight in ascending order
30, 30, 32, 35, 38, 40, 41, 43.
Here \[n=8\]
\[\therefore \] Median weight
\[=\frac{1}{2}\left\{ \frac{\text{nth}\,\text{term}\,\text{value}}{\text{2}}+\left( \frac{n}{2}+1 \right)\text{th}\,\text{term}\,\text{value} \right\}\]
\[=\frac{1}{2}\left\{ \frac{8}{2}+\left( \frac{8}{2}+1 \right) \right\}\]
\[=\frac{1}{2}\]{4th term value + 5th term value}
\[=\frac{1}{2}\times (35+38)=\frac{1}{2}\times 73=36.5\,kg\]
Example: We have the following observations in ascending order of magnitude:
\[11,\,\,12,\,\,14,\,\,18,\,\,x+2,\,\,x+4,\,\,30,\,\,32,\,\,35,\,\,41\]
If the median is 24 find the value of\[x.\]
Solution:
Here n = 10 (even)
\[\text{Median}=\frac{\frac{n}{2}\text{th observation}+\left( \frac{n}{2}+1 \right)\text{th observation}}{2}\]
\[24=\frac{\frac{10}{2}\text{th observation}+\left( \frac{10}{2}+1 \right)\text{th observation}}{2}\]
\[24=\frac{\text{5th observation + 6th observation}}{2}\]
\[24=\frac{(x+2)+(x+4)}{2}\]
\[\Rightarrow \,\,24=\frac{2x+6}{2}\Rightarrow 24=x+3\]
\[\Rightarrow \,\,x=21\]
BAR GRAPHS
Two hundred students of 6th and 7th class were asked to name their favorite colour so as to decide upon what should be the colour of their school building. The results are shown in the following table. Represent the given data on a bar graph.
43
19
55
49
34
Answer the following questions with the help of the bar graph:
(i) Which is the most preferred colour and which is the least preferred?
(ii) How many colours are there in all? What are they?
Choose a suitable scale as follows:
Start the scale at 0. The greatest value in the data is 55, so end the scale at a value greater than 55, such as 60.
Use equal divisions along the axes, such as increments of 10. You know that all the bars would lie between 0 and 60. We choose the scale such that the length between 0 and 60 is neither too long nor too small. Here we take 1 unit for 10 students.
We then draw and label the graph as shown.
From the bar graph we conclude that
(i) Blue is the most preferred colour (Because the bar representing Blue is the tallest).
(ii) Green is the least preferred colour. (Because the bar representing Green is the shortest).
(iii) There are five colours. They are Red, Green, Blue, Yellow and Orange. (These are observed on the horizontal line)
A mathematics teacher wants to see whether the new technique of teaching she applied after the quarterly test was effective or not. She takes the scores of the 5 weakest children in the quarterly test (out of 25) and in the half yearly test (out of 25):
10
15
12
20
9
15
18
16
21
15
Since there seems to be a marked improvement in most of the students, the teacher decides that she should continue to use the new technique of teaching.
HISTOGRAM Example 1:
The frequency distribution of weights (in kg) of 40 persons of a locality is given.
Weight (in kg)
40–45
45–50
50–55
55–60
60–65
Frequency
4
12
13
6
5
Solution:
Construct a histogram to represent the above data.
PROBABILITY
There are situations in our life that are certain to happen, some that are impossible and some that may or may not happen. The situation that may or may not happen not happen has chance of happening.
We know that when a coin is thrown, it was possible outcomes, Head or Tail. For a dice we have 6 possible outcomes. We also know from experience that for a coin, Head or Tail is equally likely to be obtained. We say that the probability of getting Head or Tail is equal and is\[\frac{1}{2}\]for each.
For a dice, probability of getting either of 1, 2, 3, 4, 5, or 6 is equal. That is, for a dice there are 6 equally likely possible outcomes. We say each of 1, 2, 3, 4, 5, 6 has one-sixth\[\left( \frac{1}{6} \right)\]probability.
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The Weak Tensor Product of X⊗Y
Definition: Let $X$ and $Y$ be normed linear spaces. The Weak Tensor Norm (or Injective Tensor Norm) on $X \otimes Y$ is the function $w : X \otimes Y \to [0, \infty)$ defined for all $u = \sum_{i=1}^{m} x_i \otimes y_i \in X \otimes Y$ by $\displaystyle{w(u) = \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{i=1}^{m} f(x_i)g(y_i) \right | \right \}}$. The Weak Tensor Product (or Injective Tensor Product) of $X$ and $Y$ denoted $X \otimes_w Y$ is the completion of $(X \otimes Y, w)$ in $\mathrm{BL}(X^*, Y^*, \mathbf{F})$. Sometimes the weak/injective tensor norm is denoted by $\epsilon(x)$ and sometimes the weak/injective tensor product of $X$ and $Y$ is denoted by $X \widehat{\otimes}_{\epsilon} Y$.
The following proposition tells us what the weak tensor product of $x \otimes y$ ($x \in X$, $y \in Y$) is.
Proposition 1: Let $X$ and $Y$ be normed linear spaces. Then $w(x \otimes y) = \| x \| \| y \|$ for all $x \in X$ and for all $y \in Y$. Proof:Let $x \in X$, $y \in Y$. Then:
Before we move on, we should indeed verify that the weak tensor norm is a norm on $X \otimes Y$.
Proposition 1: Let $X$ and $Y$ be normed linear spaces. Then the weak tensor norm on $X \otimes Y$ is a norm. Proof:There are three things to prove. 1. Showing that $w(u) = 0$ if and only if $u = 0$:Suppose that $w(u) = 0$. Then So for all $f \in X^*$ and $g \in Y^*$ with $\| f \| \leq 1$, $\| g \| \leq 1$ we have that $\left | \sum_{i=1} (x_i \otimes y_i)(f, g) \right | \left | u(f, g) \right | = 0$. So $u(f, g) = 0$ for all $f \in X^*$, $g \in Y^*$ with $\| f \| \leq 1$, $\| g \| \leq 1$. Note that if $f \in X^*$ with $\| f \| > 1$ and $g \in Y^*$ with $\| g \| > 1$ then from above we have that: Since $\| f \|, \| g \| > 1$, the above equation shows us that $u(f, g) = 0$ for all $f \in X^*$ and all $g \in Y^*$. Therefore $u = 0$. On the otherhand, suppose that $u = 0$. Then $u = 0_X \otimes 0_Y$ and is such that $w(u) = w(0_X \otimes 0_Y) = \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ 0_X \otimes 0_Y (f, g) \right \} = 0$. 2. Showing that $w(\alpha u) = |\alpha| w(u)$:Let $u \in X \otimes Y$ with $u = \sum_{i=1}^{m} x_i \otimes y_i$ and let $\alpha \in \mathbf{F}$ so that $\alpha u = \alpha \sum_{i=1}^{m} x_i \otimes y_i$. Then: 3. Showing that $w(u + v) \leq w(u) + w(v)$:Let $u, v \in X \otimes Y$ with $u = \sum_{i=1}^{m} x_i \otimes y_i$ and $v = \sum_{j=1}^{n} x_j' \otimes y_j'$. Then $u + v = \sum_{i=1}^{m} x_i \otimes y_i + \sum_{j=1}^{n} x_j' \otimes y_j'$ and: So we conclude that $w$ is a norm on $X \otimes Y$. $\blacksquare$ |
Is there some analogous of the Central limit theorem for discrete uniforms and discrete normal distributions?
To be more specific, let's say we have identical and independent random random variables $U_i$ uniform on a given integer interval $[0, k]\cap \mathbb{Z}$, for some $k \in \mathbb{N}$, $k \ge 2$.
Then, we can define the distribution $S_n := \sum_{i=1}^n U_i$.
Let $f_{c, s}(x) = \exp(-(x-c)/(2s^2))$, then the discrete Gaussian distribution centered on $c$ with variance $s^2$, which I will denote by $G_{c, s^2}$, assigns to each $x \in \mathbb{Z}$ the probability $$Pr[G_{c,s} = x] = \frac{f_{c, s}(x)}{\sum_{\forall y \in \mathbb{Z}}f_{c, s}(y)}.$$ That is, the "mass" of $x$ normalized by the "mass" of the whole integers.
I would like to know if there is some result that relates $S_n$ with the discrete normal distribution. For example, will $S_n$ converge to $G_{c, s^2}$ as $n$ increases?
I know that the mean of each $U$ is $\mu := k/2$ and the variance is $\sigma^2 := (k+1)^2 / 12$, so I made some tests using $G_{c, s^2}$ with $c = n\mu$ and $s = \sqrt{n}\sigma$, but it seems that $S_n$ does not converge. But it is hard to tell by the graphs... |
Answer
79 total shark attacks this year (16 more)
Work Step by Step
We know that this year there was an increase in shark attacks by $25.4\%$ (out of 63 last year). We can calculate this amount with a ratio: $\displaystyle \frac{25.4}{100}=\frac{x}{63}$ Cross-multiply: $25.4*63=x*100$ $1600.2=100x$ $x=16.002$ $x\approx 16$ Thus this year there were 16 more shark attacks. Therefore, the total was: $16+63=79$ |
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TeXing
I'm keen on turning up articles of EoM into better appearance by rewriting formulas and math symbols in TeX.
Now there are
366 (out of 15,890) articles with Category:TeX done tag.
$\quad \rightarrow \quad$ $\sum_{n=1}^{\infty}n!z^n$ Just type $\sum_{n=1}^{\infty}n!z^n$. Today You may look at Category:TeX wanted. How to Cite This Entry:
Nikita2.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Nikita2&oldid=29691 |
Given a Quantum field theory, for a scalar field $\phi$ with generic action $S[\phi]$, we have the generating functional $$Z[J] = e^{iW[J]} = \frac{\int \mathcal{D}\phi e^{i(S[\phi]+\int d^4x J(x)\phi(x))}} {\int \mathcal{D}\phi e^{iS[\phi]}}.$$
The one-point function in the presence of a source $J$ is.
$$\phi_{cl}(x) = \langle \Omega | \phi(x) | \Omega \rangle_J = {\delta\over\delta J}W[J] = \frac{\int \mathcal{D}\phi \ \phi(x)e^{i(S[\phi]+\int d^4x J(x)\phi(x))}} {\int \mathcal{D}\phi \ e^{i(S[\phi]+\int d^4x J(x)\phi(x))}}.$$
The effective Action is defined as the Legendre transform of $W$
$$\Gamma[\phi_{cl}]= W[J] -\int d^4y J(y)\phi_{cl}(y),$$ where $J$ is understood as a function of $\phi_{cl}$.
That means we have to invert the relation $$\phi_{cl}(x) = {\delta\over\delta J}W[J]$$ to $J = J(\phi_{cl})$.
How do we know that the inverse $J = J(\phi_{cl})$ exists? And does the inverse exist for every $\phi_{cl}$? Why? |
In electromagnetism, which is an abelian gauge theory, we have the nice fact that all of the components of $$ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu $$ are gauge invariant quantities. We can equivalently talk about the field in terms of $\mathbf{A}_{\perp}$ and $\mathbf{E}_{||}$, the solenoidal part of the vector potential and the irrotational part of the electric field (respectively), as being a compete set of gauge invariant quantities that fully determines the state of the field everywhere without degeneracy.
When we move to non-abelian Yang-Mills theory the field strength components \begin{align} F_{\mu\nu} & = \frac{i}{g}\left[D_\mu,\, D_\nu\right] \\ & = \partial_\mu A_\nu - \partial_\nu A_\mu - ig \left[A_\mu,\, A_\nu\right] \\ & = T^a \left(\partial_\mu A^a_\nu - \partial_\nu A^a_\mu +gf^{abc} A^b_\mu A^c_\nu\right) \end{align} become matrices in the group space ($T^a$ are generators of transformations of the group). Thus the elements of the field strength indexed by space-time, $F_{\mu\nu}$, are no longer gauge invariant but gauge covariant — transforming as a rank two tensor in whatever representation of the gauge group the the generators are in.
When I want to know the group invariant parts of a matrix I think "eigenvalues", at least when the group is defined by preserving an inner product on some vector space (as $\operatorname{SU}(N)$ is). In other words, it seems to me like the physical information will be contained in the gauge invariant parts of $F_{\mu\nu}$ and finding the eigenvalues sounds like the way to go about it.
Has this been done? It seems unlikely that a single gauge transformation could simultaneously diagonalize all of the $F_{\mu\nu}$ at any given point in space-time, what are the consequences of this (if it's true)? Finally, would the eigenvalues of $F_{\mu\nu}$ also be independent of which representation the $T^a$ are in (i.e. would finding the eigenvalues in the defining representation give the same result as any other representation), assuming the representations all have the same normalization condition (e.g. $\operatorname{Tr}(T^a T^b)= \delta_{ab} / 2$)? |
In the Wilsonian picture of renormalization, a quantum field theory is defined to have degrees of freedom only up to an energy scale $\Lambda$. The results of low-energy experiments shouldn't change as we lower $\Lambda$, as long as we change the parameters in the action to compensate for the loss of the high-energy modes. Quantitatively, we have $$S_{\Lambda'}[\phi] = S_{\Lambda}[\phi] + \log \int_{\tilde{\phi} \in (\Lambda', \Lambda)} \mathcal{D} \phi \, e^{iS_{\Lambda}[\phi]}$$ which expresses the fact that the partition function is invariant, $$\int_{\tilde{\phi} \in (0, \Lambda')} \mathcal{D}\phi\, e^{i S_{\Lambda'}[\phi]} = \int_{\tilde{\phi} \in (0, \Lambda)} \mathcal{D}\phi\, e^{i S_{\Lambda}[\phi]}.$$ Then we have scale-dependent couplings $g(\Lambda)$.
On the other hand, in 'continuum RG' techniques such as dimensional regularization, there is no high-energy cutoff. Instead, we perform some mathematical trick to assign finite values to the unbounded integrals. Sometimes, but not all of the time, the trick we use will introduce an arbitrary mass scale $\mu$, yielding scale-dependent couplings $g(\mu)$.
There are lots of questions on this site about how these two approaches are related, with the usual answer being that there's no relation whatsoever. But I suspect the two views are really the same!
Consider some loop integral in an unregularized theory, $$I = \int d^4 p\, (\text{stuff}) \sim \int_0^\infty p^3\, dp \, (\text{stuff)}.$$ In the Wilsonian picture, the integral would be finite by a hard cutoff, $$I \sim \int_0^\Lambda p^3\, dp \, (\text{stuff)}.$$ In dimensional regularization, we instead work in arbitrary dimension, giving $$I \sim \int_0^\infty p^{d-1} dp \,(\text{stuff}).$$ Then if we take $d < 4$, there are more degrees of freedom with lower energies, and less degrees of freedom with higher energies. If we say we have equality at energy $\mu$ (and this must be true, by dimensional analysis), then dimensional regularization is just doing a smoother version of the hard Wilsonian cutoff, with $\mu \sim \Lambda$!
To clarify, here's a sketch of how the number of degree of freedom at some energy changes in each of these schemes.
Is this picture valid? Are all continuum RG methods secretly Wilsonian? Are there any references where people discuss these ideas? |
Dense and Nowhere Dense Sets in Topological Spaces Examples 1
Recall from the Dense and Nowhere Dense Sets in a Topological Space page that if $(X, \tau)$ is a topological space than a set $A \subseteq X$ is said to be dense in $X$ if the intersection of $A$ with any $U \in \tau$ is nonempty, that is, for all $U \in \tau$:(1)
Furthermore, we said that $A$ is nowhere dense in $X$ if the interior of the closure of $A$ is empty, i.e.,:(2)
We will now look at some example exercises regarding dense and nowhere dense sets in topological spaces.
Example 1 Consider the topological space $(\mathbb{R}^2, \tau)$ where $\tau$ is the usual topology formed by open disks. Consider the subset $A = \{ (x, y) \in \mathbb{R}^2 : x \in \mathbb{N}, y \in \mathbb{R} \} \subseteq \mathbb{R}^2$. Is this set dense? Is this set nowhere dense?
We claim that the set $A$ is NOT dense. To show this, we must find an open neighbourhood $U$ of $\mathbb{R}^2$ whose intersection with $A$ is empty.
Consider the following open disk in $\mathbb{R}^2$:(3)
This is a disk centered at the point $\left ( \frac{1}{2}, 0 \right \}$ and with radius $\frac{1}{2}$ that does not intersect $A$ as seen in the diagram below:
Therefore $A \cap U = \emptyset$ so $A$ is not dense.
So, let's determine whether $A$ is nowhere dense. Note that $A$ itself is a closed set since $A^c$ is open. Therefore $\bar{A} = A$. Furthermore, $\mathrm{int}(\bar{A}) = \emptyset$, so $A$ is nowhere dense.
Example 2 Consider the topological space $(\mathbb{R}^2, \tau)$ where $\tau$ is the usual topology formed by open disks. Consider the subset $A = \{ (x, y) \in \mathbb{R}^2 : xy \neq 0 \} \subseteq \mathbb{R}^2$. Is this set dense? Is this set nowhere dense?
Note that $A$ is the set of points $(x, y) \in \mathbb{R}^2$ which satisfy $xy \neq 0$. Note that $xy \neq 0$ if and only if $x \neq 0$ and $y \neq 0$. If $x = 0$ then the point $(x, y) = (0, y)$ lies on the $y$-axis, and if $y = 0$ then the point $(x, y) = (x, 0)$ lies on the $x$-axis. Therefore $A$ is equal to $\mathbb{R}^2$ with the $x$-axis and $y$-axis removed.
We claim that this set is dense. To show this, let $U$ be any open disk in $\mathbb{R}^2$. If this disk does not intersect the $x$ or $y$ axes then $U$ is fully contained in $A$ and so $A \cap U \neq \emptyset$. If not, then this open disk intersects either axis at either one point or an infinite number of points. In either case, the open disk will still contain elements of $A$ and so $A \cap U \neq \emptyset$ for every open disk $U$ in $\mathbb{R}^2$.
Since these open disks form a basis of the topology on $\mathbb{R}^2$ we see that the intersection of $A$ with any open set in $\mathbb{R}^2$ is nonempty and set $A$ is a dense set.
Furthermore, this set is clearly not nowhere dense. Note that $A$ is an open set and that $\bar{A} = \mathbb{R}^2$. Furthermore, $\mathrm{int}(\bar{A}) = \mathbb{R}^2 \neq \emptyset$. |
This note is based on Professor Noam Elkies’ talk at Carnegie Mellon University on December 2, 2014.
A classical mathematical analysis problem, also known as the Basel problem, first posed by Pietro Mengoli in 1644 and solved by Leonhard Euler in 1735, asks the precise sum in closed form of the infinite series \sum_{n=1}^\infty \frac{1}{n^2} = \lim_{n\to\infty}\left(\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2}\right).
Here we present an attack to the Basel problem using the Poisson summation formula.
Poisson Summation Formula
The Poisson summation formula states that \sum_{n\in\mathbb{Z}}f(n) = \sum_{k\in\mathbb{Z}}\hat{f}(k), where f: \mathbb{R}\to\mathbb{R} is a “nice” function and \hat{f}(\omega):=\int_\mathbb{R}f(x)e^{-2\pi i\omega x}dx is the Fourier transformation of f.
For the sake of completeness, we are going to prove the Poisson summation formula and specify how “nice” the function should be for the formula to work.
Proof. Let R:\mathbb{R}\to\mathbb{R} be defined by R(t) = \sum_{n\in\mathbb{Z}}f(n+t). Because R(t+m) = R(t) for all m\in\mathbb{Z}, and so R is a periodic function, we can expand R(t) in the Fourier series R(t) = \sum_{k\in\mathbb{Z}}a_ke^{2\pi i kt}, where \begin{aligned}a_k & = \int_0^1R(t)e^{-2\pi ikt}dt \\ & = \int_0^1\sum_{n\in\mathbb{Z}}f(n+t)e^{-2\pi ikt}dt \\ & = \sum_{n\in\mathbb{Z}}\int_0^1f(n+t)e^{-2\pi ikt}dt \\ & = \sum_{n\in\mathbb{Z}}\int_n^{n+1}f(t)e^{-2\pi ikt}dt \\ & = \int_\mathbb{R}f(t)e^{-2\pi ikt}dt.\end{aligned} which equals the value at -k of the Fourier transform \hat{f} of f. We conclude that \sum_{n\in\mathbb{Z}}f(n) = R(0)=\sum_{k\in\mathbb{Z}}a_k = \sum_{k\in\mathbb{Z}}\hat{f}(k). [qed] Remark. In order to use Fourier series for R, we need R\in C^1 to ensure that its Fourier series expansion \sum_{n\in\mathbb{Z}}a_ke^{2\pi ikt} converges to R(t) pointwise (and uniformly in fact). This allows us to evaluate R at 0 and equate it with \sum_{k\in\mathbb{Z}} a_k. On the other hand, we switched the integration and the infinite summation \int_0^1\sum_{n\in\mathbb{Z}}f(n+t)e^{-2\pi ikt}dt = \sum_{n\in\mathbb{Z}}\int_0^1f(n+t)e^{-2\pi ikt}dt, in the computation. The Fubini–Tonelli theorem says that this is legit if \int_0^1\sum_{n\in\mathbb{Z}}\left|f(n+t)e^{-2\pi ikt}\right|dt = \int_0^1\sum_{n\in\mathbb{Z}}\left|f(n+t)\right|dt < \infty or \sum_{n\in\mathbb{Z}}\int_0^1\left|f(n+t)e^{-2\pi ikt}\right|dt = \sum_{n\in\mathbb{Z}}\int_n^{n+1}|f(x)|dx = \int_\mathbb{R}|f(x)|dx < \infty. Application to Basel Problem
Take f(x) = \frac{1}{x^2+c^2}, where c > 0 is a constant. For such f, R(t) = \sum_{n\in\mathbb{Z}}f(n+t) defined above converges uniformly. As each f(n+t) is continuous, so is R. As \sum_{n\in\mathbb{Z}}f'(n+t) = \sum_{n\in\mathbb{Z}}\frac{-2(t+n)}{((t+n)+c^2)^2} converges uniformly, we have that R'(t) = \sum_{n\in\mathbb{Z}}f'(n+t), and moreover, as f'(n+t) is continuous, so is R'. Now R is continuously differentiable, and so we can use Fourier series for R. On the other hand, since f is absolutely integrable, we can switch integral with summation. Therefore we can apply the Poisson summation formula.
Its Fourier transform is, by Cauchy’s integral formula, \hat{f}(\omega) = \int_\mathbb{R}\frac{1}{x^2+c^2}e^{-2\pi i\omega x}dx = \frac{\pi}{c}e^{-2\pi c|\omega|}. By the Poisson summation formula, we obtain \sum_{n\in\mathbb{Z}}\frac{1}{n^2+c^2} = \sum_{k\in\mathbb{Z}}\frac{\pi}{c}e^{-2\pi c|k|} = \frac{\pi}{c}\frac{e^{2\pi c}+1}{e^{2\pi c}-1}, and so \sum_{n=1}^\infty \frac{1}{n^2+c^2} = \frac{1}{2}\left(\frac{\pi}{c}\frac{e^{2\pi c}+1}{e^{2\pi c}-1}-\frac{1}{c^2}\right).
Sending c to 0^+, the left hand side converges to \sum_{n=1}^\infty\frac{1}{n^2} and the right hand side is equal to\begin{aligned} & \frac{1}{2}\left(\frac{\pi}{c}\frac{2+2\pi c+2\pi^2 c^2 + \ldots}{2\pi c + 2\pi^2 c^2 + 4\pi^3c^3/3 + \ldots}-\frac{1}{c^2}\right) \\ = & \frac{1}{2}\left(\frac{(2 + 2\pi c+2\pi^2c^2 + \ldots) - (2 + 2\pi c + 4\pi^2c^2/3 + \ldots)}{2c^2 + 2\pi c^3 + 4\pi^2c^4/3 + \ldots}\right) \\ = & \frac{1}{2} \frac{2\pi^2c^2 - 4\pi^2c^2/3 + \ldots}{2c^2 + \ldots} \end{aligned}
and converges to \frac{\pi^2}{6} as c\to 0^+. |
I know that we should express complex numbers generally in the standard form
$$a+bi:a,b\in\mathbb{R}$$
Like $4+5i-2=2+5i$.
But how do I express complex numbers like $e^{-i\pi/2}$ or $i+e^{2\pi i}$?
Thank you!
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I know that we should express complex numbers generally in the standard form
$$a+bi:a,b\in\mathbb{R}$$
Like $4+5i-2=2+5i$.
But how do I express complex numbers like $e^{-i\pi/2}$ or $i+e^{2\pi i}$?
Thank you!
Hint Use the Euler's Formula$$ e^{i\theta}=\cos(\theta)+i\sin(\theta).$$For exemple, $$e^{-i\pi/2} =\cos(-\pi/2)+i\sin(-\pi/2)=0+i\cdot (-1)=-i $$For reverse process $x+iy= r\cdot e^{i\theta}$ use the formulas$$r=\sqrt[\,2]{x^2+y^2}\qquad \mbox{and } \qquad \tan(\theta)=\frac{y}{x},\quad -\frac{\pi}{2}<\theta<\frac{\pi}{2}$$
Well, every complex number has a modulus and an angle which corresponds to $r, \theta$ in $re^{i\theta}$. In short, for a complex number $z=a+ib$, we have $r=\sqrt{a^2+b^2}$, $\theta=\tan^{-1}\frac{b}{a}$. To change it back, all you have to do is to use $a=r\cos\theta, b=r\sin\theta$.
In your question, you can choose to change either the polar form into the Cartesian or vice versa and you will get your answer.
Recall Euler's Formula:
$e^{ix}=\cos x + i \sin x$.
By applying this formula to the given problems we obtain, |
This is a side question which is more motivated by teaching than research.
First, I am trying to convince myself that sequences appear before series (as numerical approximations to "interesting" quantities; on the other hand, decimal expansions -- especially infinite -- are more likely to be series).
Secondly, is it natural for sequences to be placed prior to series in a calculus course?
So, which one is more original, a sequence or a series?
After-dinner edit. We define a sequence to be... a function mapping the positive numbers to a set(?). We define a series to be...a formal infinite sum $\sum_{n=1}^\infty a_n$(?). Tell me what is your way to "define" these two guys, I do not believe they are very related.
There are no doubts that it is easier to define
convergence of series via convergence of sequences, but it does not imply their "primogeniture".The notion of Cauchy sequence is an elegant way to build the apparatus of not only sequences but also of real numbers; as such it canserve as a definition of series: a series is a formal infinite sum $\sum_{n=1}^\infty a_n$, and it is called a convergent series iffor any $\epsilon>0$ there exists an $N=N(\epsilon)$ such that for any $m>n>N$ the sum $|a_n+\dots+a_m|<\epsilon$. The real numbersthen are nothing but representatives of equivalence classes of convergent series. (I have no desire here to expand all the details.)A sequence $b_n$ is convergent when the corresponding series $\sum_{n=1}^\infty a_n$ where $a_1=b_1$ and $a_n=b_n-b_{n-1}$ for $n\ge2$converges. It would be honest to say that, besides the trivialities like "algebra of limits", the techniques for investigating convergenceof series are quite independent from that of sequences. And it does not sound impossible to do series prior to sequences.
Historically, all these convergence/divergence issues were purely intuitive for both sequences and series, and they both were on the market for many centuries. I ask whether their exists an overwhelming historical support to the notion of sequence to lead. |
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Let's look at some examples of feasibility relations!
Feasibility relations work between preorders, but for simplicity suppose we have two posets \(X\) and \(Y\). We can draw them using Hasse diagrams:
Here an arrow means that one element is less than or equal to another: for example, the arrow \(S \to W\) means that \(S \le W\). But we don't bother to draw all possible inequalities as arrows, just the bare minimum. For example, obviously \(S \le S\) by reflexivity, but we don't bother to draw arrows from each element to itself. Also \(S \le N\) follows from \(S \le E\) and \(E \le N\) by transitivity, but we don't bother to draw arrows that follow from others using transitivity. This reduces clutter.
(Usually in a Hasse diagram we draw bigger elements near the top, but notice that \(e \in Y\) is not bigger than the other elements of \(Y\). In fact it's neither \(\ge\) or \(\le\) any other elements of \(Y\) - it's just floating in space all by itself. That's perfectly allowed in a poset.)
Now, we saw that a
feasibility relation from \(X\) to \(Y\) is a special sort of relation from \(X\) to \(Y\). We can think of a relation from \(X\) to \(Y\) as a function \(\Phi\) for which \(\Phi(x,y)\) is either \(\text{true}\) or \(\text{false}\) for each pair of elements \( x \in X, y \in Y\). Then a feasibility relation is a relation such that:
If \(\Phi(x,y) = \text{true}\) and \(x' \le x\) then \(\Phi(x',y) = \text{true}\).
If \(\Phi(x,y) = \text{true}\) and \(y \le y'\) then \(\Phi(x,y') = \text{true}\).
Fong and Spivak have a cute trick for drawing feasibility relations: when they draw a blue dashed arrow from \(x \in X\) to \(y \in Y\) it means \(\Phi(x,y) = \text{true}\). But again, they leave out blue dashed arrows that would follow from rules 1 and 2, to reduce clutter!
Let's do an example:
So, we see \(\Phi(E,b) = \text{true}\). But we can use the two rules to draw further conclusions from this:
Since \(\Phi(E,b) = \text{true}\) and \(S \le E\) then \(\Phi(S,b) = \text{true}\), by rule 1.
Since \(\Phi(S,b) = \text{true}\) and \(b \le d\) then \(\Phi(S,d) = \text{true}\), by rule 2.
and so on.
Puzzle 171. Is \(\Phi(E,c) = \text{true}\) ? Puzzle 172. Is \(\Phi(E,e) = \text{true}\)?
I hope you get the idea! We can think of the arrows in our Hasse diagrams as
one-way streets going between cities in two countries, \(X\) and \(Y\). And we can think of the blue dashed arrows as one-way plane flights from cities in \(X\) to cities in \(Y\). Then \(\Phi(x,y) = \text{true}\) if we can get from \(x \in X\) to \(y \in Y\) using any combination of streets and plane flights!
That's one reason \(\Phi\) is called a feasibility relation.
What's cool is that rules 1 and 2 can also be expressed by saying
$$ \Phi : X^{\text{op}} \times Y \to \mathbf{Bool} $$is a monotone function. And it's especially cool that we need the '\(\text{op}\)' over the \(X\). Make sure you understand that: the \(\text{op}\) over the \(X\) but not the \(Y\) is why we can drive
to an airport in \(X\), then take a plane, then drive from an airport in \(Y\).
Here are some ways to lots of feasibility relations. Suppose \(X\) and \(Y\) are preorders.
Puzzle 173. Suppose \(f : X \to Y \) is a monotone function from \(X\) to \(Y\). Prove that there is a feasibility relation \(\Phi\) from \(X\) to \(Y\) given by
$$ \Phi(x,y) \text{ if and only if } f(x) \le y .$$
Puzzle 174. Suppose \(g: Y \to X \) is a monotone function from \(Y\) to \(X\). Prove that there is a feasibility relation \(\Psi\) from \(X\) to \(Y\) given by
$$ \Psi(x,y) \text{ if and only if } x \le g(y) .$$
Puzzle 175. Suppose \(f : X \to Y\) and \(g : Y \to X\) are monotone functions, and use them to build feasibility relations \(\Phi\) and \(\Psi\) as in the previous two puzzles. When is
$$ \Phi = \Psi ? $$
To read other lectures go here. |
Short answer
He's basically making a parallel between a forward variance trade and a futures trade. In both cases you should have that the underlying quotes are martingales in the absence of arbitrage.
Long(er) answer
Under the physical measure $\Bbb{P}$, an arbitrage is a (self-financing) trading strategy $V$ - or rather the value of a portfolio implementing this strategy - for which there exists a time $T > 0$ such that$$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{P}-\text{a.s. and } \Bbb{P}(V_T \ne 0) > 0$$
Suppose you define an
equivalent probability measure $\Bbb{Q}\equiv\Bbb{P}$. Since by definition, both measures agree on null events, our arbitrage definition translates to$$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{Q}-\text{a.s. and } \Bbb{Q}(V_T \ne 0) > 0 \tag{A}$$
Notice that if $\Bbb{Q}$ is further a
martingale measure, that is if $(V_t)_{t\geq0}$ emerges as a $\Bbb{Q}$-martingale:$$ V_0 = \Bbb{E}_0^\Bbb{Q} [ V_T ] $$then $(A)$ will never happen. This explains the central role of equivalent martingale measures in arbitrage pricing theory.
Putting that back into context, you've managed to identify a (self-financing) strategy (i.e. buying and selling forward variance swaps), which at no cost ($V_t=0$), allows you to earn a quantity $$V_{t'} = (T_2-T_1) \left( \hat{\sigma}_{VS,T_1T_2}^2(t') - \hat{\sigma}_{VS,T_1T_2}^2(t)\right)$$
Based on what we've said earlier, in the absence of arbitrage, there should exist a measure $\Bbb{Q} \equiv \Bbb{P}$ such that$$ \Bbb{E}^\Bbb{Q}_{t}[V_{t'}] = V_t$$hence, using the definitions of $V_t$ and $V_{t'}$,$$ \Bbb{E}^\Bbb{Q}_{t}\left[ \hat{\sigma}_{VS,T_1T_2}^2(t') \right] = \hat{\sigma}_{VS,T_1T_2}^2(t) $$hence forward variance swap quotes are martingales. Assuming a
continuous paths process (= in a diffusive setting), by the martingale representation theorem we should then have$$ \hat{\sigma}_{VS,T_1T_2}^2(t) = ... dW_t^\Bbb{Q} $$hence no pricing drift under $\Bbb{Q}$. |
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Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV
(Springer, 2015-01-10)
The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (|y|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ...
Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV
(Springer Berlin Heidelberg, 2015-04-09)
The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of pion, kaon and proton production in proton–proton collisions at √s = 7 TeV
(Springer, 2015-05-27)
The measurement of primary π±, K±, p and p¯¯¯ production at mid-rapidity (|y|< 0.5) in proton–proton collisions at s√ = 7 TeV performed with a large ion collider experiment at the large hadron collider (LHC) is reported. ...
Two-pion femtoscopy in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(American Physical Society, 2015-03)
We report the results of the femtoscopic analysis of pairs of identical pions measured in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV. Femtoscopic radii are determined as a function of event multiplicity and pair ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV
(American Physical Society, 2015-06)
The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ...
Centrality dependence of high-$p_{\rm T}$ D meson suppression in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Springer, 2015-11)
The nuclear modification factor, $R_{\rm AA}$, of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$ and ${\rm D^{*+}}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at a centre-of-mass ...
K*(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2015-02)
The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ... |
A qbit is a two-element vector:
$|\psi\rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$
where $\alpha, \beta \in \mathbb{C}$ and $|\alpha|^2 + |\beta|^2 = 1$, a property called the 2-norm.
We have two important qbit values which we associate with the classical bits 0 and 1:
$|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $|1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
A qbit is in superposition when it is in any state other than $|0\rangle$ or $|1\rangle$. When people say superposition means the qbit "is both 0 and 1 at the same time", what they mean is it is a
linear combination of 0 and 1:
$|\psi\rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \alpha\begin{bmatrix} 1 \\ 0 \end{bmatrix} + \beta\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \alpha|0\rangle + \beta|1\rangle$
We operate on qbits with quantum logic gates, which must always be equivalent in action to unitary matrices. For example, here's the bitflip or "not" operator:
$X|0\rangle = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = |1\rangle$
We represent the state of multiple qbits through their tensor product:
$\begin{bmatrix} a \\ b \end{bmatrix} \otimes \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix}$
Note that the size of the product state vector grows exponentially with the number of qbits:
$\begin{bmatrix} a \\ b \end{bmatrix} \otimes \begin{bmatrix} c \\ d \end{bmatrix} \otimes \begin{bmatrix} e \\ f \end{bmatrix} = \begin{bmatrix} ace \\ acf \\ ade \\ adf \\ bce \\ bcf \\ bde \\ bdf \end{bmatrix}$
Ordinarily this doesn't matter very much, but sometimes qbits become entangled, which means their product state cannot be factored into the tensor product of individual qbits:
$\begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix} \ne \begin{bmatrix} a \\ b \end{bmatrix} \otimes \begin{bmatrix} c \\ d \end{bmatrix}$
When this happens, applying a quantum logic gate to $n$ qbits (which takes the form of a $2^n \times 2^n$ unitary matrix) can manipulate $2^n$ pieces of information (the elements of the product state, called amplitudes) in one shot. However, there's a bottleneck: you can only ever get $n$ bits of information out of an $n$-qbit system (through measurement), even though the $n$-qbit system contains $2^n$ pieces of information. This bottleneck means we cannot "solve problems by trying all possible solutions simultaneously" or some other such nonsense you might have read, and quantum computers can only provide speedup in very specific problem domains.
I gave a lecture where I expand on these concepts here. |
Weak Convergence in Hilbert Spaces
Recall that if $X$ is a normed linear space then a sequence of points $(x_n)$ in $X$ is said to weakly converge to $x \in X$ if for every $f \in X^*$ we have that:(1)
Recall from The Riesz Representation Theorem for Hilbert Spaces page that if $H$ is a Hilbert space then $f_g : H \to \mathbb{R}$ defined for each $h \in H$ by $f_g(h) = \langle h, g \rangle$ is bounded linear functional on $H$. Furthermore, the Riesz representation theorem for Hilbert spaces says that every bounded linear functional on $H$ is of this form. That is, if $f \in H^*$ then there exists a $g \in H$ such that for every $h \in H$;(2)
Therefore, we have the following characterization for weak convergence in a Hilbert space.
Proposition 1: Let $H$ be a Hilbert space. Then a sequence $(x_n)$ in $H$ weakly converges to $x \in H$ if for all $g \in H$, $\displaystyle{\lim_{n \to \infty} \langle x_n, g \rangle = \langle x, g \rangle}$. |
Criterion for a Quotient Group to be Abelian
Criterion for a Quotient Group to be Abelian
Theorem 1: Let $G$ be a group and let $H$ be a normal subgroup of $G$. Then $G / H$ is abelian if and only if for all $g_1, g_2 \in G$ we have that $[g_1, g_2] := g_1^{-1}g_2^{-1}g_1g_2\in H$.
The element $[g_1, g_2] = g_1^{-1}g_2^{-1}g_1g_2$ is called the commutator of $g_1$ and $g_2$. Thus Theorem 1 says that if $G$ is a group and $H$ is a normal subgroup of $G$ then $G/H$ is abelian if and only if $H$ contains all commutators of pairs of elements from $G$. (1) Proof: $\Rightarrow$ Suppose that $G/H$ is abelian. Let $g_1, g_2 \in G$. Then $g_1H, g_2H \in G/H$, and since $G/H$ is abelian we have that:
\begin{align} \quad (g_1H)(g_2H) &= (g_2H)(g_1H) \\ \end{align}
(2) Since $(g_1H)(g_2H) = (g_1g_2H)$ and since $(g_2H)(g_1H) = (g_2g_1H)$, we have from above that:
\begin{align} \quad g_1g_2H &= g_2g_1H \\ \quad g_2^{-1}g_1g_2H &= g_1H \\ g_1^{-1}g_2^{-1}g_1g_2H &= H \end{align}
Since $H$ is a group, $e \in H$. So $g_1^{-1}g_2^{-1}g_1g_2 \in H$, which holds true for all $g_1, g_2 \in G$. (3) $\Leftarrow$ Suppose that $g_1^{-1}g_2^{-1}g_1g_2 \in H$ for all $g_1, g_2 \in G$. Let $g_1H, g_2H \in G/H$. Then $(g_2g_1)^{-1}g_1g_2 \in H$ for all $g_1, g_2 \in G$. Therefore:
\begin{align} \quad (g_1g_2H) = (g_2g_1H) \end{align}(4)
\begin{align} \quad (g_1H)(g_2H) = (g_2H)(g_1H) \end{align}
Since this holds for all $g_1H, g_2H \in G/H$ we conclude that $G/H$ is abelian. $\blacksquare$
Corollary 2: Let $G$ be a group and let $H$ be a normal subgroup of $G$. If $G$ is abelian then $G/H$ is abelian. Proof: Suppose that $G$ is abelian. Then $g_1g_2 = g_1g_2$ for all $g_1, g_2 \in G$. So $g_1^{-1}g_2^{-1}g_1g_2 = e \in H$ for all $g_1, g_2 \in G$. So by Theorem 1 above we see that $G/H$ is an abelian group. $\blacksquare$ |
The Absolute Value of a Complex Number
Given any complex number $z$, we can compute its absolute value as described in the following definition:
Definition: Let $z = a + bi$ be a complex number. Then the Absolute Value of $z$ denoted $\mid z \mid = \sqrt{a^2 + b^2} = \sqrt{\Re (z) ^2 + \Im (z)^2}$.
For example, consider the complex number $z = 2 + 3i$. Then $\mid z \mid = \sqrt{2^2 + 3^3} = \sqrt{13}$. Geometrically, the absolute value of a complex number represents the length of vector representing that complex number on a 2-dimensional grid with a real axis and imaginary axis:
Let's now look at some properties regarding the absolute value of a complex number.
Theorem 1: Let $z, z' \in \mathbb{C}$ be complex numbers $z = a + bi$ and $z' = a' + b'i$. Then: a) $z \bar{z} = \mid z \mid^2$ where $\bar{z} = a - bi$ is the Complex Conjugate of $z$. b) $\mid zz' \mid = \mid z \mid \mid z' \mid$. Proof a)$z\bar{z} = (a + bi)(a -bi) = a^2 - abi + abi + b^2 = a^2 + b^2 = \mid z \mid^2$. Proof b)We will not prove part b)as it is rather lengthy, but the reader is advised to construct a proof. $\blacksquare$ |
I have a problem that sounds like this:
Find the limit $$\lim_{x\rightarrow 0} \frac{14\tan(6x)-84x}{6x^3}$$ using Maclaurin series, and don't forget the importance of big O notation.
I have tried to find the Maclaurin series in different ways, but I always end up with the wrong answer. And I don't know how to use the big O notation in a helpful way here.
Thank you in advance. |
The Annals of Applied Probability Ann. Appl. Probab. Volume 2, Number 2 (1992), 481-502. The Tail of the Convolution of Densities and its Application to a Model of HIV-Latency Time Abstract
Let $p(x)$ and $q(x)$ be density functions and let $(p \ast q)(x)$ be their convolution. Define $w(x) = -(d/dx)\log q(x) \text{and} v(x) = -(d/dx)\log p(x).$ Under the hypothesis of the regular oscillation of the functions $w$ and $v$, the asymptotic form of $(p \ast q)(x)$, for $x \rightarrow \infty$, is obtained. The results are applied to a model previously introduced by the author for the estimation of the distribution of HIV latency time.
Article information Source Ann. Appl. Probab., Volume 2, Number 2 (1992), 481-502. Dates First available in Project Euclid: 19 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aoap/1177005712 Digital Object Identifier doi:10.1214/aoap/1177005712 Mathematical Reviews number (MathSciNet) MR1161063 Zentralblatt MATH identifier 0752.62014 JSTOR links.jstor.org Citation
Berman, Simeon M. The Tail of the Convolution of Densities and its Application to a Model of HIV-Latency Time. Ann. Appl. Probab. 2 (1992), no. 2, 481--502. doi:10.1214/aoap/1177005712. https://projecteuclid.org/euclid.aoap/1177005712 |
Existence and uniqueness theorems for 1st-order ODE
The general 1st-order initial value problem (IVP) is $$\begin{equation}\tag{*}y’=F(x,y), y(x_0) = y_0.\end{equation}$$ We are interested in the following questions:
Under what conditions can we be sure that a solution to (*) exists? Under what conditions can we be sure that there is a unique solution to (*)?
Here are the answers.
Theorem (Existence and uniqueness). Suppose that F(x,y) is a continuous function defined in some region $$R = (x_0-\delta, x_0+\delta)\times (y_0-\epsilon, y_0+\epsilon)$$ containing the point (x_0, y_0). Then there exists \delta_1 > 0 so that a solution y=f(x) to (*) is defined for x\in (x_0-\delta_1, x_0+\delta_1). Suppose, furthermore, that \frac{\partial F}{\partial y}(x,y) is a continuous function defined on R. Then there exists \delta_2 > 0 so that the solution is the unique solution to (*) for x\in (x_0-\delta_2, x_0 + \delta_2). Example 1 Consider the IVP y'=x-y+1, y(1)=2. In this case, both the F(x,y)=x-y+1 and \frac{\partial F}{\partial y}(x,y)=-1 are defined and continuous at all points. The theorem guarantees that a solution to the ODE uniquely exists in some open interval centered at 1. In fact, an explicit solution to this equation is y(x) = x+e^{1-x}. This solution exists for all x. Example 2 Consider the IVP y' = 1+y^2, y(0)=0. Both F(x,y)=1+y^2 and \frac{\partial F}{\partial y}(x,y)=2y are defined and continuous at all points, so by the theorem we can conclude that a unique solution exists in some open interval centered at 0. By separating variables and integrating, we derive a solution to this equation y(x)=\tan x. Remember that a solution to a DE must be a continuous function! In order for this function to be considered as a solution to this IVP, we must restrict the domain to x\in (-\pi/2,\pi/2). Example 3 Consider the IVP y' = 2y/x, y(x_0)=y_0. In this example, F(x,y) = 2y/x and \frac{\partial F}{\partial y}(x,y)=2/x. Both functions are defined for all x\neq 0, so the theorem tells us that for each x_0 \neq 0 there exists a unique solution in an open interval around x_0. By separating variables and integrating, we derive solutions y(x) = Cx^2 for some constant C. Notice that all of theses solutions pass through (0,0). So the IVP y'=2y/x, y(0)=0 has infinitely many solutions, but the IVP y'=2y/x, y(0)=y_0\neq 0 has no solutions. |
Commentary Open Access Published: Notes on the result of solutions of the equilibrium equations Boundary Value Problems volume 2018, Article number: 57 (2018) Article metrics
423 Accesses
Abstract
In this short note, we correct some expressions obtained by Wang et al. (Bound. Value Probl. 2015:230, 2015). The corrected expressions will be useful for evaluating the boundary behaviors of solutions of modified equilibrium equations with finite mass subject. Moreover, the correction of Theorem 2.1 is also given.
Introduction
The origin of our work lies in Wang et al. [1]. In [1], they investigated slow equilibrium equations with finite mass subject to a homogeneous Neumann-type boundary condition. As an application, the existence of solutions for Laplace equations with a Neumann-type boundary condition was also proved, which has recently been used to study the Cauchy problem of Laplace equation by Wang [2].
However, there exist some misprints and erroneous expressions in [1]. Firstly, we correct some misprints in Sect. 2. Then we correct erroneous expressions in Sect. 3. The corrected versions will be useful for evaluating the boundary behaviors of solutions of the equilibrium equations with finite mass subject. Finally, we correct Theorem 2.1 in Sect. 4. The present notation and terminology is the same as in [1].
Some misprints
We are indebted to the anonymous reviewer for pointing out to us that the following should also be corrected in [1].
(I) A correct version of Abstract reads as follows.
The aim of this paper is to study the models of rotating stars with prescribed angular velocity. We prove that it can be formulated as a variational problem. As an application, we are also concerned with the existence of equilibrium solution.
(II) \({\mathbb{R}}^{4}\) and \(x_{4}\) should be written as \({\mathbb{R}}^{3}\) and \(x_{3}\), respectively.
(III) Introduction: instead of “3-D”, there should be “4-D”.
(IV) Some main references [1, 2, 3] should be corrected as follows:
[1] Auchmuty, G, Beals, R: Variational solutions of some nonlinear free boundary problems. Arch. Ration. Mech. Anal. 43, 255–271 (1971)
[2] Li, Y: On Uniformly Rotating Stars. Arch. Ration. Mech. Anal. 115, 367–393 (1991)
[3] Deng, Y, Yang, T: Multiplicity of stationary solutions to the Euler–Poisson equations. J. Differ. Equ. 231, 252–289 (2006)
(V) On page 2, line 12: instead of “
P:”, there should be “\(P_{1}\):”. Corrected expressions
We find that [1, inequality (2.4)] is not correct and should be modified as (the sign before the function “\((\frac{M_{1}}{M} )^{5/3}\)” should be “+”)
These corrections will be useful for the readers who want to use [1, Theorem 2.1] to evaluate the boundary behavior of solutions of the equilibrium equations with finite mass subject.
Corrected Theorem 2.1
A correction of Theorem 2.1 in [1] reads as follows.
Theorem 2.1 Let \(P_{1}\) hold. Let \((\rho_{n})_{n=1}^{\infty}\in{\mathcal{A}}_{M}\) be a minimizing sequence of F. Then there exists a subsequence, still denoted by \((\rho_{n} )_{n=1}^{\infty }\), and a sequence of translations \(T\rho_{n}:=\rho_{n} (\cdot+a_{n} e_{3} ) \), where \(a_{n} \) are constants, and \(e_{3}=(0,0,1)\), such that and \(T\rho_{n} \rightharpoonup\rho_{0} \) weakly in \(L^{\frac{4}{3}}({\mathbb{R}}^{3})\). For the induced potentials, we have \(\nabla\Phi_{T\rho_{n} }\rightharpoonup \nabla\Phi_{\rho_{0} } \) weakly in \(L^{2}({\mathbb{R}}^{3})\). Proof
Define
for \(l,m=1,2,3\).
Let \(\rho=\rho_{1} + \rho_{2} + \rho_{3}\), where \(\rho_{1}=\chi _{B_{R_{1}}}\rho\), \(\rho_{2}=\chi_{B_{R_{1},R_{2}}}\rho\), and \(\rho_{3}=\chi_{B_{R_{2}}}\rho\). So we have
Choosing \(R_{2} >2R_{1}\), we have
Next, we estimate \(I_{12}\) and \(I_{23}\):
If we define \(M_{l} =\int\rho_{l} \,dx\), then it is easy to see that \(M=M_{1}+ M_{2} +M_{3}\).
The remaining proofs are carried out in the same way as for Theorem 2.1 in [1], except that instead of the erroneous expressions (2.4) and (2.8), we have to use their corrected versions given in Sect. 2. □
Conclusions
In this note, we corrected some expressions obtained by Wang et al. [1]. The corrected expressions will be useful for evaluating the boundary behavior of solutions of the equilibrium equations with finite mass subject. Moreover, the correction of Theorem 2.1 was also given.
References 1.
Wang, J., Pu, J., Huang, B., Shi, G.: Boundary value behaviors for solutions of the equilibrium equations with angular velocity. Bound. Value Probl.
2015, 230 (2015) 2.
Wang, Y.: A regularization method for the Cauchy problem of Laplace equation. Acta Anal. Funct. Appl.
19(2), 199–205 (2017) Acknowledgements
The authors would like to thank Professor D. Simms for bringing to our attention that the expressions in [1] are not true in general. The authors are also grateful to the anonymous reviewer for his valuable observation.
Funding
This work was supported by FONDECYT (No. 11130619).
Ethics declarations Competing interests
The authors declare that they have no competing interests.
Additional information Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. |
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In this chapter we learned about left and right adjoints, and about joins and meets. At first they seemed like two rather different pairs of concepts. But then we learned some deep relationships between them. Briefly:
Left adjoints preserve joins, and monotone functions that preserve enough joins are left adjoints.
Right adjoints preserve meets, and monotone functions that preserve enough meets are right adjoints.
Today we'll conclude our discussion of Chapter 1 with two more bombshells:
Joins
are left adjoints, and meets are right adjoints.
Left adjoints are right adjoints seen upside-down, and joins are meets seen upside-down.
This is a good example of how category theory works. You learn a bunch of concepts, but then you learn more and more facts relating them, which unify your understanding... until finally all these concepts collapse down like the core of a giant star, releasing a supernova of insight that transforms how you see the world!
Let me start by reviewing what we've already seen. To keep things simple let me state these facts just for posets, not the more general preorders. Everything can be generalized to preorders.
In Lecture 6 we saw that given a left adjoint \( f : A \to B\), we can compute its right adjoint using joins:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$ Similarly, given a right adjoint \( g : B \to A \) between posets, we can compute its left adjoint using meets:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ In Lecture 16 we saw that left adjoints preserve all joins, while right adjoints preserve all meets.
Then came the big surprise: if \( A \) has all joins and a monotone function \( f : A \to B \) preserves all joins, then \( f \) is a left adjoint! But if you examine the proof, you'l see we don't really need \( A \) to have
all joins: it's enough that all the joins in this formula exist:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$Similarly, if \(B\) has all meets and a monotone function \(g : B \to A \) preserves all meets, then \( f \) is a right adjoint! But again, we don't need \( B \) to have
all meets: it's enough that all the meets in this formula exist:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ Now for the first of today's bombshells: joins are left adjoints and meets are right adjoints. I'll state this for binary joins and meets, but it generalizes.
Suppose \(A\) is a poset with all binary joins. Then we get a function
$$ \vee : A \times A \to A $$ sending any pair \( (a,a') \in A\) to the join \(a \vee a'\). But we can make \(A \times A\) into a poset as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Then \( \vee : A \times A \to A\) becomes a monotone map, since you can check that
$$ a \le a' \textrm{ and } b \le b' \textrm{ implies } a \vee b \le a' \vee b'. $$And you can show that \( \vee : A \times A \to A \) is the left adjoint of another monotone function, the
diagonal
$$ \Delta : A \to A \times A $$sending any \(a \in A\) to the pair \( (a,a) \). This diagonal function is also called
duplication, since it duplicates any element of \(A\).
Why is \( \vee \) the left adjoint of \( \Delta \)? If you unravel what this means using all the definitions, it amounts to this fact:
$$ a \vee a' \le b \textrm{ if and only if } a \le b \textrm{ and } a' \le b . $$ Note that we're applying \( \vee \) to \( (a,a') \) in the expression at left here, and applying \( \Delta \) to \( b \) in the expression at the right. So, this fact says that \( \vee \) the left adjoint of \( \Delta \).
Puzzle 45. Prove that \( a \le a' \) and \( b \le b' \) imply \( a \vee b \le a' \vee b' \). Also prove that \( a \vee a' \le b \) if and only if \( a \le b \) and \( a' \le b \).
A similar argument shows that meets are really right adjoints! If \( A \) is a poset with all binary meets, we get a monotone function
$$ \wedge : A \times A \to A $$that's the
right adjoint of \( \Delta \). This is just a clever way of saying
$$ a \le b \textrm{ and } a \le b' \textrm{ if and only if } a \le b \wedge b' $$ which is also easy to check.
Puzzle 46. State and prove similar facts for joins and meets of any number of elements in a poset - possibly an infinite number.
All this is very beautiful, but you'll notice that all facts come in pairs: one for left adjoints and one for right adjoints. We can squeeze out this redundancy by noticing that every preorder has an "opposite", where "greater than" and "less than" trade places! It's like a mirror world where up is down, big is small, true is false, and so on.
Definition. Given a preorder \( (A , \le) \) there is a preorder called its opposite, \( (A, \ge) \). Here we define \( \ge \) by
$$ a \ge a' \textrm{ if and only if } a' \le a $$ for all \( a, a' \in A \). We call the opposite preorder\( A^{\textrm{op}} \) for short.
I can't believe I've gone this far without ever mentioning \( \ge \). Now we finally have really good reason.
Puzzle 47. Show that the opposite of a preorder really is a preorder, and the opposite of a poset is a poset. Puzzle 48. Show that the opposite of the opposite of \( A \) is \( A \) again. Puzzle 49. Show that the join of any subset of \( A \), if it exists, is the meet of that subset in \( A^{\textrm{op}} \). Puzzle 50. Show that any monotone function \(f : A \to B \) gives a monotone function \( f : A^{\textrm{op}} \to B^{\textrm{op}} \): the same function, but preserving \( \ge \) rather than \( \le \). Puzzle 51. Show that \(f : A \to B \) is the left adjoint of \(g : B \to A \) if and only if \(f : A^{\textrm{op}} \to B^{\textrm{op}} \) is the right adjoint of \( g: B^{\textrm{op}} \to A^{\textrm{ op }}\).
So, we've taken our whole course so far and "folded it in half", reducing every fact about meets to a fact about joins, and every fact about right adjoints to a fact about left adjoints... or vice versa! This idea, so important in category theory, is called
duality. In its simplest form, it says that things come on opposite pairs, and there's a symmetry that switches these opposite pairs. Taken to its extreme, it says that everything is built out of the interplay between opposite pairs.
Once you start looking you can find duality everywhere, from ancient Chinese philosophy:
to modern computers:
But duality has been studied very deeply in category theory: I'm just skimming the surface here. In particular, we haven't gotten into the connection between adjoints and duality!
This is the end of my lectures on Chapter 1. There's more in this chapter that we didn't cover, so now it's time for us to go through all the exercises. |
Table of Contents
Connected and Disconnected Sets In Topological Spaces
Recall from the Connected and Disconnected Topological Spaces page that a topological space $X$ is said to be disconnected if there exists open sets $A, B \subset X$ such that $A, B \neq \emptyset$, $A \cap B = \emptyset$ and:(1)
In such cases, we cal $\{ A, B \}$ a separation of $X$. Furthermore, the topological space $X$ is said to be connected if it is not disconnected.
We now turn our attention to defining connected and disconnected sets in a topological space.
Definition: Let $X$ be a topological space and let $A \subseteq X$. $A$ is said to be Connected if it is a connected topological space with respect to the subspace topology on $A$. Similarly, $A$ is said to be Disconnected if it is a disconnected topological space with respect to the subspace topology on $A$.
For example, consider the topological space $\mathbb{R}$ with the usual topology. Let $A = (0, 1) \cup [2, 3]$. Then the set $A$ (with the subspace topology) is a disconnected topological space.
Let $B = (0, 1)$ and $C = [2, 3]$. Clearly $B, C \subset A$, $B, C \neq \emptyset$, $B \cap C = \emptyset$, and $A = B \cup C$. So, we only need to show that $B$ and $C$ are open in $A$.
Since $A$ has the subspace topology, consider the open sets $(0, 1)$ and $\left ( \frac{3}{2}, 4 \right )$ in $\mathbb{R}$. When we intersect these open sets with $A$ we see that:(2)
Therefore $B$ and $C$ are open in $A$. This shows that $\{ B, C \}$ is a separation of $A$ and that $A$ is a disconnected set. |
Indirect adjusted comparisons in BE [General Statistics]
yep, you are all right. Nobody’s definition of “pragmatic” is correct. IMHO, here it should be called – following Wolfgang Pauli’s footprints – “not even wrong”.
Pooling variances without weighting by the sample size is just false.
Quoting Shuanghe’s reference, Section Data analysis:
In line with recommendations by Gwaza
et al., a pragmatic method based on t test was used […]. This method has been shown to give comparable results as the homoscedastic method. et al.* really wrote/recommended is this:
The [pragmatic] method based on a combination of both (SE calculation of the heteroscedastic method and degrees of freedom of the homocedastic method), which is the simplest numerical calculation, gives results similar to those obtained with the heteroscedastic approach, showing that the contribution of the different method of calculation of the degrees of freedom is negligible.
In conclusion, the homoscedastic method is recommended when performing adjusted indirect comparisons of bioequivalence studies—unless a clear difference in variances is observed in the data, which is unlikely.
Gwaza drives me nuts. What is “SE calculation of the heteroscedastic method”? There is only one correct way to pool variances. No, it’s not \(SE_d^2 = SE_1^2 + SE_2^2\). The homo- and heteroscedastic methods differ only in the degrees of freedom.
homoscedastic \(\nu = n_1+n_2-2\) heteroscedastic \(\nu \approx \frac{\left (\frac{s_{1}^{2}}{n_1}+\frac{s_{1}^{2}}{n_2}\right )^2}{\frac{s_{1}^{4}}{n_1(n_1-1)}+\frac{s_{1}^{4}}{n_2(n_2-1)}}\)
Comes handy cause most people don’t read beyond the abstract (men on a mission?). When I rank the width of the CIs given in the two tables:
The pragmatic method lies between the hetero- and homoscedastic methods. The pragmatic method is much closer to the hetero- than to the homoscedastic method.
If I use the crappy formula I end up with CIs which are not slightly tighter than the ones of the heteroscedastic method – as expected – but slightly
wider. Puzzle not solved.
Since I’m dealing with HVD(P)s, where the CV jumps around like crazy between studies I prefer the heteroscedastic approach.
» OT again:
Easy.
» Thanks a lot for implementing MathJax. Equations are looking much better. one puzzle though: command \LaTeX does do what it supposed to do.
There is not \LaTex in MathJax. See there for examples (including a link to the MathJax site).
Gwaza L, Gordon J, Welink J, Potthast H, Hansson H, Stahl M, García-Arieta A. Statistical approaches to indirectly compare bioequivalence between generics: a comparison of methodologies employing artemether / lumefantrine 20/120 mg tablets as prequalified by WHO.Eur J Clin Pharmacol. 2012;68(12):1611–8. doi:10.1007/s00228-012-1396-1.
Cheers,
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. ☼
Science Quotes
Complete thread: Indirect adjusted comparisons in BE Helmut 2019-07-21 15:20 [General Statistics] Pragmatic assumptions mittyri 2019-07-22 01:46 Indirect adjusted comparisons in BE Shuanghe 2019-07-22 09:53 |
Tsukuba Journal of Mathematics Tsukuba J. Math. Volume 12, Number 1 (1988), 241-248. Existence of all the asymptotic $\lambda$-th means for certain arithmetical convolutions Abstract
Let $E$ designate either of the classical error terms for the summatory functions of the arithmetical functions $\phi(n)/n$ and $\sigma(n)/n$ ($\phi$ is Euler's function and $\sigma$ the divisor function). By following an idea of Codecà's [3] and by refining some of his estimates we prove that $|E|$ has asymptotic $\lambda$-th order means for all positive real numbers $\lambda$. We also prove that $E$ has asymptotic $k$-th order means for all positive integers $k$, and that this mean is zero whenever $k$ is odd. The results obtained can be applied to functions other than $E$ as well, such as the functions $P$ and $Q$ of Hardy and Littlewood[8], or the divisor functions $G_{-1,k}$[9].
Article information Source Tsukuba J. Math., Volume 12, Number 1 (1988), 241-248. Dates First available in Project Euclid: 30 May 2017 Permanent link to this document https://projecteuclid.org/euclid.tkbjm/1496160645 Digital Object Identifier doi:10.21099/tkbjm/1496160645 Mathematical Reviews number (MathSciNet) MR949911 Zentralblatt MATH identifier 0661.10056 Citation
Y-F.S, Petermann. Existence of all the asymptotic $\lambda$-th means for certain arithmetical convolutions. Tsukuba J. Math. 12 (1988), no. 1, 241--248. doi:10.21099/tkbjm/1496160645. https://projecteuclid.org/euclid.tkbjm/1496160645 |
COMMERCIAL MATHEMATICS
Category : 6th Class
Learning Objective PERCENTAGE
The term percent means "for every hundred". A fraction whose denominator is 100 is called percentage and the numerator of the fraction is called the rate percent. Thus, when we say a man made a profit of 20 percent we mean to say that he gained Rs 20 for every hundred rupees he invested in the business, i.e., 20/100 rupees for each Rupee. The abbreviation of percent is p.c. and it is generally denoted by %.
VALUE OF PERCENTAGE
Value of percentage always depends on the quantity to which it refers consider the statement:
“65% of the students in this class are boys". From the context, it is understood that boys form 65% of the total number of students in the class. To know the value of 65%, the value of the total number of student should be mown.
If the total number of students is 200, then, the number of boys \[=\frac{200\times 65}{100}=130;\] It can also be written as \[(200)\times (0.65)=130\].
Note that the expressions 6%, 63%, 72%, 155% etc. do not have any value intrinsic to themselves. Their values depend on the quantities to which they refer.
COMPARING PERCENTAGES
Which of the three 25%, 5% and 125% is the largest?
If should be remembered that no comparison can be made about the above percentages, because they do not have intrinsic values. If 25% refers to 25% of 10,000 then its value is 0.25 x 10,000 = 2,500 and if 75% of 100, its value is 0.75 x 100 =75.
And so we can conclude that 25% of 10,000 > 75% of 100.
Note: Percentages can be compared only when the quantities they refer to are known. IMPORTANT RESULTS
1. To express a percentage as a fraction divide it by \[100\Rightarrow \] a percentage = 1/100.
Example:
Express the following as fraction
(a) 25%
(b) \[33\frac{1}{3}%\]
Sol.
(a) \[25%\,=\frac{25}{10}\left[ \text{since}\,\text{ }\!\!%\!\!\text{ }\,\text{means}\,\frac{1}{100} \right]=\frac{1}{4}\]
(b) \[33\frac{1}{3}%\,=\frac{100}{3}%\,=\frac{100}{3\times 100}=\frac{1}{3}\]
2. To express a fraction as a percent multiply it by \[100\,\frac{a}{b}\,=\left[ \left( \frac{a}{b}\times 100 \right) \right]\,%\]
Example:
Express \[\frac{1}{8}\] as a percentage.
Sol. \[\frac{1}{8}=\frac{1}{8}\times 100%\] \[=\frac{100}{8}%=\frac{25}{2}%\,=12\frac{1}{2}%\]
3. To express percentage as a decimal we remove the symbol% and shift the decimal point by two places the left.
Example:
Express \[6\frac{1}{2}%\] as a decimal.
Sol.
\[6\frac{1}{2}%=\frac{13}{2}%\,=6.5%\,=\frac{65}{100}=0.065\]
4. To express decimal as a percentage we shift the decimal point by two places to the right and write number obtained with the symbol % or simply we multiply the decimal with 100.
Example:
0.345 as a percentage.
Sol. \[0.345\times 100%\]
= 34.5%
If A is R% of a given number N, then \[N=\frac{A\times 100}{R}\]
Example:
25% of a number is 80. What is the number?
Sol.
Let the number be X.
According to the given condition
\[\frac{25}{100}\,\times \,X=80\Rightarrow \,X=\frac{80\times 100}{25}=320\].
PROFIT AND LOSS COST PRICE (CP)
The price for which an article is bought is called its cost price.
SELLING PRICE (SP)
The price at which an article is sold is called its selling price.
PROFIT (GAIN)
The difference between the selling price and cost price is called the profit. For profit, selling price should be greater than cost price.
LOSS
The difference between the cost price and the selling price is called the loss. When cost price is greater than the selling price there is a loss.
Wit and loss is generally represented as a percent of the cost price, unless otherwise stated.
FORMULAE
Example:
A man buys a radio for Rs 600 and sells it at a gain of 25%, what will be the selling price for him?
Sol.
C.P = Rs 600 and Gain (Profit) % = 25%
\[S.P=\left( \frac{100+\text{Profit}%}{100} \right)\times C.P\]
\[=\left( \frac{100+25}{100} \right)\]
\[=\frac{125}{100}\times 600\]
= Rs 70
Example:
Find the cost price of good, sells at Rs 160, having Loss% = 20%
Sol.
S.P = Rs 160% = 20%
\[C.P=\frac{S.P\times 100}{100-Loss%}\]
\[=\frac{160\times 100}{80}\]
= Rs 200
Cost price of goods = Rs 200.
SIMPLE INTEREST
Suppose Vishal borrows money from a bank for his higher studies, then at the end of the specified period, he has to pay the money he borrowed and some additional money for the privilege of having used the bank money
Now, we can define the following terms:
The money borrowed is called the Principal ‘p’;
The additional money paid is called the Interest 'SI;
The total money paid is called the Amount 'A’
Formula
1. Amount = Principal + Interests A =P+S.I. 2. Simple Interest
\[=\frac{\text{Principal}\,\text{ }\!\!\times\!\!\text{ Time}\,\text{ }\!\!\times\!\!\text{ Rate}}{\text{100}}\,\Rightarrow \,S.I.=\frac{P\times T\times R}{100}\]
3. Principal
\[=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Time }\!\!\times\!\!\text{ Rate}}\,\Rightarrow \,P=\frac{100\times S.I.}{T\times R}\]
4. Rate \[=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Principle}\,\times \text{Time}}\,\Rightarrow \,R=\frac{100\times S.I.}{P\times T}\] 5. Time \[=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Principle}\,\text{ }\!\!\times\!\!\text{ Rate}}\,\Rightarrow \,T=\frac{100\times S.I.}{P\times R}\] Example
Find (i) the interest, (ii) the amount on Rs 5000 at 5% per annum S.I. for 4 years.
Sol.
(i) Given P = Rs 5000, T = 4 years, R = 5%
We know \[S.I.=\frac{P\times T\times R}{100}\]\[\therefore \,\,S.I.\,=\frac{5000\times 4\times 5}{100}\]
Hence, S.I. =Rs 1000.
(ii) Amount = P + S.I.
\[\therefore \] Amount = Rs 5000 + Rs 1000 = Rs 6000.
Example
What sum of money will amount to Rs 1400 at the rate of 8% per annum S.I. in 5 years?
Sol. Let the principal be Rs 100
100x5x8
\[\therefore \] S.I. on Rs 100 for 5 years at 8% \[=\frac{100\times 5\times 8}{100}=5\times 8=\text{Rs}40\]
\[\therefore \] Amount = P + S.I.
=Rs (100+40)=Rs 140.
If the amount is Rs 140, then Principal = Rs 100
If the amount is Rs 1400, then Principal
\[=\frac{100\times 1400}{140}\]
\[\therefore \] Principal = Rs 1000.
Example
In what time will Rs 450 amount to Rs 540 at 5% per annum S.I.?
Sol. Given A = Rs 540, P = Rs 450, R = 5%, T = Rs
\[\therefore \] S.I. = A - P = Rs 540 - Rs 450 == Rs 90
\[\therefore \] \[T=\frac{100\times S.I.}{P\times R}\]
\[\therefore \] \[T=\frac{100\times 90}{450\times 5}=4\]
\[\therefore \] Time = 4 years.
Example
In how many years will a sum of money double itself at 10% per annum S.I.?
Sol. Let the principle be Rs 'x’
\[\therefore \] Amount == 2x
\[\therefore \] S.I. = A – P
\[\therefore \] S.I. \[=2x-x=x\]
\[T=\frac{100\times S.I.}{P\times R}\]
\[\therefore \] \[T=\frac{100\times x}{x\times 10}\,=10\]
\[\therefore \] Time = 10 years.
Note:
1. If a sum of money doubles, means S.I. = 2P - P = P
2. If a sum of money triple, means S.I. = 3P - P = 2P
Example
If Rs 800 amounts to Rs 960 in 4 years, find the rate percent per annum S.I.
Sol. Given: P = Rs 800, A, = Rs 960
\[\therefore \] S.I. = Rs 960 – Rs 800 = Rs 160
\[R=\frac{100\times S.I.}{P\times T}\,=\frac{100\times 160}{800\times }=5\]
\[\therefore \] Rate == 5%.
AVERAGE OR MEAN
If \[{{X}_{1}},{{X}_{2}},\,{{X}_{3}}....{{X}_{n}}\] be n observations then their arithmetic mean is given by:
\[\overline{X}=\frac{{{X}_{1}}+{{X}_{2}}+......{{X}_{n}}}{n}\]
Example
Find the mean height if the heights of 5 persons are 144cm, 153 cm, 150cm, 158 cm and 155 cm respectively
Sol.
Mean Height \[=\frac{144+153+150+158+155}{5}\]
\[=\frac{760}{5}\,=152\,cm\]
Example
Find the mean of the first 10 odd numbers.
Sol.
First ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
So mean
\[(\overline{x})=\frac{1+3+5+9+11+13+15+17+19}{10}\]
\[=\frac{100}{10}=10\]
SPEED
The speed of a body is defined as the distance covered by it in unit time.
Speed \[=\frac{\text{Distance}}{\text{Time}}\] distance is constant
Time \[=\frac{\text{Distance}}{\text{Speed}};\] time is constant
Distance = Speed x Time; Speed is constant
UNITS OF MEASUREMENT AVERAGE SPEED
Total distance travelled
Average speed \[=\frac{\text{Total}\,\text{distance}\,\text{travelled}}{\text{Total}\,\text{time}\,\text{take}}\]
Example
Find the Speed of train travels 90 km in 2 hours.
Sol.
\[Speed\,=\frac{\text{Distance}}{\text{Time}}\]
Speed \[=\frac{90}{2}\]
Speed = 45 Km/hr
Example
Find the average speed of a person if he goes 5 km from point A to B and then goes 3 km from point B to C, he takes 2 hours to cover the distance from point A to C.
Sol.
Total distance covered by man = (5 + 3) km = 8 km.
Average speed
\[=\frac{\text{Total}\,\text{distance}}{\text{Total}\,\text{time}\,\text{take}}=\frac{8}{2\,}=4\,km/hr\]
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In this chapter we learned about left and right adjoints, and about joins and meets. At first they seemed like two rather different pairs of concepts. But then we learned some deep relationships between them. Briefly:
Left adjoints preserve joins, and monotone functions that preserve enough joins are left adjoints.
Right adjoints preserve meets, and monotone functions that preserve enough meets are right adjoints.
Today we'll conclude our discussion of Chapter 1 with two more bombshells:
Joins
are left adjoints, and meets are right adjoints.
Left adjoints are right adjoints seen upside-down, and joins are meets seen upside-down.
This is a good example of how category theory works. You learn a bunch of concepts, but then you learn more and more facts relating them, which unify your understanding... until finally all these concepts collapse down like the core of a giant star, releasing a supernova of insight that transforms how you see the world!
Let me start by reviewing what we've already seen. To keep things simple let me state these facts just for posets, not the more general preorders. Everything can be generalized to preorders.
In Lecture 6 we saw that given a left adjoint \( f : A \to B\), we can compute its right adjoint using joins:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$ Similarly, given a right adjoint \( g : B \to A \) between posets, we can compute its left adjoint using meets:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ In Lecture 16 we saw that left adjoints preserve all joins, while right adjoints preserve all meets.
Then came the big surprise: if \( A \) has all joins and a monotone function \( f : A \to B \) preserves all joins, then \( f \) is a left adjoint! But if you examine the proof, you'l see we don't really need \( A \) to have
all joins: it's enough that all the joins in this formula exist:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$Similarly, if \(B\) has all meets and a monotone function \(g : B \to A \) preserves all meets, then \( f \) is a right adjoint! But again, we don't need \( B \) to have
all meets: it's enough that all the meets in this formula exist:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ Now for the first of today's bombshells: joins are left adjoints and meets are right adjoints. I'll state this for binary joins and meets, but it generalizes.
Suppose \(A\) is a poset with all binary joins. Then we get a function
$$ \vee : A \times A \to A $$ sending any pair \( (a,a') \in A\) to the join \(a \vee a'\). But we can make \(A \times A\) into a poset as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Then \( \vee : A \times A \to A\) becomes a monotone map, since you can check that
$$ a \le a' \textrm{ and } b \le b' \textrm{ implies } a \vee b \le a' \vee b'. $$And you can show that \( \vee : A \times A \to A \) is the left adjoint of another monotone function, the
diagonal
$$ \Delta : A \to A \times A $$sending any \(a \in A\) to the pair \( (a,a) \). This diagonal function is also called
duplication, since it duplicates any element of \(A\).
Why is \( \vee \) the left adjoint of \( \Delta \)? If you unravel what this means using all the definitions, it amounts to this fact:
$$ a \vee a' \le b \textrm{ if and only if } a \le b \textrm{ and } a' \le b . $$ Note that we're applying \( \vee \) to \( (a,a') \) in the expression at left here, and applying \( \Delta \) to \( b \) in the expression at the right. So, this fact says that \( \vee \) the left adjoint of \( \Delta \).
Puzzle 45. Prove that \( a \le a' \) and \( b \le b' \) imply \( a \vee b \le a' \vee b' \). Also prove that \( a \vee a' \le b \) if and only if \( a \le b \) and \( a' \le b \).
A similar argument shows that meets are really right adjoints! If \( A \) is a poset with all binary meets, we get a monotone function
$$ \wedge : A \times A \to A $$that's the
right adjoint of \( \Delta \). This is just a clever way of saying
$$ a \le b \textrm{ and } a \le b' \textrm{ if and only if } a \le b \wedge b' $$ which is also easy to check.
Puzzle 46. State and prove similar facts for joins and meets of any number of elements in a poset - possibly an infinite number.
All this is very beautiful, but you'll notice that all facts come in pairs: one for left adjoints and one for right adjoints. We can squeeze out this redundancy by noticing that every preorder has an "opposite", where "greater than" and "less than" trade places! It's like a mirror world where up is down, big is small, true is false, and so on.
Definition. Given a preorder \( (A , \le) \) there is a preorder called its opposite, \( (A, \ge) \). Here we define \( \ge \) by
$$ a \ge a' \textrm{ if and only if } a' \le a $$ for all \( a, a' \in A \). We call the opposite preorder\( A^{\textrm{op}} \) for short.
I can't believe I've gone this far without ever mentioning \( \ge \). Now we finally have really good reason.
Puzzle 47. Show that the opposite of a preorder really is a preorder, and the opposite of a poset is a poset. Puzzle 48. Show that the opposite of the opposite of \( A \) is \( A \) again. Puzzle 49. Show that the join of any subset of \( A \), if it exists, is the meet of that subset in \( A^{\textrm{op}} \). Puzzle 50. Show that any monotone function \(f : A \to B \) gives a monotone function \( f : A^{\textrm{op}} \to B^{\textrm{op}} \): the same function, but preserving \( \ge \) rather than \( \le \). Puzzle 51. Show that \(f : A \to B \) is the left adjoint of \(g : B \to A \) if and only if \(f : A^{\textrm{op}} \to B^{\textrm{op}} \) is the right adjoint of \( g: B^{\textrm{op}} \to A^{\textrm{ op }}\).
So, we've taken our whole course so far and "folded it in half", reducing every fact about meets to a fact about joins, and every fact about right adjoints to a fact about left adjoints... or vice versa! This idea, so important in category theory, is called
duality. In its simplest form, it says that things come on opposite pairs, and there's a symmetry that switches these opposite pairs. Taken to its extreme, it says that everything is built out of the interplay between opposite pairs.
Once you start looking you can find duality everywhere, from ancient Chinese philosophy:
to modern computers:
But duality has been studied very deeply in category theory: I'm just skimming the surface here. In particular, we haven't gotten into the connection between adjoints and duality!
This is the end of my lectures on Chapter 1. There's more in this chapter that we didn't cover, so now it's time for us to go through all the exercises. |
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Isomorphisms are very important in mathematics, and we can no longer put off talking about them. Intuitively, two objects are 'isomorphic' if they look the same. Category theory makes this precise and shifts the emphasis to the 'isomorphism' - the
way in which we match up these two objects, to see that they look the same.
For example, any two of these squares look the same after you rotate and/or reflect them:
An isomorphism between two of these squares is a
process of rotating and/or reflecting the first so it looks just like the second.
As the name suggests, an isomorphism is a kind of morphism. Briefly, it's a morphism that you can 'undo'. It's a morphism that has an inverse:
Definition. Given a morphism \(f : x \to y\) in a category \(\mathcal{C}\), an inverse of \(f\) is a morphism \(g: y \to x\) such that
and
I'm saying that \(g\) is 'an' inverse of \(f\) because in principle there could be more than one! But in fact, any morphism has at most one inverse, so we can talk about 'the' inverse of \(f\) if it exists, and we call it \(f^{-1}\).
Puzzle 140. Prove that any morphism has at most one inverse. Puzzle 141. Give an example of a morphism in some category that has more than one left inverse. Puzzle 142. Give an example of a morphism in some category that has more than one right inverse.
Now we're ready for isomorphisms!
Definition. A morphism \(f : x \to y\) is an isomorphism if it has an inverse. Definition. Two objects \(x,y\) in a category \(\mathcal{C}\) are isomorphic if there exists an isomorphism \(f : x \to y\).
Let's see some examples! The most important example for us now is a 'natural isomorphism', since we need those for our databases. But let's start off with something easier. Take your favorite categories and see what the isomorphisms in them are like!
What's an isomorphism in the category \(\mathbf{3}\)? Remember, this is a free category on a graph:
The morphisms in \(\mathbf{3}\) are paths in this graph. We've got one path of length 2:
$$ f_2 \circ f_1 : v_1 \to v_3 $$ two paths of length 1:
$$ f_1 : v_1 \to v_2, \quad f_2 : v_2 \to v_3 $$ and - don't forget - three paths of length 0. These are the identity morphisms:
$$ 1_{v_1} : v_1 \to v_1, \quad 1_{v_2} : v_2 \to v_2, \quad 1_{v_3} : v_3 \to v_3.$$ If you think about how composition works in this category you'll see that the only isomorphisms are the identity morphisms. Why? Because there's no way to compose two morphisms and get an identity morphism unless they're both that identity morphism!
In intuitive terms, we can only move from left to right in this category, not backwards, so we can only 'undo' a morphism if it doesn't do anything at all - i.e., it's an identity morphism.
We can generalize this observation. The key is that \(\mathbf{3}\) is a poset. Remember, in our new way of thinking a
preorder is a category where for any two objects \(x\) and \(y\) there is at most one morphism \(f : x \to y\), in which case we can write \(x \le y\). A poset is a preorder where if there's a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(x = y\). In other words, if \(x \le y\) and \(y \le x\) then \(x = y\). Puzzle 143. Show that if a category \(\mathcal{C}\) is a preorder, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(g\) is the inverse of \(f\), so \(x\) and \(y\) are isomorphic. Puzzle 144. Show that if a category \(\mathcal{C}\) is a poset, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then both \(f\) and \(g\) are identity morphisms, so \(x = y\).
Puzzle 144 says that in a poset, the only isomorphisms are identities.
Isomorphisms are a lot more interesting in the category \(\mathbf{Set}\). Remember, this is the category where objects are sets and morphisms are functions.
Puzzle 145. Show that every isomorphism in \(\mathbf{Set}\) is a bijection, that is, a function that is one-to-one and onto. Puzzle 146. Show that every bijection is an isomorphism in \(\mathbf{Set}\).
So, in \(\mathbf{Set}\) the isomorphisms are the bijections! So, there are lots of them.
One more example:
Definition. If \(\mathcal{C}\) and \(\mathcal{D}\) are categories, then an isomorphism in \(\mathcal{D}^\mathcal{C}\) is called a natural isomorphism.
This name makes sense! The objects in the so-called 'functor category' \(\mathcal{D}^\mathcal{C}\) are functors from \(\mathcal{C}\) to \(\mathcal{D}\), and the morphisms between these are natural transformations. So, the
isomorphisms deserve to be called 'natural isomorphisms'.
But what are they like?
Given functors \(F, G: \mathcal{C} \to \mathcal{D}\), a natural transformation \(\alpha : F \to G\) is a choice of morphism
$$ \alpha_x : F(x) \to G(x) $$ for each object \(x\) in \(\mathcal{C}\), such that for each morphism \(f : x \to y\) this naturality square commutes:
Suppose \(\alpha\) is an isomorphism. This says that it has an inverse \(\beta: G \to F\). This \(beta\) will be a choice of morphism
$$ \beta_x : G(x) \to F(x) $$ for each \(x\), making a bunch of naturality squares commute. But saying that \(\beta\) is the inverse of \(\alpha\) means that
$$ \beta \circ \alpha = 1_F \quad \textrm{ and } \alpha \circ \beta = 1_G .$$ If you remember how we compose natural transformations, you'll see this means
$$ \beta_x \circ \alpha_x = 1_{F(x)} \quad \textrm{ and } \alpha_x \circ \beta_x = 1_{G(x)} $$ for all \(x\). So, for each \(x\), \(\beta_x\) is the inverse of \(\alpha_x\).
In short: if \(\alpha\) is a natural isomorphism then \(\alpha\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\).
But the converse is true, too! It takes a
little more work to prove, but not much. So, I'll leave it as a puzzle. Puzzle 147. Show that if \(\alpha : F \Rightarrow G\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\), then \(\alpha\) is a natural isomorphism.
Doing this will help you understand natural isomorphisms. But you also need examples!
Puzzle 148. Create a category \(\mathcal{C}\) as the free category on a graph. Give an example of two functors \(F, G : \mathcal{C} \to \mathbf{Set}\) and a natural isomorphism \(\alpha: F \Rightarrow G\). Think of \(\mathcal{C}\) as a database schema, and \(F,G\) as two databases built using this schema. In what way does the natural isomorphism between \(F\) and \(G\) make these databases 'the same'. They're not necessarily equal!
We should talk about this. |
On page 101 of our textbook, we see a theorem that looks something like the following:
A set of vectors like this that spans a space and is linearly independence is called a basis.
The theorem is a bit hard to wrap one's head around but, often, the description of an algorithm like this is easier to understand if you program it. In fact, you really can't tell a computer how to do something until you understand it pretty well yourself! Let's give it a try.
Here's an example to play with lifted right off of page 89 from our textbook. Of course, this is live code so you can feel free to change it to explore other examples. Once the matrix is defined, we'll go head and reduce it to row echelon form using the
rref method.
First, let's identify the sets $D$ and $F$ that index the pivot columns and free columns respectively.
This might not be quite what we expected because Python (and therefore Sage) uses a zero-based index, rather than a one-based index. Thus, if there are 6 columns, they are indexed 0, 1, 2, 3, 4, 5. Regardless, our working has three free columns; thus, the null space should have dimension three.
Now, here is the definition of
z taken straight from the theorem and translated to Python.
Seems to work, at least!
Let's try to pick this apart a bit to get a better understanding. We'll begin by focusing on the case $i\in F$ where $(\vec{z}_j)_i$ is zero or one, effectively ignoring the last part. $$ (\vec{z_1}_j)_i = \begin{cases} 1 & \text{if } i \in F, i = f_j \\ 0 & \text{if } i \in F, i \neq f_j \\ \color{#eee}-b_{k,f_j} & \color{#eee}\text{if } i \in D, i=d_k \end{cases} $$ We might denote this first step by $\vec{z_1}$ and we can implement it by just suppressing the last part as follows:
We can now see a nice way to interpret this. For each vector, we've got to determine one of the free index locations to place a 1; we'll place a 0 in the other free index locations. For the $j^{\text{th}}$ vector, find the free index in the $j^{\text{th}}$ location and place the 1 there.
In our working example, the free indices (counting from one, like a normal person) are 3, 5, and 6. Thus, our first vector has a 1 in the $3^{\text{rd}}$ spot, our second vector has a 1 in the $5^{\text{th}}$ spot, and our third vector has a 1 in the $6^{\text{th}}$ spot.
Note that this portion alone guarantees linear independence of the vectors.
Finally, we'd like to get a grip on why this set of vectors spans the null space. To do so, let's first consider how we find the null space by translating the reduced row echelon form of the matrix to a linear system. That is, we take our example matrix: $$ \left(\begin{array}{rrrrrr} 1 & 0 & 2 & 0 & -1 & 2 \\ 0 & 1 & 1 & 0 & 3 & -1 \\ 0 & 0 & 0 & 1 & 4 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$ and translate it to the homeogenous system $$ \begin{align} x_1 + 2x_3 - x_5 + 2x_6 &= 0 \\ x_2 + x_3 + 3x_5 - x_6 &= 0 \\ x_4 + 4x_5 - 2x_6 &= 0 \end{align}. $$ Solving for the pivot variables $x_1$, $x_2$, and $x_4$ in terms of the free variables $x_3$, $x_5$, and $x_6$, we find $$ \begin{align} x_1 &= -2x_3 + x_5 - 2x_6 \\ x_2 &= -x_3 - 3x_5 + x_6 \\ x_4 &= -4x_5 + 2x_6 \end{align}. $$ From there, it's pretty easy to see the general form of a solution: $$ x_3 \left(\begin{matrix} -2 \\ -1 \\ \color{lightgray}1 \\ 0 \\ \color{lightgray}0 \\ \color{lightgray}0 \end{matrix}\right) + x_5 \left(\begin{matrix} 1 \\ -3 \\ \color{lightgray}0 \\ -4 \\ \color{lightgray}1 \\ \color{lightgray}0 \end{matrix}\right) + x_6 \left(\begin{matrix} -2 \\ 1 \\ \color{lightgray}0 \\ 2 \\ \color{lightgray}0 \\ \color{lightgray}1 \end{matrix}\right) $$ Note that each vector corresponds naturally to the free variable that appears as its scalar multiple - just as in the formula for $(\vec{z}_j)_i$ in our algorithm. Also note that entries corresponding to the free indices have been lightly grayed. This should allow us to distinguish between the free indices and the pivot indices fairly easily.
For the free indices, we have $$ (\vec{z}_j)_i = \begin{cases} 1 & \text{if } i \in F, i = f_j \\ 0 & \text{if } i \in F, i \neq f_j \\ \color{#ddd}-b_{k,f_j} & \color{#ddd}\text{if } i \in D, i=d_k \end{cases}. $$ Note that $j$ tells us which spanning vector we are constructing; we want the $i^{\text{th}}$ elment to be $1$ precisely when $i$ is the $j^{\text{th}}$ free index. For example, when $j=3$ we should examine our $3^{\text{rd}}$ element of $F=\{3,5,6\}$, which is 6. Thus, the $6^{\text{th}}$ element of the third vector should be 1. The other free index positions (3 and 5) should contain zeros.
For the pivot indices, we have $$ (\vec{z}_j)_i = \begin{cases} \color{#ddd}1 & \color{#ddd}\text{if } i \in F, i = f_j \\ \color{#ddd}0 & \color{#ddd}\text{if } i \in F, i \neq f_j \\ -b_{k,f_j} & \text{if } i \in D, i=d_k \end{cases}. $$ Again, $j$ indicates which spanning vector we are constructing so it makes sense that we should look in the column corresponding to the $j^{\text{th}}$ free variable. Now, however, we are solving an equation to find the contribution from the $j^{\text{th}}$ free variable to one of the pivot variables. That's where the $f_j$ comes from in the $-b_{k,f_j}$. To contribute to the $k^{\text{th}}$ pivot variable, we should be working in the $k^{\text{th}}$ row of the reduced row echelon form of the matrix. That's where the $k$ comes from in the $-b_{k,f_j}$.
In our working example, the second vector corresponds to the second free index, which is 5. Thus, the terms we see in that vector should come from the $5^{\text{th}}$ column of our matrix. If we are interested in, say, the third pivot variable, which is 4, we should look in the third row of the matrix. That's why we see the $-4$ in the fourth position of the second vector. |
Euler Differential Equations
Another type of second order differential equation that we can solve are known as
Euler Differential Equations. What's particularly nice about Euler differential equations is that we can take a second order linear homogenous differential equation with non-constant coefficients and use a substitution to transform it into a second order linear homogenous differential equation with constant coefficients. We define Euler differential equations below.
Definition: A second order differential equation in the form $t^2 \frac{d^2 y}{dt^2} + \alpha t \frac{dy}{dt} + \beta y = 0$, $t > 0$ where $\alpha, \beta \in \mathbb{R}$ is called an Euler Differential Equation.
In order to transform an Euler differential equation into a second order linear homogenous differential equation with constant coefficients, we will let $x = \ln t$. We will compute $\frac{dy}{dt}$ and $\frac{d^2y}{dt^2}$ as follows:(1)
We will now plug this into our differential equation to get that:(3)
Thus we have a second order linear homogenous differential equation with constant coefficients, namely, $1$, $\alpha - 1$, and $\beta$. We know how to solve differential equations of this type! Note that if $y = y_1(x)$ and $y = y_2(x)$ form a fundamental set of solutions to $\frac{d^2 y}{dx^2} + (\alpha - 1) \frac{dy}{dx} + \beta y = 0$ then $y = y_1(\ln t)$ and $y = y_2(\ln t)$ (which are obtained by reintroducing the substitution $x = \ln t$) form a fundamental set of solutions to the Euler equation $t^2 \frac{d^2 y}{dt^2} + \alpha t \frac{dy}{dt} + \beta y = 0$.
Let's now look at an example of solving an Euler differential equation.
Example 1 Solve the Euler differential equation $t^2 \frac{d^2y}{dt^2} + 4t \frac{dy}{dt} + 2y = 0$ for $t > 0$.
Note that the second order linear homogenous differential equation above is indeed an Euler differential equation with $\alpha = 4$ and $\beta = 2$. Therefore we can immediately let $x = \ln t$ and rewrite this Euler differential equation as:(4)
The characteristic equation for this differential equation is $r^2 + 3r + 2 = 0$. This can easily be factored as $(r + 2)(r + 1) = 0$ and so we have two distinct real roots, $r_1 = -2$ and $r_2 = -1$. Therefore the general solution to this differential equation is:(5)
Making the substitution $x = \ln t$ back and we get a solution to the Euler differential equation:(6) |
The Boundary of a Set in a Topological Space
Definition: Let $(X, \tau)$ be a topological space and $A \subseteq X$. A point $x \in X$ is said to be a Boundary Point of $A$ if $x$ is in the closure of $A$ but not in the interior of $A$, i.e., $x \in \bar{A} \setminus \mathrm{int} (A)$. The set of all boundary points of $A$ is called the Boundary of $A$ and is denoted $\partial A = \bar{A} \setminus \mathrm{int} (A)$. Equivalently, $x \in \partial A$ if every $U \in \tau$ with $x \in U$ intersects $A$ and $A^c = X \setminus A$ nontrivially. Another equivalent definition for the boundary of $A$ is the set of all points $x \in X$ such that every open neighbourhood of $x$ contains at least one point of $A$ and at least one point of $X \setminus A$. I.e., $x \in \partial A$ if and only if for every open neighbourhood $U$ of $x$ we have that $A \cap U \neq \emptyset$ and $(X \setminus A) \cap U \neq \emptyset$.
Lemma 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $\overline{X \setminus A} = X \setminus \mathrm{int}(A)$. Proof:Let $x \in \overline{X \setminus A}$. Then $x$ is in the closure of $X \setminus A$. Then for every $U \in \tau$ with $x \in U$ we have that: So there does NOT exist an open neighbourhood of $x$ that is fully contained in $A$. Thus, $x \not \in \mathrm{int}(A)$, i.e., $x \in X \setminus \mathrm{int}(A)$. So $\overline{X \setminus A} \subseteq X \setminus \mathrm{int}(A)$. Now let $x \in X \setminus \mathrm{int}(A)$. Then $x \not \in \mathrm{int}(A)$. So for every open neighbourhood $U$ of $x$ we have that $U \not \subseteq A$. So $U \cap (X \setminus A) \neq \emptyset$ for every open neighbourhood $U$ of $x$. Thus $x \in \overline{X \setminus A}$. So $\overline{X \setminus A} \supseteq X \setminus \mathrm{int}(A)$. So we conclude that:
Proposition 2: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $\partial A = \overline{A} \cap \overline{X \setminus A}$. Proof:By definition, $\partial A = \overline{A} \setminus \mathrm{int}(A)$, or equivalently: By lemma 1 we have that:
Corollary 3: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $\partial A$ is closed. Proof:By proposition 2, $\partial A$ can be written as an intersection of two closed sets and so $\partial A$ is closed. $\blacksquare$.
Corollary 4: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $\partial A = \partial (X \setminus A)$. Proof:By proposition 2 we have that: And since $X \setminus (X \setminus A) = A$, we also have by proposition that: Comparing the two above expressions yields $\partial A = \partial (X \setminus A)$. $\blacksquare$. Example 1
Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology on $\mathbb{R}$ of open intervals and consider the set $A = [0, 1) \subset \mathbb{R}$. The closure of $A$ is:(7)
Furthermore, the interior of $A$ is:(8)
Hence we see that the boundary of $A$ is as expected:(9)
Example 2
For another example, consider the set $B = [0, 1) \cup (2, 3) \subset \mathbb{R}$. The closure of $A$ is:(10)
The interior of $B$ is:(11)
Hence we see that the boundary of $B$ is:(12)
Example 3
For a third example, consider the set $X = \mathbb{R}^2$ with the the usual topology $\tau$ containing open disks with positive radii. Let $A = [0, 1) \times [0, 1) \subseteq \mathbb{R}^2$. Then $A$ can be depicted as illustrated:
Then the boundary of $A$, $\partial A$ is therefore the set of points illustrated in the image below: |
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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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Now showing items 1-1 of 1
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
Dense and Nowhere Dense Sets in a Topological Space Dense Sets in a Topological Space
Definition: Let $(X, \tau)$ be a topological space. The set $A \subseteq X$ is said to be Dense in $X$ if the intersection of every nonempty open set with $A$ is nonempty, that is, $A \cap U \neq \emptyset$ for all $U \in \tau \setminus \{ \emptyset \}$.
Given any topological space $(X, \tau)$ it is important to note that $X$ is dense in $X$ because every $U \in \tau$ is such that $U \subseteq X$, and so $X \cap U = U \neq \emptyset$ for all $U \in \tau \setminus \{ \emptyset \}$.
For another example, consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology of open intervals. Then the set of rational numbers $\mathbb{Q} \subset \mathbb{R}$ is dense in $\mathbb{R}$. If not, then there exists an $U \in \tau \setminus \{ \emptyset \}$ such that $\mathbb{Q} \cap U = \emptyset$.
Since $U \in \tau$ we have that $(a, b) \subseteq U$ for some open interval $(a, b)$ with $a, b \in \mathbb{R}$ and $a < b$. Suppose that $\mathbb{Q} \setminus U = \emptyset$. Then we must also have that:(1)
The intersection above implies that there exists no rational numbers in the interval $(a, b)$, i.e., there exists no $q \in \mathbb{Q}$ such that $a < q < b$. But this is a contradiction since for all $a, b \in \mathbb{R}$ with $a < b$ there ALWAYS exists a rational number $q \in \mathbb{Q}$ such that $a < q < b$, i.e., $q \in (a, b)$. So $\mathbb{Q} \cap (a, b) \neq \emptyset$ for all $U \in \tau \setminus \{ \emptyset \}$. Thus, $\mathbb{Q}$ is dense in $\mathbb{R}$.
We will now look at a very important theorem which will give us a way to determine whether a set $A \subseteq X$ is dense in $X$ or not.
Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is dense in $X$ if and only if $\bar{A} = X$. Proof:$\Rightarrow$ Suppose that $A$ is dense in $X$. Then for all $U \in \tau \setminus \{ \emptyset \}$ we have that $A \cap U = \emptyset$. Clearly $\bar{A} \subseteq X$ so we only need to show that $X \subseteq \bar{A}$. Nowhere Dense Sets in a Topological Space
Definition: Let $(X, \tau)$ be a topological space. A set $A \subseteq X$ is said to be Nowhere Dense in $X$ if the interior of the closure of $A$ is empty, that is, $\mathrm{int} (\bar{A}) = \emptyset$.
For example, consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usually topology of open intervals on $\mathbb{R}$, and consider the set of integers $\mathbb{Z}$. The closure of $\mathbb{Z}$, $\bar{\mathbb{Z}}$ is the smallest closed set containing $\mathbb{Z}$. The smallest closed set containing $\mathbb{Z}$ is $\mathbb{Z}$ since $\mathbb{Z}^c$ is open as $\mathbb{Z}^c$ is an arbitrary union of open sets:(2)
So what is the interior of $\bar{\mathbb{Z}} = \mathbb{Z}$? It is the largest open set contained in $\bar{\mathbb{Z}} = \mathbb{Z}$. All open sets of $\mathbb{R}$ with respect to this topology $\tau$ are either the empty set, an open interval, a union of open intervals, or the whole set (the union of all open intervals). But no open intervals are contained in $\mathbb{Z}$ and so:(3)
Therefore $\mathbb{Z}$ is a nowhere dense set in $\mathbb{R}$ with respect to the usual topology $\tau$ on $\mathbb{R}$. |
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The ISO 217:1995 standard defines the RA and SRA paper formats.
These paper series are untrimmed raw paper. RA stands for “raw format A” and SRA stands for “supplementary raw format A”. The RA and SRA formats are slightly larger than the corresponding A series formats. This allows bleed (ink to the edge) on printed material that will be later cut down to size. These paper sheets will after printing and binding be cut to match the A format.
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Relational model, PostgreSQL, International Electrotechnical Commission, Prolog, Ibm
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Common Connector Criterion for Connectedness of Unions of Topological Subspaces
Recall from the Common Point Criterion for Connectedness of Unions of Topological Subspaces page that if $X$ is a topological space and $\{ A_i \}_{i \in I}$ is an arbitrary collection of connected topological subspaces then if $\displaystyle{\bigcap_{i \in I} A_i \neq \emptyset}$, i.e., there exists a point in common between all of the topological subspaces $A_i$, then the union $\displaystyle{\bigcap_{i \in I} A_i}$ is also a connected space.
The result stated above is sufficient to show that the union of topological subspaces is connected, however, it is not necessary. For example, if we consider the topological space $\mathbb{R}$ with the usual topology and the subspaces $A_n = [n, n + 1]$, then the collection $\{ A_n \}_{n \in \mathbb{N}} = \{ [1, 2], [2, 3], [3, 4], ... \}$ is such that $\displaystyle{\bigcap_{i \in I} A_n = \emptyset}$. However, the union $\displaystyle{\bigcup_{n = 1}^{\infty} A_n = [0, \infty)}$ is connected.
We now develop a similar result to determine whether the union of a collection of topological spaces is connected or not.
Theorem 1: Let $X$ be a topological space and let $\{ A_i \}_{i \in I}$ be an arbitrary collection of connected topological subspaces. If $A_0$ is another connected topological subspace and $A_0 \cap A_i \neq \emptyset$ for all $i \in I$ then $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is connected. Proof:Suppose instead that $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is disconnected. Then there exists a clopen set $\displaystyle{B \subset A_0 \cup \bigcup_{i \in I} A_i}$ such that $B \neq \emptyset$ and $\displaystyle{B \neq A_0 \cup \bigcup_{i \in I} A_i}$. Let $j \in I \cup \{ 0 \}$ and suppose that $B \cap A_j \neq \emptyset$ (and/or $B \cap A_0 \neq \emptyset$. We have already established that then $A_j \subseteq B$ (and/or $A_0 \subseteq B$) otherwise we could form a separation of $A_j$ (and/or $A_0$) which contradicts $A_j$ (or $A_0$) from being connected. Hence we see that for some $J \subseteq I \cup \{ 0 \}$ that then: There are two cases to consider: If $A_0 \subseteq B$, then $B \cap A_i \neq \emptyset$ for all $i \in I$. So $\displaystyle{B = A_0 \cup \bigcup_{i \in I} A_i}$ which contradicts $B$ being a proper subset of $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$. If $A_0 \not \subseteq B$ then this implies that $A_0 \cap B = \emptyset$. But $A_0 \cap A_i \neq \emptyset$ for all $i \in I$, and since $B$ is a union of $A_i$s we have that this is a contradiction. In both cases we see that the assumption that $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is disconnected leads to contradictions. Therefore, if $X$ is a topological space, $\{ A_i \}_{i \in I}$ is an arbitrary collection of connected topological subspaces, and $A_0$ is another connected topological subspace such that $A_0 \cap A_i \neq \emptyset$ for all $i \in I$, then $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is also connected. $\blacksquare$ |
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Physics Letters B, ISSN 0370-2693, 01/2018, Volume 776, Issue C, pp. 430 - 439
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Boundary controllability for a one-dimensional heat equation with a singular inverse-square potential
Facultad Ingeniería, Universidad de Deusto, Avda Universidades 24, 48007 Bilbao, Basque Country, Spain
$\begin{align*} u_t-u_{xx}-\frac{\mu}{x^2}u = 0, \;\;\; (x, t)\in(0, 1)\times(0, T).\end{align*}$
$\mu<1/4$
$f\in H^1(0, T)$ x= 0. This result is obtained employing the moment method by Fattorini and Russell. Mathematics Subject Classification:35K05, 35K67, 93B05, 93B07. Citation:Umberto Biccari. Boundary controllability for a one-dimensional heat equation with a singular inverse-square potential. Mathematical Control & Related Fields, 2019, 9 (1) : 191-219. doi: 10.3934/mcrf.2019011
References:
[1]
F. Araruna, E. Fernández-Cara and M. Santos,
Stackelberg-Nash exact controllability for linear and semilinear parabolic equations,
[2] [3] [4]
U. Biccari and E. Zuazua,
Null controllability for a heat equation with a singular inverse-square potential involving the distance to the boundary function,
[5] [6] [7]
P. Cannarsa, P. Martinez and J. Vancostenoble,
[8]
P. Cannarsa, P. Martinez and J. Vancostenoble,
The cost of controlling weakly degenerate parabolic equations by boundary controls,
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P. Cannarsa, P. Martinez and J. Vancostenoble, The cost of controlling strongly degenerate parabolic equations,
[11]
P. Cannarsa, J. Tort and M. Yamamoto,
Unique continuation and approximate controllability for a degenerate parabolic equation,
[12]
C. Cazacu,
Schrödinger operators with boundary singularities: Hardy inequality, Pohozaev identity and controllability results,
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S. Ervedoza,
Control and stabilization properties for a singular heat equation with an inverse-square potential,
[15]
H. O. Fattorini and D. L. Russell,
Exact controllability theorems for linear parabolic equations in one space dimension,
[16]
H. O. Fattorini and D. L. Russell,
Uniform bounds on biorthogonal functions for real exponentials with an application to the control theory of parabolic equations,
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E. Fernández-Cara and E. Zuazua, Null and approximate controllability for weakly blowing up semilinear heat equations, In
[20]
J. Garcia Azorero and I. Peral Alonso,
Hardy inequalities and some critical elliptic and parabolic problems,
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V. Hernández-Santamaría and L. de Teresa,
Robust Stackelberg controllability for linear and semilinear heat equations,
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V. Komornik and P. Loreti,
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N. N. Lebedev and R. A. Silverman,
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J. Vancostenoble,
Improved Hardy-Poincaré inequalities and sharp Carleman estimates for degenerate/singular parabolic problems,
[31]
J. Vancostenoble and E. Zuazua,
Null controllability for the heat equation with singular inverse-square potentials,
[32]
J. Vancostenoble and E. Zuazua,
Hardy inequalities, observability, and control for the wave and Schrödinger equations with singular potentials,
[33]
J. L. Vázquez and E. Zuazua,
The Hardy inequality and the asymptotic behaviour of the heat equation with an inverse-square potential,
[34]
G. N. Watson,
show all references
References:
[1]
F. Araruna, E. Fernández-Cara and M. Santos,
Stackelberg-Nash exact controllability for linear and semilinear parabolic equations,
[2] [3] [4]
U. Biccari and E. Zuazua,
Null controllability for a heat equation with a singular inverse-square potential involving the distance to the boundary function,
[5] [6] [7]
P. Cannarsa, P. Martinez and J. Vancostenoble,
[8]
P. Cannarsa, P. Martinez and J. Vancostenoble,
The cost of controlling weakly degenerate parabolic equations by boundary controls,
[9] [10]
P. Cannarsa, P. Martinez and J. Vancostenoble, The cost of controlling strongly degenerate parabolic equations,
[11]
P. Cannarsa, J. Tort and M. Yamamoto,
Unique continuation and approximate controllability for a degenerate parabolic equation,
[12]
C. Cazacu,
Schrödinger operators with boundary singularities: Hardy inequality, Pohozaev identity and controllability results,
[13] [14]
S. Ervedoza,
Control and stabilization properties for a singular heat equation with an inverse-square potential,
[15]
H. O. Fattorini and D. L. Russell,
Exact controllability theorems for linear parabolic equations in one space dimension,
[16]
H. O. Fattorini and D. L. Russell,
Uniform bounds on biorthogonal functions for real exponentials with an application to the control theory of parabolic equations,
[17] [18] [19]
E. Fernández-Cara and E. Zuazua, Null and approximate controllability for weakly blowing up semilinear heat equations, In
[20]
J. Garcia Azorero and I. Peral Alonso,
Hardy inequalities and some critical elliptic and parabolic problems,
[21] [22] [23]
V. Hernández-Santamaría and L. de Teresa,
Robust Stackelberg controllability for linear and semilinear heat equations,
[24]
V. Komornik and P. Loreti,
[25] [26]
N. N. Lebedev and R. A. Silverman,
[27] [28]
P. Martinez and J. Vancostenoble, The cost of boundary controllability for a parabolic equation with inverse square potential, Submitted.Google Scholar
[29] [30]
J. Vancostenoble,
Improved Hardy-Poincaré inequalities and sharp Carleman estimates for degenerate/singular parabolic problems,
[31]
J. Vancostenoble and E. Zuazua,
Null controllability for the heat equation with singular inverse-square potentials,
[32]
J. Vancostenoble and E. Zuazua,
Hardy inequalities, observability, and control for the wave and Schrödinger equations with singular potentials,
[33]
J. L. Vázquez and E. Zuazua,
The Hardy inequality and the asymptotic behaviour of the heat equation with an inverse-square potential,
[34]
G. N. Watson,
[1]
Víctor Hernández-Santamaría, Liliana Peralta.
Some remarks on the Robust Stackelberg controllability for the heat equation with controls on the boundary.
[2]
Patrick Martinez, Judith Vancostenoble.
The cost of boundary controllability for a parabolic equation with inverse square potential.
[3]
Xiumei Deng, Jun Zhou.
Global existence and blow-up of solutions to a semilinear heat equation with singular potential and logarithmic nonlinearity.
[4] [5] [6] [7] [8] [9]
Laurence Cherfils, Stefania Gatti, Alain Miranville.
A doubly nonlinear parabolic equation with a singular potential.
[10]
Kazuhiro Ishige, Y. Kabeya.
Hot spots for the two dimensional heat equation with
a rapidly decaying negative potential.
[11]
Kazuhiro Ishige, Asato Mukai.
Large time behavior of solutions of the heat equation with inverse square potential.
[12] [13]
Bopeng Rao, Laila Toufayli, Ali Wehbe.
Stability and controllability of a wave equation with dynamical boundary control.
[14]
Jamel Ben Amara, Hedi Bouzidi.
Exact boundary controllability for the Boussinesq equation with variable coefficients.
[15]
Abdelhakim Belghazi, Ferroudja Smadhi, Nawel Zaidi, Ouahiba Zair.
Carleman inequalities for the two-dimensional heat equation in singular domains.
[16]
David Gómez-Castro, Juan Luis Vázquez.
The fractional Schrödinger equation with singular potential and measure data.
[17]
Ning-An Lai, Jinglei Zhao.
Potential well and exact boundary controllability for radial semilinear wave equations on Schwarzschild spacetime.
[18]
Mi-Ho Giga, Yoshikazu Giga, Takeshi Ohtsuka, Noriaki Umeda.
On behavior of signs for the heat equation and a diffusion method for data separation.
[19]
Arnaud Heibig, Mohand Moussaoui.
Exact controllability of the wave equation for domains with slits and for mixed boundary conditions.
[20]
Abdelmouhcene Sengouga.
Exact boundary observability and controllability of the wave equation in an interval with two moving endpoints.
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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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